Replacing character dates with Date values-CodePudding

I have some data where the final column has dates (extracted from file name where they are stored originally), and for all intents and purposes looks like this:

Person <- c("Person A", "Person B", "Person C", "Person A", "Person B", "Person C")
School <- c("School A", "School B", "School C", "School D", "School E", "School F" )
Date <- c("Aug 2019", "Oct 2019","Sep 2019","Nov 2019","Dec 2019","Jul 2019")
df <- data.frame(Person, School, Date)

The dates come from the file name, I added the file name as a column and then use the following to extract the date value as it appears in the example above.

mutate(Date = substr(File, 44, 51))

From this point I need to convert it from a character "Aug 2019" to a date of any format, so I can create some timeseries, I'll only use month/day on the X axis anyway and there are no days specified in the data, it's all month/year.

I can see that R recognises it as a date when I parse it using

parse_date_time(data3$Date, orders = c("bY"))

and it gives the result

> parse_date_time(df$Date, orders = c("bY"))
[1] "2019-08-01 UTC" "2019-10-01 UTC" "2019-09-01 UTC" "2019-11-01 UTC" "2019-12-01 UTC" "2019-07-01 UTC"

I am unfortunately drawing a blank at getting these into the table in R, the furthest I got was the following, the mutate line being what I use to extract the date from the file name.

df2 <- head(df, 4) %>% 
  mutate(Date = substr(Date, 10, 18)) %>% 
  replace(df2$Date,  parse_date_time(df2$Date, orders = c("bY")))

However it is creating new columns instead of replacing the date in the right column. I also tried using mutate instead of replace but that didn't appear to do anything.

Grateful for any advice on this.

CodePudding user response：

The replace is not needed i.e. after the substr, just wrap the parse_date_time within mutate to convert the column

library(dplyr)
library(lubridate)
df2 <- df %>%
   mutate(Date = parse_date_time(substr(Date, 10, 18), orders = c("bY")))