Deleting multiple elements from a list

Question

Is it possible to delete multiple elements from a list at the same time  If I want to delete elements at index 0 and 2  and try something like del somelist 0   followed by del somelist 2   the second statement will actually delete somelist 3    I suppose I could always delete the higher numbered elements first but I m hoping there is a better way

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some list remove some list max i  j      Avoids sorting cost and having to explicitly copy list

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I can actually think of two ways to do it    slice the list like  this deletes the 1st 3rd and 8th elements   somelist   somelist 1 2  somelist 3 7  somelist 8   do that in place  but one at a time   somelist pop 2  somelist pop 0

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Importing it only for this reason might be overkill  but if you happen to be using pandas anyway  then the solution is simple and straightforward   import pandas as pd stuff   pd Series   a   b   a   c   a   d    less stuff   stuff stuff     a      define any condition here   results   b   c   d

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To generalize the comment from  sth  Item deletion in any class  that implements abc MutableSequence  and in list in particular  is done via   delitem   magic method  This method works similar to   getitem    meaning it can accept either an integer or a slice  Here is an example   class MyList list       def   delitem   self  item           if isinstance item  slice               for i in range  item indices len self                     self i     null          else              self item     null    l   MyList range 10   print l  del l 5 8  print l    This will output   0  1  2  3  4  5  6  7  8  9   0  1  2  3  4   null    null    null   8  9

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Remove method will causes a lot of shift of list elements  I think is better to make a copy       new list      for el in obj my list     if condition is true el         new list append el  del obj my list obj my list   new list

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As a specialisation of Greg s answer  you can even use extended slice syntax  eg  If you wanted to delete items 0 and 2    gt  gt  gt  a   0  1  2  3  4   gt  gt  gt  del a 0 3 2   gt  gt  gt  a  1  3  4    This doesn t cover any arbitrary selection  of course  but it can certainly work for deleting any two items

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I wanted to a way to compare the different solutions that made it easy to turn the knobs   First I generated my data   import random  N   16   1024 x   range N  random shuffle x  y   random sample range N   N   10    Then I defined my functions   def list set value list  index list       index list   set index list      result    value for index  value in enumerate value list  if index not in index list      return result  def list del value list  index list       for index in sorted index list  reverse True           del value list index    def list pop value list  index list       for index in sorted index list  reverse True           value list pop index    Then I used timeit to compare the solutions   import timeit from collections import OrderedDict  M   1000 setup    from   main   import x  y  list set  list del  list pop  statement dict   OrderedDict         overhead     a   x             set    a   x     list set a  y           del    a   x     list del a  y           pop    a   x     list pop a  y         overhead   None result dict   OrderedDict   for name  statement in statement dict iteritems        result   timeit timeit statement  number M  setup setup      if overhead is None          overhead   result     else          result   result - overhead         result dict name    result  for name  result in result dict iteritems        print   s    7 3f     name  result    Output  set     1 711 del     3 450 pop     3 618   So the generator with the indices in a set was the winner  And del is slightly faster then pop

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This has been mentioned  but somehow nobody managed to actually get it right   On O n  solution would be   indices    0  2  somelist    i for j  i in enumerate somelist  if j not in indices    This is really close to SilentGhost s version  but adds two braces

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An alternative list comprehension method that uses list index values   stuff     a    b    c    d    e    f    woof   index    0  3  6  new    i for i in stuff if stuff index i  not in index    This returns     b    c    e    f

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You can do that way on a dict  not on a list  In a list elements are in sequence  In a dict they depend only on the index   Simple code just to explain it by doing    gt  gt  gt  lst     a   b   c    gt  gt  gt  dct    0   a   1   b   2  c    gt  gt  gt  lst 0   a   gt  gt  gt  dct 0   a   gt  gt  gt  del lst 0   gt  gt  gt  del dct 0   gt  gt  gt  lst 0   b   gt  gt  gt  dct 0  Traceback  most recent call last     File   lt pyshell 19 gt    line 1  in  lt module gt      dct 0  KeyError  0  gt  gt  gt  dct 1   b   gt  gt  gt  lst 1   c    A way to  convert  a list in a dict is    gt  gt  gt  dct       gt  gt  gt  for i in xrange 0 len lst    dct i    lst i    The inverse is   lst    dct i  for i in sorted dct keys        Anyway I think it s better to start deleting from the higher index as you said

