以下python代码应该清楚我想要完成的任务。
# 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]
请告诉我是否可以使用更紧凑的代码来执行此操作。
谢谢!
答案 0 :(得分:2)
使用sum
和列表理解。
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))]
答案 1 :(得分:2)
[sum([x.count(1) for x in L[:i]]) for i in range(1, len(L) + 1)]
应该做你想做的事。