R中的这个非数字矩阵范围错误是什么?

What is this non numeric matrix extent error in R?

我正在尝试将一个函数应用到我的列表中,但是 returns 这个错误

"non numeric matrix extent error"

这是我的代码

错误发生在最后几行 该代码直到最后都运行良好,因此,我无法绘制我的图表 我在网上搜索过,但找不到任何有用的东西,而且我看不出代码有什么问题



#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()

#Question 4
nmean <- 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)
}
sequentialMeans <- lapply(lyst,nmean)

pdf("Question4.pdf",width=15,height=10)
for (i in 1:7)
{
  for (j in 1:7)
  {
    plot(y=sequentialMeans[[i]][,j],x=1:10000,xlab="n value",ylab="Values", main=paste("Alpha-Lambda:",colnames(lyst[[i]])[j]),type="l")
  }
}
dev.off()


您的代码的问题在于 nmean 函数的输入数据格式与行

相符
nmean <- 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)
}

是一个矩阵,您希望为其提供以下行中指定的伽马分布值向量

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)
  }
}

对于具有向量类型的 x,函数 ncol(x)nrow(x)return NULL。此外,也无法应用 ncol(x)

如果你想保存你的方法,你需要考虑将数据转换为矩阵格式,或者使用矢量格式,但使用矢量兼容函数 length(x) 作为矢量的长度和names(lyst) 的名字。


更新:

注释中的代码有效,但您必须更改 lapply 语句,因为您现在有一个可以直接用作 nmean 函数输入的矩阵。以下代码用于生成 sampleMeans 并避免您问题的原始错误消息。为了减少运行时间,它只需要 100 个样本。

#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(100,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 = TRUE) 
colnames(lyst) = c("100-0.1","100-0.5","100-1","100-2","100-5","100-10","100-100")
#Question 4
nmean <- 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)
}
sequentialMeans <- nmean(lyst)

另请注意,您需要调整Q4的代码,即剧情生成。以下代码有效。

pdf("Question4.pdf",width=15,height=10)
    for (i in 1:7)
    {
      for (j in 1:7)
      {
        plot(y=sequentialMeans[,j],x=1:700,xlab="n value",ylab="Values", main=paste("Alpha-Lambda:",colnames(lyst[,j]),type="l"))
      }
    }
    dev.off()

如果有帮助请告诉我。