Finding the index of an item in a list

Question

Given a list   foo    bar    baz   and an item in the list  bar   how do I get its index  1  in Python

User · Answer

If you are going to find an index once then using "index" method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect. Using bisect module on my machine is about 20 times faster than using index method.

Here is an example of code using Python 3.8 and above syntax:

import bisect
from timeit import timeit

def bisect_search(container, value):
    return (
      index 
      if (index := bisect.bisect_left(container, value)) < len(container) 
      and container[index] == value else -1
    )

data = list(range(1000))
# value to search
value = 666

# times to test
ttt = 1000

t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)

print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")

Output:

t1=0.0400, t2=0.0020, diffs t1/t2=19.60

User · Answer

Let   s give the name lst to the list that you have  One can convert the list lst to a numpy array  And  then use numpy where to get the index of the chosen item in the list  Following is the way in which you will implement it   import numpy as np  lst     foo    bar    baz     lst     list  data type print np where  np array lst      bar   0  0    gt  gt  gt  1

User · Answer

Finding index of item x in list L   idx   L index x  if  x in L  else -1

User · Answer

Getting all the occurrences and the position of one or more  identical  items in a list  With enumerate alist  you can store the first element  n  that is the index of the list when the element x is equal to what you look for    gt  gt  gt  alist     foo    spam    egg    foo    gt  gt  gt  foo indexes    n for n x in enumerate alist  if x   foo    gt  gt  gt  foo indexes  0  3   gt  gt  gt    Let s make our function findindex  This function takes the item and the list as arguments and return the position of the item in the list  like we saw before   def indexlist item2find  list or string      Returns all indexes of an item in a list or a string    return  n for n item in enumerate list or string  if item  item2find   print indexlist  1    010101010        Output     1  3  5  7    Simple  for n  i in enumerate  1  2  3  4  1        if i    1          print n    Output   0 4

User · Answer

index   returns the first index of value         index                L index value   start   stop    -  integer -- return first index of value   def all indices value  qlist       indices          idx   -1     while True          try              idx   qlist index value  idx 1              indices append idx          except ValueError              break     return indices  all indices  foo     foo   bar   baz   foo

User · Answer

For one comparable   Throws ValueError if nothing is found some list     foo    bar    baz   index  baz     some list    2  Custom predicate some list    item1  item2  item3     Throws StopIteration if nothing is found    unless  you provide a second parameter to  next  index of value you like   next      i for i  item in enumerate some list      if item matches your criteria     Finding index of all items by predicate index of staff members         i for i  user in enumerate users      if user is staff

User · Answer

a     foo   bar   baz   bar   any   much    indexes    index for index in range len a   if a index      bar

User · Answer

If performance is of concern   It is mentioned in numerous answers that the built-in method of list index item  method is an O n  algorithm  It is fine if you need to perform this once  But if you need to access the indices of elements a number of times  it makes more sense to first create a dictionary  O n   of item-index pairs  and then access the index at O 1  every time you need it   If you are sure that the items in your list are never repeated  you can easily   myList     foo    bar    baz      Create the dictionary myDict   dict  e i  for i e in enumerate myList      Lookup myDict  bar     Returns 1   myDict get  blah   if you don t want an error to be raised if element not found    If you may have duplicate elements  and need to return all of their indices   from collections import defaultdict as dd myList     foo    bar    bar    baz    foo      Create the dictionary myDict   dd list  for i e in enumerate myList       myDict e  append i     Lookup myDict  foo     Returns  0  4

User · Answer

All of the proposed functions here reproduce inherent language behavior but obscure what s going on    i for i in range len mylist   if mylist i   myterm     get the indices   each for each in mylist if each  myterm                get the items  mylist index myterm  if myterm in mylist else None      get the first index and fail quietly   Why write a function with exception handling if the language provides the methods to do what you want itself

User · Answer

You have to set a condition to check if the element you re searching is in the list  if  your element  in mylist      print mylist index  your element   else      print None

User · Answer

There is a more functional answer to this   list filter lambda x  x 1    bar  enumerate   foo    bar    baz    bar    baz    bar    a    b    c        More generic form   def get index of lst  element       return list map lambda x  x 0            list filter lambda x  x 1   element  enumerate lst

User · Answer

using dictionary   where process the list first and then add the index to it   from collections import defaultdict  index dict   defaultdict list      word list      foo   bar   baz   bar   any    foo    much    for word index in range len word list         index dict word list word index   append word index   word index to find    foo         print index dict word index to find      output     0  5

