如何从应用方法中获取map的第二个参数

Question

我有以下数据集：

data = {"C1":[[(3, 5), (6, 8), (9-10)], [(0, 2), (5, 7), (9, 10)], [], [(1, 11)], [(0, 7), (8, 10)], [(5, 6)], [(0, 1)]]}
dt = pd.DataFrame(data)
print(dt)

看起来像：

0       [(3, 5), (6, 8), (9,10)]
1       [(0, 2), (5, 7), (9, 10)]
2       []
3       [(1, 11)]
4       [(0, 7), (8, 10)]
5       [(5, 6)]
6       [(0, 1)]

我想得到每个元组的长度(元组的第二个元素减去第一个元素(。

我最喜欢的输出是类似的东西

0       [(3, 5), (6, 8), (9,10)]        [2,2,1]
1       [(0, 2), (5, 7), (9,10)]        [2,2,1]
2                         []            []
3                  [(1, 11)]            [10]
4          [(0, 7), (8, 10)]            [7,2]
5                   [(5, 6)]            [1]
6                   [(0, 1)]            [1]

我目前正在使用这个代码：

dt['C2] = dt['C1'].apply(list(map(lambda x: x[1]-x[0])))

它给出以下错误：

map() must have at least two arguments

既然我使用的是apply方法，我希望map的第二个参数能自动从apply中获得，为什么没有发生？

Answer 1

给.apply()的lambda被单独应用于列中的每一行。所以你可以把理解列表放进去，做你想做的事：

data = {"C1":[[(3, 5), (6, 8), (9, 10)], [(0, 2), (5, 7), (9, 10)], [], [(1, 11)], [(0, 7), (8, 10)], [(5, 6)], [(0, 1)]]}
dt = pd.DataFrame(data)
print(dt)
>>> dt['C2'] = dt['C1'].apply(lambda lst: [tup[1] - tup[0] for tup in lst])
>>> dt
C1         C2
0  [(3, 5), (6, 8), (9, 10)]  [2, 2, 1]
1  [(0, 2), (5, 7), (9, 10)]  [2, 2, 1]
2                         []         []
3                  [(1, 11)]       [10]
4          [(0, 7), (8, 10)]     [7, 2]
5                   [(5, 6)]        [1]
6                   [(0, 1)]        [1]