how to correctly modify the iterator of a loop in python from within the loop

Question

what I basically need is to check every element of a list and if some criteria fit I want to remove it from the list.

So for example let\'s say that

list         


        

Answer 1

If you are ok with creating a copy of the list you can do it like this (list comprehension):

[s for s in list if s != 'b' and s != 'c']

Answer 2

Its better not to reinvent things which are already available. Use filter functions and lambda in these cases. Its more pythonic and looks cleaner.

filter(lambda x:x not in ['b','c'],['a','b','c','d','e'])

alternatively you can use list comprehension

[x for x in ['a','b','c','d','e'] if x not in ['b','c']]

Answer 3

The easier way is to use a copy of the list - it can be done with a slice that extends "from the beginning" to the "end" of the list, like this:

for s in list[:]:
    if s=='b' or s=='c':
        list.remove(s)

You have considered this, and this is simple enough to be in your code, unless this list is really big, and in a critical part of the code (like, in the main loop of an action game). In that case, I sometimes use the following idiom:

to_remove = []
for index, s in enumerate(list):
    if s == "b" or s == "c":
         to_remove.append(index)

for index in reversed(to_remove):
    del list[index]

Of course you can resort to a while loop instead:

index = 0
while index < len(list):
   if s == "b" or s == "c":
       del list[index]
       continue
   index += 1

Answer 4

This is exactly what itertools.ifilter is designed for.

from itertools import ifilter

ifilter(lambda x: x not in ['b', 'c'], ['a', 'b', 'c', 'd', 'e'])

will give you back a generator for your list. If you actually need a list, you can create it using one of the standard techniques for converting a generator to a list:

list(ifilter(lambda x: x not in ['b', 'c'], ['a', 'b', 'c', 'd', 'e']))

or

[x for x in ifilter(lambda x: x not in ['b', 'c'], ['a', 'b', 'c', 'd', 'e'])]