从numpy的矩阵构建一套巨蟒 [英] Constructing a python set from a numpy matrix

问题描述

我试图执行以下

>> from numpy import *
>> x = array([[3,2,3],[4,4,4]])
>> y = set(x)
TypeError: unhashable type: 'numpy.ndarray'

我怎么能轻松高效地从numpy的阵列创建一组？

How can I easily and efficiently create a set from a numpy array?

推荐答案

如果你想要一组元素，这里是另一个，可能是更快的方式：

If you want a set of the elements, here is another, probably faster way:

y = set(x.flatten())

PS： x.flat ， x.flatten之间进行比较（）和<$ C $之后C> x.ravel（） A 10x100阵列上，我发现他们都在大致相同的速度执行。对于一个3x3的阵列，最快的版本是迭代器版本：

PS: after performing comparisons between x.flat, x.flatten(), and x.ravel() on a 10x100 array, I found out that they all perform at about the same speed. For a 3x3 array, the fastest version is the iterator version:

y = set(x.flat)

我会建议，因为它是较少的内存便宜的版本（它扩展了很好的数组的大小）。

which I would recommend because it is the less memory expensive version (it scales up well with the size of the array).

PS ：还有一个numpy的功能，做类似的事情：

PS: There is also a NumPy function that does something similar:

y = numpy.unique(x)

这确实产生numpy的阵列相同的元素为集（x.flat），但作为一个numpy的数组。这是非常快的（几乎10倍快），但如果你需要一个设置，然后做设置（numpy.unique（X））比其他程序慢一点（建造了一套带有一个大的开销）。

This does produce a NumPy array with the same element as set(x.flat), but as a NumPy array. This is very fast (almost 10 times faster), but if you need a set, then doing set(numpy.unique(x)) is a bit slower than the other procedures (building a set comes with a large overhead).

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