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If you re deleting multiple non-adjacent items  then what you describe is the best way  and yes  be sure to start from the highest index    If your items are adjacent  you can use the slice assignment syntax   a 2 10

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Importing it only for this reason might be overkill  but if you happen to be using pandas anyway  then the solution is simple and straightforward   import pandas as pd stuff   pd Series   a   b   a   c   a   d    less stuff   stuff stuff     a      define any condition here   results   b   c   d

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None of the answers offered so far performs the deletion in place in O n  on the length of the list for an arbitrary number of indices to delete  so here s my version   def multi delete the list  indices       assert type indices  in  set  frozenset    indices must be a set or frozenset      offset   0     for i in range len the list            if i in indices              offset    1         elif offset              the list i - offset    the list i      if offset          del the list -offset      Example  a    0  1  2  3  4  5  6  7  multi delete a   1  2  4  6  7   print a     prints  0  3  5

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You can do that way on a dict  not on a list  In a list elements are in sequence  In a dict they depend only on the index   Simple code just to explain it by doing    gt  gt  gt  lst     a   b   c    gt  gt  gt  dct    0   a   1   b   2  c    gt  gt  gt  lst 0   a   gt  gt  gt  dct 0   a   gt  gt  gt  del lst 0   gt  gt  gt  del dct 0   gt  gt  gt  lst 0   b   gt  gt  gt  dct 0  Traceback  most recent call last     File   lt pyshell 19 gt    line 1  in  lt module gt      dct 0  KeyError  0  gt  gt  gt  dct 1   b   gt  gt  gt  lst 1   c    A way to  convert  a list in a dict is    gt  gt  gt  dct       gt  gt  gt  for i in xrange 0 len lst    dct i    lst i    The inverse is   lst    dct i  for i in sorted dct keys        Anyway I think it s better to start deleting from the higher index as you said

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For the indices 0 and 2 from listA   for x in  2 0   listA pop x    For some random indices to remove from listA    indices  5 3 2 7 0   for x in sorted indices    -1   listA pop x

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Another implementation of the idea of removing from the highest index   for i in range len yourlist -1  -1  -1       del yourlist i

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I can actually think of two ways to do it    slice the list like  this deletes the 1st 3rd and 8th elements   somelist   somelist 1 2  somelist 3 7  somelist 8   do that in place  but one at a time   somelist pop 2  somelist pop 0

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Remove method will causes a lot of shift of list elements  I think is better to make a copy       new list      for el in obj my list     if condition is true el         new list append el  del obj my list obj my list   new list

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You can use enumerate and remove the values whose index matches the indices you want to remove  indices   0  2 somelist    i for j  i in enumerate somelist  if j not in indices

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You can use numpy delete as follows   import numpy as np a     a    l   3 14  42   u   I    0  2  np delete a  I  tolist     Returns    l    42    u     If you don t mind ending up with a numpy array at the end  you can leave out the  tolist    You should see some pretty major speed improvements  too  making this a more scalable solution  I haven t benchmarked it  but numpy operations are compiled code written in either C or Fortran

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You can use remove  too   delete from somelist      for i in  int 0   int 2         delete from somelist append somelist i   for j in delete from somelist       newlist   somelist remove j

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You can use enumerate and remove the values whose index matches the indices you want to remove  indices   0  2 somelist    i for j  i in enumerate somelist  if j not in indices

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technically  the answer is NO it is not possible to delete two objects AT THE SAME TIME  However  it IS possible to delete two objects in one line of beautiful python   del  foo  bar   foo  baz      will recusrively delete foo  bar    then foo  baz

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If you re deleting multiple non-adjacent items  then what you describe is the best way  and yes  be sure to start from the highest index    If your items are adjacent  you can use the slice assignment syntax   a 2 10

