从 label python 列表创建关联矩阵的快速方法？

Question

I have an array y, len(y) = M that contains values from 0 -> N . For example, with N = 3 :

y = [0, 2, 0, 1, 2, 1, 0, 2]

Incidence matrix A is defined as followed:

• Size MxM
• A(i,j) = 1 if y(i) == y(j)
• A(i,j) = 0 if y(i) != y(j)

A simple algorithm would be:

def incidence(y):
    M = len(y)
    A = np.zeros((M,M))
    for i in range(M):
        for j in range(M):
            if y[i]==y[j]:
                A[i,j] = 1
    return A

But this is very slow. Is there any way to do this faster? Using list comprehension or vectorization, for example.

Answer 1

You can take advantage of numpy broadcasting to gain some efficiency here over our python by simply asking if y equals its transpose:

import numpy as np

y = np.array([1, 2, 1, 0, 0, 1, 2])

def mat_me(y):
    return (y == y.reshape(-1, 1)).astype(int)

mat_me(y)

which produces:

array([[1, 0, 1, 0, 0, 1, 0],
       [0, 1, 0, 0, 0, 0, 1],
       [1, 0, 1, 0, 0, 1, 0],
       [0, 0, 0, 1, 1, 0, 0],
       [0, 0, 0, 1, 1, 0, 0],
       [1, 0, 1, 0, 0, 1, 0],
       [0, 1, 0, 0, 0, 0, 1]])

for comparison:

y = np.random.choice([1, 2, 3], size=3000)

def mat_me_py(y):
    return (y == y.reshape([-1, 1])).astype(int)

%timeit mat_me_py(y)  
# 28.6 ms ± 1.11 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

vs.

y = np.random.choice([1, 2, 3], size=3000)
y = list(y)

def mat_me_py(y):
    return [[int(a == b) for a in y] for b in y]

%timeit mat_me_py(y)
# 4.16 s ± 213 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

The difference will become very pronounced on larger lists.