R中具有不同输出名称的多个文件进程

Question

我有一个 csv 文件的数字数据集（比如 50 个文件）：crasha、crashabd、crashd、... 我编写了一个函数来对单个数据进行一些更改和分析。我想要一个动态的输出名称。例如，我想将 newcrasha、newcrashabd、newcrashd 和……作为输出 csv 文件。确实，我想获取导入文件的名称并将其用作输出文件名？ 例如：

filenames <- list.files(path = "D:/health/car crash/", pattern = "csv",full.names = TRUE)

analyze <- function(filename) {

  # Input is character string of a csv file.
  crash <- read.csv(file = filename, header = TRUE)

    #merg and summation (crashcounter and NUMBER_INJURED)

newcrash<-crash %>% group_by(COLLISION_DATE) %>% summarise(crashcounter = sum(crashcounter), NUMBER_INJURED = sum(NUMBER_INJURED))
   write.csv( newcrash, "D://health//car crash// newcrash.csv", row.names = FALSE)
 }
filenames <- filenames[1:50]
for (f in filenames) {
  analyze(f)
}

感谢您的帮助

Answer 1

按照@mhovd 的建议试试这个：

filename <- list.files(path = "D:/health/car crash/", pattern = "csv",full.names = TRUE)

analyze <- function(filename) {
  
  # Input is character string of a csv file.
  crash <- read.csv(file = filename, header = TRUE)
  
  #merg and summation (crashcounter and NUMBER_INJURED)
  newcrash<-crash %>% group_by(COLLISION_DATE) %>% summarise(crashcounter = sum(crashcounter), NUMBER_INJURED = sum(NUMBER_INJURED))
  new.name <- paste0("D:/health/car crash/new",basename(tools::file_path_sans_ext(filename)),".csv")
  write.csv( newcrash, file=new.name, row.names = FALSE)
}

lapply(filename[1:50], analyze)