如果col1和col2是第一和第二列名，则最简单的解决方案是：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

另一个有^{}或列表理解的解决方案：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 fcols = [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

最快的是列表理解：df.columns = [col+'_x' if col != 'col1' and col != 'col2' else col for col in df.columns]

计时：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 loopdf1 = 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))