python - Pandas 数据框中元素和子集的总长度

Question

我如何计算数据框中的总元素数(包括子集)并将结果放入新列？

import pandas as pd
x = pd.Series([[1, (2,5,6)], [2, (3,4)], [3, 4], [(5,6), (7,8,9)]], \
              index=range(1, len(x)+1))
df = pd.DataFrame({'A': x})

我尝试使用以下代码，但它在每一行中给出 2:

df['Length'] = df['A'].apply(len)

print(df)

                         A  Length
    1       [1, (2, 5, 6)]       2
    2          [2, (3, 4)]       2
    3               [3, 4]       2
    4  [(5, 6), (7, 8, 9)]       2

然而，我想要得到的是:

                         A  Length
    1       [1, (2, 5, 6)]       4
    2          [2, (3, 4)]       3
    3               [3, 4]       2
    4  [(5, 6), (7, 8, 9)]       5

谢谢

Answer 1

给定:

import pandas as pd
x = pd.Series([[1, (2,5,6)], [2, (3,4)], [3, 4], [(5,6), (7,8,9)]])
df = pd.DataFrame({'A': x}) 

您可以编写一个递归生成器，为每个不可迭代的嵌套元素生成 1。沿着这些线的东西:

import collections 

def glen(LoS):
    def iselement(e):
        return not(isinstance(e, collections.Iterable) and not isinstance(e, str))
    for el in LoS:
        if iselement(el):
            yield 1
        else:
            for sub in glen(el): yield sub    

df['Length'] = df['A'].apply(lambda e: sum(glen(e)))

产量:

>>> df
                     A  Length
0       [1, (2, 5, 6)]       4
1          [2, (3, 4)]       3
2               [3, 4]       2
3  [(5, 6), (7, 8, 9)]       5

这将适用于 Python 2 或 3。对于 Python 3.3 或更高版本，您可以使用 yield from 来替换循环:

def glen(LoS):
    def iselement(e):
        return not(isinstance(e, collections.Iterable) and not isinstance(e, str))
    for el in LoS:
        if iselement(el):
            yield 1
        else:
            yield from glen(el)