擅长:python、mysql、java
<p><strong>方法1:</strong>这是一种利用<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html" rel="nofollow noreferrer">^{<cd1>}</a>和{a2}-</p>
<pre><code>K = np.arange(n)[:,None]
mask = (i == K) | (j == K)
out = np.dot(mask,a)
</code></pre>
<p><strong>方法2:</strong>对于列数较少的情况,我们可以使用<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html" rel="nofollow noreferrer">^{<cd3>}</a>对每列进行基于bin的求和,如下-</p>
^{pr2}$