[英]If the value of pandas series is a list, how to get a subList of each element?
使用兩個Pandas系列: series1和series2 ,我願意制作series3 。 series1的每個值都是一個列表, series2的每個值都是series1的對應索引。
>>> 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...
我的解決方法1:從事實和系列1 系列2的長度相等,我可以做一個for循環迭代系列1和計算類似series1.ix[i][series2.ix[i]]
,並作出新系列( series3 )保存結果。
我的解決方案2 :使用df = pd_concat([series1, series2])
生成dataFrame df,並創建一個新列(使用apply函數的行方式操作 - 例如,df ['series3'] = df.apply(lambda x: subList(x),axis = 1)。
但是,我認為上面兩種解決方案並不是實現我想要的方式。 如果你建議更整潔的解決方案,我將不勝感激
如果您希望避免創建中間pd.DataFrame
,並且只需要新的pd.Series
,則可以在map
對象上使用pd.Series
構造函數。 所以給出:
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
如果要在不創建中間容器的情況下修改S1
,可以使用for循環:
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
你基本上可以連接指定wich軸的系列(0 =行,1列),更好的是相同的長度
series3=pd.concat([series2, series1], axis=1).reset_index()
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