User · Answer

Finding the index of an item given a list containing it in Python For a list   quot foo quot    quot bar quot    quot baz quot   and an item in the list  quot bar quot   what s the cleanest way to get its index  1  in Python   Well  sure  there s the index method  which returns the index of the first occurrence   gt  gt  gt  l     quot foo quot    quot bar quot    quot baz quot    gt  gt  gt  l index  bar   1  There are a couple of issues with this method   if the value isn t in the list  you ll get a ValueError if more than one of the value is in the list  you only get the index for the first one  No values If the value could be missing  you need to catch the ValueError  You can do so with a reusable definition like this  def index a list  value       try          return a list index value      except ValueError          return None  And use it like this   gt  gt  gt  print index l   quux    None  gt  gt  gt  print index l   bar    1  And the downside of this is that you will probably have a check for if the returned value is or is not None  result   index a list  value  if result is not None      do something result   More than one value in the list If you could have more occurrences  you ll not get complete information with list index   gt  gt  gt  l append  bar    gt  gt  gt  l   foo    bar    baz    bar    gt  gt  gt  l index  bar                  nothing at index 3  1  You might enumerate into a list comprehension the indexes   gt  gt  gt   index for index  v in enumerate l  if v     bar    1  3   gt  gt  gt   index for index  v in enumerate l  if v     boink       If you have no occurrences  you can check for that with boolean check of the result  or just do nothing if you loop over the results  indexes    index for index  v in enumerate l  if v     boink   for index in indexes      do something index   Better data munging with pandas If you have pandas  you can easily get this information with a Series object   gt  gt  gt  import pandas as pd  gt  gt  gt  series   pd Series l   gt  gt  gt  series 0    foo 1    bar 2    baz 3    bar dtype  object  A comparison check will return a series of booleans   gt  gt  gt  series     bar  0    False 1     True 2    False 3     True dtype  bool  Pass that series of booleans to the series via subscript notation  and you get just the matching members   gt  gt  gt  series series     bar   1    bar 3    bar dtype  object  If you want just the indexes  the index attribute returns a series of integers   gt  gt  gt  series series     bar   index Int64Index  1  3   dtype  int64    And if you want them in a list or tuple  just pass them to the constructor   gt  gt  gt  list series series     bar   index   1  3   Yes  you could use a list comprehension with enumerate too  but that s just not as elegant  in my opinion - you re doing tests for equality in Python  instead of letting builtin code written in C handle it   gt  gt  gt   i for i  value in enumerate l  if value     bar    1  3   Is this an XY problem   The XY problem is asking about your attempted solution rather than your actual problem   Why do you think you need the index given an element in a list  If you already know the value  why do you care where it is in a list  If the value isn t there  catching the ValueError is rather verbose - and I prefer to avoid that  I m usually iterating over the list anyways  so I ll usually keep a pointer to any interesting information  getting the index with enumerate  If you re munging data  you should probably be using pandas - which has far more elegant tools than the pure Python workarounds I ve shown  I do not recall needing list index  myself  However  I have looked through the Python standard library  and I see some excellent uses for it  There are many  many uses for it in idlelib  for GUI and text parsing  The keyword module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming  In Lib mailbox py it seems to be using it like an ordered mapping  key list key list index old     new  and del key list key list index key    In Lib http cookiejar py  seems to be used to get the next month  mon   MONTHS LOWER index mon lower    1  In Lib tarfile py similar to distutils to get a slice up to an item  members   members  members index tarinfo    In Lib pickletools py  numtopop   before index markobject   What these usages seem to have in common is that they seem to operate on lists of constrained sizes  important because of O n  lookup time for list index   and they re mostly used in parsing  and UI in the case of Idle   While there are use-cases for it  they are fairly uncommon  If you find yourself looking for this answer  ask yourself if what you re doing is the most direct usage of the tools provided by the language for your use-case

User · Answer

The majority of answers explain how to find a single index  but their methods do not return multiple indexes if the item is in the list multiple times  Use enumerate     for i  j in enumerate   foo    bar    baz         if j     bar           print i    The index   function only returns the first occurrence  while enumerate   returns all occurrences   As a list comprehension    i for i  j in enumerate   foo    bar    baz    if j     bar       Here s also another small solution with itertools count    which is pretty much the same approach as enumerate    from itertools import izip as zip  count   izip for maximum efficiency  i for i  j in zip count      foo    bar    baz    if j     bar     This is more efficient for larger lists than using enumerate       python -m timeit -s  from itertools import izip as zip  count    i for i  j in zip count      foo    bar    baz   500  if j     bar    10000 loops  best of 3  174 usec per loop   python -m timeit   i for i  j in enumerate   foo    bar    baz   500  if j     bar    10000 loops  best of 3  196 usec per loop

User · Answer

Python index   method throws an error if the item was not found  So instead you can make it similar to the indexOf   function of JavaScript which returns -1 if the item was not found   try      index   array index  search keyword   except ValueError      index   -1

