Question

我尝试使用此注释https://web.stanford.edu/class/stats202/content/lec9.pdf（第34页）

从头开始实施LDA

为了检查我的实现，我将我的先验，组均值和线性判别系数与MASS库中的lda()函数进行比较。我的先验和组意味着与lda()生成的值匹配。但是，我的系数不同。这是玩具示例：

group1 = replicate(3, rnorm(10, mean = 1))
group2 = replicate(3, rnorm(15, mean = 2)) 
x = rbind(group1, group2)
colnames(x) = c(1, 2, 3)
y = matrix(rep(1, 10), ncol = 1)
y = rbind(y, matrix(rep(2, 15), ncol = 1))
colnames(y) = 'y'

library(MASS)
xy = cbind(x, y)
lda.fit = lda(y ~ ., as.data.frame(xy))

LDA <- function(x, y) {
    group1_index = which( y == 1 )
    group2_index = which( y == 2 )
    #priors:
    prior_group1 = length(group1_index) / length(y)
    prior_group2 = length(group2_index) / length(y)
    print("Prior probabilities of groups:")
    print(c(prior_group1, prior_group2))
    #means:
    mean_group1 = colMeans(x[group1_index, ])
    mean_group2 = colMeans(x[group2_index, ])
    print("Group means:")
    print(rbind(mean_group1, mean_group2))
    #discriminant coefficients:
    x[group1_index, ] = sweep(x[group1_index, ], 2, mean_group1, "-")
    x[group2_index, ] = sweep(x[group2_index, ], 2, mean_group2, "-")
    sigma = solve(cov(x))
    disc_coeff = sigma %*% mean_group1 
    print("Coefficients of linear discriminants:")
    print(disc_coeff)
}

LDA(x, y)

因此，我计算偏差的向量并估计协方差矩阵，如幻灯片中所述。以下是我的实施的输出和lda()：

> LDA(x, y)
[1] "Prior probabilities of groups:"
[1] 0.4 0.6

[1] "Group means:"
                    1         2        3
mean_group1 0.9886488 0.7906502 1.228568
mean_group2 1.9180531 1.9603046 2.175800

[1] "Coefficients of linear discriminants:"
      [,1]
1 2.449038
2 2.687901
3 3.108110

> lda.fit
Call:
lda(y ~ ., data = as.data.frame(xy))

Prior probabilities of groups:
  1   2 
0.4 0.6 

Group means:
        `1`       `2`      `3`
1 0.9886488 0.7906502 1.228568
2 1.9180531 1.9603046 2.175800

Coefficients of linear discriminants:
          LD1
`1` 0.7204303
`2` 1.0612863
`3` 1.0114053

我试着查看lda()函数的源代码，然而，对于新手R程序员来说这是非常复杂的。

有人可以建议是否应该以不同的方式计算线性判别系数？或者，lda()可能进行了一些后期处理，我没有在实现中加入？

实施线性判别分析

0 个答案: