python – 二维矩阵：查找和删除作为其他列的子集的列

发布时间：2020-12-20 13:12:27 所属栏目：Python 来源：网络整理

导读：我有一个问题,我想识别和删除逻辑矩阵中的列是其他列的子集.即[1,1]是[1,1,1]的子集;但[1,0]和[0,1]都不是彼此的子集.我写了一段快速代码来识别作为子集的列,它使用一对嵌套for循环进行(n ^ 2-n)/ 2次检查. import numpy as npA = np.array([[1,1],[0,0],[1,

我有一个问题,我想识别和删除逻辑矩阵中的列是其他列的子集.即[1,1]是[1,1,1]的子集;但[1,0]和[0,1]都不是彼此的子集.我写了一段快速代码来识别作为子集的列,它使用一对嵌套for循环进行(n ^ 2-n)/ 2次检查.

import numpy as np
A = np.array([[1,1],[0,0],[1,0]])
rows,cols = A.shape
columns = [True]*cols
for i in range(cols):
    for j in range(i+1,cols):
        diff = A[:,i]-A[:,j]
        if all(diff >= 0):
            print "%d is a subset of %d" % (j,i)
            columns[j] = False
        elif all(diff <= 0):
            print "%d is a subset of %d" % (i,j)
            columns[i] = False
B = A[:,columns]

解决方案应该是

>>> print B
[[1 0 0]
 [0 1 1]
 [1 1 0]
 [1 0 1]
 [1 0 1]
 [1 0 0]
 [0 1 1]
 [0 1 0]]

对于大型矩阵,我确信有一种方法可以更快地完成这项工作.一个想法就是在我去的时候消除子列,所以我不会检查已知为子集的列.另一个想法是对此进行矢量化,因此不要进行O(n ^ 2)运算.谢谢.

解决方法

这是使用 NumPy broadcasting的另一种方法 –

A[:,~((np.triu(((A[:,:,None] - A[:,None,:])>=0).all(0),1)).any(0))]

下面列出了详细的评论说明 –

# Perform elementwise subtractions keeping the alignment along the columns
sub = A[:,:]

# Look for >=0 subtractions as they indicate non-subset criteria
mask3D = sub>=0

# Check if all elements along each column satisfy that criteria giving us a 2D
# mask which represent the relationship between all columns against each other
# for the non subset criteria
mask2D = mask3D.all(0)

# Finally get the valid column mask by checking for all columns in the 2D mas
# that have at least one element in a column san the diagonal elements.
# Index into input array with it for the final output.
colmask = ~(np.triu(mask2D,1).any(0))
out = A[:,colmask]

（编辑：李大同）

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