查找列表的前N个元素，直到满足条件

Question

I have a list of objects, and I want to take the beginning of the list up until the first object that meets a simple condition (like imp[1] == 'conversion' when imp is some element in the list).

An easy way would be: initialize a new list, iterate through the original list, and at each step append the current element and check the condition on the current element. If the condition is not satisfied then continue, and if it is satisfied then break.

new_list = []
for ele in old_list:
    new_list.append(ele)
    if condish(ele):
        break

But this seems inefficient in memory, runtime, and code (the big three!).

Answer 1

You can try this:

for idx, el in enumerate(your_list):
    if satisfies_condition(el):
        return your_list[:idx]

which will save you the cost of creating a new list in memory.

Or you can use itertools.takewhile

return list(itertools.takewhile(not_condition, your_list))

Answer 2

There is itertools.takewhile which should satisfy your need. Of course, here you need to negate the condition in your original post so that you break when the condition is met...

itertools.takewhile(lambda ele: not condish(ele), old_list)

If you want more flexibility and control (eg you also want to take the first element that doesn't meet the condition), it might be worth considering a generator function:

def take_until(iterable):
    for item in iterable
        yield item
        if condition(item):
           return

then you use it like this:

for item in take_until(old_list):
    ...

This avoids building a list that you don't really need/want and gives you an iterable instead.