将世纪添加到 2 位数的年份

Question

I currently have a df that looks like

        STA YR MO DA MAX       date
58716 33013 43  3 11  60 0043-03-11
58717 33013 43  3 12  55 0043-03-12
58718 33013 43  3 13  63 0043-03-13
58719 33013 43  3 14  50 0043-03-14
58720 33013 43  3 15  58 0043-03-15
58721 33013 43  3 16  63 0043-03-16

I did df$date <- as.Date(with(df, paste(YR, MO, DA,sep="-")), "%Y-%m-%d") as you can see to get the date column, but clearly because there's no '19' in front of the year column, the year in the date comes out wacky. These are all 19xx dates. What would be a good way to fix this?

Answer 1

Try

df$date <- as.Date(with(df, paste(1900+YR, MO, DA,sep="-")), "%Y-%m-%d")

Answer 2

You should use %y since you have two digit year.

df$date <- as.Date(with(df, paste(YR, MO, DA,sep="-")), "%y-%m-%d")

However, this doesn't solve your problem since anything less than 69 is prefixed with 20 in 2 digit-years so 43 becomes 2043.

If you know that all your years are in the form of 19XX , you can do

df$date <- as.Date(with(df, sprintf('19%d-%d-%d', YR, MO, DA)))

Answer 3

If your years contain a mixture of 2-digit years from more than one century, then this code converts them all into valid dates in the past (no future dates).

dates_y2Y <- function(y,m,d) {
  library(stringr)
  y <- stringr::str_pad(y, width=2, pad="0")
  m <- stringr::str_pad(m, width=2, pad="0")
  d <- stringr::str_pad(d, width=2, pad="0")

  toyear <- format(Sys.Date(), "%y")
  tomnth <- format(Sys.Date(), "%m")
  today  <- format(Sys.Date(), "%d")

  as.Date(
    ifelse(y<toyear | y==toyear & m<tomnth | y==toyear & m==tomnth & d<=today,
           as.Date(paste(y,m,d,sep="-"), format="%y-%m-%d"),
           as.Date(paste(paste0("19",y),m,d,sep="-"), format="%Y-%m-%d"))
    , origin="1970-01-01")
}

df$date <- dates_y2Y(df$YR, df$MO, df$DA)
df

---

    STA YR MO DA       date
1 33013 23  1 31 1923-01-31
2 33013 43  2 30       <NA>
3 33013 63  5  5 1963-05-05
4 33013 83  7 27 1983-07-27
5 33013  3 12  9 2003-12-09
6 33013 20  4 21 2020-04-21
7 33013 20  4 22 1920-04-22

---

Data :

df <- structure(list(STA = c(33013L, 33013L, 33013L, 33013L, 33013L, 
33013L, 33013L), YR = c(23L, 43L, 63L, 83L, 3L, 20L, 20L), MO = c(1L, 
2L, 5L, 7L, 12L, 4L, 4L), DA = c(31L, 30L, 5L, 27L, 9L, 21L, 
22L), date = structure(c(-17137, NA, -2433, 4955, 12395, 18373, 
-18151), class = "Date")), row.names = c(NA, -7L), class = "data.frame")

Answer 4

another solution

library(lubridate)
df %>% 
  mutate(date = make_date(year = 1900 + YR, month = MO, day = DA))

Answer 5

Another option with sprintf

df$date <- as.Date(do.call(sprintf, c(f = '19%d-%d-%d', df[2:4])))

---

Or with unite

library(dplyr)
library(tidyr)
library(stringr)
df %>%
  mutate(YR = str_c('19', YR)) %>%
  unite(date, YR, MO, DA, sep="-", remove = FALSE) %>%
  mutate(date = as.Date(date))