R 从文件名中存在数据的list.files中选择特定项(即条件项) game\u name
这里是对我用来提取多个文件的解决方案的改编。它根据R 从文件名中存在数据的list.files中选择特定项(即条件项) game\u name,r,R,这里是对我用来提取多个文件的解决方案的改编。它根据file.info()创建时间提取最近的文件。类似于此: game_name <- "fungame" day_from <- 7 # train from day_to <- 30 # predict to directory <- paste0("/home/rstudio-doug/analysis/radhoc/ltv_models/models/", game_name) list.files(director
file.info()
创建时间提取最近的文件。类似于此:
game_name <- "fungame"
day_from <- 7 # train from
day_to <- 30 # predict to
directory <- paste0("/home/rstudio-doug/analysis/radhoc/ltv_models/models/", game_name)
list.files(directory)
fungame_20200201_day_7_to_day_30.rds
fungame_20200221_day_7_to_day_30.rds
fungame_20200222_day_7_to_day_30.rds
fungame_20200201_day_7_to_day_60.rds
fungame_20200221_day_7_to_day_60.rds
fungame_20200222_day_7_to_day_90.rds
文件如果它是基于更新的,一个选项是out谢谢你的建议!当我这样做时,一切都是NA:file.info(list.files(directory))size isdir mode mtime ctime atime uid gid uname grname fungame 20200204\u day\u 7到day\u 30 NA NA fungame 20200205\u day\u 7到day\u 30 NA NA fungame\u 20200206\u day\u 7到day\u 30 NA NA NA
这可能是因为您没有使用全名=正确
。例如,如果您执行file.info(c(“A”、“B”))
操作,您将得到相同的结果(假设“A”和“B”不在您的工作目录中)。不过我猜。将ful.names=TRUE
添加到list.files()
命令是否有帮助?
files <- list.files(path_to_files, pattern = "day_7_to_day_30.rds$", full.names = TRUE)
files[which.max(file.info(files)$ctime)]