Question

我想计算一个向量与Python中数组的每一行之间的Pearson相关系数（假设为numpy和/或scipy）。由于实际数据阵列的大小和存储器约束，将不可能使用标准相关矩阵计算功能。这是我的天真实施：

import numpy as np
import scipy.stats as sps

np.random.seed(0)

def correlateOneWithMany(one, many):
    """Return Pearson's correlation coef of 'one' with each row of 'many'."""
    pr_arr = np.zeros((many.shape[0], 2), dtype=np.float64)
    pr_arr[:] = np.nan
    for row_num in np.arange(many.shape[0]):
        pr_arr[row_num, :] = sps.pearsonr(one, many[row_num, :])
    return pr_arr

obs, varz = 10 ** 3, 500
X = np.random.uniform(size=(obs, varz))

pr = correlateOneWithMany(X[0, :], X)

%timeit correlateOneWithMany(X[0, :], X)
# 10 loops, best of 3: 38.9 ms per loop

任何关于加速这一点的想法都将非常感谢！

Answer 1

模块scipy.spatial.distance实现“相关距离”，它只是减去相关系数的一个。您可以使用函数cdist来计算一对多距离，并通过从1中减去结果来获得相关系数。

以下是您的脚本的修改版本，其中包括使用cdist计算相关系数：

import numpy as np
import scipy.stats as sps
from scipy.spatial.distance import cdist

np.random.seed(0)

def correlateOneWithMany(one, many):
    """Return Pearson's correlation coef of 'one' with each row of 'many'."""
    pr_arr = np.zeros((many.shape[0], 2), dtype=np.float64)
    pr_arr[:] = np.nan
    for row_num in np.arange(many.shape[0]):
        pr_arr[row_num, :] = sps.pearsonr(one, many[row_num, :])
    return pr_arr

obs, varz = 10 ** 3, 500
X = np.random.uniform(size=(obs, varz))

pr = correlateOneWithMany(X[0, :], X)

c = 1 - cdist(X[0:1, :], X, metric='correlation')[0]

print(np.allclose(c, pr[:, 0]))

定时：

In [133]: %timeit correlateOneWithMany(X[0, :], X)
10 loops, best of 3: 37.7 ms per loop

In [134]: %timeit 1 - cdist(X[0:1, :], X, metric='correlation')[0]
1000 loops, best of 3: 1.11 ms per loop

加速Python中的一对多关联计算

1 个答案: