R 中的這個非數字矩陣范圍錯誤是什么？

Question

我試圖將 function 應用到我的列表中，但它返回此錯誤

“非數字矩陣范圍錯誤”

這是我的代碼

錯誤發生在最后幾行代碼可以正常工作到最后，因此，我無法 plot 我在網上搜索過但找不到任何有用的圖表，我看不出代碼有什么問題

#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()

Answer 1

您的代碼的問題是nmean function 的輸入數據格式根據行

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)
}

是一個矩陣，你想給它提供伽馬分布值的向量，如以下行中所指定

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)
  }
}

對於具有向量類型的x ， function ncol ncol(x)和nrow(x)返回NULL 。 此外， ncol(x)也不可能應用。

如果您想保存您的方法，您需要考慮將數據轉換為矩陣格式，或者使用向量格式，但使用與向量兼容的函數length(x)作為向量的長度，使用names(lyst)作為向量的長度名字。

---

更新：

注釋中的代碼有效，但您必須更改lapply語句，因為您現在有一個矩陣，您可以直接將其nmean function 的輸入。 以下代碼用於生成sampleMeans並避免您問題的原始錯誤消息。 為了減少運行時間，它只需要 100 個樣本。

#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)

另請注意，您需要調整 Q4 的代碼，即 plot 代。 以下代碼有效。

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()

讓我知道這是否有幫助。