为什么 NumPy 没有复制 J 秩的概念？
最近有一篇帖子提到，当使用形状大于2的数组时，numpy的使用并不容易。[1]中提到的问题之一是，不能使用Python循环，因为它们速度较慢。在J语言中，可以通过秩的概念来解决例如100个方程，这是一个简单的例子：
```
a=: 2 2 $ 1 1 1 _1
b=: 10 2 $ ? 20 # 10
solutions =: b %. "(1 _) a
```
这段代码解决了方程组 a * v_i = b_i，针对十个随机向量。我认为可以在numpy中开发类似的概念。语法“(1 _)”表示从左侧操作数中取行，并对a的所有行（_表示无限）应用求解（在J中是%.）。在这个例子中，系统是 x+y=y0，x-y=y1。
因此，我建议可以使用类似于 numpy.linalg.solve(a, b, rank=(1, inf)) 的语法。
[1] https://news.ycombinator.com/item?id=43996431
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Recently there was a post [1] about numpy being not easy to work with when using arrays of shape greater than 2.<p><pre><code> One of the problems mentioned in [1] is that you can not use python loops because they are slow. In J you can solve for example 100 equations using the rank concept, this is a simple example:
a=: 2 2 $ 1 1 1 _1
b=: 10 2 $ ? 20 # 10
solutions =: b %. "(1 _) a
</code></pre>
That code solve the systems a * v_i = b_i for ten random vectors. I think a similar concept could be developed in numpy. The syntax "(1 _)" indicates to take the rows from the left operator and all (_ is infinite) of a apply solve (that is %. in J). In this case the system is x+y=y0, x-y=y1.<p>So I would suggest somthing like numpy.linalg.solve(a,b,rank=(1,inf))<p>[1] https://news.ycombinator.com/item?id=43996431