remove a sublist from a list in python

Question

Hi I am looking for a way to remove a sublist from a list. something like this:

a=range(1,10)  
a.remove([2,3,7])  
print a  
a=[1,4,5,6,8,9]  

Answer 1

>>> a=range(1,10)
>>> [x for x in a if x not in [2,3,7]]
[1, 4, 5, 6, 8, 9]

Answer 2

a=range(1,10)
b = filter(lambda x:x not in set([2,3,7]), a)
print b
b = [1, 4, 5, 6, 8, 9]

update:

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]

Answer 3

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.

Answer 4

The simplest way is

>>> a=range(1,10)
>>> for x in [2,3,7]:
...  a.remove(x)
... 
>>> 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

>>> removeset = set([2,3,7])
>>> 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

>>> removeset = set([2,3,7])
>>> a[:]=[x for x in a if x not in removeset]

Answer 5

>>> a=range(1,10)
>>> for i in [2,3,7]: a.remove(i)
...
>>> a
[1, 4, 5, 6, 8, 9]

>>> a=range(1,10)
>>> b=map(a.remove,[2,3,7])
>>> a
[1, 4, 5, 6, 8, 9]