期望最大化算法的数字例子

本文介绍了期望最大化算法的数字例子的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

谁能提供EM算法的一个简单的数字例子，因为我不知道给出的公式？一个非常简单的有4个或5个笛卡尔坐标会完全做到。

Could anyone provide a simple numeric example of the EM algorithm as I am not sure about the formulas given? A really simple one with 4 or 5 Cartesian coordinates would perfectly do.

推荐答案

你看这个： en.wikibooks/wiki/Data_Mining_Algorithms_In_R/Clustering/Expectation_Maximization_(EM)#A_simple_example

我也写在（编辑）一个简单的例子R A年前，不幸的是我无法找到它。我会稍后再试找到它。

I had also written a simple example in (edit)R a year ago, unfortunately I am unable to locate it. I'll try again to find it later.

编辑：这是 -

EM <- function() { ### Read file, get necessary cols dataFile <- read.csv("wine.csv", head = FALSE, sep = ",") sl <- dataFile[, 2] #sw <- dataFile[, 3] #pl <- dataFile[, 3] #pw <- dataFile[, 4] class <- dataFile[, 5] N <- length(sl) pi1 <- 0.5 ### Init ### rand1 <- floor(runif(1) * N) rand2 <- floor(runif(1) * N) mu1 <- sl[rand1] mu2 <- sl[rand2] mean1 <- sum(sl)/N sigma1 <- sum( (sl - mean1) ** 2) / N sigma2 <- sigma1 print(mu1) print(mu2) print(sigma1) print(sigma2) COUNTLIM <- 10 count <- 1 prevmu1 <- 0.0; prevmu2 <- 0.0; prevsigma1 <- 0.0; prevsigma2 <- 0.0; gamma <- array(0, length(sl)) while (count <= COUNTLIM) { gamma <- pi1 * dnorm(sl, mu2, sigma2)/ ( (1 - pi1) * dnorm(sl, mu1, sigma1) + pi1 * dnorm(sl, mu2, sigma2)) mu1 <- sum((1 - gamma) * sl) / sum(1 - gamma) mu2 <- sum((gamma) * sl) / sum(gamma) sigma1 <- sum((1 - gamma) * (sl - mu1) ** 2)/sum(1 - gamma) sigma2 <- sum((gamma) * (sl - mu2) ** 2)/sum(gamma) pi1 <- sum(gamma)/N print(c(mu1, mu2, sigma1, sigma2, pi1)) if (count == 1) { prevmu1 <- mu1; prevmu2 <- mu2; prevsigma1 <- sigma1; prevsigma2 <- sigma2; } else { val <- ((prevmu1 - mu1)*2 + (prevmu2 - mu2)*2 + (prevsigma1 - sigma1)*2 + (prevsigma2 - sigma2)*2) ** 0.5; print(c("val: " , val)) if (val <= 1) { break; } } count <- count + 1 } print(mu1) print(mu2) print(sigma1) print(sigma2) }