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As a function   def multi delete list    args       indexes   sorted list args   reverse True      for index in indexes          del list  index      return list    Runs in n log n  time  which should make it the fastest correct solution yet

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As a specialisation of Greg s answer  you can even use extended slice syntax  eg  If you wanted to delete items 0 and 2    gt  gt  gt  a   0  1  2  3  4   gt  gt  gt  del a 0 3 2   gt  gt  gt  a  1  3  4    This doesn t cover any arbitrary selection  of course  but it can certainly work for deleting any two items

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So  you essentially want to delete multiple elements in one pass   In that case  the position of the next element to delete will be offset by however many were deleted previously   Our goal is to delete all the vowels  which are precomputed to be indices 1  4  and 7   Note that its important the to delete indices are in ascending order  otherwise it won t work   to delete    1  4  7  target   list  hello world   for offset  index in enumerate to delete     index -  offset   del target index    It d be a more complicated if you wanted to delete the elements in any order   IMO  sorting to delete might be easier than figuring out when you should or shouldn t subtract from index

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Here is an alternative  that does not use enumerate   to create tuples  as in SilentGhost s original answer    This seems more readable to me   Maybe I d feel differently if I was in the habit of using enumerate   CAVEAT  I have not tested performance of the two approaches     Returns a new list   lst  is not modified  def delete by indices lst  indices       indices as set   set indices      return   lst i  for i in xrange len lst   if i not in indices as set     NOTE  Python 2 7 syntax  For Python 3  xrange    range   Usage   lst     11 x for x in xrange 10    somelist   delete by indices  lst   0  4  5     somelist    11  22  33  66  77  88  99      --- BONUS ---  Delete multiple values from a list  That is  we have the values we want to delete     Returns a new list   lst  is not modified  def delete  by values lst  values       values as set   set values      return   x for x in lst if x not in values as set     Usage   somelist   delete  by values  lst   0  44  55      somelist    11  22  33  66  77  88  99    This is the same answer as before  but this time we supplied the VALUES to be deleted  0  44  55

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You can use enumerate and remove the values whose index matches the indices you want to remove  indices   0  2 somelist    i for j  i in enumerate somelist  if j not in indices

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I m a total beginner in Python  and my programming at the moment is crude and dirty to say the least  but my solution was to use a combination of the basic commands I learnt in early tutorials   some list    1 2 3 4 5 6 7 8 10  rem    0 5 7   for i in rem      some list i          mark for deletion  for i in range 0  some list count            some list remove        remove print some list   Obviously  because of having to choose a  mark-for-deletion  character  this has its limitations    As for the performance as the size of the list scales  I m sure that my solution is sub-optimal   However  it s straightforward  which I hope appeals to other beginners  and will work in simple cases where some list is of a well-known format  e g   always numeric

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How about one of these  I m very new to Python  but they seem ok    ocean basin     a    Atlantic    Pacific    Indian    a    a    a   for i in range 1   ocean basin count  a     1        ocean basin remove  a   print ocean basin      Atlantic    Pacific    Indian    ob     a    b   4  5  Atlantic    Pacific    Indian    a    a   4   a   remove     a    b   4  5  ob    i for i in ob if i not in  remove   print ob      Atlantic    Pacific    Indian

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For some reason I don t like any of the answers here  Yes  they work  but strictly speaking most of them aren t deleting elements in a list  are they   But making a copy and then replacing the original one with the edited copy    Why not just delete the higher index first   Is there a reason for this  I would just do   for i in sorted indices  reverse True       del somelist i    If you really don t want to delete items backwards  then I guess you should just deincrement the indices values which are greater than the last deleted index  can t really use the same index since you re having a different list  or use a copy of the list  which wouldn t be  deleting  but replacing the original with an edited copy    Am I missing something here  any reason to NOT delete in the reverse order

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As a specialisation of Greg s answer  you can even use extended slice syntax  eg  If you wanted to delete items 0 and 2    gt  gt  gt  a   0  1  2  3  4   gt  gt  gt  del a 0 3 2   gt  gt  gt  a  1  3  4    This doesn t cover any arbitrary selection  of course  but it can certainly work for deleting any two items

