How to check if all elements of a list match a condition?

Question

I have a list that contains many sub-lists of 3 elements each, like:

my_list = [["a", "b", 0], ["c", "d", 0], ["e", "f", 0], .....]

The last element of each sub-list is a sort of flag, which is initially 0 for each sub-list. As my algorithm progresses, I want to check whether this flag is 0 for at least one element. Currently I use a while loop, like so:

def check(list_):
    for item in list_:
        if item[2] == 0:
            return True
    return False

The overall algorithm loops as long as that condition is satisfied, and sets some of the flags in each iteration:

while check(my_list):
    for item in my_list:
        if condition:
            item[2] = 1
        else:
            do_sth()

Because it causes problems to remove elements from the list while iterating over it, I use these flags to keep track of elements that have already been processed.

How can I simplify or speed up the code?

---

_{See also Pythonic way of checking if a condition holds for any element of a list for checking the condition for any element. Keep in mind that "any" and "all" checks are related through De Morgan's law, just as "or" and "and" are related.}

_{Existing answers here use the built-in function all to do the iteration. See How do Python's any and all functions work? for an explanation of all and its counterpart, any.}

_{If the condition you want to check is "is found in another container", see How to check if all of the following items are in a list? and its counterpart, How to check if one of the following items is in a list?. Using any and all will work, but more efficient solutions are possible.}

Answer 1

The best answer here is to use all(), which is the builtin for this situation. We combine this with a generator expression to produce the result you want cleanly and efficiently. For example:

>>> items = [[1, 2, 0], [1, 2, 0], [1, 2, 0]]
>>> all(flag == 0 for (_, _, flag) in items)
True
>>> items = [[1, 2, 0], [1, 2, 1], [1, 2, 0]]
>>> all(flag == 0 for (_, _, flag) in items)
False

Note that all(flag == 0 for (_, _, flag) in items) is directly equivalent to all(item[2] == 0 for item in items), it's just a little nicer to read in this case.

And, for the filter example, a list comprehension (of course, you could use a generator expression where appropriate):

>>> [x for x in items if x[2] == 0]
[[1, 2, 0], [1, 2, 0]]

If you want to check at least one element is 0, the better option is to use any() which is more readable:

>>> any(flag == 0 for (_, _, flag) in items)
True