How to get item s position in a list

Question

I am iterating over a list and I want to print out the index of the item if it meets a certain condition  How would I do this   Example     testlist    1 2 3 5 3 1 2 1 6  for item in testlist      if item    1          print position

User · Accepted Answer

Hmmm   There was an answer with a list comprehension here  but it s disappeared   Here     i for i x in enumerate testlist  if x    1    Example    gt  gt  gt  testlist  1  2  3  5  3  1  2  1  6   gt  gt  gt   i for i x in enumerate testlist  if x    1   0  5  7    Update   Okay  you want a generator expression  we ll have a generator expression   Here s the list comprehension again  in a for loop    gt  gt  gt  for i in  i for i x in enumerate testlist  if x    1           print i      0 5 7   Now we ll construct a generator      gt  gt  gt   i for i x in enumerate testlist  if x    1   lt generator object at 0x6b508 gt   gt  gt  gt  for i in  i for i x in enumerate testlist  if x    1           print i      0 5 7   and niftily enough  we can assign that to a variable  and use it from there      gt  gt  gt  gen    i for i x in enumerate testlist  if x    1   gt  gt  gt  for i in gen  print i      0 5 7   And to think I used to write FORTRAN

User · Answer

Just to illustrate complete example along with the input list which has searies1  example  input list 0   in which you want to do a lookup of series2  example  input list 1   and get indexes of series2 if it exists in series1  Note  Your certain condition will go in lambda expression if conditions are simple input list     1 2 3 4 5 6 7   1 3 7   series1   input list 0  series2   input list 1  idx list   list map lambda item  series1 index item  if item in series1 else None  series2   print idx list   output   0  2  6

User · Answer

Why complicate things   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

If your list got large enough and you only expected to find the value in a sparse number of indices  consider that this code could execute much faster because you don t have to iterate every value in the list   lookingFor   1 i   0 index   0 try    while i  lt  len testlist       index   testlist index lookingFor i      i   index   1     print index except ValueError   testlist index   cannot find lookingFor   pass   If you expect to find the value a lot you should probably just append  index  to a list and print the list at the end to save time per iteration

User · Answer

for i in xrange len testlist      if testlist i     1      print i   xrange instead of range as requested  see comments

User · Answer

If your list got large enough and you only expected to find the value in a sparse number of indices  consider that this code could execute much faster because you don t have to iterate every value in the list   lookingFor   1 i   0 index   0 try    while i  lt  len testlist       index   testlist index lookingFor i      i   index   1     print index except ValueError   testlist index   cannot find lookingFor   pass   If you expect to find the value a lot you should probably just append  index  to a list and print the list at the end to save time per iteration

User · Answer

I think that it might be useful to use the curselection   method from thte Tkinter library   from Tkinter import    listbox curselection     This method works on Tkinter listbox widgets  so you ll need to construct one of them instead of a list   This will return a position like this     0     although later versions of Tkinter may return a list of ints instead   Which is for the first position and the number will change according to the item position   For more information  see this page  http   effbot org tkinterbook listbox htm  Greetings

User · Answer

for i in xrange len testlist      if testlist i     1      print i   xrange instead of range as requested  see comments

User · Answer

I think that it might be useful to use the curselection   method from thte Tkinter library   from Tkinter import    listbox curselection     This method works on Tkinter listbox widgets  so you ll need to construct one of them instead of a list   This will return a position like this     0     although later versions of Tkinter may return a list of ints instead   Which is for the first position and the number will change according to the item position   For more information  see this page  http   effbot org tkinterbook listbox htm  Greetings

User · Answer

Here is another way to do this   try     id   testlist index  1      print testlist id  except ValueError     print  Not Found

User · Answer

for i in xrange len testlist      if testlist i     1      print i   xrange instead of range as requested  see comments

User · Answer

What about the following   print testlist index element    If you are not sure whether the element to look for is actually in the list  you can add a preliminary check  like  if element in testlist      print testlist index element    or  print testlist index element  if element in testlist else None    or the  pythonic way   which I don t like so much because code is less clear  but sometimes is more efficient   try      print testlist index element  except ValueError      pass

User · Answer

Use enumerate   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

Use enumerate   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

Use enumerate   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

testlist    1 2 3 5 3 1 2 1 6  for id  value in enumerate testlist       if id    1          print testlist id    I guess that it s exacly what you want   -   id  will be always the index of the values on the list

User · Answer

If your list got large enough and you only expected to find the value in a sparse number of indices  consider that this code could execute much faster because you don t have to iterate every value in the list   lookingFor   1 i   0 index   0 try    while i  lt  len testlist       index   testlist index lookingFor i      i   index   1     print index except ValueError   testlist index   cannot find lookingFor   pass   If you expect to find the value a lot you should probably just append  index  to a list and print the list at the end to save time per iteration

User · Answer

Use enumerate   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

x for x in range len testlist   if testlist x   1

User · Answer

Why complicate things   testlist    1 2 3 5 3 1 2 1 6  for position  item in enumerate testlist       if item    1          print position

User · Answer

Try the below   testlist    1 2 3 5 3 1 2 1 6      position 0 for i in testlist     if i    1        print position     position position 1

User · Answer

for i in xrange len testlist      if testlist i     1      print i   xrange instead of range as requested  see comments

User · Answer

Here is another way to do this   try     id   testlist index  1      print testlist id  except ValueError     print  Not Found

User · Answer

If your list got large enough and you only expected to find the value in a sparse number of indices  consider that this code could execute much faster because you don t have to iterate every value in the list   lookingFor   1 i   0 index   0 try    while i  lt  len testlist       index   testlist index lookingFor i      i   index   1     print index except ValueError   testlist index   cannot find lookingFor   pass   If you expect to find the value a lot you should probably just append  index  to a list and print the list at the end to save time per iteration

User · Answer

Try the below   testlist    1 2 3 5 3 1 2 1 6      position 0 for i in testlist     if i    1        print position     position position 1

User · Answer

testlist    1 2 3 5 3 1 2 1 6  for id  value in enumerate testlist       if id    1          print testlist id    I guess that it s exacly what you want   -   id  will be always the index of the values on the list

User · Answer

What about the following   print testlist index element    If you are not sure whether the element to look for is actually in the list  you can add a preliminary check  like  if element in testlist      print testlist index element    or  print testlist index element  if element in testlist else None    or the  pythonic way   which I don t like so much because code is less clear  but sometimes is more efficient   try      print testlist index element  except ValueError      pass

User · Answer

x for x in range len testlist   if testlist x   1

User · Answer

Just to illustrate complete example along with the input list which has searies1  example  input list 0   in which you want to do a lookup of series2  example  input list 1   and get indexes of series2 if it exists in series1  Note  Your certain condition will go in lambda expression if conditions are simple input list     1 2 3 4 5 6 7   1 3 7   series1   input list 0  series2   input list 1  idx list   list map lambda item  series1 index item  if item in series1 else None  series2   print idx list   output   0  2  6

[python] How to get item's position in a list?

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