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You can use this logic   my list     word   yes   no   nice    c  b for i b in enumerate my list  if not i in  0 2 3    print c

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To generalize the comment from  sth  Item deletion in any class  that implements abc MutableSequence  and in list in particular  is done via   delitem   magic method  This method works similar to   getitem    meaning it can accept either an integer or a slice  Here is an example   class MyList list       def   delitem   self  item           if isinstance item  slice               for i in range  item indices len self                     self i     null          else              self item     null    l   MyList range 10   print l  del l 5 8  print l    This will output   0  1  2  3  4  5  6  7  8  9   0  1  2  3  4   null    null    null   8  9

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You can use remove  too   delete from somelist      for i in  int 0   int 2         delete from somelist append somelist i   for j in delete from somelist       newlist   somelist remove j

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here is another method which removes the elements in place  also if your list is really long  it is faster    gt  gt  gt  a   range 10   gt  gt  gt  remove    0 4 5   gt  gt  gt  from collections import deque  gt  gt  gt  deque  list pop a  i  for i in sorted remove  reverse True    maxlen 0    gt  gt  gt  timeit timeit   i for j  i in enumerate a  if j not in remove    setup  import random remove  random randrange 100000  for i in range 100    a   range 100000    number 1  0 1704120635986328   gt  gt  gt  timeit timeit  deque  list pop a  i  for i in sorted remove  reverse True    maxlen 0    setup  from collections import deque import random remove  random randrange 100000  for i in range 100    a   range 100000    number 1  0 004853963851928711

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So  you essentially want to delete multiple elements in one pass   In that case  the position of the next element to delete will be offset by however many were deleted previously   Our goal is to delete all the vowels  which are precomputed to be indices 1  4  and 7   Note that its important the to delete indices are in ascending order  otherwise it won t work   to delete    1  4  7  target   list  hello world   for offset  index in enumerate to delete     index -  offset   del target index    It d be a more complicated if you wanted to delete the elements in any order   IMO  sorting to delete might be easier than figuring out when you should or shouldn t subtract from index

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As a function   def multi delete list    args       indexes   sorted list args   reverse True      for index in indexes          del list  index      return list    Runs in n log n  time  which should make it the fastest correct solution yet

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As a specialisation of Greg s answer  you can even use extended slice syntax  eg  If you wanted to delete items 0 and 2    gt  gt  gt  a   0  1  2  3  4   gt  gt  gt  del a 0 3 2   gt  gt  gt  a  1  3  4    This doesn t cover any arbitrary selection  of course  but it can certainly work for deleting any two items

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You can use enumerate and remove the values whose index matches the indices you want to remove  indices   0  2 somelist    i for j  i in enumerate somelist  if j not in indices

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l     a   b   a   c   a   d   to remove    1  3   l i  for i in range 0  len l   if i not in to remove     It s basically the same as the top voted answer  just a different way of writing it  Note that using l index   is not a good idea  because it can t handle duplicated elements in a list

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You may want to simply use np delete  list indices    0  2  original list    0  1  2  3  new list   np delete original list  list indices   Output array  1  3    Here  the first argument is the original list  the second is the index or a list of indices you want to delete  There is a third argument which you can use in the case of having ndarrays  axis  0 for rows and 1 for columns in case of ndarrays

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You can do that way on a dict  not on a list  In a list elements are in sequence  In a dict they depend only on the index   Simple code just to explain it by doing    gt  gt  gt  lst     a   b   c    gt  gt  gt  dct    0   a   1   b   2  c    gt  gt  gt  lst 0   a   gt  gt  gt  dct 0   a   gt  gt  gt  del lst 0   gt  gt  gt  del dct 0   gt  gt  gt  lst 0   b   gt  gt  gt  dct 0  Traceback  most recent call last     File   lt pyshell 19 gt    line 1  in  lt module gt      dct 0  KeyError  0  gt  gt  gt  dct 1   b   gt  gt  gt  lst 1   c    A way to  convert  a list in a dict is    gt  gt  gt  dct       gt  gt  gt  for i in xrange 0 len lst    dct i    lst i    The inverse is   lst    dct i  for i in sorted dct keys        Anyway I think it s better to start deleting from the higher index as you said

