Remove all occurrences of a value from a list

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In Python remove   will remove the first occurrence of value in a list   How to remove all occurrences of a value from a list   This is what I have in mind    gt  gt  gt  remove values from list  1  2  3  4  2  2  3   2   1  3  4  3

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Repeating the solution of the first post in a more abstract way    gt  gt  gt  x    1  2  3  4  2  2  3   gt  gt  gt  while 2 in x  x remove 2   gt  gt  gt  x  1  3  4  3

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Numpy approach and timings against a list array with 1 000 000 elements   Timings   In  10   a shape Out 10    1000000    In  13   len lst  Out 13   1000000  In  18    timeit a a    2  100 loops  best of 3  2 94 ms per loop  In  19    timeit  x for x in lst if x    2  10 loops  best of 3  79 7 ms per loop   Conclusion  numpy is 27 times faster  on my notebook  compared to list comprehension approach  PS if you want to convert your regular Python list lst to numpy array   arr   np array lst    Setup   import numpy as np a   np random randint 0  1000  10  6   In  10   a shape Out 10    1000000    In  12   lst   a tolist    In  13   len lst  Out 13   1000000   Check   In  14   a a    2  shape Out 14    998949    In  15   len  x for x in lst if x    2   Out 15   998949

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You can use slice assignment if the original list must be modified  while still using an efficient list comprehension  or generator expression     gt  gt  gt  x    1  2  3  4  2  2  3   gt  gt  gt  x       value for value in x if value    2   gt  gt  gt  x  1  3  4  3

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What s wrong with   Motor   1   2   2   For i in Motor         If i      2          Print i  Print motor    Using anaconda

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you can do this while 2 in x         x remove 2   better solution with list comprehension x     i for i in x if i  2

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hello      h    e    l    l    o         w    o    r    l    d    chech every item for a match for item in range len hello -1        if hello item            if there is a match  rebuild the list with the list before the item   the list after the item          hello   hello  item    hello  item   1   print hello     h    e    l    l    o    w    o    r    l    d

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for i in range a count            a remove        Much simpler I believe

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At the cost of readability  I think this version is slightly faster as it doesn t force the while to reexamine the list  thus doing exactly the same work remove has to do anyway   x    1  2  3  4  2  2  3  def remove values from list the list  val       for i in range the list count val            the list remove val   remove values from list x  2   print x

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p  2 3 4 4 4  p clear   print p       Only with Python 3

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You can use a list comprehension   def remove values from list the list  val      return  value for value in the list if value    val   x    1  2  3  4  2  2  3  x   remove values from list x  2  print x    1  3  4  3

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We can also do in-place remove all using either del or pop   import random  def remove values from list lst  target       if type lst     list          return lst      i   0     while i  lt  len lst           if lst i     target              lst pop i     length decreased by 1 already         else              i    1      return lst  remove values from list None  2  remove values from list     2  remove values from list  1  2  3  4  2  2  3   2  lst   remove values from list  random randrange 0  10  for x in range 1000000    2  print len lst       Now for the efficiency   In  21    timeit -n1 -r1 x   random randrange 0 10  1 loop  best of 1  43 5 us per loop  In  22    timeit -n1 -r1 lst    random randrange 0  10  for x in range 1000000   g1 loop  best of 1  660 ms per loop  In  23    timeit -n1 -r1 lst   remove values from list  random randrange 0  10  for x in range 1000000              random randrange 0 10   1 loop  best of 1  11 5 s per loop  In  27    timeit -n1 -r1 x   random randrange 0 10   lst    a for a in  random randrange 0  10  for x in           range 1000000   if x    a  1 loop  best of 1  710 ms per loop    As we see that in-place version remove values from list   does not require any extra memory  but it does take so much more time to run    11 seconds for inplace remove values 710 milli seconds for list comprehensions  which allocates a new list in memory

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See the simple solution   gt  gt  gt   i for i in x if i    2    This will return a list having all elements of x without 2

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a    1  2  2  3  1  to remove   1 a    i for i in a if i    to remove  print a       Perhaps not the most pythonic but still the easiest for me haha

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I believe this is probably faster than any other way if you don t care about the lists order  if you do take care about the final order store the indexes from the original and resort by that   category ids sort   ones last index   category ids count  1   del category ids 0 ones last index

