<p>这是一个矢量化的方法-</p>
<pre><code>def vectorized_app(a,n):
M,N = a.shape
idx = np.arange(a.shape[0])[:,None] - np.arange(n+1)
out = a[idx.ravel(),:].reshape(-1,N*(n+1))
out[N*(np.arange(1,M+1))[:,None] <= np.arange(N*(n+1))] = 0
return out
</code></pre>
<p>样本运行-</p>
<pre><code>In [255]: a
Out[255]:
array([[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9],
[10, 11, 12],
[13, 14, 15],
[16, 17, 18]])
In [256]: vectorized_app(a,3)
Out[256]:
array([[ 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 4, 5, 6, 1, 2, 3, 0, 0, 0, 0, 0, 0],
[ 7, 8, 9, 4, 5, 6, 1, 2, 3, 0, 0, 0],
[10, 11, 12, 7, 8, 9, 4, 5, 6, 1, 2, 3],
[13, 14, 15, 10, 11, 12, 7, 8, 9, 4, 5, 6],
[16, 17, 18, 13, 14, 15, 10, 11, 12, 7, 8, 9]])
</code></pre>
<p>运行时测试-</p>
<p>我正在计时<a href="https://stackoverflow.com/a/38595430/3293881">^{<cd1>}</a>,并在问题中发布的样本的<code>100x</code>放大版本上使用本文中列出的向量化方法:</p>
<pre><code>In [246]: a = np.random.randint(0,9,(600,200))
In [247]: n = 200
In [248]: %timeit np.column_stack(mypad(a, i) for i in range(n + 1))
1 loops, best of 3: 748 ms per loop
In [249]: %timeit vectorized_app(a,n)
1 loops, best of 3: 224 ms per loop
</code></pre>