Question

正如标题所示，我想做上述事情。示例如下：

library(stringr)
library(XLConnect)

df <- data.frame(do.call("rbind", lapply(1:10, function(i) rnorm(10))))
df.list <- rep(list(df), 10)
names(df.list) <- paste("DataFrame", str_pad(1:length(df.list), 2, pad = "0"), sep = "")

df.list.workbook <- loadWorkbook("df.list.workbook.xlsx", create = TRUE)
lapply(1:length(df.list), function(i) createSheet(df.list.workbook, name = names(df.list[i])))
lapply(df.list[1:length(df.list)], function(i) writeWorksheet(df.list.workbook, i, sheet = names(i)))

最后一行是它引发错误的地方：

Error: IllegalArgumentException (Java): Sheet index (-1) is out of range (0..9)

要解决此问题，我尝试了：

lapply(df.list[1:length(df.list)], function(i) print(names(i)))

并意识到列的名称正被传递给工作表变量。任何想法如何克服这个？

Answer 1

您需要lapply而不是列表：

lapply(seq_along(df.list), function(i) writeWorksheet(df.list.workbook, df.list[[i]], sheet = names(df.list)[i]))

Answer 2

XLConnect在许多地方被矢量化。在您的情况下，这适用于createSheet和writeWorksheet，因此您可以写：

df.list.workbook <- loadWorkbook("df.list.workbook.xlsx", create = TRUE)
createSheet(df.list.workbook, name = names(df.list))
writeWorksheet(df.list.workbook, data = df.list, sheet = names(df.list))
saveWorkbook(df.list.workbook)

使用列表名称作为使用XLConnect的工作表的名称

2 个答案: