[英]pandas - get values from multindex columns
我有以下数据框df:
H,Nu,City,Code,Code2
0.965392,15,Madrid,es,es
0.920614,15,Madrid,it,es
0.726219,16,Madrid,tn,es
0.739119,17,Madrid,fr,es
0.789923,55,Dublin,mt,en
0.699239,57,Dublin,en,en
0.890462,68,Dublin,ar,en
0.746863,68,Dublin,pt,en
0.789923,55,Milano,it,it
0.699239,57,Milano,es,it
0.890462,68,Milano,ar,it
0.746863,68,Milano,pt,it
我想为每个City
添加一个新列HCode
,其H
值对应于Code2
字符串映射的Code
,因此结果数据帧显示为:
H,Nu,City,Code,Code2,HCode
0.965392,15,Madrid,es,es,0.965392
0.920614,15,Madrid,it,es,0.965392
0.726219,16,Madrid,tn,es,0.965392
0.739119,17,Madrid,fr,es,0.965392
0.789923,55,Dublin,mt,en,0.699239
0.699239,57,Dublin,en,en,0.699239
0.890462,68,Dublin,ar,en,0.699239
0.746863,68,Dublin,pt,en,0.699239
0.789923,55,Milano,it,it,0.789923
0.699239,57,Milano,es,it,0.789923
0.890462,68,Milano,ar,it,0.789923
0.746863,68,Milano,pt,it,0.789923
到目前为止,我尝试按City和Code2分组,但没有结果。
您可以groupby
在“城市”和“代码2”,拨打first
就这个问题和复位导致以下指标:
In [172]:
gp = df.groupby(['City','Code2'])['H'].first().reset_index()
gp
Out[172]:
City Code2 H
0 Dublin en 0.789923
1 Madrid es 0.965392
2 Milano it 0.789923
然后在原始df上执行左合并,然后选择'H_y'列,该名称来自以下事实:各列发生冲突并ffill
以下条件:
In [173]:
df['HCode'] = df.merge(gp, left_on=['City', 'Code'], right_on=['City', 'Code2'], how='left')['H_y'].ffill()
df
Out[173]:
H Nu City Code Code2 HCode
0 0.965392 15 Madrid es es 0.965392
1 0.920614 15 Madrid it es 0.965392
2 0.726219 16 Madrid tn es 0.965392
3 0.739119 17 Madrid fr es 0.965392
4 0.789923 55 Dublin mt en 0.965392
5 0.699239 57 Dublin en en 0.789923
6 0.890462 68 Dublin ar en 0.789923
7 0.746863 68 Dublin pt en 0.789923
8 0.789923 55 Milano it it 0.789923
9 0.699239 57 Milano es it 0.789923
10 0.890462 68 Milano ar it 0.789923
11 0.746863 68 Milano pt it 0.789923
merge
结果以显示产生的结果:
In [165]:
df.merge(gp, left_on=['City', 'Code'], right_on=['City', 'Code2'])['H_y']
Out[165]:
0 0.965392
1 0.789923
2 0.789923
Name: H_y, dtype: float64
编辑
好的,IIUC可以像以前一样进行分组,但是可以过滤“ Code2”等于“ Code”的组,然后将其合并为:
In [200]:
gp = df.groupby('City')
mask = gp.apply(lambda x: x['Code2'] == x['Code'])
lookup = df.loc[mask[mask].reset_index(level=0).index]
lookup
Out[200]:
H Nu City Code Code2
5 0.699239 57 Dublin en en
0 0.965392 15 Madrid es es
8 0.789923 55 Milano it it
In [202]:
df['HCode'] = df.merge(lookup, left_on=['City', 'Code'], right_on=['City', 'Code2'], how='left')['H_y'].ffill()
df
Out[202]:
H Nu City Code Code2 HCode
0 0.965392 15 Madrid es es 0.965392
1 0.920614 15 Madrid it es 0.965392
2 0.726219 16 Madrid tn es 0.965392
3 0.739119 17 Madrid fr es 0.965392
4 0.789923 55 Dublin mt en 0.965392
5 0.699239 57 Dublin en en 0.699239
6 0.890462 68 Dublin ar en 0.699239
7 0.746863 68 Dublin pt en 0.699239
8 0.789923 55 Milano it it 0.789923
9 0.699239 57 Milano es it 0.789923
10 0.890462 68 Milano ar it 0.789923
11 0.746863 68 Milano pt it 0.789923
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