使用两个Pandas系列：series1和series2,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的对应索引.
>>> 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
我期待的是：
>>> 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：
从series1和series2的长度相等的事实来看,我可以使for循环迭代series1并计算类似series1.ix [i] [series2.ix [i]]的东西并制作一个新的系列(series3)来保存结果.
我的解决方案2：
使用df = pd_concat([series1,series2])生成dataFrame df,并创建一个新列(使用apply函数的行方式操作 – 例如,df [‘series3’] = df.apply(lambda x：subList(x) ,轴= 1).
但是,我认为上面两种解决方案并不是实现我想要的方式.如果你建议更整洁的解决方案,我将不胜感激
解决方法:
如果您希望避免创建中间pd.DataFrame,并且只是想要一个新的pd.Series,则可以在地图对象上使用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
如果要在不创建中间容器的情况下修改S1,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
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