# 背景介紹 通常我們不會(huì)在Pandas中主動(dòng)設(shè)置多層索引,但是如果一個(gè)字段做多個(gè)不同的聚合運(yùn)算, 比如sum, max這樣形成的Column Level是有層次的,這樣閱讀非常方便,但是對(duì)編程定位比較麻煩. # 數(shù)據(jù)準(zhǔn)備
import pandas as pdimport numpy as npdf = pd.DataFrame(np.arange(0, 14).reshape(7,2),columns =['a','b'] )df.a = df.a %3df['who'] = 'Bob'df.loc[df.a%4==0,'who'] = 'Alice'
| a | b | who | |
|---|---|---|---|
| 0 | 0 | 1 | Alice |
| 1 | 2 | 3 | Bob |
| 2 | 1 | 5 | Bob |
| 3 | 0 | 7 | Alice |
| 4 | 2 | 9 | Bob |
| 5 | 1 | 11 | Bob |
| 6 | 0 | 13 | Alice |
# 對(duì)一個(gè)字段同時(shí)用3個(gè)聚合函數(shù)
gp1 = df.groupby('who').agg({'b':[sum,np.max, np.min], 'a':sum})gp1| b | a | |||
|---|---|---|---|---|
| sum | amax | amin | sum | |
| who | ||||
| Alice | 8.0 | 7.0 | 1.0 | 0 |
| Bob | 28.0 | 11.0 | 3.0 | 6 |
索引是有層次的,虛要通過(guò)下面這種方式,個(gè)人感覺(jué)不是很方便.下面介紹2種方法來(lái)解決這個(gè)問(wèn)題
#有層次的索引訪(fǎng)問(wèn)方法gp1.loc['Bob', ('b', 'sum')]28.0
# 直接去除一層
gp2 = gp1.copy(deep=True)gp2.columns = gp1.columns.droplevel(0)gp2
| sum | amax | amin | sum | |
|---|---|---|---|---|
| who | ||||
| Alice | 8.0 | 7.0 | 1.0 | 0 |
| Bob | 28.0 | 11.0 | 3.0 | 6 |
# 把2層合并到一層
gp3 = gp1.copy(deep=True)gp3.columns = ["_".join(x) for x in gp3.columns.ravel()]gp3
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