矩阵所有成对行的相关系数和p值

问题描述

我有一个矩阵data，其中包含 m 行和 n 列.我曾经使用 :

I have a matrix data with m rows and n columns. I used to compute the correlation coefficients between all pairs of rows using np.corrcoef:

import numpy as np
data = np.array([[0, 1, -1], [0, -1, 1]])
np.corrcoef(data)

现在我也想看看这些系数的p值. np.corrcoef不提供这些； scipy.stats.pearsonr 可以.但是，scipy.stats.pearsonr不接受输入矩阵.

Now I would also like to have a look at the p-values of these coefficients. np.corrcoef doesn't provide these; scipy.stats.pearsonr does. However, scipy.stats.pearsonr does not accept a matrix on input.

有没有一种快速的方法来计算所有成对的行的系数和p值(例如，到达两个 m 乘 m 矩阵，其中一个相关系数，另一个具有相应的p值)而不必手动遍历所有对?

Is there a quick way how to compute both the coefficient and the p-value for all pairs of rows (arriving e.g. at two m by m matrices, one with correlation coefficients, the other with corresponding p-values) without having to manually go through all pairs?

推荐答案

我今天遇到了同样的问题.

I have encountered the same problem today.

经过一个半小时的搜索，我在numpy/scipy库中找不到任何代码可以帮助我做到这一点.

After half an hour of googling, I can't find any code in numpy/scipy library can help me do this.

所以我写了自己的 corrcoef

import numpy as np
from scipy.stats import pearsonr, betai

def corrcoef(matrix):
    r = np.corrcoef(matrix)
    rf = r[np.triu_indices(r.shape[0], 1)]
    df = matrix.shape[1] - 2
    ts = rf * rf * (df / (1 - rf * rf))
    pf = betai(0.5 * df, 0.5, df / (df + ts))
    p = np.zeros(shape=r.shape)
    p[np.triu_indices(p.shape[0], 1)] = pf
    p[np.tril_indices(p.shape[0], -1)] = p.T[np.tril_indices(p.shape[0], -1)]
    p[np.diag_indices(p.shape[0])] = np.ones(p.shape[0])
    return r, p

def corrcoef_loop(matrix):
    rows, cols = matrix.shape[0], matrix.shape[1]
    r = np.ones(shape=(rows, rows))
    p = np.ones(shape=(rows, rows))
    for i in range(rows):
        for j in range(i+1, rows):
            r_, p_ = pearsonr(matrix[i], matrix[j])
            r[i, j] = r[j, i] = r_
            p[i, j] = p[j, i] = p_
    return r, p

第一个版本使用np.corrcoef的结果，然后基于corrcoef矩阵的上三角值计算p值.

The first version use the result of np.corrcoef, and then calculate p-value based on triangle-upper values of corrcoef matrix.

第二个循环版本只是遍历行，请手动执行pearsonr.

The second loop version just iterating over rows, do pearsonr manually.

def test_corrcoef():
    a = np.array([
        [1, 2, 3, 4],
        [1, 3, 1, 4],
        [8, 3, 8, 5],
        [2, 3, 2, 1]])

    r1, p1 = corrcoef(a)
    r2, p2 = corrcoef_loop(a)

    assert np.allclose(r1, r2)
    assert np.allclose(p1, p2)

测试通过，它们是相同的.

The test passed, they are the same.

def test_timing():
    import time
    a = np.random.randn(100, 2500)

    def timing(func, *args, **kwargs):
        t0 = time.time()
        loops = 10
        for _ in range(loops):
            func(*args, **kwargs)
        print('{} takes {} seconds loops={}'.format(
            func.__name__, time.time() - t0, loops))

    timing(corrcoef, a)
    timing(corrcoef_loop, a)


if __name__ == '__main__':
    test_corrcoef()
    test_timing()

在Macbook上针对100x2500矩阵的性能

The performance on my Macbook against 100x2500 matrix

corrcoef_loop需要7.585600137710571秒的循环次数= 10

corrcoef_loop takes 7.585600137710571 seconds loops=10

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