我正在尝试创建一个图,其中每个我都有一个密度图和一个直方图。对于这个例子,i = 1..3
我遇到的问题是创建要传递给grid.arrange的列表。但是,我这样做似乎以某种方式重复自己。
DF:
x1 x2 x3
1 108.28 17.05 1484.10
2 152.36 16.59 750.33
3 95.04 10.91 766.42
4 65.45 14.14 1110.46
5 62.97 9.52 1031.29
6 263.99 25.33 195.26
7 265.19 18.54 193.83
8 285.06 15.73 191.11
9 92.01 8.10 1175.16
10 165.68 11.13 211.15
X <- df
mu.X <- colMeans(X)
cov.X <- cov(X)
eg <- eigen(cov.X)
myprinboot = function(
X,
iter = 10000,
alpha = 0.05,
prettyPlot = T
){
# Find the dimensions of X
nrX <- dim(X)[1]
nx <- dim(X)[2]
# Make matrices of suitable sizes to hold the booted parameter estimates
# lambda
# each cov matrix will have nx lambdas
lambda.mat <- matrix(NA, nr = nx, nc = iter)
# e vectors nx components each and one vector per eigen value
# Each cov matrix will therefore produce a nx X nx matrix of components
Y.mat <- matrix(NA, nr = nx, nc = iter * nx)
# For loop to fill the matrices created above
for (i in 1:iter)
{
# ind will contain random integers used to make random samples of the X matrix
# Must use number of rows nrX to index
ind <- sample(1:nrX,nrX,replace=TRUE)
# eigen will produce lambdas in decreasing order of size
# make an object then remove extract the list entries using $
eigvalvec <- eigen(cov(X[ind,]))
lambda.mat[,i] <- eigvalvec$values
colstart <- 1 + nx * (i - 1)
colend <- colstart + nx - 1
Y.mat[,colstart:colend] = eigvalvec$vectors
}
if(prettyPlot){
p <- list()
i <- 0
for(j in 1:(2*nx))
{
if (j %% 2 == 0){
p[[j]] <- ggplot(NULL, aes(lambda.mat[i,])) +
geom_histogram(color = 'black', fill = 'green', alpha = .5) +
xlab(substitute(lambda[i])) +
ggtitle(substitute(paste("Histogram of the pc variance ", lambda[i])))
} else {
i <- i + 1
p[[j]] <- ggplot(NULL, aes(lambda.mat[i,])) +
geom_density(fill = 'blue', alpha = .5) +
xlab((substitute(lambda[i]))) +
ggtitle(substitute(paste("Density plot of the pc variance ", lambda[i])))
}
do.call(grid.arrange, p)
}
do.call(grid.arrange, p)
} else {
layout(matrix(1:(2*nx),nr=nx,nc=2,byrow=TRUE))
for(i in 1:nx)
{
plot(density(lambda.mat[i,]),xlab=substitute(lambda[i]),
main=substitute(paste("Density plot of the pc variance ", lambda[i])
))
hist(lambda.mat[i,],xlab=substitute(lambda[i]),
main=substitute(paste("Histogram of the pc variance ", lambda[i])))
}
}
library(rgl)
plot3d(t(lambda.mat))
list(lambda.mat = lambda.mat, Y.mat = Y.mat)
}
pc <- myprinboot(X = Y, iter=1000, alpha=0.5)
任何人都知道我做错了什么或这是不可能的?
答案 0 :(得分:0)
我不明白你的代码Jay,因为它似乎做了很多事情并使用了base和ggplot绘图,但是如果你想要的是为每个j创建一个组合的直方图和密度图,为什么不循环在j和内部,对于j循环做这样的事情:
d&lt; - 创建密度图,使其仅依赖于j
h&lt; - 创建直方图,使其仅依赖于j
p [[j]]&lt; - grid.arrange(d,h,ncol = 2)
然后,当你离开循环时,你将有一个对象p,它包含一个绘图列表,每个绘图由密度图和直方图组合而成。
然后你可以使用cowplot包(在安装之后)做这样的事情:
cowplot :: plot_grid(plotlist = p,ncol = 2)
其中可能需要更改列数。有关绘制图表列表的其他方法,请参见此处:How do I arrange a variable list of plots using grid.arrange?
我对你的问题知之甚少,无法理解你为什么以不同的方式对待j even和j的情况。但潜在的想法应该与我在这里建议的相同。
答案 1 :(得分:0)
我最终得到了如下工作。
getHist <- function(x, i){
lam <- paste('$\\lambda_', i, '$', sep='')
p <- qplot(x[i,],
geom="histogram",
fill = I('green'),
color = I('black'),
alpha = I(.5),
main=TeX(paste("Histogram of the pc variance ", lam, sep='')),
xlab=TeX(lam),
ylab="Count",
show.legend=F)
return(p)
}
getDens <- function(x, i){
lam <- paste('$\\lambda_', i, '$', sep='')
p <- qplot(x[i,],
geom="density",
fill = I('blue'),
alpha = I(.5),
main=TeX(paste("Density plot of the pc variance ", lam, sep='')),
xlab=TeX(lam),
ylab="Density",
show.legend=F)
return(p)
}
fp <- lapply(1:3, function(x) arrangeGrob(getHist(lambda.mat, x), getDens(lambda.mat, x), ncol=2))
print(marrangeGrob(fp, nrow = 3, ncol=1, top = textGrob("Lambda.mat Histogram and Density Plot",gp=gpar(fontsize=18))))