R 将 excel 与不兼容的列组合

Question

我正在将多个 excel 文件与一对多工作表合并。他们每个人都有不同的专栏。我只对将工作表与地址信息结合起来感兴趣。如果工作表没有地址信息，我需要在生成的组合文件中记下它。如果我遇到问题，其中一个工作表有蔬菜但没有地址，另一个有地址信息..我正在使用下面的代码将它们放在一起。在我开始工作后，我会清理标准化它们并将它们放在一起。

dir_path <- "C:/temp/ConsigneeList/stuff4/"         # target directory where the xlsx files are located. 
re_file <- list.files(dir_path, pattern=".xls*")    # regex pattern to match the file name format, in this case 'test1.xlsx', 'test2.xlsx' etc.

read_sheets <- function(dir_path, file){
  xls_file <- paste0(dir_path, file)
  xls_file %>%
    excel_sheets() %>%
    set_names() %>%
    map_df(read_excel, path = xls_file, .id = 'sheet_name') %>% 
    mutate(file_name = file) %>% 
    select(file_name, sheet_name, everything())
}

number_of_excel_files<-length(file.list)
mybiggerlist<-vector('list',number_of_excel_files)
for(file in 1:length(mybiggerlist)) {

  mybiggerlist[[file]]<-read_sheets(dir_path, file.list[file])
  
}

我收到错误：错误：无法合并 Customer Quick REF$Order No 和 CH Belt$Order No 。我尝试使用 %>% mutate_all(as.character) 因为列本质上应该都是字符。关于如何解决这个问题有什么想法吗？或者，有没有一种方法可以跳过导入有问题的数据并生成一行来指示该工作表存在问题？谢谢！

Answer 1

尝试这样的事情：

dir_path <- "C:/temp/ConsigneeList/stuff4/"         # target directory where the xlsx files are located. 
re_file <- list.files(dir_path, pattern=".xls*")    # regex pattern to match the file name format, in this case 'test1.xlsx', 'test2.xlsx' etc.

read_sheets <- function(dir_path, file){
  xls_file <- paste0(dir_path, file)
  sheets <- xls_file %>%
    excel_sheets() %>%
    set_names() %>% ## not really sure if this is doing anything?
    map(read_excel, path = xls_file)
    
    # Now we have all the sheets in a list. 
    # Time to figure out which ones to combine
    # Use purrr::keep to only keep sheets that meet some condition
    # I just put in a wild guess, edit the test so that only sheets
    # you want are kept
    sheets <- purrr::keep(sheets, ~ "Address" %in% names(.))    
    
    bind_rows(sheets, .id = 'sheet_name') %>%
      mutate(file_name = file) %>% 
      select(file_name, sheet_name, everything())
}

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