How to find first elements in list that match a condition?

user_14263566 0
Viewed 164

Say I have a list, l, which contains [5, 10, 15, 16, 20]. I want to get the first elements that are all divisible by 5 (so the first three elements), without the last 20. How do I do this? I've done:

done = False
i = 0
while not done and i < len(l):
    if l[i] % 5 == 0:
        answer.append(source[i])
    else:
        done = True
    i += 1

However, this seems inefficient. Is there any better way to do this?

python-3.x list python
6 Answers

The itertools.takewhile function seems like what you want:

from itertools import takewhile
list(takewhile(lambda x: x % 5 == 0, [5, 10, 15, 16, 20]))

This returns [5, 10, 15].

user_50449500

Your idea is actually an efficient answer because it avoids processing more elements than needed. @SamEstep's answer should be the accepted answer because it uses itertools exactly as intended, but regardless, the correct setup for your code would be the following:

l = [5, 10, 15, 16, 20]
result = []
for elem in l:
    if elem % 5 == 0:
        result.append(elem)
    else:
        break
# [5, 10, 15]

Control Structures

As a side note: a while loop that just indexes a list is discouraged, because it makes your intentions less explicit, and it makes your code more prone to bugs, as you have to handle the indexing yourself. A better way to write it would be to write:

for i in range(len(l)):

which gives you an index for each element in the list. However, if all you're going to do is index the list, then you can just iterate through each item in the list, which adds readability (this is what I've done above):

for elem in l:

If you need both the index and the element, use enumerate:

for idx, elem in enumerate(l):

This will make your code more readable, and less bug-prone.

user_116598810

in the else code block just use break will be fine

user_101312420

Say you have myList and you want to find items in it that can be divided by n:

myList = [1,2,10,12,5,20]
ansList = []
ansCount = 0
for i in myList:
    if i % n == 0:
        # condition is true, do whaterver you want with "i"; here I added it to a new list(ansList)
        ansList.append(i)
        ansCount += 1
        # you can also break the loop as your wish
        if ansCount > 2:
            break
user_132895750

It's not really inefficient, but probably a clearer way would be -

l = [5, 10, 15, 16, 20]
result = []
for item in l:
    if item % 5 == 0:
        result.append(item)
    else:
        break

this gives [5, 10, 15]

user_59119720

you can use something like this

mylist = [5, 10, 15, 16, 20]
result = []
for item in mylist:
    if item % 5 != 0:
        break
    result.append(item)
print(result)
user_162082630
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