Remove all values within one list from another list

Question

I am looking for a way to remove all values within a list from another list   Something like this   a   range 1 10    a remove  2 3 7     print a   a    1 4 5 6 8 9

User · Answer

a   range 1 10  itemsToRemove   set  2  3  7   b   filter lambda x  x not in itemsToRemove  a    or  b    x for x in a if x not in itemsToRemove    Don t create the set inside the lambda or inside the comprehension  If you do  it ll be recreated on every iteration  defeating the point of using a set at all

User · Answer

gt  gt  gt  a range 1 10   gt  gt  gt  for i in  2 3 7   a remove i       gt  gt  gt  a  1  4  5  6  8  9    gt  gt  gt  a range 1 10   gt  gt  gt  b map a remove  2 3 7    gt  gt  gt  a  1  4  5  6  8  9

User · Answer

gt  gt  gt  a   range 1  10   gt  gt  gt   x for x in a if x not in  2  3  7    1  4  5  6  8  9

User · Answer

If you don t have repeated values  you could use set difference   x   set range 10   y   x - set  2  3  7     y   set  0  1  4  5  6  8  9     and then convert back to list  if needed

User · Answer

I was looking for fast way to do the subject  so I made some experiments with suggested ways  And I was surprised by results  so I want to share it with you   Experiments were done using pythonbenchmark tool and with   a   range 1 50000    Source list b   range 1 15000    Items to remove   Results    def comprehension a  b        return  x for x in a if x not in b    5 tries  average time 12 8 sec  def filter function a  b       return filter lambda x  x not in b  a    5 tries  average time 12 6 sec  def modification a b       for x in b          try              a remove x          except ValueError              pass     return a   5 tries  average time 0 27 sec  def set approach a b       return list set a -set b     5 tries  average time 0 0057 sec  Also I made another measurement with bigger inputs size for the last two functions  a   range 1 500000  b   range 1 100000    And the results   For modification  remove method  - average time is 252 seconds For set approach - average time is 0 75 seconds  So you can see that approach with sets is significantly faster than others  Yes  it doesn t keep similar items  but if you don t need it  - it s for you  And there is almost no difference between list comprehension and using filter function  Using  remove  is  50 times faster  but it modifies source list  And the best choice is using sets - it s more than 1000 times faster than list comprehension

User · Answer

The simplest way is   gt  gt  gt  a   range 1  10   gt  gt  gt  for x in  2  3  7        a remove x        gt  gt  gt  a  1  4  5  6  8  9    One possible problem here is that each time you call remove    all the items are shuffled down the list to fill the hole  So if a grows very large this will end up being quite slow   This way builds a brand new list  The advantage is that we avoid all the shuffling of the first approach   gt  gt  gt  removeset   set  2  3  7    gt  gt  gt  a    x for x in a if x not in removeset    If you want to modify a in place  just one small change is required   gt  gt  gt  removeset   set  2  3  7    gt  gt  gt  a       x for x in a if x not in removeset

User · Answer

Others have suggested ways to make newlist after filtering e g   newl    x for x in l if x not in  2 3 7     or  newl   filter lambda x  x not in  2 3 7   l     but from your question it looks you want in-place modification for that you can do this  this will also be much much faster if original list is long and items to be removed less  l   range 1 10  for o in set  2 3 7 11        try          l remove o      except ValueError          pass  print l   output       1  4  5  6  8  9   I am checking for ValueError exception so it works even if items are not in orginal list   Also if you do not need in-place modification solution by S Mark is simpler

[python] Remove all values within one list from another list?

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