User · Answer

gt  gt  gt    quot foo quot    quot bar quot    quot baz quot   index  quot bar quot   1  Reference  Data Structures  gt  More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked  index is a rather weak component of the list API  and I can t remember the last time I used it in anger  It s been pointed out to me in the comments that because this answer is heavily referenced  it should be made more complete  Some caveats about list index follow  It is probably worth initially taking a look at the documentation for it   list index x   start   end     Return zero-based index in the list of the first item whose value is equal to x  Raises a ValueError if there is no such item  The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list  The returned index is computed relative to the beginning of the full sequence rather than the start argument   Linear time-complexity in list length An index call checks every element of the list in order  until it finds a match  If your list is long  and you don t know roughly where in the list it occurs  this search could become a bottleneck  In that case  you should consider a different data structure  Note that if you know roughly where to find the match  you can give index a hint  For instance  in this snippet  l index 999 999  999 990  1 000 000  is roughly five orders of magnitude faster than straight l index 999 999   because the former only has to search 10 entries  while the latter searches a million   gt  gt  gt  import timeit  gt  gt  gt  timeit timeit  l index 999 999    setup  l   list range 0  1 000 000     number 1000  9 356267921015387  gt  gt  gt  timeit timeit  l index 999 999  999 990  1 000 000    setup  l   list range 0  1 000 000     number 1000  0 0004404920036904514    Only returns the index of the first match to its argument A call to index searches through the list in order until it finds a match  and stops there  If you expect to need indices of more matches  you should use a list comprehension  or generator expression   gt  gt  gt   1  1  index 1  0  gt  gt  gt   i for i  e in enumerate  1  2  1   if e    1   0  2   gt  gt  gt  g    i for i  e in enumerate  1  2  1   if e    1   gt  gt  gt  next g  0  gt  gt  gt  next g  2  Most places where I once would have used index  I now use a list comprehension or generator expression because they re more generalizable  So if you re considering reaching for index  take a look at these excellent Python features  Throws if element not present in list A call to index results in a ValueError if the item s not present   gt  gt  gt   1  1  index 2  Traceback  most recent call last     File  quot  lt stdin gt  quot   line 1  in  lt module gt  ValueError  2 is not in list  If the item might not be present in the list  you should either  Check for it first with item in my list  clean  readable approach   or Wrap the index call in a try except block which catches ValueError  probably faster  at least when the list to search is long  and the item is usually present

User · Answer

This solution is not as powerful as others  but if you re a beginner and only know about forloops it s still possible to find the first index of an item while avoiding the ValueError   def find element p t       i   0     for e in p          if e    t              return i         else              i   1     return -1

User · Answer

And now  for something completely different           like confirming the existence of the item before getting the index   The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list   It works with strings as well   def indices l  val          Always returns a list containing the indices of val in the list        retval          last   0     while val in l last                i   l last   index val              retval append last   i              last    i   1        return retval  l     bar   foo   bar   baz   bar   bar   q    bar  print indices l q  print indices l  bat   print indices  abcdaababb   a     When pasted into an interactive python window   Python 2 7 6  v2 7 6 3a1db0d2747e  Nov 10 2013  00 42 54    GCC 4 2 1  Apple Inc  build 5666   dot 3   on darwin Type  help    copyright    credits  or  license  for more information   gt  gt  gt  def indices the list  val              Always returns a list containing the indices of val in the list            retval              last   0         while val in the list last                    i   the list last   index val                  retval append last   i                  last    i   1            return retval       gt  gt  gt  l     bar   foo   bar   baz   bar   bar    gt  gt  gt  q    bar   gt  gt  gt  print indices l q   0  2  4  5   gt  gt  gt  print indices l  bat       gt  gt  gt  print indices  abcdaababb   a    0  4  5  7   gt  gt  gt     Update  After another year of heads-down python development  I m a bit embarrassed by my original answer  so to set the record straight  one can certainly use the above code  however  the much more idiomatic way to get the same behavior would be to use list comprehension  along with the enumerate   function     Something like this     def indices l  val          Always returns a list containing the indices of val in the list        return  index for index  value in enumerate l  if value    val   l     bar   foo   bar   baz   bar   bar   q    bar  print indices l q  print indices l  bat   print indices  abcdaababb   a     Which  when pasted into an interactive python window yields   Python 2 7 14  Anaconda  Inc    default  Dec  7 2017  11 07 58    GCC 4 2 1 Compatible Clang 4 0 1  tags RELEASE 401 final   on darwin Type  help    copyright    credits  or  license  for more information   gt  gt  gt  def indices l  val              Always returns a list containing the indices of val in the list            return  index for index  value in enumerate l  if value    val        gt  gt  gt  l     bar   foo   bar   baz   bar   bar    gt  gt  gt  q    bar   gt  gt  gt  print indices l q   0  2  4  5   gt  gt  gt  print indices l  bat       gt  gt  gt  print indices  abcdaababb   a    0  4  5  7   gt  gt  gt     And now  after reviewing this question and all the answers  I realize that this is exactly what FMc suggested in his earlier answer   At the time I originally answered this question  I didn t even see that answer  because I didn t understand it   I hope that my somewhat more verbose example will aid understanding     If the single line of code above still doesn t make sense to you  I highly recommend you Google  python list comprehension  and take a few minutes to familiarize yourself   It s just one of the many powerful features that make it a joy to use Python to develop code

User · Answer

This solution is not as powerful as others  but if you re a beginner and only know about forloops it s still possible to find the first index of an item while avoiding the ValueError   def find element p t       i   0     for e in p          if e    t              return i         else              i   1     return -1

