<p>所以我有一个</p>
<pre><code>df = read_excel(...)
</code></pre>
<p>循环确实有效:</p>
<pre><code>for i, row in df.iterrows(): #loop through rows
a = df[df.columns].SignalName[i] #column "SignalName" of row i, is read
b = (row[7]) #column "Bus-Signalname" of row i, taken primitively=hardcoded
</code></pre>
<p>访问a是正常的,如何用excel表中动态找到/定位的“Bus Signalname”元素替换硬编码的b=(行[7])。有哪些方法可以做到这一点</p>
<pre><code> b = df[df.columns].Bus-Signalname[i]
</code></pre>
<p>不起作用</p>