为特定列添加前缀 Dataframe

我有 DataFrame 这样的 :


col1 col2 col3 col4 col5 col6 col7 col8

0 5345 rrf rrf rrf rrf rrf rrf

1 2527 erfr erfr erfr erfr erfr erfr

2 2727 f f f f f f


我想重命名所有列，但不是

col1

和

col2

.

所以我试图制作一个循环


print/df.columns/

 for col in df.columns:

 if col != 'col1' and col != 'col2':

 col.rename = str/col/ + '_x'


但这不是很好 efficient...it 不起作用 !
2021-03-20 添加评论
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没有找到相关结果

    已邀请:

    詹大官人

    赞同来自:

    您可以使用该方法 DataFrame.rename//


    new_names = [/i,i+'_x'/ for i in df.iloc[:, 2:].columns.values]
    
df.rename/columns = dict/new_names/, inplace=True/
    2021-03-20 0 0
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    知食

    赞同来自:

    最简单的解决方案
    col1

    和
    col2

    - 这些是第一个和第二列的名称：


    df.columns = df.columns[:2].union/df.columns[2:] + '_x'/
    
print /df/
    
 col1 col2 col3_x col4_x col5_x col6_x col7_x col8_x
    
0 0 5345 rrf rrf rrf rrf rrf rrf
    
1 1 2527 erfr erfr erfr erfr erfr erfr
    
2 2 2727 f f f f f f


    另一个决定。
    http://pandas.pydata.org/panda ... .html
    或理解列表：


    cols = df.columns[~df.columns.isin/['col1','col2']/]
    
print /cols/
    
['col3', 'col4', 'col5', 'col6', 'col7', 'col8']
    

    
df.rename/columns = dict/zip/cols, cols + '_x'//, inplace=True/
    

    
print /df/
    

    
 col1 col2 col3_x col4_x col5_x col6_x col7_x col8_x
    
0 0 5345 rrf rrf rrf rrf rrf rrf
    
1 1 2527 erfr erfr erfr erfr erfr erfr
    
2 2 2727 f f f f f f



    cols = [col for col in df.columns if col not in ['col1', 'col2']]
    
print /cols/
    
['col3', 'col4', 'col5', 'col6', 'col7', 'col8']
    

    
df.rename/columns = dict/zip/cols, cols + '_x'//, inplace=True/
    

    
print /df/
    

    
 col1 col2 col3_x col4_x col5_x col6_x col7_x col8_x
    
0 0 5345 rrf rrf rrf rrf rrf rrf
    
1 1 2527 erfr erfr erfr erfr erfr erfr
    
2 2 2727 f f f f f f


    最快的是对列表的理解：


    df.columns = [col+'_x' if col != 'col1' and col != 'col2' else col for col in df.columns]


    Tayyumi

    :


    In [350]: %timeit /akot/df//
    
1000 loops, best of 3: 387 µs per loop
    

    
In [351]: %timeit /jez/df1//
    
The slowest run took 4.12 times longer than the fastest. This could mean that an intermediate result is being cached.
    
10000 loops, best of 3: 207 µs per loop
    

    
In [363]: %timeit /jez3/df2//
    
The slowest run took 6.41 times longer than the fastest. This could mean that an intermediate result is being cached.
    
10000 loops, best of 3: 75.7 µs per loop



    df1 = df.copy//
    
df2 = df.copy//
    

    
def jez/df/:
    
 df.columns = df.columns[:2].union/df.columns[2:] + '_x'/
    
 return df
    

    
def akot/df/:
    
 new_names = [/i,i+'_x'/ for i in df.iloc[:, 2:].columns.values]
    
 df.rename/columns = dict/new_names/, inplace=True/
    
 return df
    

    

    
def jez3/df/:
    
 df.columns = [col + '_x' if col != 'col1' and col != 'col2' else col for col in df.columns]
    
 return df
    

    

    
print /akot/df//
    
print /jez/df1//
    
print /jez2/df1//
    2021-03-20 0 0
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    帅驴

    赞同来自:

    您可以使用
    http://pandas.pydata.org/panda ... .html
    有模板 regex 用于过滤感兴趣 cols, 然后在帮助下
    zip

    建造 dict 并将其转移为 arg 在
    http://pandas.pydata.org/panda ... .html
    :


    In [94]:
    
cols = df.columns[~df.columns.str.contains/'col1|col2'/]
    
df.rename/columns = dict/zip/cols, cols + '_x'//, inplace=True/
    
df
    

    
Out[94]:
    
 col1 col2 col3_x col4_x col5_x col6_x col7_x col8_x
    
0 0 5345 rrf rrf rrf rrf rrf rrf
    
1 1 2527 erfr erfr erfr erfr erfr erfr
    
2 2 2727 f f f f f f


    所以这是使用
    str.contains

    过滤列将返回不重合的列，因此列的顺序无关紧要
    2021-03-20 0 0
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