User · Answer

And now  for something completely different           like confirming the existence of the item before getting the index   The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list   It works with strings as well   def indices l  val          Always returns a list containing the indices of val in the list        retval          last   0     while val in l last                i   l last   index val              retval append last   i              last    i   1        return retval  l     bar   foo   bar   baz   bar   bar   q    bar  print indices l q  print indices l  bat   print indices  abcdaababb   a     When pasted into an interactive python window   Python 2 7 6  v2 7 6 3a1db0d2747e  Nov 10 2013  00 42 54    GCC 4 2 1  Apple Inc  build 5666   dot 3   on darwin Type  help    copyright    credits  or  license  for more information   gt  gt  gt  def indices the list  val              Always returns a list containing the indices of val in the list            retval              last   0         while val in the list last                    i   the list last   index val                  retval append last   i                  last    i   1            return retval       gt  gt  gt  l     bar   foo   bar   baz   bar   bar    gt  gt  gt  q    bar   gt  gt  gt  print indices l q   0  2  4  5   gt  gt  gt  print indices l  bat       gt  gt  gt  print indices  abcdaababb   a    0  4  5  7   gt  gt  gt     Update  After another year of heads-down python development  I m a bit embarrassed by my original answer  so to set the record straight  one can certainly use the above code  however  the much more idiomatic way to get the same behavior would be to use list comprehension  along with the enumerate   function     Something like this     def indices l  val          Always returns a list containing the indices of val in the list        return  index for index  value in enumerate l  if value    val   l     bar   foo   bar   baz   bar   bar   q    bar  print indices l q  print indices l  bat   print indices  abcdaababb   a     Which  when pasted into an interactive python window yields   Python 2 7 14  Anaconda  Inc    default  Dec  7 2017  11 07 58    GCC 4 2 1 Compatible Clang 4 0 1  tags RELEASE 401 final   on darwin Type  help    copyright    credits  or  license  for more information   gt  gt  gt  def indices l  val              Always returns a list containing the indices of val in the list            return  index for index  value in enumerate l  if value    val        gt  gt  gt  l     bar   foo   bar   baz   bar   bar    gt  gt  gt  q    bar   gt  gt  gt  print indices l q   0  2  4  5   gt  gt  gt  print indices l  bat       gt  gt  gt  print indices  abcdaababb   a    0  4  5  7   gt  gt  gt     And now  after reviewing this question and all the answers  I realize that this is exactly what FMc suggested in his earlier answer   At the time I originally answered this question  I didn t even see that answer  because I didn t understand it   I hope that my somewhat more verbose example will aid understanding     If the single line of code above still doesn t make sense to you  I highly recommend you Google  python list comprehension  and take a few minutes to familiarize yourself   It s just one of the many powerful features that make it a joy to use Python to develop code

User · Answer

Let   s give the name lst to the list that you have  One can convert the list lst to a numpy array  And  then use numpy where to get the index of the chosen item in the list  Following is the way in which you will implement it   import numpy as np  lst     foo    bar    baz     lst     list  data type print np where  np array lst      bar   0  0    gt  gt  gt  1

User · Answer

Since Python lists are zero-based  we can use the zip built-in function as follows    gt  gt  gt   i for i j in zip range len haystack    haystack  if j     needle      where  haystack  is the list in question and  needle  is the item to look for    Note  Here we are iterating using i to get the indexes  but if we need rather to focus on the items we can switch to j

User · Answer

If you want all indexes  then you can use NumPy   import numpy as np  array    1  2  1  3  4  5  1  item   1 np array   np array array  item index   np where np array  item  print item index   Out   array  0  2  6   dtype int64      It is clear  readable solution

User · Answer

Finding index of item x in list L   idx   L index x  if  x in L  else -1

User · Answer

As indicated by  TerryA  many answers discuss how to find one index   more itertools is a third-party library with tools to locate multiple indices within an iterable    Given  import more itertools as mit   iterable     foo    bar    baz    ham    foo    bar    baz     Code  Find indices of multiple observations   list mit locate iterable  lambda x  x     bar       1  5    Test multiple items   list mit locate iterable  lambda x  x in   bar    ham        1  3  5    See also more options with more itertools locate   Install via  gt  pip install more itertools

User · Answer

All indexes with the zip function   get indexes   lambda x  xs   i for  y  i  in zip xs  range len xs    if x    y   print get indexes 2   1  2  3  4  5  6  3  2  3  2   print get indexes  f    xsfhhttytffsafweef

User · Answer

Python index   method throws an error if the item was not found  So instead you can make it similar to the indexOf   function of JavaScript which returns -1 if the item was not found   try      index   array index  search keyword   except ValueError      index   -1

User · Answer

All of the proposed functions here reproduce inherent language behavior but obscure what s going on    i for i in range len mylist   if mylist i   myterm     get the indices   each for each in mylist if each  myterm                get the items  mylist index myterm  if myterm in mylist else None      get the first index and fail quietly   Why write a function with exception handling if the language provides the methods to do what you want itself

User · Answer

One thing that is really helpful in learning Python is to use the interactive help function    gt  gt  gt  help   foo    bar    baz    Help on list object   class list object               index              L index value   start   stop    - gt  integer -- return first index of value      which will often lead you to the method you are looking for

User · Answer

The majority of answers explain how to find a single index  but their methods do not return multiple indexes if the item is in the list multiple times  Use enumerate     for i  j in enumerate   foo    bar    baz         if j     bar           print i    The index   function only returns the first occurrence  while enumerate   returns all occurrences   As a list comprehension    i for i  j in enumerate   foo    bar    baz    if j     bar       Here s also another small solution with itertools count    which is pretty much the same approach as enumerate    from itertools import izip as zip  count   izip for maximum efficiency  i for i  j in zip count      foo    bar    baz    if j     bar     This is more efficient for larger lists than using enumerate       python -m timeit -s  from itertools import izip as zip  count    i for i  j in zip count      foo    bar    baz   500  if j     bar    10000 loops  best of 3  174 usec per loop   python -m timeit   i for i  j in enumerate   foo    bar    baz   500  if j     bar    10000 loops  best of 3  196 usec per loop

