gini
entropy
计算 importance，比较和模型生成权重的一致性 import numpy as npdef split_gain(gini, n, gini1, n1, gini2, n2, t): return (n*gini - n1*gini1 - n2*gini2)*1.0/t# ginix3_gain = split_gain(0.6667, 150, 0, 50, 0.5, 100, 150) + split_gain(0.5, 100, 0.168, 54, 0.0425, 46, 150)x2_gain = split_gain(0.168, 54, 0.0408, 48, 0.4444, 6, 150) + split_gain(0.0425, 46, 0.4444, 3, 0, 43, 150)x = np.array([x2_gain, x3_gain])x = x / np.sum(x)print('gini:', x)# [ 0.05389858 0.94610142] computed# [ 0.05393633 0.94606367] sklearnx3_gain = split_gain(1.585, 150, 0, 50, 1, 100, 150) + split_gain(1, 100, 0.4451, 54, 0.1511, 46, 150)x2_gain = split_gain(0.4451, 54, 0.1461, 48, 0.9183, 6, 150) + split_gain(0.1511, 46, 0.9183, 3, 0, 43, 150)x = np.array([x2_gain, x3_gain])x = x / np.sum(x)print('entropy:', x)# [ 0.07060873 0.92939127] computed# [ 0.07060267 0.92939733] sklearn 总结 计算特征 对不存度减少的贡献，同时考虑 节点的样本量 对于某节点计算（**criterion可为gini或entropy**） 父节点 有样本量$n_0$，criterion为${c}_0$ 子节点1有样本量$n_1$，criterion为${c}_1$ 子节点2有样本量$n_2$，criterion为${c}_2$ 总样本个数为$T$ $gain = left(n_0*{c}_0 -n_1*{c}_1-n_2*{c}_2 right)/T$
转载于:https://www.cnblogs.com/porco/p/8440768.html