擅长:python、mysql、java
<p><strong>更新:</strong></p>
<pre><code>counts = complaints['Provider'].value_counts()
counts[counts == 1]
</code></pre>
<p>显示小于或等于3的“计数”:</p>
<pre><code>counts[counts <= 3]
</code></pre>
<p><strong>旧答案:</strong></p>
<p>你可以这样做:</p>
<pre><code>complaints['Provider'].value_counts().nsmallest(1)
</code></pre>
<p>或者您可以使用<code>iloc</code>定位器,这可能要快一点:</p>
<pre><code>complaints['Provider'].value_counts().iloc[-1]
</code></pre>