4

次の python コードは、私が達成したいことを明確にする必要があります。

# Say I have the following list and I want to keep count of 1's while going through each nested list
L = [[1,1,0,0,0,1],[1,1,0,0,0,0],[0,0,0,0,0,1],[1,1,1,1,1,1]]

# So I'd like to get a list containing [3, 5, 6, 12]
# I tried to find a compact way of doing this by mapping this list into a another list like such 
T = [L[:i].count(1) for i in range(len(L))]
# >>> [0, 0, 0, 0]


# But this doesn't work so how to count occurances for nested lists?
# Is there a compact way of doing this (maybe with Lambda functions)?

# I'd like to avoid using a function like:
def Nested_Count():
    Result = []
    count = 0
    for i in L:
        count += i.count(1)
        Result.append(count)
    return Result
# >>> [3, 5, 6, 12]

これを行うためのよりコンパクトなコードが可能かどうか教えてください。

ありがとう！

4

2 に答える 2

2

使用sumとリスト内包表記。

L = [[1,1,0,0,0,1],[1,1,0,0,0,0],[0,0,0,0,0,1],[1,1,1,1,1,1]]
L2 = [sum(x) for x in L]
T = [sum(L2[:x+1]) for x in xrange(len(L2))]
于 2013-09-07T22:25:57.523 に答える
2 
[sum([x.count(1) for x in L[:i]]) for i in range(1, len(L) + 1)]

あなたがしたいことをするべきです。

于 2013-09-07T22:27:21.070 に答える