擅长:python、mysql、java
<p>正如<strong>@jojo</strong>的回答,但是使用熊猫:</p>
<pre><code>df.A = df.A.where(df.A > 0.5, (1/df.A)*-1)
</code></pre>
<p>或者</p>
<pre><code>df.A.where(df.A > 0.5, (1/df.A)*-1, inplace=True) # this should be faster
</code></pre>
<p>。其中docstring:</p>
<blockquote>
<p>Definition: df.A.where(self, cond, other=nan, inplace=False,
axis=None, level=None, try_cast=False, raise_on_error=True) </p>
<p>Docstring:
Return an object of same shape as self and whose corresponding entries
are from self where cond is True and otherwise are from other.</p>
</blockquote>