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l     a   b   a   c   a   d   to remove    1  3   l i  for i in range 0  len l   if i not in to remove     It s basically the same as the top voted answer  just a different way of writing it  Note that using l index   is not a good idea  because it can t handle duplicated elements in a list

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An alternative list comprehension method that uses list index values   stuff     a    b    c    d    e    f    woof   index    0  3  6  new    i for i in stuff if stuff index i  not in index    This returns     b    c    e    f

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How about one of these  I m very new to Python  but they seem ok    ocean basin     a    Atlantic    Pacific    Indian    a    a    a   for i in range 1   ocean basin count  a     1        ocean basin remove  a   print ocean basin      Atlantic    Pacific    Indian    ob     a    b   4  5  Atlantic    Pacific    Indian    a    a   4   a   remove     a    b   4  5  ob    i for i in ob if i not in  remove   print ob      Atlantic    Pacific    Indian

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I put it all together into a list diff function that simply takes two lists as inputs and returns their difference  while preserving the original order of the first list   def list diff list a  list b  verbose False          returns a difference of list a and list b        preserving the original order  unlike set-based solutions        get indices of elements to be excluded from list a     excl ind    i for i  x in enumerate list a  if x in list b      if verbose          print excl ind         filter out the excluded indices  producing a new list      new list    i for i in list a if list a index i  not in excl ind      if verbose          print new list       return new list    Sample usage   my list     a    b    c    d    e    f    woof     index    0  3  6     define excluded names list excl names list     woof    c    list diff my list  excl names list   gt  gt    a    b    d    e    f

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I m a total beginner in Python  and my programming at the moment is crude and dirty to say the least  but my solution was to use a combination of the basic commands I learnt in early tutorials   some list    1 2 3 4 5 6 7 8 10  rem    0 5 7   for i in rem      some list i          mark for deletion  for i in range 0  some list count            some list remove        remove print some list   Obviously  because of having to choose a  mark-for-deletion  character  this has its limitations    As for the performance as the size of the list scales  I m sure that my solution is sub-optimal   However  it s straightforward  which I hope appeals to other beginners  and will work in simple cases where some list is of a well-known format  e g   always numeric

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I can actually think of two ways to do it    slice the list like  this deletes the 1st 3rd and 8th elements   somelist   somelist 1 2  somelist 3 7  somelist 8   do that in place  but one at a time   somelist pop 2  somelist pop 0

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If you re deleting multiple non-adjacent items  then what you describe is the best way  and yes  be sure to start from the highest index    If your items are adjacent  you can use the slice assignment syntax   a 2 10

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I can actually think of two ways to do it    slice the list like  this deletes the 1st 3rd and 8th elements   somelist   somelist 1 2  somelist 3 7  somelist 8   do that in place  but one at a time   somelist pop 2  somelist pop 0

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You can use this logic   my list     word   yes   no   nice    c  b for i b in enumerate my list  if not i in  0 2 3    print c

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I put it all together into a list diff function that simply takes two lists as inputs and returns their difference  while preserving the original order of the first list   def list diff list a  list b  verbose False          returns a difference of list a and list b        preserving the original order  unlike set-based solutions        get indices of elements to be excluded from list a     excl ind    i for i  x in enumerate list a  if x in list b      if verbose          print excl ind         filter out the excluded indices  producing a new list      new list    i for i in list a if list a index i  not in excl ind      if verbose          print new list       return new list    Sample usage   my list     a    b    c    d    e    f    woof     index    0  3  6     define excluded names list excl names list     woof    c    list diff my list  excl names list   gt  gt    a    b    d    e    f

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You may want to simply use np delete  list indices    0  2  original list    0  1  2  3  new list   np delete original list  list indices   Output array  1  3    Here  the first argument is the original list  the second is the index or a list of indices you want to delete  There is a third argument which you can use in the case of having ndarrays  axis  0 for rows and 1 for columns in case of ndarrays

