在大型数据框中转换为日期时，如何为缺少日期和月份的混乱日期数据指定日期和月份

Question

I have a large date frame of over 100k rows. The date column contains dates in multiple formats such as "%m/%d/%Y" , "%Y-%m" , "%Y" , and "%Y-%m-%d" . I can convert these all to dates with parse_date_time() from lubridate .

dates <- c("05/10/1983","8/17/2014","1953-12","1975","2001-06-17")

parse_date_time(dates, orders = c("%m/%d/%Y","%Y-%m","%Y","%Y-%m-%d"))

[1] "1983-05-10 UTC" "2014-08-17 UTC" "1953-12-01 UTC" "1975-01-01 UTC" "2001-06-17 UTC"

But as you can see, this sets dates with missing day to the first of the month and dates with missing month and day to the first of the year. How can I set those to the 15th and June 15th, respectively?

Answer 1

Use nchar to check the dates vector and paste what is missing.

library(lubridate)

dates <- c("05/10/1983","8/17/2014","1953-12","1975","2001-06-17")


dates <- ifelse(nchar(dates) == 4, paste(dates, "06-15", sep = "-"),
             ifelse(nchar(dates) == 7, paste(dates, 15, sep = "-"), dates))
dates
#[1] "05/10/1983" "8/17/2014"  "1953-12-15" "1975-06-15"
#[5] "2001-06-17"

parse_date_time(dates, orders = c("%m/%d/%Y","%Y-%m","%Y","%Y-%m-%d"))
#[1] "1983-05-10 UTC" "2014-08-17 UTC" "1953-12-15 UTC"
#[4] "1975-06-15 UTC" "2001-06-17 UTC"

Another solution would be to use an index vector, also based on nchar .

n <- nchar(dates)
dates[n == 4] <- paste(dates[n == 4], "06-15", sep = "-")
dates[n == 7] <- paste(dates[n == 7], "15", sep = "-")

dates
#[1] "05/10/1983" "8/17/2014"  "1953-12-15" "1975-06-15"
#[5] "2001-06-17"

As you can see, the result is the same as with ifelse .

Answer 2

Here's another way of doing that - based on orders :

library(lubridate)
dates <- c("05/10/1983","8/17/2014","1953-12","1975","2001-06-17")

parseDates <- function(x, orders = c('mdY', 'dmY', 'Ymd', 'Y', 'Ym')){
  fmts <- guess_formats(x, orders = orders)
  dte <- parse_date_time(x, orders = fmts[1], tz = 'UTC')
  if(!grepl('m', fmts[1]) ){
    dte <- dte + days(165)
    return(dte)
  }
  if(!grepl('d', fmts[1]) ){
    dte <- dte + days(14)
  }
  return(dte)
}

output

> parseDates(dates[4])
[1] "1975-06-15 UTC"
> parseDates(dates[3])
[1] "1953-12-15 UTC"

This way for different date formats you only need to change the orders argument while the rest is done using lubridate .

Hope this is helpful!