Fastest or most idiomatic way to remove object from list of objects in Python

Fastest or most idiomatic way to remove object from list of objects in Python

Suppose I have a list L of unknown objects, O1 to On, and I want to remove another object reference M which may refer to one of the objects in L, I've managed to do it using:

L = [ O1, O2, ... On]    ...    L = [ j for j in L if j not in [ M ] ]  

which is lovely and idiomatic... but I'm having to do it a lot, and I wonder if there's not another more idiomatic way, or if there is not a faster way.

The important point is that the list of objects is unknown, and may or may not include the object to be excluded. I want to avoid extending or subclassing the objects where possible.

Answer by Francesco for Fastest or most idiomatic way to remove object from list of objects in Python

What about the built in filter function?

>>> l = [1,2,3,4,5]  >>> f = [4]  >>> filter(lambda x: x not in f, l)  [1, 2, 3, 5]  

or in python3

>>> list(filter(lambda x: x not in f, l))  [1, 2, 3, 5]  

Answer by piRSquared for Fastest or most idiomatic way to remove object from list of objects in Python

Use try/except wrapped in a recursive function recursion takes care of potential multiple M's

def tremove(L, M):      try:          L.remove(M)          return tremove(L, M)      except:          return L    tremove(L, M)  

Answer by Muhammad Tahir for Fastest or most idiomatic way to remove object from list of objects in Python

list.remove seems to be the fastest way, with list comprehension as the second fastest and filter at last.

Here are the timeit results

In: python -m timeit '[x for x in [1,2,3,4,5] if x not in [4]]'  Out: 1000000 loops, best of 3: 0.285 usec per loop    In: python -m timeit '[x for x in [1,2,3,4,5] if x != 4]'  Out: 1000000 loops, best of 3: 0.254 usec per loop    In: python -m timeit 'filter(lambda x: x not in [4], [1,2,3,4,5])'  Out: 1000000 loops, best of 3: 0.577 usec per loop    In: python -m timeit 'filter(lambda x: x != 4, [1,2,3,4,5])'  Out: 1000000 loops, best of 3: 0.569 usec per loop    In: python -m timeit '[1,2,3,4,5].remove(4)'  Out: 10000000 loops, best of 3: 0.132 usec per loop  

Answer by MSeifert for Fastest or most idiomatic way to remove object from list of objects in Python

If there are multiple occurences of your value then probably a while-loop with remove is needed:

L = [1,2,3,4,5]  while True:      try:          L.remove(4)      except:          break  

It is a bit slower (due to the exception handling and multiple iterations over the list) than the list comprehension:

[ j for j in L if j != 4 ]  

but both do work fine. If you want to exclude multiple values then you should use the list-comprehension:

M = [1, 4]  [ j for j in L if j not in M ]  

because the try / except will be nested and the list comprehension only needs to traverse the list once.

Answer by timgeb for Fastest or most idiomatic way to remove object from list of objects in Python

Here's an idea that let's you make O(1) containment checks for hashable items. It should be dramatically faster for long lists M with lots of hashables.

class MixedBag(object):      def __init__(self, *args):          self.hashed = set()          self.nothashed = []            for x in args:              self.add(x)        def add(self, x):          try:              self.hashed.add(x)          except TypeError:              self.nothashed.append(x)        def __contains__(self, x):          try:              return x in self.hashed          except TypeError:              return x in self.nothashed      L = [[1,2,3], 4, '5', {6}]  M = [[1,2,3], '5', {4}]    mix = MixedBag(*M)  L = [x for x in L if x not in mix]  print(L) # [4, set([6])]  

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