Python：比较不同列表的长度并返回最长的

Question

我想比较5个列表的长度，并返回最长的列表。

我不知道... 也许是这样的：

a = [1,1,1,1]
b = [1]
c = [1,1,1]
d = [1,1,1,1,1]
e = [1,1,1,1,1]

L = [a,b,c,d,e]

def compare(lists):
    counter = count()
    previous = lists[0]
    group_index = next(counter)
    for value in lists:
        if len(value) >= len(previous):
                ...

结果应为“ e”。

Answer 1

下面是py3

a = [1,1,1,1]
b = [1]
c = [1,1,1]
d = [1,1,1,1,1]
e = [1,1,1,1,2]
f = [1,1,1,1,1,2]
L = [a,f,b,c,d,e]

def compare(lists):
    previous = lists[0]
    for value in lists:
        if len(value) >= len(previous):
            previous=value
    return previous
Rsss=compare(L)
print(Rsss)
print([ k for k,v in locals().items() if v == Rsss][0])

python 2将项目更改为迭代项

print（）进行打印