擅长:python、mysql、java
<p>我肯定会选择<code>numexpr</code>。我不确定<code>numexpr</code>是否可以处理索引,但我敢打赌以下(或类似的)方法会起作用:</p>
<pre><code>import numexpr as ne
yold = self.oldYD[ind]
y = self.YD[ind]
xold = self.oldXD[ind]
x = self.XD[ind]
YD_zero = ne.evaluate("yold - ((yold - y) * xold)/(xold - x)")
</code></pre>