Question

我想计算两个矩阵A和B的马修相关系数。循环遍历A的列，并计算该列和矩阵B的所有2000行的MCC，然后取最大索引。代码是：

import numpy as np
import pandas as pd
from sklearn.metrics import matthews_corrcoef as mcc

A = pd.read_csv('A.csv', squeeze=True)
B = pd.read_csv('B.csv', squeeze=True)

ind = {}
for col in A:
   ind[col] = np.argmax(list(mcc(B.iloc[i], A[col]) for i in range(2000)))
   print(ind[col])

我的问题是，这确实花费很长时间（每列一秒钟）。我看到R中几乎相同的代码运行得更快（例如5秒）。怎么会这样？我可以改善我的Python代码吗？

R代码：

A <- as.matrix(read.csv(file='A.csv'))
B <- t(as.matrix(read.csv(file='B.csv', check.names = FALSE)))
library('mccr')
C <- rep(NA, ncol(A))
for (query in 1:ncol(A)) {
    mcc <- sapply(1:ncol(B), function(i) 
           mccr(A[, query], B[, i]))
    C[query] <- which.max(mcc)
}

Answer 1

也许在 python 中使用 numpy 和 dot 产品试试这个

def compute_mcc(true_labels, pred_labels):
    """Compute matthew's correlation coefficient.

    :param true_labels: 2D integer array (features x samples)
    :param pred_labels: 2D integer array (features x samples)
    :return: mcc (samples1 x samples2)
    """
    # prep inputs for confusion matrix calculations
    pred_labels_1 = pred_labels == 1; pred_labels_0 = pred_labels == 0
    true_labels_1 = true_labels == 1; true_labels_0 = true_labels == 0
    
    # dot product of binary matrices
    confusion_dot = lambda a,b: np.dot(a.T.astype(int), b.astype(int)).T
    TP = confusion_dot(pred_labels_1, true_labels_1)
    TN = confusion_dot(pred_labels_0, true_labels_0)
    FP = confusion_dot(pred_labels_1, true_labels_0)
    FN = confusion_dot(pred_labels_0, true_labels_1)

    mcc = (TP * TN) - (FP * FN)
    denom = np.sqrt((TP + FP) * (TP + FN) * (TN + FP) * (TN + FN))
    
    # avoid dividing by 0
    denom[denom == 0] = 1

    return mcc / denom

计算矩阵的Matthew相关系数需要太长时间

1 个答案: