python - 二维矩阵 : Finding and deleting columns that are subsets of other columns

Question

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

import numpy as np
A = np.array([[1, 0, 0, 0, 0, 1],
              [0, 1, 1, 1, 1, 0],
              [1, 0, 1, 0, 1, 1],
              [1, 1, 0, 1, 0, 1],
              [1, 1, 0, 1, 0, 0],
              [1, 0, 0, 0, 0, 0],
              [0, 0, 1, 1, 1, 0],
              [0, 0, 1, 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]]

不过，对于大型矩阵，我确信有一种方法可以更快地完成此操作。一种想法是在我进行时消除子集列，这样我就不会检查已知是子集的列。另一个想法是对其进行矢量化，这样就没有 O(n^2) 操作。谢谢。

Answer 1

由于我实际处理的 A 矩阵是 5000x5000 并且密度大约为 4% 的稀疏矩阵，我决定尝试结合 Python 的“集合”对象的稀疏矩阵方法。总的来说，它比我原来的解决方案快得多，但我觉得我从矩阵 A 到集合列表 D 的过程并没有那么快。任何关于如何更好地做到这一点的想法都将受到赞赏。

解决方案

import numpy as np

A = np.array([[1, 0, 0, 0, 0, 1],
              [0, 1, 1, 1, 1, 0],
              [1, 0, 1, 0, 1, 1],
              [1, 1, 0, 1, 0, 1],
              [1, 1, 0, 1, 0, 0],
              [1, 0, 0, 0, 0, 0],
              [0, 0, 1, 1, 1, 0],
              [0, 0, 1, 0, 1, 0]])

rows,cols = A.shape
drops = np.zeros(cols).astype(bool)

# sparse nonzero elements
C = np.nonzero(A)

# create a list of sets containing the indices of non-zero elements of each column
D = [set() for j in range(cols)]
for i in range(len(C[0])):
    D[C[1][i]].add(C[0][i])

# find subsets, ignoring columns that are known to already be subsets
for i in range(cols):
    if drops[i]==True:
        continue
    col1 = D[i]
    for j in range(i+1,cols):
        col2 = D[j]
        if col2.issubset(col1):
            # I tried `if drops[j]==True: continue` here, but that was slower
            print "%d is a subset of %d" % (j, i)
            drops[j] = True
        elif col1.issubset(col2):
            print "%d is a subset of %d" % (i, j)
            drops[i] = True
            break

B = A[:, ~drops]
print B