比较NumPy数组，使NaNs比较相等 [英] Comparing NumPy arrays so that NaNs compare equal

问题描述

有一个惯用的方法来比较两个NumPy数组，它们会将NaN彼此相等（但不等于NaN以外的任何其他）。

Is there an idiomatic way to compare two NumPy arrays that would treat NaNs as being equal to each other (but not equal to anything other than a NaN).

例如，我想要以下两个数组进行比较：

For example, I want the following two arrays to compare equal:

np.array([1.0, np.NAN, 2.0])
np.array([1.0, np.NAN, 2.0])

和以下两个数组来比较不等号：

and the following two arrays to compare unequal:

np.array([1.0, np.NAN, 2.0])
np.array([1.0, 0.0, 2.0])

以下操作可以实现：

np.all((a == b) | (np.isnan(a) & np.isnan(b)))

但它很笨重，并创建所有这些中间数组。

but it's clunky and creates all those intermediate arrays.

有一种方法更容易眼睛，使用内存？

Is there a way that's easier on the eye and makes better use of memory?

PS如果它有帮助，数组已知具有相同的形状和dtype。

P.S. If it helps, the arrays are known to have the same shape and dtype.

推荐答案

如果你真的关心内存使用有非常大的数组），那么你应该使用numexpr和下面的表达式将为你工作：

If you really care about memory use (e.g. have very large arrays), then you should use numexpr and the following expression will work for you:

np.all(numexpr.evaluate('(a==b)|((a!=a)&(b!=b))'))

我测试了它在长度为3e8的非常大的数组上，代码在我的机器上具有相同的性能

I've tested it on very big arrays with length of 3e8, and the code has the same performance on my machine as

np.all(a==b)

的内存

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