User · Answer

You have to set a condition to check if the element you re searching is in the list  if  your element  in mylist      print mylist index  your element   else      print None

User · Answer

One thing that is really helpful in learning Python is to use the interactive help function    gt  gt  gt  help   foo    bar    baz    Help on list object   class list object               index              L index value   start   stop    - gt  integer -- return first index of value      which will often lead you to the method you are looking for

User · Answer

For those coming from another language like me  maybe with a simple loop it s easier to understand and use it   mylist     foo    bar    baz    bar   newlist   enumerate mylist  for index  item in newlist    if item     bar       print index  item    I am thankful for So what exactly does enumerate do   That helped me to understand

User · Answer

Getting all the occurrences and the position of one or more  identical  items in a list  With enumerate alist  you can store the first element  n  that is the index of the list when the element x is equal to what you look for    gt  gt  gt  alist     foo    spam    egg    foo    gt  gt  gt  foo indexes    n for n x in enumerate alist  if x   foo    gt  gt  gt  foo indexes  0  3   gt  gt  gt    Let s make our function findindex  This function takes the item and the list as arguments and return the position of the item in the list  like we saw before   def indexlist item2find  list or string      Returns all indexes of an item in a list or a string    return  n for n item in enumerate list or string  if item  item2find   print indexlist  1    010101010        Output     1  3  5  7    Simple  for n  i in enumerate  1  2  3  4  1        if i    1          print n    Output   0 4

User · Answer

If you want all indexes  then you can use NumPy   import numpy as np  array    1  2  1  3  4  5  1  item   1 np array   np array array  item index   np where np array  item  print item index   Out   array  0  2  6   dtype int64      It is clear  readable solution

User · Answer

using dictionary   where process the list first and then add the index to it   from collections import defaultdict  index dict   defaultdict list      word list      foo   bar   baz   bar   any    foo    much    for word index in range len word list         index dict word list word index   append word index   word index to find    foo         print index dict word index to find      output     0  5

User · Answer

Simply you can go with  a      hand    head      phone    wallet      lost    stock    b     phone    lost    res     x 0  for x in a  index y  for y in b

User · Answer

If performance is of concern   It is mentioned in numerous answers that the built-in method of list index item  method is an O n  algorithm  It is fine if you need to perform this once  But if you need to access the indices of elements a number of times  it makes more sense to first create a dictionary  O n   of item-index pairs  and then access the index at O 1  every time you need it   If you are sure that the items in your list are never repeated  you can easily   myList     foo    bar    baz      Create the dictionary myDict   dict  e i  for i e in enumerate myList      Lookup myDict  bar     Returns 1   myDict get  blah   if you don t want an error to be raised if element not found    If you may have duplicate elements  and need to return all of their indices   from collections import defaultdict as dd myList     foo    bar    bar    baz    foo      Create the dictionary myDict   dd list  for i e in enumerate myList       myDict e  append i     Lookup myDict  foo     Returns  0  4

User · Answer

There is a chance that that value may not be present so to avoid this ValueError  we can check if that actually exists in the list   list      quot foo quot    quot bar quot    quot baz quot    item to find    quot foo quot   if item to find in list        index   list index item to find        print  quot Index of the item is  quot    str index   else      print  quot That word does not exist quot

User · Answer

All indexes with the zip function   get indexes   lambda x  xs   i for  y  i  in zip xs  range len xs    if x    y   print get indexes 2   1  2  3  4  5  6  3  2  3  2   print get indexes  f    xsfhhttytffsafweef

User · Answer

Another option   gt  gt  gt  a     red    blue    green    red    gt  gt  gt  b    red   gt  gt  gt  offset   0   gt  gt  gt  indices   list    gt  gt  gt  for i in range a count b            indices append a index b offset           offset   indices -1  1       gt  gt  gt  indices  0  3   gt  gt  gt

User · Answer

For one comparable   Throws ValueError if nothing is found some list     foo    bar    baz   index  baz     some list    2  Custom predicate some list    item1  item2  item3     Throws StopIteration if nothing is found    unless  you provide a second parameter to  next  index of value you like   next      i for i  item in enumerate some list      if item matches your criteria     Finding index of all items by predicate index of staff members         i for i  user in enumerate users      if user is staff

