根据行逐步对Numpy Python矩阵进行排序 [英] Sort a Numpy Python matrix progressively according to rows
问题描述
我四处搜寻并试图找到似乎简单的问题的解决方案,但没有提出任何建议.问题是逐步根据矩阵的列对矩阵进行排序.因此,如果我有一个类似numpy的矩阵:
I have searched around and tried to find a solution to what seems to be a simple problem, but have come up with nothing. The problem is to sort a matrix based on its columns, progressively. So, if I have a numpy matrix like:
import numpy as np
X=np.matrix([[0,0,1,2],[0,0,1,1],[0,0,0,4],[0,0,0,3],[0,1,2,5]])
print(X)
[[0 0 1 2]
[0 0 1 1]
[0 0 0 4]
[0 0 0 3]
[0 1 2 5]]
我想根据第一列对其进行排序,然后对第二,第三等进行排序,以获得类似以下结果:
I would like to sort it based on the first column, then the second, the third, and so on, to get a result like:
Xsorted=np.matrix([[0,0,0,3],[0,0,0,4],[0,0,1,1],[0,0,1,2],[0,1,2,5]])
print(Xsorted)
[[0,0,0,3]
[0,0,0,4]
[0,0,1,1]
[0,0,1,2]
[0,1,2,5]]
虽然我认为可以通过命名列和所有类似的东西来对矩阵进行排序,但我更希望有一种排序方法,该方法与矩阵的大小无关.我正在使用Python 3.4,如果那很重要的话.
While I think it is possible to sort a matrix like this by naming the columns and all that, I would prefer to have a method for sorting that doesn't depend so much on how big the matrix is. I am using Python 3.4, if that is important.
任何帮助将不胜感激!
推荐答案
它不会特别快,但是您始终可以将行转换为元组,然后使用Python的sort:
It's not going to be particularly fast, but you can always convert your rows to tuples, then use Python's sort:
np.matrix(sorted(map(tuple, X.A)))
您也可以使用np.lexsort
,如此答案对尽管您应该使用实际数据进行测试以确保:
The lexsort approach appears to be faster, though you should test with your actual data to make sure:
In [20]: X = np.matrix(np.random.randint(10, size=(100,100)))
In [21]: %timeit np.matrix(sorted(map(tuple, X.A)))
100 loops, best of 3: 2.23 ms per loop
In [22]: %timeit X[np.lexsort(X.T[::-1])]
1000 loops, best of 3: 1.22 ms per loop
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