擅长:python、mysql、java
<p>看起来循环是不可避免的,因为您必须替换/删除子字符串。在这种情况下,列表理解可能会派上用场:</p>
<pre><code>%%timeit
df.apply(lambda x: x['col3'].replace(x['col1'], ''), axis=1)
# 767 µs ± 24.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
</code></pre>
<p>而</p>
<pre><code>%%timeit
[a.replace(b,'') for a,b in zip(df['col3'], df['col1'])]
# 24.4 µs ± 3.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
</code></pre>