How to find all occurrences of an element in a list

Question

index   will give the first occurrence of an item in a list  Is there a neat trick which returns all indices in a list for an element

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If you are using Python 2  you can achieve the same functionality with this   f   lambda my list  value filter lambda x  my list x     value  range len my list      Where my list is the list you want to get the indexes of  and value is the value searched  Usage   f some list  some element

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Using a for-loop   Answers with enumerate and a list comprehension are more pythonic  not necessarily faster  however  this answer is aimed at students who may not be allowed to use some of those built-in functions  create an empty list  indices create the loop with for i in range len x     which essentially iterates through a list of index locations  0  1  2  3       len x -1  in the loop  add any i  where x i  is a match to value  to indices  x i  accesses the list by index    def get indices x  list  value  int  - gt  list      indices   list       for i in range len x            if x i     value              indices append i      return indices  n    1  2  3  -50  -60  0  6  9  -60  -60  print get indices n  -60     gt  gt  gt   4  8  9    The functions  get indices  are implemented with type hints   In this case  the list  n  is a bunch of ints  therefore we search for value  also defined as an int    Using a while-loop and  index   With  index  use try-except for error handling  because a ValueError will occur if value is not in the list   def get indices x  list  value  int  - gt  list      indices   list       i   0     while True          try                find an occurrence of value and update i to that index             i   x index value  i                add i to the list             indices append i                advance i by 1             i    1         except ValueError as e              break     return indices  print get indices n  -60    gt  gt  gt   4  8  9

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One more solution sorry if duplicates  for all occurrences   values    1 2 3 1 2 4 5 6 3 2 1  map lambda val   val   i for i in xrange len values   if values i     val    values

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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

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There is an answer using np where to find the indices of a single value This solution will use np where and np unique to find the indices of all unique elements in the list  Converting the list to an array  and using np where is 6 8x faster than any list-comprehension for finding all indices of a single element  Faster solutions using numpy can be found in Get a list of all indices of repeated elements in a numpy array  import numpy as np import random    to create test list    create sample list random seed 365  l    random choice   s1    s2    s3    s4    for   in range 20      convert the list to an array for use with these numpy methods a   np array l     create a dict of each unique entry and the associated indices idx    v  np where a    v  0  tolist   for v in np unique a      print idx    s1    7  9  10  11  17     s2    1  3  6  8  14  18  19     s3    0  2  13  16     s4    4  5  12  15      find a single element with  idx   np where a     s1    print idx   out    array   7   9  10  11  17   dtype int64      timeit  Find indices of a single element in a 2M element list with 4 unique elements    create 2M element list random seed 365  l    random choice   s1    s2    s3    s4    for   in range 2000000      create array a   np array l     np where  timeit np where a     s1    out   25 9 ms    827   s per loop  mean    std  dev  of 7 runs  10 loops each     list-comprehension  timeit  i for i  x in enumerate l  if x     quot s1 quot    out   175 ms    2 73 ms per loop  mean    std  dev  of 7 runs  10 loops each

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more itertools locate finds indices for all items that satisfy a condition     from more itertools import locate   list locate  0  1  1  0  1  0  0       1  2  4   list locate   a    b    c    b    lambda x  x     b       1  3    more itertools is a third-party library  gt  pip install more itertools

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If you need to search for all element s positions between certain indices  you can state them    i for i x in enumerate  1 2 3 2   if x  2  amp  2 lt   i  lt  3    - gt   3

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A solution using list index   def indices lst  element       result          offset   -1     while True          try              offset   lst index element  offset 1          except ValueError              return result         result append offset    It s much faster than the list comprehension with enumerate  for large lists  It is also much slower than the numpy solution if you already have the array  otherwise the cost of converting outweighs the speed gain  tested on integer lists with 100  1000 and 10000 elements    NOTE  A note of caution based on Chris Rands  comment  this solution is faster than the list comprehension if the results are sufficiently sparse  but if the list has many instances of the element that is being searched  more than  15  of the list  on a test with a list of 1000 integers   the list comprehension is faster

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You can use a list comprehension   indices    i for i  x in enumerate my list  if x     whatever

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While not a solution for lists directly  numpy really shines for this sort of thing   import numpy as np values   np array  1 2 3 1 2 4 5 6 3 2 1   searchval   3 ii   np where values    searchval  0    returns   ii    gt array  2  8     This can be significantly faster for lists  arrays  with a large number of elements vs some of the other solutions

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You can create a defaultdict  from collections import defaultdict d1   defaultdict int         defaults to 0 values for keys unq   set lst1                 lst1    1  2  2  3  4  1  2  7  for each in unq        d1 each    lst1 count each  else        print d1

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Here is a time performance comparison between using np where vs list comprehension  Seems like np where is faster on average    np where start times      end times      for i in range 10000       start   time time       start times append start      temp list   np array  1 2 3 3 5       ixs   np where temp list  3  0  tolist       end   time time       end times append end  print  quot Took on average    seconds quot  format      np mean end times -np mean start times     Took on average 3 81469726562e-06 seconds    list comprehension start times      end times      for i in range 10000       start   time time       start times append start      temp list   np array  1 2 3 3 5       ixs    i for i in range len temp list   if temp list i   3      end   time time       end times append end  print  quot Took on average    seconds quot  format      np mean end times -np mean start times     Took on average 4 05311584473e-06 seconds

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Or Use range  python 3    l  i for i in range len lst   if lst i    something        For  python 2    l  i for i in xrange len lst   if lst i    something        And then  both cases    print l    Is as expected

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occurrences   lambda s  lst   i for i e in enumerate lst  if e    s  list occurrences 1   1 2 3 1         0  3

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How about   In  1   l  1 2 3 4 3 2 5 6 7   In  2    i for i val in enumerate l  if val  3  Out 2    2  4

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Using filter   in python2    gt  gt  gt  q     Yeehaw    Yeehaw    Googol    B9    Googol    NSM    B9    NSM    Dont Ask    Googol    gt  gt  gt  filter lambda i  q i    Googol   range len q     2  4  9

[python] How to find all occurrences of an element in a list

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