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None of the answers offered so far performs the deletion in place in O n  on the length of the list for an arbitrary number of indices to delete  so here s my version   def multi delete the list  indices       assert type indices  in  set  frozenset    indices must be a set or frozenset      offset   0     for i in range len the list            if i in indices              offset    1         elif offset              the list i - offset    the list i      if offset          del the list -offset      Example  a    0  1  2  3  4  5  6  7  multi delete a   1  2  4  6  7   print a     prints  0  3  5

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Another implementation of the idea of removing from the highest index   for i in range len yourlist -1  -1  -1       del yourlist i

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Here is an alternative  that does not use enumerate   to create tuples  as in SilentGhost s original answer    This seems more readable to me   Maybe I d feel differently if I was in the habit of using enumerate   CAVEAT  I have not tested performance of the two approaches     Returns a new list   lst  is not modified  def delete by indices lst  indices       indices as set   set indices      return   lst i  for i in xrange len lst   if i not in indices as set     NOTE  Python 2 7 syntax  For Python 3  xrange    range   Usage   lst     11 x for x in xrange 10    somelist   delete by indices  lst   0  4  5     somelist    11  22  33  66  77  88  99      --- BONUS ---  Delete multiple values from a list  That is  we have the values we want to delete     Returns a new list   lst  is not modified  def delete  by values lst  values       values as set   set values      return   x for x in lst if x not in values as set     Usage   somelist   delete  by values  lst   0  44  55      somelist    11  22  33  66  77  88  99    This is the same answer as before  but this time we supplied the VALUES to be deleted  0  44  55

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This has been mentioned  but somehow nobody managed to actually get it right   On O n  solution would be   indices    0  2  somelist    i for j  i in enumerate somelist  if j not in indices    This is really close to SilentGhost s version  but adds two braces

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technically  the answer is NO it is not possible to delete two objects AT THE SAME TIME  However  it IS possible to delete two objects in one line of beautiful python   del  foo  bar   foo  baz      will recusrively delete foo  bar    then foo  baz

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So  you essentially want to delete multiple elements in one pass   In that case  the position of the next element to delete will be offset by however many were deleted previously   Our goal is to delete all the vowels  which are precomputed to be indices 1  4  and 7   Note that its important the to delete indices are in ascending order  otherwise it won t work   to delete    1  4  7  target   list  hello world   for offset  index in enumerate to delete     index -  offset   del target index    It d be a more complicated if you wanted to delete the elements in any order   IMO  sorting to delete might be easier than figuring out when you should or shouldn t subtract from index

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I wanted to a way to compare the different solutions that made it easy to turn the knobs   First I generated my data   import random  N   16   1024 x   range N  random shuffle x  y   random sample range N   N   10    Then I defined my functions   def list set value list  index list       index list   set index list      result    value for index  value in enumerate value list  if index not in index list      return result  def list del value list  index list       for index in sorted index list  reverse True           del value list index    def list pop value list  index list       for index in sorted index list  reverse True           value list pop index    Then I used timeit to compare the solutions   import timeit from collections import OrderedDict  M   1000 setup    from   main   import x  y  list set  list del  list pop  statement dict   OrderedDict         overhead     a   x             set    a   x     list set a  y           del    a   x     list del a  y           pop    a   x     list pop a  y         overhead   None result dict   OrderedDict   for name  statement in statement dict iteritems        result   timeit timeit statement  number M  setup setup      if overhead is None          overhead   result     else          result   result - overhead         result dict name    result  for name  result in result dict iteritems        print   s    7 3f     name  result    Output  set     1 711 del     3 450 pop     3 618   So the generator with the indices in a set was the winner  And del is slightly faster then pop

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we can do this by use of a for loop iterating over the indexes after sorting the index list in descending order  mylist  66 25  333  1  4  6  7  8  56  8769  65  indexes   4 6 indexes   sorted indexes  reverse True  for i in index      mylist pop i  print mylist

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some list remove some list max i  j      Avoids sorting cost and having to explicitly copy list