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Finding the index of an item given a list containing it in Python For a list   quot foo quot    quot bar quot    quot baz quot   and an item in the list  quot bar quot   what s the cleanest way to get its index  1  in Python   Well  sure  there s the index method  which returns the index of the first occurrence   gt  gt  gt  l     quot foo quot    quot bar quot    quot baz quot    gt  gt  gt  l index  bar   1  There are a couple of issues with this method   if the value isn t in the list  you ll get a ValueError if more than one of the value is in the list  you only get the index for the first one  No values If the value could be missing  you need to catch the ValueError  You can do so with a reusable definition like this  def index a list  value       try          return a list index value      except ValueError          return None  And use it like this   gt  gt  gt  print index l   quux    None  gt  gt  gt  print index l   bar    1  And the downside of this is that you will probably have a check for if the returned value is or is not None  result   index a list  value  if result is not None      do something result   More than one value in the list If you could have more occurrences  you ll not get complete information with list index   gt  gt  gt  l append  bar    gt  gt  gt  l   foo    bar    baz    bar    gt  gt  gt  l index  bar                  nothing at index 3  1  You might enumerate into a list comprehension the indexes   gt  gt  gt   index for index  v in enumerate l  if v     bar    1  3   gt  gt  gt   index for index  v in enumerate l  if v     boink       If you have no occurrences  you can check for that with boolean check of the result  or just do nothing if you loop over the results  indexes    index for index  v in enumerate l  if v     boink   for index in indexes      do something index   Better data munging with pandas If you have pandas  you can easily get this information with a Series object   gt  gt  gt  import pandas as pd  gt  gt  gt  series   pd Series l   gt  gt  gt  series 0    foo 1    bar 2    baz 3    bar dtype  object  A comparison check will return a series of booleans   gt  gt  gt  series     bar  0    False 1     True 2    False 3     True dtype  bool  Pass that series of booleans to the series via subscript notation  and you get just the matching members   gt  gt  gt  series series     bar   1    bar 3    bar dtype  object  If you want just the indexes  the index attribute returns a series of integers   gt  gt  gt  series series     bar   index Int64Index  1  3   dtype  int64    And if you want them in a list or tuple  just pass them to the constructor   gt  gt  gt  list series series     bar   index   1  3   Yes  you could use a list comprehension with enumerate too  but that s just not as elegant  in my opinion - you re doing tests for equality in Python  instead of letting builtin code written in C handle it   gt  gt  gt   i for i  value in enumerate l  if value     bar    1  3   Is this an XY problem   The XY problem is asking about your attempted solution rather than your actual problem   Why do you think you need the index given an element in a list  If you already know the value  why do you care where it is in a list  If the value isn t there  catching the ValueError is rather verbose - and I prefer to avoid that  I m usually iterating over the list anyways  so I ll usually keep a pointer to any interesting information  getting the index with enumerate  If you re munging data  you should probably be using pandas - which has far more elegant tools than the pure Python workarounds I ve shown  I do not recall needing list index  myself  However  I have looked through the Python standard library  and I see some excellent uses for it  There are many  many uses for it in idlelib  for GUI and text parsing  The keyword module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming  In Lib mailbox py it seems to be using it like an ordered mapping  key list key list index old     new  and del key list key list index key    In Lib http cookiejar py  seems to be used to get the next month  mon   MONTHS LOWER index mon lower    1  In Lib tarfile py similar to distutils to get a slice up to an item  members   members  members index tarinfo    In Lib pickletools py  numtopop   before index markobject   What these usages seem to have in common is that they seem to operate on lists of constrained sizes  important because of O n  lookup time for list index   and they re mostly used in parsing  and UI in the case of Idle   While there are use-cases for it  they are fairly uncommon  If you find yourself looking for this answer  ask yourself if what you re doing is the most direct usage of the tools provided by the language for your use-case

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One thing that is really helpful in learning Python is to use the interactive help function    gt  gt  gt  help   foo    bar    baz    Help on list object   class list object               index              L index value   start   stop    - gt  integer -- return first index of value      which will often lead you to the method you are looking for

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gt  gt  gt    quot foo quot    quot bar quot    quot baz quot   index  quot bar quot   1  Reference  Data Structures  gt  More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked  index is a rather weak component of the list API  and I can t remember the last time I used it in anger  It s been pointed out to me in the comments that because this answer is heavily referenced  it should be made more complete  Some caveats about list index follow  It is probably worth initially taking a look at the documentation for it   list index x   start   end     Return zero-based index in the list of the first item whose value is equal to x  Raises a ValueError if there is no such item  The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list  The returned index is computed relative to the beginning of the full sequence rather than the start argument   Linear time-complexity in list length An index call checks every element of the list in order  until it finds a match  If your list is long  and you don t know roughly where in the list it occurs  this search could become a bottleneck  In that case  you should consider a different data structure  Note that if you know roughly where to find the match  you can give index a hint  For instance  in this snippet  l index 999 999  999 990  1 000 000  is roughly five orders of magnitude faster than straight l index 999 999   because the former only has to search 10 entries  while the latter searches a million   gt  gt  gt  import timeit  gt  gt  gt  timeit timeit  l index 999 999    setup  l   list range 0  1 000 000     number 1000  9 356267921015387  gt  gt  gt  timeit timeit  l index 999 999  999 990  1 000 000    setup  l   list range 0  1 000 000     number 1000  0 0004404920036904514    Only returns the index of the first match to its argument A call to index searches through the list in order until it finds a match  and stops there  If you expect to need indices of more matches  you should use a list comprehension  or generator expression   gt  gt  gt   1  1  index 1  0  gt  gt  gt   i for i  e in enumerate  1  2  1   if e    1   0  2   gt  gt  gt  g    i for i  e in enumerate  1  2  1   if e    1   gt  gt  gt  next g  0  gt  gt  gt  next g  2  Most places where I once would have used index  I now use a list comprehension or generator expression because they re more generalizable  So if you re considering reaching for index  take a look at these excellent Python features  Throws if element not present in list A call to index results in a ValueError if the item s not present   gt  gt  gt   1  1  index 2  Traceback  most recent call last     File  quot  lt stdin gt  quot   line 1  in  lt module gt  ValueError  2 is not in list  If the item might not be present in the list  you should either  Check for it first with item in my list  clean  readable approach   or Wrap the index call in a try except block which catches ValueError  probably faster  at least when the list to search is long  and the item is usually present

