Python: Removing list element while iterating over list

I’m iterating over a list of elements in Python, do some action on it, and then remove them if they meet certain criteria.

for element in somelist:
    do_action(element)
    if check(element):
        remove_element_from_list

What should I use in place of remove_element?
I have seen similar questions asked, but notice the presence of the do_action part that is to be executed for all elements and thus eliminates the solution of using filters.

Answers:

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Method 1

You could always iterate over a copy of the list, leaving you free to modify the original:

for item in list(somelist):
  ...
  somelist.remove(item)

Method 2

To meet these criteria: modify original list in situ, no list copies, only one pass, works, a traditional solution is to iterate backwards:

for i in xrange(len(somelist) - 1, -1, -1):
    element = somelist[i]
    do_action(element)
    if check(element):
        del somelist[i]

Bonus: Doesn’t do len(somelist) on each iteration. Works on any version of Python (at least as far back as 1.5.2) … s/xrange/range/ for 3.X.

Update: If you want to iterate forwards, it’s possible, just trickier and uglier:

i = 0
n = len(somelist)
while i < n:
    element = somelist[i]
    do_action(element)
    if check(element):
        del somelist[i]
        n = n - 1
    else:
        i = i + 1

Method 3

List comp:

results = [x for x in (do_action(element) for element in somelist) if check(element)]

Method 4

for element in somelist:
    do_action(element)
somelist[:] = (x for x in somelist if not check(x))

If you really need to do it in one pass without copying the list

i=0
while i < len(somelist):
    element = somelist[i] 
    do_action(element)
    if check(element):
        del somelist[i]
    else:
        i+=1

Method 5

You can still use filter, moving to an outside function the element modification (iterating just once)

def do_the_magic(x):
    do_action(x)
    return check(x)

# you can get a different filtered list
filter(do_the_magic,yourList)

# or have it modified in place (as suggested by Steven Rumbalski, see comment)
yourList[:] = itertools.ifilter(do_the_magic, yourList)

Method 6

Another way of doing so is:

while i<len(your_list):
    if #condition :
        del your_list[i]
    else:
        i+=1

So, you delete the elements side by side while checking

Method 7

You can make a generator that returns everything that isn’t removed:

def newlist(somelist):
    for element in somelist:
        do_action(element)
        if not check(element):
            yield element

Method 8

Why not rewrite it to be

for element in somelist: 
   do_action(element)  

if check(element): 
    remove_element_from_list

See this question for how to remove from the list, though it looks like you’ve already seen that
Remove items from a list while iterating

Another option is to do this if you really want to keep this the same

newlist = [] 
for element in somelist: 
   do_action(element)  

   if not check(element): 
      newlst.append(element)

Method 9

Not exactly in-place, but some idea to do it:

a = ['a', 'b']

def inplace(a):
    c = []
    while len(a) > 0:
        e = a.pop(0)
        if e == 'b':
            c.append(e)
    a.extend(c)

You can extend the function to call you filter in the condition.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0