[英]How to group Pandas data frame by column with regex match
我有以下數據框:
import pandas as pd
df = pd.DataFrame({'id':['a','b','c','d','e'],
'XX_111_S5_R12_001_Mobile_05':[-14,-90,-90,-96,-91],
'YY_222_S00_R12_001_1-999_13':[-103,0,-110,-114,-114],
'ZZ_111_S00_R12_001_1-999_13':[1,2.3,3,5,6],
})
df.set_index('id',inplace=True)
df
看起來像這樣:
Out[6]:
XX_111_S5_R12_001_Mobile_05 YY_222_S00_R12_001_1-999_13 ZZ_111_S00_R12_001_1-999_13
id
a -14 -103 1.0
b -90 0 2.3
c -90 -110 3.0
d -96 -114 5.0
e -91 -114 6.0
我想要做的是根據以下正則表達式對列進行分組:
\w+_\w+_\w+_\d+_([\w\d-]+)_\d+
所以最終它被Mobile
和1-999
分組。
有什么辦法呢。 我嘗試了這個,但未能將它們分組:
import re
grouped = df.groupby(lambda x: re.search("\w+_\w+_\w+_\d+_([\w\d-]+)_\d+", x).group(), axis=1)
for name, group in grouped:
print name
print group
哪個印刷品:
XX_111_S5_R12_001_Mobile_05
YY_222_S00_R12_001_1-999_13
ZZ_111_S00_R12_001_1-999_13
我們想要的是name
打印到:
Mobile
1-999
1-999
並且group
打印相應的數據框。
您可以在列上使用.str.extract
,以便為您的groupby
提取子字符串 :
# Performing the groupby.
pat = '\w+_\w+_\w+_\d+_([\w\d-]+)_\d+'
grouped = df.groupby(df.columns.str.extract(pat, expand=False), axis=1)
# Showing group information.
for name, group in grouped:
print name
print group, '\n'
返回預期的組:
1-999
YY_222_S00_R12_001_1-999_13 ZZ_111_S00_R12_001_1-999_13
id
a -103 1.0
b 0 2.3
c -110 3.0
d -114 5.0
e -114 6.0
Mobile
XX_111_S5_R12_001_Mobile_05
id
a -14
b -90
c -90
d -96
e -91
分組后,將新數據幀的索引設置為[re.findall(r'\\w+_\\w+_\\w+_\\d+_([\\w\\d-]+)_\\d+', col)[0] for col in df.columns]
( ['Mobile', '1-999', '1-999']
)。
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