用分隔符pandas python拆分列
发布于 2021-01-29 19:35:05
我有一个小样本数据:
import pandas as pd
df = {'ID': [3009, 129,119,120,121,122,130,3014,266,849,174,844 ],
'V': ['IGHV7-B*01','IGHV7-B*01','IGHV6-A*01','GHV6-A*01','IGHV6-A*01','IGHV6-A*01','IGHV4-L*03','IGHV4-L*03','IGHV5-A*01','IGHV5-A*04','IGHV6-A*02','IGHV6-A*02'],
'Prob': [1,1,0.8,0.8056,0.9,0.805 ,1,1,0.997,0.401,1,1]}
df = pd.DataFrame(df)
好像
df
Out[25]:
ID Prob V
0 3009 1.0000 IGHV7-B*01
1 129 1.0000 IGHV7-B*01
2 119 0.8000 IGHV6-A*01
3 120 0.8056 IGHV6-A*01
4 121 0.9000 IGHV6-A*01
5 122 0.8050 IGHV6-A*01
6 130 1.0000 IGHV4-L*03
7 3014 1.0000 IGHV4-L*03
8 266 0.9970 IGHV5-A*01
9 849 0.4010 IGHV5-A*04
10 174 1.0000 IGHV6-A*02
11 844 1.0000 IGHV6-A*02
我想用’-‘分隔符分隔列’V’并将其移至另一个名为’allele’的列
Out[25]:
ID Prob V allele
0 3009 1.0000 IGHV7 B*01
1 129 1.0000 IGHV7 B*01
2 119 0.8000 IGHV6 A*01
3 120 0.8056 IGHV6 A*01
4 121 0.9000 IGHV6 A*01
5 122 0.8050 IGHV6 A*01
6 130 1.0000 IGHV4 L*03
7 3014 1.0000 IGHV4 L*03
8 266 0.9970 IGHV5 A*01
9 849 0.4010 IGHV5 A*04
10 174 1.0000 IGHV6 A*02
11 844 1.0000 IGHV6 A*02
到目前为止,我尝试过的代码不完整,无法正常工作:
df1 = pd.DataFrame()
df1[['V']] = pd.DataFrame([ x.split('-') for x in df['V'].tolist() ])
要么
df.add(Series, axis='columns', level = None, fill_value = None)
newdata = df.DataFrame({'V':df['V'].iloc[::2].values, 'Allele': df['V'].iloc[1::2].values})
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1 个回答
-
str.split
与vectoried一起使用expand=True
:In [42]: df[['V','allele']] = df['V'].str.split('-',expand=True) df Out[42]: ID Prob V allele 0 3009 1.0000 IGHV7 B*01 1 129 1.0000 IGHV7 B*01 2 119 0.8000 IGHV6 A*01 3 120 0.8056 GHV6 A*01 4 121 0.9000 IGHV6 A*01 5 122 0.8050 IGHV6 A*01 6 130 1.0000 IGHV4 L*03 7 3014 1.0000 IGHV4 L*03 8 266 0.9970 IGHV5 A*01 9 849 0.4010 IGHV5 A*04 10 174 1.0000 IGHV6 A*02 11 844 1.0000 IGHV6 A*02