我有一组CSV文件。我想打包它们并将数据导出到包含多个工作表的单个Excel文件。我将CSV文件作为一组数据框读入。
我的问题是如何在openxlsx中构建命令，我可以手动完成，但我有一个列表构造问题。具体来说，如何将数据框添加为命名列表的子组件，然后作为参数传递给write.xlsx()
好的，我首先在磁盘上列出CSV文件并在内存中生成一组数据帧...
# Generate a list of csv files on disk and shorten names...
filePath <- "../02benchmark/results/results_20170330/"
filePattern <- "*.csv"
fileListwithPath = list.files(path = filePath, pattern = filePattern, full.names = TRUE)
fileList = list.files(path = filePath, pattern = filePattern, full.names = FALSE)
datasets <- gsub("*.csv$", "", fileList)
datasets <- gsub("sample_", "S", datasets)
datasets
# Now generate the dataframes for each csv file...
list2env(
lapply(setNames(fileListwithPath, make.names(datasets)),
read.csv), envir = .GlobalEnv)
示例输出：
dput(datasets)
c("S10000_R3.3.2_201703301839", "S10000_T4.3.0_201703301843",
"S20000_R3.3.2_201703301826", "S20000_T4.3.0_201703301832", "S280000_R3.3.2_201704020847",
"S280000_T4.3.0_201704021100", "S290000_R3.3.2_201704020447",
"S290000_T4.3.0_201704020702", "S30000_R3.3.2_201703301803",
"S30000_T4.3.0_201703301817", "S310000_R3.3.2_201704012331",
"S310000_T4.3.0_201704020242", "S320000_R3.3.2_201704011827",
"S320000_T4.3.0_201704012128", "S330000_R3.3.2_201704011304",
"S330000_T4.3.0_201704011546", "S340000_R3.3.2_201704010652",
"S340000_T4.3.0_201704011010", "S350000_R3.3.2_201704010020",
"S350000_T4.3.0_201704010404", "S360000_R3.3.2_201703311819",
"S360000_T4.3.0_201703312134", "S370000_R3.3.2_201703310914",
"S370000_T4.3.0_201703311301", "S380000_R3.3.2_201703310134",
"S380000_T4.3.0_201703310509", "S390000_R3.3.2_201703301846",
"S390000_T4.3.0_201703302252", "S40000_R3.3.2_201703301738",
"S40000_T4.3.0_201703301752", "S50000_R3.3.2_201703301707", "S50000_T4.3.0_201703301724",
"S60000_R3.3.2_201703301624", "S60000_T4.3.0_201703301647", "S70000_R3.3.2_201703301535",
"S70000_T4.3.0_201703301602", "S80000_R3.3.2_201703301430", "S80000_T4.3.0_201703301508",
"S90000_R3.3.2_201703301324", "S90000_T4.3.0_201703301400")
wb <- createWorkbook()
saveWorkbook(wb, 'output.xlsx')
lapply(names(myList), function(x) write.xlsx(myList[[x]], 'output.xlsx', sheetName=x, append=TRUE))
问题是我可以手动创建列表结构并可以确认它是否有效但是我似乎无法自动构建列表。
myList <- sapply(datasets,function(x) NULL)
names(myList)
str(myList)
myList$S10000_R3.3.2_201703301839 <- eval(S10000_R3.3.2_201703301839)
从而：
> str(myList)
List of 40
$ S10000_R3.3.2_201703301839 :'data.frame': 43 obs. of 4 variables:
..$ function.: Factor w/ 42 levels "DF add random number vector",..: 30 25 38 42 36 39 40 29 26 22 ...
..$ user : num [1:43] 2.144 0.263 0.024 0.068 0.008 ...
..$ system : num [1:43] 0.63 0.065 0.001 0.004 0 ...
..$ elapsed : num [1:43] 12.274 1.104 0.047 0.115 0.009 ...
$ S10000_T4.3.0_201703301843 : NULL
$ S20000_R3.3.2_201703301826 : NULL
...
myList <- lapply( myList, function(x) eval(x) )
拉普利在这里做错了什么？上面的lapply（）不会遍历列表并将数据框附加到名称列表条目。
i.e. myList$S10000_R3.3.2_201703301839 <- eval(S10000_R3.3.2_201703301839)
> str(myList)
List of 40
$ S10000_R3.3.2_201703301839 :'data.frame': 43 obs. of 4 variables:
..$ function.: Factor w/ 42 levels "DF add random number vector",..: 30 25 38 42 36 39 40 29 26 22 ...
..$ user : num [1:43] 2.144 0.263 0.024 0.068 0.008 ...
..$ system : num [1:43] 0.63 0.065 0.001 0.004 0 ...
..$ elapsed : num [1:43] 12.274 1.104 0.047 0.115 0.009 ...
$ S10000_T4.3.0_201703301843 : NULL
$ S20000_R3.3.2_201703301826 : NULL
...
我错过了什么？所有的帮助感激不尽。是的，我很确定我错过了一些明显的东西......但是......我很难过。
我没有你的数据框，所以我无法测试，但下面的代码类似于我需要读取和写入Excel文件时使用的方法。下面的代码使用xlsx包，因为这是我熟悉的，但希望你可以调整它，如果你需要使用openxlsx。
library(xlsx)
首先，将文件读入列表。像这样的东西：
filePath <- "../02benchmark/results/results_20170330/"
filePattern <- "*.csv"
fileListwithPath = list.files(path = filePath,
pattern = filePattern,
full.names = TRUE)
fileList = list.files(path = filePath, pattern = filePattern, full.names = FALSE)
fileListwithPath = setNames( fileListwithPath,
list.files(path = filePath, pattern = filePattern))
df.list = lapply(fileListwithPath, read.csv)
# Now we rename the List Names for use in worksheets...
# Remove .csv and sample_ prefix used in filenames...
# Reult in workbook S<size>_<R version>_<date>
names(df.list) <- gsub("\\.csv$","", names(df.list))
names(df.list) <- gsub("sample_","S", names(df.list))
您现在有一个列表，其中每个元素都是一个数据框，每个元素的名称都是文件的名称。现在，让我们将每个数据框写入同一Excel工作簿中的不同工作表，然后将该文件另存为xlsx文件：
wb = createWorkbook()
lapply( names(df.list),
function(df) {
sheet = createSheet(wb, df)
addDataFrame(df.list[[df]], sheet = sheet, row.names = FALSE)
} )
saveWorkbook(wb, "My_workbook.xlsx")
我已将读取和写入csv文件分开以进行说明，但您可以将它们组合成一个函数，该函数读取每个单独的csv文件并将其写入单个Excel工作簿中的新工作表。
这是openxlsx的解决方案：
## create data;
dataframes <- split(iris, iris$Species)
# create workbook
wb <- createWorkbook()
#Iterate the same way as PavoDive, slightly different (creating an anonymous function inside Map())
Map(function(data, nameofsheet){
addWorksheet(wb, nameofsheet)
writeData(wb, nameofsheet, data)
}, dataframes, names(dataframes))
## Save workbook to excel file
saveWorkbook(wb, file = "file.xlsx", overwrite = TRUE)
..但是，openxlsx也可以使用write.xlsx，所以你可以给对象提供你的数据帧列表和文件路径，而openxlsx足够聪明，可以完成其余的...上面的解决方案与Map（ ）是否要以特定方式格式化工作表。