User · Answer

To get all indexes   indexes    i for i x in enumerate xs  if x     foo

User · Answer

One thing that is really helpful in learning Python is to use the interactive help function    gt  gt  gt  help   foo    bar    baz    Help on list object   class list object               index              L index value   start   stop    - gt  integer -- return first index of value      which will often lead you to the method you are looking for

User · Answer

A problem will arise if the element is not in the list  This function handles the issue     if element is found it returns index of element else returns None  def find element in list element  list element       try          index element   list element index element          return index element     except ValueError          return None

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A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry    gt  gt  gt  a     foo   bar   baz   bar   any    foo    much    gt  gt  gt  l   dict zip set a   map lambda y   i for i z in enumerate a  if z is y    set a      gt  gt  gt  l  foo    0  5   gt  gt  gt  l   much    6   gt  gt  gt  l   baz    2    foo    0  5    bar    1  3    any    4    much    6    gt  gt  gt     You could also use this as a one liner to get all indices for a single entry  There are no guarantees for efficiency  though I did use set a  to reduce the number of times the lambda is called

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A problem will arise if the element is not in the list  This function handles the issue     if element is found it returns index of element else returns None  def find element in list element  list element       try          index element   list element index element          return index element     except ValueError          return None

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There is a more functional answer to this   list filter lambda x  x 1    bar  enumerate   foo    bar    baz    bar    baz    bar    a    b    c        More generic form   def get index of lst  element       return list map lambda x  x 0            list filter lambda x  x 1   element  enumerate lst

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index   returns the first index of value         index                L index value   start   stop    -  integer -- return first index of value   def all indices value  qlist       indices          idx   -1     while True          try              idx   qlist index value  idx 1              indices append idx          except ValueError              break     return indices  all indices  foo     foo   bar   baz   foo

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There is a chance that that value may not be present so to avoid this ValueError  we can check if that actually exists in the list   list      quot foo quot    quot bar quot    quot baz quot    item to find    quot foo quot   if item to find in list        index   list index item to find        print  quot Index of the item is  quot    str index   else      print  quot That word does not exist quot

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For those coming from another language like me  maybe with a simple loop it s easier to understand and use it   mylist     foo    bar    baz    bar   newlist   enumerate mylist  for index  item in newlist    if item     bar       print index  item    I am thankful for So what exactly does enumerate do   That helped me to understand

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To get all indexes   indexes    i for i x in enumerate xs  if x     foo

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Another option   gt  gt  gt  a     red    blue    green    red    gt  gt  gt  b    red   gt  gt  gt  offset   0   gt  gt  gt  indices   list    gt  gt  gt  for i in range a count b            indices append a index b offset           offset   indices -1  1       gt  gt  gt  indices  0  3   gt  gt  gt

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name   bar  list      foo   1     bar   2     baz   3   new list    for item in list      new list append item 0   print new list  try      location  new list index name  except      location -1 print  location    This accounts for if the string is not in the list too  if it isn t in the list then location   -1

User · Answer

If you are going to find an index once then using "index" method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect. Using bisect module on my machine is about 20 times faster than using index method.

Here is an example of code using Python 3.8 and above syntax:

import bisect
from timeit import timeit

def bisect_search(container, value):
    return (
      index 
      if (index := bisect.bisect_left(container, value)) < len(container) 
      and container[index] == value else -1
    )

data = list(range(1000))
# value to search
value = 666

# times to test
ttt = 1000

t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)

print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")

Output:

t1=0.0400, t2=0.0020, diffs t1/t2=19.60

User · Answer

A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry    gt  gt  gt  a     foo   bar   baz   bar   any    foo    much    gt  gt  gt  l   dict zip set a   map lambda y   i for i z in enumerate a  if z is y    set a      gt  gt  gt  l  foo    0  5   gt  gt  gt  l   much    6   gt  gt  gt  l   baz    2    foo    0  5    bar    1  3    any    4    much    6    gt  gt  gt     You could also use this as a one liner to get all indices for a single entry  There are no guarantees for efficiency  though I did use set a  to reduce the number of times the lambda is called

