如何更改我的 x 和 y 值以适合

Question

这是我的代码 所以问题现在代码有效，但只有在行中 lyst[[elementname]] <- rgamma(10000,i,j) 10000 设置为 100 并在行中 plot(y=sample_mean_q4[,j],x=1:7000,xlab="n value",ylab="Values", main=paste("Alpha-Lambda:",colnames(lyst[,j]),type="l")) x=1:1000 的值设置为 x=1:700 否则我会收到错误“xy.coords(x, y, xlabel, ylabel, log) 中的错误： 'x' 和 'y' 长度不同"

我需要值为 10000 和 x=1=1:1000

#Question 1
set.seed(10000)

v <- c(0.1,0.5,1,2,5,10,100)

lyst <- list()

for(i in v)
{
  for(j in v)
  {
    elementname <- paste0(as.character(i),"-",as.character(j))
    print(elementname)
    lyst[[elementname]] <- rgamma(10000,i,j)
  }
}
#Question 2
pdf("Question2.pdf",width = 20, height = 10)
par(mfcol=c(7,7))
for(x in names(lyst))
{
  hist(lyst[[x]],
       xlab = "Value",
       main = paste("Alpha-Lambda:",x))
}
dev.off()

#Question 3
theoretical_mean <- matrix(ncol=7,nrow=7,dimnames=list(as.character(v), as.character(v)))
theoretical_var <- matrix(ncol=7,nrow=7,dimnames=list(as.character(v), as.character(v)))
for (i in 1:7)
{
  for (j in 1:7)
  {
    theoretical_mean[j,i] <- as.character(v[i]/v[j])
    theoretical_var[j,i] <- as.character(v[i]/(v[j]^2))
  }
}

sample_mean <-lapply(lyst, mean)
sample_mean <- as.data.frame(matrix(unlist(sample_mean),nrow = 7, ncol = 7, byrow = T))
sample_mean <- round(sample_mean,digits = 3)
sample_mean <- data.matrix(sample_mean, rownames.force = NA)

sample_var <-lapply(lyst, var)

sample_var <- as.data.frame(matrix(unlist(sample_var),nrow = 7, ncol = 7, byrow = T))
sample_var <- round(sample_var,digits = 3)
sample_var <- data.matrix(sample_var, rownames.force = NA)

theor_sample_mean <- matrix(paste(theoretical_mean, sample_mean, sep=" - "),nrow=7,dimnames = dimnames(theoretical_var))
theor_sample_var <- matrix(paste(theoretical_var, sample_var, sep=" - "),nrow=7,dimnames= dimnames(theoretical_var))

sink("Q3.txt")
cat("Theoretical Mean vs. Sample Mean:\n")
print(as.table(theor_sample_mean))
cat("\n")
cat("Theoretical Variance vs. Sample Variance:\n")
print(as.table(theor_sample_var))
sink()

lyst = matrix(unlist(lyst), ncol = 7, byrow = F) 
colnames(lyst) = c("100-0.1","100-0.5","100-1","100-2","100-5","100-10","100-100")
#Question 4
q4mean <- function(x)
{
  m <- matrix(nrow=nrow(x))
  for (j in 1:ncol(x))
  {
    v <- c()
    for(i in 1:nrow(x))
    {
      v <- c(v,mean(x[1:i,j]))
    }
    m <- cbind(m,v)
  }
  m <- m[,-1]
  colnames(m) <- colnames(x)
  rownames(m) <- NULL
  return(m)
}

sample_mean_q4 <- q4mean(lyst)
pdf("Question4.pdf",width=15,height=10)
for (i in 1:7)
{
  for (j in 1:7)
  {
    plot(y=sample_mean_q4[,j],x=1:1000,xlab="n value",ylab="Values", main=paste("Alpha-Lambda:",colnames(lyst[,j]),type="l"))
  }
}
dev.off()

Answer 1

请不要在一篇文章中添加很多问题。确定问题并在每个帖子中添加一个。您的代码中的错误与问题 4 有关。

# Select names of the column to be included in the plot
selected_cols <- c("100-0.1","100-0.5","100-1","100-2","100-5","100-10","100-100")

# this is equivalent to: `lyst = matrix(unlist(lyst), ncol = 7, byrow = F) `
# I used abc as name instead of lyst
abc <- sapply(selected_cols, function(x) lyst[[x]])   

sample_mean_q4 <- q4mean(abc)

# draw plot
for(x in selected_cols){
  pdf(paste0("Question4", "_", x, ".pdf"),width=15,height=10)
  plot( x=1:1000, y = sample_mean_q4[1:1000, x], 
        xlab="n value", ylab="Values", main=bquote(alpha - lambda == .(x) ), type="l")
  dev.off()
}

数据：

q4mean <- function(x)
{
  m <- matrix(nrow=nrow(x))
  for (j in 1:ncol(x)) {
    v <- c()
    for(i in 1:nrow(x)){
      v <- c(v,mean(x[1:i,j]))
    }
    m <- cbind(m,v)
  }
  m <- m[,-1]
  colnames(m) <- colnames(x)
  rownames(m) <- NULL
  return(m)
}

set.seed(10000)
v <- c(0.1,0.5,1,2,5,10,100)
lyst <- list()
for(i in v){
  for(j in v){
    elementname <- paste0(as.character(i),"-",as.character(j))
    print(elementname)
    lyst[[elementname]] <- rgamma(10000,i,j)
  }
}

Answer 2

你知道 R 的回收功能吗？你试过把问题分解成更小的块吗？