Count consecutive ones in a binary list

There is a list consisting of zeroes and ones. I want to find out the length of the longest streak of ones. Is there a better solution?

def consecutive_one(data):
   one_list = []
   size = 0
   for num in data:
       if num == 1:
           one_list.append(num)
       elif num == 0 and size < len(one_list):
           size = len(one_list)
           one_list = []

    return size

if __name__ == '__main__':
    data = [0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0]
    print(consecutive_one(data))

Both your and janos' implementations are broken:

data = [1, 0] * 10000
consecutive_one(data)
#>>> 140

This is because you don't always reset after seeing a 0. Going from janos', you should have

longest = 0
current = 0
for num in data:
    if num == 1:
        current += 1
    else:
        longest = max(longest, current)
        current = 0

return max(longest, current)

and equivalent for the original.

You'll find that this functionality is largely provided by itertools.groupby, though:

from itertools import groupby

def len_iter(items):
    return sum(1 for _ in items)

def consecutive_one(data):
    return max(len_iter(run) for val, run in groupby(data) if val)

StackExchange Code Review Q#138550, answer score: 10