我有以下数据框:
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
看起来像这样:
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
我想要做的是根据以下正则表达式对列进行分组:
\w+_\w+_\w+_\d+_([\w\d-]+)_\d+
所以最终它被Mobile和1-999分组.
有什么办法呢.我尝试了这个,但未能将它们分组:
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
哪个印刷品:
XX_111_S5_R12_001_Mobile_05
YY_222_S00_R12_001_1-999_13
ZZ_111_S00_R12_001_1-999_13
我们想要的是名字打印到:
Mobile
1-999
1-999
并且组打印相应的数据框.
最佳答案 您可以在列上使用
.str.extract
,以便为您的groupby使用
extract substrings:
# 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'
返回预期的组:
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