如果pandasseries的值是一个list,如何得到每个元素的subList?

If the value of pandas series is a list, how to get a subList of each element?

使用两个Pandas系列:series1series2,我愿意制作series3 series1的每个值都是一个列表,series2的每个值都是series1对应的一个索引。

>>> print(series1)

0      [481, 12, 11, 220, 24, 24, 645, 153, 15, 13, 6...
1      [64, 80, 79, 147, 14, 20, 56, 288, 12, 208, 26...
4      [5, 6, 152, 31, 295, 127, 711, 5, 271, 291, 11...
5          [363, 121, 727, 249, 483, 122, 241, 494, 555]
7      [112, 20, 41, 9, 104, 131, 26, 298, 65, 214, 1...
9      [129, 797, 19, 151, 448, 47, 19, 106, 299, 144...
11     [72, 35, 25, 200, 122, 5, 75, 30, 208, 24, 14,...
18     [137, 339, 71, 14, 19, 54, 61, 15, 73, 104, 43...



>>> print(series2)

0       0
1       3
4       1
5       6
7       4
9       5
11      7
18      2

我的期望:

>>> print(series3)

0      [481, 12, 11, 220, 24, 24, 645, 153, 15, 13, 6...
1      [147, 14, 20, 56, 288, 12, 208, 26...
4      [6, 152, 31, 295, 127, 711, 5, 271, 291, 11...
5      [241, 494, 555]
7      [104, 131, 26, 298, 65, 214, 1...
9      [47, 19, 106, 299, 144...
11     [30, 208, 24, 14,...
18     [71, 14, 19, 54, 61, 15, 73, 104, 43...

我的解决方案1: 由于 series1series2 的长度相等,我可以制作一个 for 循环来迭代 series1 并计算类似 series1.ix[i][series2.ix[i]] 的东西并制作一个新系列(series3)来保存结果。

我的方案2: 使用 df = pd_concat([series1, series2]) 生成数据帧 df,并创建一个新列(使用应用函数进行逐行操作 - 例如,df['series3'] = df.apply(lambda x: subList(x),轴=1).

但是,我认为以上两种解决方案都不是实现我想要的东西的好方法。如果您提出更简洁的解决方案,我将不胜感激!

您基本上可以连接指定轴(0=行,1 列)的系列,最好具有相同的长度

series3=pd.concat([series2, series1], axis=1).reset_index()

如果您希望避免创建中间 pd.DataFrame,而只想创建一个新的 pd.Series,您可以在 map 对象上使用 pd.Series 构造函数。所以给出:

In [6]: S1
Out[6]:
0    [481, 12, 11, 220, 24, 24, 645, 153, 15, 13, 6]
1    [64, 80, 79, 147, 14, 20, 56, 288, 12, 208, 26]
2    [5, 6, 152, 31, 295, 127, 711, 5, 271, 291, 11]
3      [363, 121, 727, 249, 483, 122, 241, 494, 555]
4    [112, 20, 41, 9, 104, 131, 26, 298, 65, 214, 1]
5    [129, 797, 19, 151, 448, 47, 19, 106, 299, 144]
6     [72, 35, 25, 200, 122, 5, 75, 30, 208, 24, 14]
7    [137, 339, 71, 14, 19, 54, 61, 15, 73, 104, 43]
dtype: object

In [7]: S2
Out[7]:
0    0
1    3
2    1
3    6
4    4
5    5
6    7
7    2
dtype: int64

你可以这样做:

In [8]: pd.Series(map(lambda x,y : x[y:], S1, S2), index=S1.index)
Out[8]:
0    [481, 12, 11, 220, 24, 24, 645, 153, 15, 13, 6]
1                [147, 14, 20, 56, 288, 12, 208, 26]
2       [6, 152, 31, 295, 127, 711, 5, 271, 291, 11]
3                                    [241, 494, 555]
4                    [104, 131, 26, 298, 65, 214, 1]
5                            [47, 19, 106, 299, 144]
6                                  [30, 208, 24, 14]
7              [71, 14, 19, 54, 61, 15, 73, 104, 43]
dtype: object

如果您想修改S1而不创建中间容器,您可以使用for循环:

In [10]: for i, x in enumerate(map(lambda x,y : x[y:], S1, S2)):
    ...:     S1.iloc[i] = x
    ...:

In [11]: S1
Out[11]:
0    [481, 12, 11, 220, 24, 24, 645, 153, 15, 13, 6]
1                [147, 14, 20, 56, 288, 12, 208, 26]
2       [6, 152, 31, 295, 127, 711, 5, 271, 291, 11]
3                                    [241, 494, 555]
4                    [104, 131, 26, 298, 65, 214, 1]
5                            [47, 19, 106, 299, 144]
6                                  [30, 208, 24, 14]
7              [71, 14, 19, 54, 61, 15, 73, 104, 43]
dtype: object