将前缀添加到Dataframe的特定列

Question

我有一个这样的DataFrame:

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,但不是重命名col1和col2.

所以我试着做一个循环

print(df.columns)
    for col in df.columns:
        if col != 'col1' and col != 'col2':
            col.rename = str(col) + '_x'

但它不是很有效......它不起作用!

Answer 1

您可以使用DataFrame.rename()方法

new_names = [(i,i+'_x') for i in df.iloc[:, 2:].columns.values]
df.rename(columns = dict(new_names), inplace=True)

Answer 2

如果col1和col2是第一列和第二列名称的最简单解决方案：

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

具有isin或列表理解的另一种解决方案：

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      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

最快的是列表理解：

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

时间：

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))

Answer 3

您可以使用str.contains正则表达式模式来过滤感兴趣的列，然后使用zip构造一个字典并将其作为参数传递给rename：

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

因此，这里使用str.contains过滤列将返回不匹配的列，因此列顺序无关紧要