User · Answer

gt  gt  gt    quot foo quot    quot bar quot    quot baz quot   index  quot bar quot   1  Reference  Data Structures  gt  More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked  index is a rather weak component of the list API  and I can t remember the last time I used it in anger  It s been pointed out to me in the comments that because this answer is heavily referenced  it should be made more complete  Some caveats about list index follow  It is probably worth initially taking a look at the documentation for it   list index x   start   end     Return zero-based index in the list of the first item whose value is equal to x  Raises a ValueError if there is no such item  The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list  The returned index is computed relative to the beginning of the full sequence rather than the start argument   Linear time-complexity in list length An index call checks every element of the list in order  until it finds a match  If your list is long  and you don t know roughly where in the list it occurs  this search could become a bottleneck  In that case  you should consider a different data structure  Note that if you know roughly where to find the match  you can give index a hint  For instance  in this snippet  l index 999 999  999 990  1 000 000  is roughly five orders of magnitude faster than straight l index 999 999   because the former only has to search 10 entries  while the latter searches a million   gt  gt  gt  import timeit  gt  gt  gt  timeit timeit  l index 999 999    setup  l   list range 0  1 000 000     number 1000  9 356267921015387  gt  gt  gt  timeit timeit  l index 999 999  999 990  1 000 000    setup  l   list range 0  1 000 000     number 1000  0 0004404920036904514    Only returns the index of the first match to its argument A call to index searches through the list in order until it finds a match  and stops there  If you expect to need indices of more matches  you should use a list comprehension  or generator expression   gt  gt  gt   1  1  index 1  0  gt  gt  gt   i for i  e in enumerate  1  2  1   if e    1   0  2   gt  gt  gt  g    i for i  e in enumerate  1  2  1   if e    1   gt  gt  gt  next g  0  gt  gt  gt  next g  2  Most places where I once would have used index  I now use a list comprehension or generator expression because they re more generalizable  So if you re considering reaching for index  take a look at these excellent Python features  Throws if element not present in list A call to index results in a ValueError if the item s not present   gt  gt  gt   1  1  index 2  Traceback  most recent call last     File  quot  lt stdin gt  quot   line 1  in  lt module gt  ValueError  2 is not in list  If the item might not be present in the list  you should either  Check for it first with item in my list  clean  readable approach   or Wrap the index call in a try except block which catches ValueError  probably faster  at least when the list to search is long  and the item is usually present

User · Answer

Simply you can go with  a      hand    head      phone    wallet      lost    stock    b     phone    lost    res     x 0  for x in a  index y  for y in b

User · Answer

a     foo   bar   baz   bar   any   much    indexes    index for index in range len a   if a index      bar

User · Answer

name   bar  list      foo   1     bar   2     baz   3   new list    for item in list      new list append item 0   print new list  try      location  new list index name  except      location -1 print  location    This accounts for if the string is not in the list too  if it isn t in the list then location   -1

User · Answer

gt  gt  gt    quot foo quot    quot bar quot    quot baz quot   index  quot bar quot   1  Reference  Data Structures  gt  More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked  index is a rather weak component of the list API  and I can t remember the last time I used it in anger  It s been pointed out to me in the comments that because this answer is heavily referenced  it should be made more complete  Some caveats about list index follow  It is probably worth initially taking a look at the documentation for it   list index x   start   end     Return zero-based index in the list of the first item whose value is equal to x  Raises a ValueError if there is no such item  The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list  The returned index is computed relative to the beginning of the full sequence rather than the start argument   Linear time-complexity in list length An index call checks every element of the list in order  until it finds a match  If your list is long  and you don t know roughly where in the list it occurs  this search could become a bottleneck  In that case  you should consider a different data structure  Note that if you know roughly where to find the match  you can give index a hint  For instance  in this snippet  l index 999 999  999 990  1 000 000  is roughly five orders of magnitude faster than straight l index 999 999   because the former only has to search 10 entries  while the latter searches a million   gt  gt  gt  import timeit  gt  gt  gt  timeit timeit  l index 999 999    setup  l   list range 0  1 000 000     number 1000  9 356267921015387  gt  gt  gt  timeit timeit  l index 999 999  999 990  1 000 000    setup  l   list range 0  1 000 000     number 1000  0 0004404920036904514    Only returns the index of the first match to its argument A call to index searches through the list in order until it finds a match  and stops there  If you expect to need indices of more matches  you should use a list comprehension  or generator expression   gt  gt  gt   1  1  index 1  0  gt  gt  gt   i for i  e in enumerate  1  2  1   if e    1   0  2   gt  gt  gt  g    i for i  e in enumerate  1  2  1   if e    1   gt  gt  gt  next g  0  gt  gt  gt  next g  2  Most places where I once would have used index  I now use a list comprehension or generator expression because they re more generalizable  So if you re considering reaching for index  take a look at these excellent Python features  Throws if element not present in list A call to index results in a ValueError if the item s not present   gt  gt  gt   1  1  index 2  Traceback  most recent call last     File  quot  lt stdin gt  quot   line 1  in  lt module gt  ValueError  2 is not in list  If the item might not be present in the list  you should either  Check for it first with item in my list  clean  readable approach   or Wrap the index call in a try except block which catches ValueError  probably faster  at least when the list to search is long  and the item is usually present

User · Answer

As indicated by  TerryA  many answers discuss how to find one index   more itertools is a third-party library with tools to locate multiple indices within an iterable    Given  import more itertools as mit   iterable     foo    bar    baz    ham    foo    bar    baz     Code  Find indices of multiple observations   list mit locate iterable  lambda x  x     bar       1  5    Test multiple items   list mit locate iterable  lambda x  x in   bar    ham        1  3  5    See also more options with more itertools locate   Install via  gt  pip install more itertools

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Since Python lists are zero-based  we can use the zip built-in function as follows    gt  gt  gt   i for i j in zip range len haystack    haystack  if j     needle      where  haystack  is the list in question and  needle  is the item to look for    Note  Here we are iterating using i to get the indexes  but if we need rather to focus on the items we can switch to j

[python] Finding the index of an item in a list

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