R中的这个非数字矩阵范围错误是啥？

Posted 2023-03-24

【中文标题】R中的这个非数字矩阵范围错误是啥？

【英文标题】：What is this non numeric matrix extent error in R?R中的这个非数字矩阵范围错误是什么？

【发布时间】：2020-07-16 21:20:11

【问题描述】：

我试图将一个函数应用到我的列表中，但它返回此错误

“非数值矩阵范围误差”

这是我的代码

错误发生在最后几行 代码到最后都可以正常工作，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，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()

【参考方案1】：

你的代码的问题是nmean函数的输入数据格式根据行

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)

是一个矩阵，您希望向它提供伽马分布值的向量，如下行中指定的那样

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)和nrow(x)return NULL。此外，ncol(x)也不能申请。

如果您想保存您的方法，您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) 来获取向量的长度，names(lyst) 来获取向量的长度名字。

---

更新：

cmets 中的代码有效，但您必须更改lapply-statement，因为您现在有一个矩阵，您可以直接将其用作nmean 函数的输入。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)

还要注意，您需要调整 Q4 的代码，即绘图生成。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()

如果这有帮助，请告诉我。

【讨论】：

类似lyst <- matrix(unlist(lyst), ncol = 7, byrow = T)

是的，这是一种可能性。但请注意，您仍需要事先提供 colnames(lyst) 才能使您的功能正常工作！

嗯，我在colnames(lyst) <- c("100-0.1","100-0.5","100-1","100-2","100-5","100-10","100-100") 中添加了但它仍然为我返回了同样的错误

@user12197328 我更新了答案以反映您的评论。

感谢您的大力回复，感谢您的帮助，但它确实修复了我收到的错误，但现在它给了我另一个错误Error in xy.coords(x, y, xlabel, ylabel, log) : object 'sequentialMeans' not found