最近在做特征选择,要考量几个特征的相关性,想找这个方法的描述,发现很难在网页上搜到。以下为整合的:
[11] 王黎明, 吴香华, 赵天良,等. 基于距离相关系数和支持向量机回归的PM_(2.5)浓度滚动统计预报方案[J]. 环境科学学报, 2017,37(4):1268-1276.(我是从这篇论文上找的,*上有更细致的,可惜我看不下去啊)
下为python程序:
原文:https://gist.github.com/satra/aa3d19a12b74e9ab7941
from scipy.spatial.distance import pdist, squareform
import numpy as np
from numbapro import jit, float32
def distcorr(X, Y):
""" Compute the distance correlation function
>>> a = [1,2,3,4,5]
>>> b = np.array([1,2,9,4,4])
>>> distcorr(a, b)
0.762676242417
"""
X = np.atleast_1d(X)
Y = np.atleast_1d(Y)
if np.prod(X.shape) == len(X):
X = X[:, None]
if np.prod(Y.shape) == len(Y):
Y = Y[:, None]
X = np.atleast_2d(X)
Y = np.atleast_2d(Y)
n = X.shape[0]
if Y.shape[0] != X.shape[0]:
raise ValueError('Number of samples must match')
a = squareform(pdist(X))
b = squareform(pdist(Y))
A = a - a.mean(axis=0)[None, :] - a.mean(axis=1)[:, None] + a.mean()
B = b - b.mean(axis=0)[None, :] - b.mean(axis=1)[:, None] + b.mean()
dcov2_xy = (A * B).sum()/float(n * n)
dcov2_xx = (A * A).sum()/float(n * n)
dcov2_yy = (B * B).sum()/float(n * n)
dcor = np.sqrt(dcov2_xy)/np.sqrt(np.sqrt(dcov2_xx) * np.sqrt(dcov2_yy))
return dcor