pandas 为 Dataframe 的特定列添加前缀
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Add prefix to specific columns of Dataframe
提问by Geo-x
I've a DataFrame like that :
我有一个这样的数据帧：
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
I would like to rename all columns but not col1and col2.
我想重命名所有列，但不重命名col1和col2。
So I tried to make a loop
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'
But it's not very efficient...it doesn't work !
但它不是很有效......它不起作用！
回答by A.Kot
You can use the DataFrame.rename()method
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)
回答by jezrael
Simpliest solution if col1and col2are first and second column names:
如果col1和col2是第一列和第二列名称，则最简单的解决方案：
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
Another solution with isinor list comprehension:
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
The fastest is list comprehension:
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]
Timings:
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))
回答by EdChum
You can use str.containswith a regex pattern to filter the cols of interest, then using zipconstruct a dict and pass this as the arg to rename:
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 dict 并将其作为 arg 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
So here using str.containsto filter the columns will return the columns that don't match so the column order is irrelevant
所以这里使用str.contains过滤列将返回不匹配的列，因此列顺序无关紧要