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As a function   def multi delete list    args       indexes   sorted list args   reverse True      for index in indexes          del list  index      return list    Runs in n log n  time  which should make it the fastest correct solution yet

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So  you essentially want to delete multiple elements in one pass   In that case  the position of the next element to delete will be offset by however many were deleted previously   Our goal is to delete all the vowels  which are precomputed to be indices 1  4  and 7   Note that its important the to delete indices are in ascending order  otherwise it won t work   to delete    1  4  7  target   list  hello world   for offset  index in enumerate to delete     index -  offset   del target index    It d be a more complicated if you wanted to delete the elements in any order   IMO  sorting to delete might be easier than figuring out when you should or shouldn t subtract from index

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we can do this by use of a for loop iterating over the indexes after sorting the index list in descending order  mylist  66 25  333  1  4  6  7  8  56  8769  65  indexes   4 6 indexes   sorted indexes  reverse True  for i in index      mylist pop i  print mylist

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If you re deleting multiple non-adjacent items  then what you describe is the best way  and yes  be sure to start from the highest index    If your items are adjacent  you can use the slice assignment syntax   a 2 10

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You can use numpy delete as follows   import numpy as np a     a    l   3 14  42   u   I    0  2  np delete a  I  tolist     Returns    l    42    u     If you don t mind ending up with a numpy array at the end  you can leave out the  tolist    You should see some pretty major speed improvements  too  making this a more scalable solution  I haven t benchmarked it  but numpy operations are compiled code written in either C or Fortran

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As a function   def multi delete list    args       indexes   sorted list args   reverse True      for index in indexes          del list  index      return list    Runs in n log n  time  which should make it the fastest correct solution yet

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For some reason I don t like any of the answers here  Yes  they work  but strictly speaking most of them aren t deleting elements in a list  are they   But making a copy and then replacing the original one with the edited copy    Why not just delete the higher index first   Is there a reason for this  I would just do   for i in sorted indices  reverse True       del somelist i    If you really don t want to delete items backwards  then I guess you should just deincrement the indices values which are greater than the last deleted index  can t really use the same index since you re having a different list  or use a copy of the list  which wouldn t be  deleting  but replacing the original with an edited copy    Am I missing something here  any reason to NOT delete in the reverse order

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For the indices 0 and 2 from listA   for x in  2 0   listA pop x    For some random indices to remove from listA    indices  5 3 2 7 0   for x in sorted indices    -1   listA pop x

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here is another method which removes the elements in place  also if your list is really long  it is faster    gt  gt  gt  a   range 10   gt  gt  gt  remove    0 4 5   gt  gt  gt  from collections import deque  gt  gt  gt  deque  list pop a  i  for i in sorted remove  reverse True    maxlen 0    gt  gt  gt  timeit timeit   i for j  i in enumerate a  if j not in remove    setup  import random remove  random randrange 100000  for i in range 100    a   range 100000    number 1  0 1704120635986328   gt  gt  gt  timeit timeit  deque  list pop a  i  for i in sorted remove  reverse True    maxlen 0    setup  from collections import deque import random remove  random randrange 100000  for i in range 100    a   range 100000    number 1  0 004853963851928711

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You can do that way on a dict  not on a list  In a list elements are in sequence  In a dict they depend only on the index   Simple code just to explain it by doing    gt  gt  gt  lst     a   b   c    gt  gt  gt  dct    0   a   1   b   2  c    gt  gt  gt  lst 0   a   gt  gt  gt  dct 0   a   gt  gt  gt  del lst 0   gt  gt  gt  del dct 0   gt  gt  gt  lst 0   b   gt  gt  gt  dct 0  Traceback  most recent call last     File   lt pyshell 19 gt    line 1  in  lt module gt      dct 0  KeyError  0  gt  gt  gt  dct 1   b   gt  gt  gt  lst 1   c    A way to  convert  a list in a dict is    gt  gt  gt  dct       gt  gt  gt  for i in xrange 0 len lst    dct i    lst i    The inverse is   lst    dct i  for i in sorted dct keys        Anyway I think it s better to start deleting from the higher index as you said

[python] Deleting multiple elements from a list

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