r - 如何将数据框的名称传递给 Excel 工作表(使用 xlsx 包)

Question

我正在尝试使用 lapply 遍历数据帧列表并对每个数据帧执行自定义函数。在函数中，我试图根据数据集的名称命名工作表(使用 xlsx)。这可能吗？参见示例:

myList <- list(DataFrame1, DataFrame2, DataFrame3, DataFrame4)

require(xlsx)
export <- createWorkbook()

lapply(myList,
       ExcelExport <- function(dataset) {

nameDF <- deparse(substitute(dataset))

# Use another function and store the output               
DF <- as.data.frame(function2(dataset)) 
# Here is where I'm having trouble naming the worksheet according to the name of the Dataframe:               
wksht <- createSheet(wb=export, sheetName = paste("Dataset is ", nameDF, sep = ""))  
               addDataFrame(x=DF, sheet = wksht)


)
# Save it to an excel file (either existing or new) under a given name
saveWorkbook(export, "Export1.xlsx")

我从 Getting the name of a data frame 中找到了 deparse(substitute())但 lapply 似乎将我的数据框重命名为 X[[i]]，然后抛出 '[' 的无效字符错误

Answer 1

这里是经过编辑的代码，允许您访问列表节点的名称。请在发布前检查您的代码，它包含一些歧义。

myList <- list(DataFrame1 = data.frame(matrix(rnorm(100), 10, 10)), 
               DataFrame2 = data.frame(matrix(rnorm(100), 10, 10)), 
               DataFrame3 = data.frame(matrix(rnorm(100), 10, 10)), 
               DataFrame4 = data.frame(matrix(rnorm(100), 10, 10)))

require(xlsx)
export <- createWorkbook()

lapply(seq_along(myList), function(i) {
  ExcelExport <- function1(myList[[i]]) # Your code was incomplete here
  # You don't have object 'ExcelExport' anywhere esle in your code
  # so this step seems useless...

  # Now you have a full access to myList inside lapply 
  nameDF <- names(myList)[i]

  # Use another function and store the output               
  DF <- as.data.frame(function2(myList[[i]])) 

  wksht <- createSheet(wb=export, sheetName = paste("Dataset is ", nameDF, sep = ""))  
  addDataFrame(x=DF, sheet = wksht) # Btw, object 'wksht' is not defined in your code

})
# Save it to an excel file (either existing or new) under a given name
saveWorkbook(export, "Export1.xlsx")