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Let   gt  gt  gt  x    1  2  3  4  2  2  3    The simplest and efficient solution as already posted before is   gt  gt  gt  x       v for v in x if v    2   gt  gt  gt  x  1  3  4  3    Another possibility which should use less memory but be slower is   gt  gt  gt  for i in range len x  - 1  -1  -1           if x i     2              x pop i     takes time   len x  - i  gt  gt  gt  x  1  3  4  3    Timing results for lists of length 1000 and 100000 with 10  matching entries  0 16 vs 0 25 ms  and 23 vs 123 ms

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Functional approach   Python 3 x   gt  gt  gt  x    1 2 3 2 2 2 3 4   gt  gt  gt  list filter  2    ne    x    1  3  3  4    or    gt  gt  gt  x    1 2 3 2 2 2 3 4   gt  gt  gt  list filter lambda a  a    2  x    1  3  3  4    Python 2 x   gt  gt  gt  x    1 2 3 2 2 2 3 4   gt  gt  gt  filter lambda a  a    2  x   1  3  3  4

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Remove all occurrences of a value from a Python list  lists    6 9 7 8 9 3 5 4 9 1 2 9 7 9 12 9 10 9 11 7  def remove values from list        for list in lists        if list  7            print list  remove values from list     Result  6 9  8 9  3  5  4 9  1  2 9  9  12 9  10 9  11  Alternatively   lists    6 9 7 8 9 3 5 4 9 1 2 9 7 9 12 9 10 9 11 7  def remove values from list remove       for list in lists        if list  remove           print list  remove values from list 7    Result  6 9  8 9  3  5  4 9  1  2 9  9  12 9  10 9  11

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About the speed   import time s time   time time    print  start  a   range 100000000  del a    print  finished in  0 2f     time time   - s time    start   finished in 3 25  s time   time time   print  start  a   range 100000000  a      print  finished in  0 2f     time time   - s time    start   finished in 2 11

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To remove all duplicate occurrences and leave one in the list   test    1  1  2  3   newlist   list set test    print newlist   1  2  3    Here is the function I ve used for Project Euler   def removeOccurrences e     return list set e

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No one has posted an optimal answer for time and space complexity  so I thought I would give it a shot  Here is a solution that removes all occurrences of a specific value without creating a new array and at an efficient time complexity  The drawback is that the elements do not maintain order   Time complexity  O n   Additional space complexity  O 1   def main        test case  1  2  3  4  2  2  3   2         1  3  3  4      test case  3  3  3   3                           test case  1  1  1   3                     1  1  1    def test case test val  remove val       remove element in place test val  remove val      print test val    def remove element in place my list  remove value       length my list   len my list      swap idx   length my list - 1      for idx in range length my list - 1  -1  -1           if my list idx     remove value              my list idx   my list swap idx    my list swap idx   my list idx              swap idx -  1      for pop idx in range length my list - swap idx - 1           my list pop     O 1  operation   if   name         main         main

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I just did this for a list  I am just a beginner  A slightly more advanced programmer can surely write a function like this   for i in range len spam        spam remove  cat       if  cat  not in spam           print  All instances of      cat      have been removed            break

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All of the answers above  apart from Martin Andersson s  create a new list without the desired items  rather than removing the items from the original list     gt  gt  gt  import random  timeit  gt  gt  gt  a   list range 5     1000  gt  gt  gt  random shuffle a    gt  gt  gt  b   a  gt  gt  gt  print b is a  True   gt  gt  gt  b    x for x in b if x    0   gt  gt  gt  print b is a  False  gt  gt  gt  b count 0  0  gt  gt  gt  a count 0  1000   gt  gt  gt  b   a  gt  gt  gt  b   filter lambda a  a    2  x   gt  gt  gt  print b is a  False   This can be important if you have other references to the list hanging around   To modify the list in place  use a method like this   gt  gt  gt  def removeall inplace x  l           for   in xrange l count x                l remove x       gt  gt  gt  removeall inplace 0  b   gt  gt  gt  b is a True  gt  gt  gt  a count 0  0   As far as speed is concerned  results on my laptop are  all on a 5000 entry list with 1000 entries removed    List comprehension -  400us Filter -  900us  remove   loop - 50ms   So the  remove loop is about 100x slower         Hmmm  maybe a different approach is needed  The fastest I ve found is using the list comprehension  but then replace the contents of the original list    gt  gt  gt  def removeall replace x  l           t    y for y in l if y    x          del l            l extend t     removeall replace   - 450us

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If you didn t have built-in filter or didn t want to use extra space and you need a linear solution     def remove all A  v       k   0     n   len A      for i in range n           if A i      v              A k    A i              k    1      A   A  k

[python] Remove all occurrences of a value from a list?

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