从 Numpy 矩阵构造 Python 集

2022-01-17 00:00:00 python numpy arrays set

问题描述

我正在尝试执行以下操作

I'm trying to execute the following

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

如何轻松高效地创建包含 Numpy 数组中所有元素的集合?

How can I easily and efficiently create a set with all the elements from the Numpy array?

解决方案

如果你想要一组元素，这里有另一种可能更快的方法:

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

y = set(x.flatten())

PS:在x.flat、x.flatten()和x.ravel()之间进行比较后 在 10x100 阵列上，我发现它们都以大致相同的速度执行.对于 3x3 数组，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)

我会推荐它，因为它是内存成本较低的版本(它可以很好地随着数组的大小而扩展).

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

PPS:还有一个 NumPy 函数可以做类似的事情:

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

y = numpy.unique(x)

这确实会生成一个与 set(x.flat) 具有相同元素的 NumPy 数组，但它是一个 NumPy 数组.这非常快(几乎快 10 倍)，但是如果你需要一个 set，那么执行 set(numpy.unique(x)) 会比另一个慢一点程序(构建一个集合需要很大的开销).

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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