如何将 integer 转换为 R 中的日期（格式 YYYY）

Question

I have a big data set, filled with dates that are in integer form, and different types of integer form (YYYYMMDD, YYYYMM and YYYY)

I just want to get all of them to YYYY.

I first tried splitting up the data frame into three data frames with the different respective date (integer) forms. And then I tried to change the data form from integer to date.

Below is me splitting up the data and then all the various things i have tried and subsequently commented out.

Opera_A <- Opera_split$Date_Format_A
Opera_B <- Opera_split$Date_Format_B
Opera_C <- Opera_split$Date_Format_C


lubridate::dmy(Opera_C$Composer_Born)
#as.Date(Opera_C$Composer_Born, "%m/%d/%y")
#Composer_Born_C <- data.frame(Composer_Born)
#Opera_C <- Opera_C %>% mutate(df,dateTime=as.Date(Composer_Born, format = "%Y-%m-%d"))
#Opera_C <- transform(Opera_C,Composer_Born=as.Date(as.character(Integer),"%Y%m%d"))
# Opera_C$Composer_Born <- as.Date(Opera_C$Composer_Born, '%Y-%m-%d')

I keep getting Error in Opera_C$Composer_Born: $ operator is invalid for atomic vectors --- do i have to turn these vectors into dfs or can i just convert them directly?

Any help much appreciated --- I'm an R beginner!

Thank you

Answer 1

> library(anytime)
> anydate(c(20220210, 202202, 2022))
[1] "2022-02-10" "2022-02-01" "2022-01-01"
> 

However, YYYYMM is not a really date: it is indeterminate as it could be any day of that month, ditto for YYYY. So anydate guesses for you here.

Answer 2

As suggested by @caldwellst in the comments you can just pull the first 4 characters if they're all in the format you indicated. Then you can use as.Date(format = "%Y) to turn into actual date format and just fill in the M and D portions. Then use lubridate::year() to pull the year info.

library(tidyverse)
library(lubridate)

c(17701216, 177012, 1770) %>% 
  as.character() %>% 
  str_sub(1, 4) %>% 
  as.Date(format = "%Y") %>% 
  year()
#> [1] 1770 1770 1770

^{Created on 2022-02-10 by the reprex package (v2.0.1)}

Answer 3

If you want them to be recognized as dates, you could do this:

date <- as.numeric(19351225)

as.Date(paste(substr(as.character(date), 1, 4),"-01-01",sep=""),format="%Y-%m-%d")

Answer 4

Assuming that you actually just want an integer indicating the years and all your dates are after the year 1000, then this should work:

years_from_date_int <- function (i) {
  digits <- ceiling(log10(i))
  divide_by <- 10 ** (digits - 4)
  floor(i / divide_by)
}

print(years_from_date_int(10210104))
# [1] 1021

It also has the advantage that I don't have to import any libraries and it may be potentially much more efficient as it doesn't have to convert the ints to strings like many of the other answers.

Quick benchmark:

set.seed(12345)
samples <- runif(1000000, min=1000, max = 20200101)

start <- Sys.time()
for (s in samples) {
  years_from_date_int(s)
}
end <- Sys.time()
print("Direct int manipulation:")
print(end - start)

start <- Sys.time()
for (s in samples) {
  years_from_date_int_string_manipulation(s)
}
end <- Sys.time()
print("With string conversion:")
print(end - start)

# [1] "Direct int manipulation:"
# Time difference of 3.928902 secs
# [1] "With string conversion:"
# Time difference of 16.86341 secs