通过分隔符pandas python拆分列
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splitting a column by delimiter pandas python
提问by Jessica
i have a small sample data:
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)
looks like
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
I want to split the column 'V' by the '-' delimiter and move it to another column named 'allele'
我想用“-”分隔符拆分“V”列并将其移动到名为“等位基因”的另一列
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
the code i have tried so far is incomplete and didn't work:
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() ])
or
或者
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})
回答by EdChum
Use vectoried str.splitwith expand=True:
使用vectoriedstr.split有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
回答by Lukas
For storing data into a new dataframe use the same approach, just with the new dataframe:
要将数据存储到新数据帧中，请使用相同的方法，只需使用新数据帧：
tmpDF = pd.DataFrame(columns=['A','B'])
tmpDF[['A','B']] = df['V'].str.split('-', expand=True)
Eventually (and more usefull for my purposes) if you would need get only a part of the string value (i.e. text before '-'), you could use .str.split(...).str[idx] like:
最终（并且对我的目的更有用）如果您只需要获取字符串值的一部分（即“-”之前的文本），您可以使用 .str.split(...).str[idx] ，例如：
df['V'] = df['V'].str.split('-').str[0]
df
ID V Prob
0 3009 IGHV7 1.0000
1 129 IGHV7 1.0000
2 119 IGHV6 0.8000
3 120 GHV6 0.8056
- splits 'V' values into list according to separator '-' and stores 1st item back to the column
- 根据分隔符“-”将“V”值拆分为列表并将第一个项目存储回列
