pandas dataframe多层索引取值
作者:互联网
import pandas as pd
import numpy as np
# 新建df数据
df = pd.DataFrame(np.random.randint(50, 100, size=(4, 4)),
columns=pd.MultiIndex.from_product(
[['math', 'physics'], ['term1', 'term2']]),
index=pd.MultiIndex.from_tuples(
[('class1', 'LiLei'), ('class2', 'HanMeiMei'),
('class2', 'LiLei'), ('class2', 'HanMeiMei')]))
df.index.names = ['class', 'name']
# >>输出df:
行索引取值
# 取外层索引为'class1'的数据
df.loc['class1']
# 同时根据多个索引筛选取值,法一:
df.loc[('class2', 'HanMeiMei')]
# 同时根据多个索引筛选取值,法二:
df.loc['class2'].loc['HanMeiMei']
# 取内层索引:
# 先交换内外层索引位置
df.swaplevel()
# 输出:
math physics
term1 term2 term1 term2
name class
LiLei class1 81 81 77 91
HanMeiMei class2 82 83 84 79
LiLei class2 78 50 81 64
HanMeiMei class2 59 94 89 52
# 再通过取外层索引的方法取值
df.swaplevel().loc['HanMeiMei']
列索引取值
df数据:
# 外层列索引:
df['math']
# 根据多层索引联合取值:
# 以下4句代码等效:
df['math','term2']
df.loc[:, ('math','term1')]
df['math']['term2']
df[('math','term1')]
# 与行索引类似,取内层索引先交换轴
df.swaplevel(axis=1)
df.swaplevel(axis=1)['term1']
标签:term1,HanMeiMei,df,取值,dataframe,索引,class2,pandas,math 来源: https://www.cnblogs.com/jaysonteng/p/13475618.html