检查两个 scipy.sparse.csr_matrix 是否相等 [英] Check if two scipy.sparse.csr_matrix are equal

问题描述

我想检查两个 csr_matrix 是相等的.

I want to check if two csr_matrix are equal.

如果我这样做:

x.__eq__(y)

我明白了:

raise ValueError("The truth value of an array with more than one "
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all().

然而，这很有效:

assert (z in x for z in y)

有没有更好的方法来做到这一点?也许使用一些 scipy 优化的函数代替?

Is there a better way to do it? maybe using some scipy optimized function instead?

非常感谢

推荐答案

我们可以假设它们的形状相同吗?

Can we assume they are the same shape?

In [202]: a=sparse.csr_matrix([[0,1],[1,0]])
In [203]: b=sparse.csr_matrix([[0,1],[1,1]])
In [204]: (a!=b).nnz==0   
Out[204]: False

这会检查不等式数组的稀疏性.

This checks the sparsity of the inequality array.

如果您尝试 a==b(至少第一次使用它)，它会给您一个效率警告.那是因为它必须测试所有这些零.它不能充分利用稀疏性.

It will give you an efficiency warning if you try a==b (at least the 1st time you use it). That's because it has to test all those zeros. It can't take much advantage of the sparsity.

您需要一个相对较新的版本才能使用这样的逻辑运算符.您是否尝试在某些 if 表达式中使用 x.__eq__(y) ，还是仅从该表达式中出错?

You need a relatively recent version to use logical operators like this. Were you trying to use x.__eq__(y) in some if expression, or did you get error from just that expression?

一般来说，您可能想先检查几个参数.相同的shape，相同的nnz，相同的dtype.你需要小心浮动.

In general you probably want to check several parameters first. Same shape, same nnz, same dtype. You need to be careful with floats.

对于密集数组 np.allclose 是测试相等性的好方法.如果稀疏数组不太大，那也不错

For dense arrays np.allclose is a good way of testing equality. And if the sparse arrays aren't too large, that might be good as well

np.allclose(a.A, b.A)

allclose 使用 all(less_equal(abs(x-y), atol + rtol * abs(y))).您可以使用 a-b，但我怀疑这也会发出效率警告.

allclose uses all(less_equal(abs(x-y), atol + rtol * abs(y))). You can use a-b, but I suspect that this too will give an efficiecy warning.

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