R中的日期值到日期格式

2 回答

• 0

  您的数据作为 factor 变量输入，可能通过 read.csv() 或未设置 StringsAsFactors=FALSE 时相关 .

  转换为角色无法修复 . 模拟示例：

  R> v <- factor(c("13-1-1990", "13-1-1991", "13-10-1959")) # your first three
  R> v
  [1] 13-1-1990  13-1-1991  13-10-1959
  Levels: 13-1-1990 13-1-1991 13-10-1959
  R> 
  R> d <- as.Date(as.character(v), format="%d-%m-%Y")
  R> d        ## key here were the as.character() and the correct format
  [1] "1990-01-13" "1991-01-13" "1959-10-13"
  R> 
  R> class(d) ## these are now Date objects
  [1] "Date"
  R> 
  R> d + 1    ## that we can compute with
  [1] "1990-01-14" "1991-01-14" "1959-10-14"
  R>
  

  回复于 2024-05-17T12:57:57+08:00
• 0

  我可以解决这个问题只是将数据类型从日期转换为字符串到数字，然后我使用“as.Date”函数来更改格式 .

  > New_His_data$BirthDate[1]
  [1] 36473
  1378 Levels: 13-1-1990 13-1-1991 13-10-1959 13-10-1973 13-10-1979 13-10-1988 13-10-1989 13-10-1994 13-11-1963 13-11-1970 ... 36473
  > as.Date(New_His_data$BirthDate[1],origin = "1899-12-30")
  Error in charToDate(x) : 
    character string is not in a standard unambiguous format
  > as.Date(strptime(New_His_data$BirthDate[1],"%d-%m-%Y"),origin ="1899-12-30")
  [1] NA
  > as.Date(as.numeric(as.character(New_His_data$BirthDate[1])),origin = "1899-12-30")
  [1] "1999-11-09"
  

  回复于 2024-05-17T12:57:57+08:00