Python: Removing list element while iterating over list [duplicate]

Answer 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)

Answer 2

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

Answer 3

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

Answer 4

List comp:

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

Answer 5

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

Answer 6

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