Access scores for observation on linear discrimina

2019-04-14 02:53发布

问题:

library(MASS)
example(lda)
plot(z)

How can I access all the points in z? I want to know the values of every point along LD1 and LD2 depending on their Sp (c,s,v).

回答1:

What you are looking for is computed as part of the predict() method of objects of class "lda" (see ?predict.lda). It is returned as component x of the object produced by predict(z):

## follow example from ?lda
Iris <- data.frame(rbind(iris3[,,1], iris3[,,2], iris3[,,3]),
                   Sp = rep(c("s","c","v"), rep(50,3)))
set.seed(1) ## remove this line if you want it to be pseudo random
train <- sample(1:150, 75)
table(Iris$Sp[train])
## your answer may differ
##  c  s  v
## 22 23 30
z <- lda(Sp ~ ., Iris, prior = c(1,1,1)/3, subset = train)

## get the whole prediction object
pred <- predict(z)
## show first few sample scores on LDs
head(z$x)

the last line shows the first few rows of the object scores on the linear discriminants

> head(pred$x)
          LD1        LD2
40  -8.334664  0.1348578
56   2.462821 -1.5758927
85   2.998319 -0.6648073
134  4.030165 -1.4724530
30  -7.511226 -0.6519301
131  6.779570 -0.8675742

These scores can be plotted like so

plot(LD2 ~ LD1, data = pred$x)

producing the following plot (for this training sample!)



回答2:

When you calling the function plot(z), you are actually calling the function plot.lda - this is an S3 method. Basically, the object z has class lda:

class(z)

We can look at the actual function that is being used:

getS3method("plot", "lda")

This turns out to be rather involved. But the key points are:

x = z
Terms <- x$terms
data <- model.frame(x)
X <- model.matrix(delete.response(Terms), data)
g <- model.response(data)
xint <- match("(Intercept)", colnames(X), nomatch = 0L)
X <- X[, -xint, drop = FALSE]
means <- colMeans(x$means)
X <- scale(X, center = means, scale = FALSE) %*% x$scaling

We can no plot as before:

plot(X[,1], X[,2])

Proviso There might well be an easier way of getting what you want - I just don't know the lda function.