<p>如果<code>col1</code>和<code>col2</code>是第一和第二列名,则最简单的解决方案是:</p>
<pre><code>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
</code></pre>
<p>另一个有<a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.isin.html" rel="nofollow">^{<cd3>}</a>或列表理解的解决方案:</p>
<pre><code>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
</code></pre>
<hr/>
<pre><code>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
</code></pre>
<p>最快的是列表理解:</p>
<pre><code>df.columns = [col+'_x' if col != 'col1' and col != 'col2' else col for col in df.columns]
</code></pre>
<p><strong>计时</strong>:</p>
<pre><code>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
</code></pre>
<hr/>
<pre><code>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))
</code></pre>