I want to combine two variables into one with a date format

Question

I have a data set with a character column for months (MONTH) and a numeric column indicating years (YEAR). In order to work with it as panel data, I ne

Answer 1

You could simplify the below, but it makes it easier to see what's going on:

library(lubridate)
library(tidyverse)

df2 <- df %>% 
  mutate(TIME = parse_date_time(paste0(MONTH, YEAR), orders = "%b%Y"),
         TIME = as.character(substr(TIME, 6, 7)),
         TIME = paste0(TIME, "-", YEAR))

This is using lubridate - the easiest way to parse dates in R IMO, dplyr from tidyverse and substr from base R.

If you want to keep the date column then just pipe in another mutate and call the new column something different.

Answer 2

If you wish to use a full-on Tidyverse solution, consider this combination of tidyr, and lubridate's parse_date_time:

library(tidyverse)
df <- tibble::tribble(
  ~STATE,      ~MONTH,      ~YEAR,   ~VALUE,
"California",     "JAN",      2018,      800,
"California",     "FEB",      2018,      780,
"California",     "MAR",      2018,      600,
"Minesota",       "JAN",      2018,      800,
"Minesota",       "FEB",      2018,      780,
"Minesota",       "MAR",      2018,      600)

df %>%
   tidyr::unite(TIME, c(MONTH, YEAR), sep = "-") %>%
   dplyr::mutate(TIME = lubridate::parse_date_time(TIME, "my"))
#> # A tibble: 6 x 3
#>   STATE      TIME                VALUE
#>   <chr>      <dttm>              <dbl>
#> 1 California 2018-01-01 00:00:00   800
#> 2 California 2018-02-01 00:00:00   780
#> 3 California 2018-03-01 00:00:00   600
#> 4 Minesota   2018-01-01 00:00:00   800
#> 5 Minesota   2018-02-01 00:00:00   780
#> 6 Minesota   2018-03-01 00:00:00   600

Also check out the following related question: Converting year and month ("yyyy-mm" format) to a date?

Answer 3

Combining Tim's response with an easy to use date package lubridate we get:

# This can handle months of JAN, FEB, ETC. Or it can handle months of 01,02,etc.
df$TIME <- lubridate::ymd(paste0(df$YEAR,df$MONTH,"01")) 

# or if you need it in MM-YYYY format:
df$TIME <- format(lubridate::ymd(paste0(df$YEAR,df$MONTH,"01")), "%m-%Y")

Answer 4

I would recommend handling this by going through bona-fide R dates, using as.Date to generate an R date, then using format to render the string output you want. Something like this:

df$TIME <- format(as.Date(paste0(df$MONTH, df$YEAR, "01"), format="%b%Y%d"), "%m-%Y")

I arbitrarily assign the first to each date in your data set, but this doesn't matter, because the call to format only includes the month and year.

Answer 5

In base R you could do something like:

transform(df,TIME = paste(sprintf('%02d',match(MONTH,toupper(month.abb))),YEAR,sep = '-'))[c(1,5,4)]
       STATE    TIME VALUE
1 California 01-2018   800
2 California 02-2018   780
3 California 03-2018   600
4        ...  NA-...   ...
5   Minesota 01-2018   800
6   Minesota 02-2018   780
7   Minesota 03-2018   600