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

Question

I have a list consisting of like 20000 lists. I use each list’s 3rd element as a flag. I want to do some operations on this list as long as at least one element’s flag is 0, it’s like:

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

In the beginning, all flags are 0. I use a while loop to check if at least one element’s flag is 0:

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

If check(my_list) returns True, then I continue working on my list:

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

Actually, I wanted to remove an element in my_list as I iterated over it, but I’m not allowed to remove items as I iterate over it.

Original my_list didn’t have flags:

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

Since I couldn’t remove elements as I iterated over it, I invented these flags. But the my_list contains many items, and while loop reads all of them at each for loop, and it consumes lots of time! Do you have any suggestions?

Asked By: alwbtc

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Source

Answer 1

You could use itertools’s takewhile like this, it will stop once a condition is met that fails your statement. The opposite method would be dropwhile

for x in itertools.takewhile(lambda x: x[2] == 0, list)
    print x

Answered By: Hedde van der Heide

Answer 2

If you want to check if any item in the list violates a condition use all:

if all([x[2] == 0 for x in lista]):
    # Will run if all elements in the list has x[2] = 0 (use not to invert if necessary)

To remove all elements not matching, use filter

# Will remove all elements where x[2] is 0
listb = filter(lambda x: x[2] != 0, listb)

Answered By: Hampus Nilsson

Answer 3

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

Answered By: Gareth Latty

Answer 4

this way is a bit more flexible than using all():

my_list = [[1, 2, 0], [1, 2, 0], [1, 2, 0]]
all_zeros = False if False in [x[2] == 0 for x in my_list] else True
any_zeros = True if True in [x[2] == 0 for x in my_list] else False

or more succinctly:

all_zeros = not False in [x[2] == 0 for x in my_list]
any_zeros = 0 in [x[2] for x in my_list]

Answered By: mulllhausen

Answer 5

Another way to use itertools.ifilter. This checks truthiness and process
(using lambda)

Sample-

for x in itertools.ifilter(lambda x: x[2] == 0, my_list):
    print x

Answered By: SIslam

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