I have monthly column value in number
df <- data.frame(Month= c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "1", "2", "3"))
I want to convert this to: Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec, Jan, Feb, Mar
Can anyone help me with this please?
Thanking in advance for your time
Not the most elegant or programmatic solution, but you could replace all the values with the mutate() function from the dplyr package:
library(dplyr) df = mutate(df, Month = case_when(Month == "1" ~ "Jan", Month == "2" ~ "Feb", Month == "3" ~ "Mar", Month == "4" ~ "Apr", Month == "5" ~ "May", Month == "6" ~ "Jun", Month == "7" ~ "Jul", Month == "8" ~ "Aug", Month == "9" ~ "Sep", Month == "10" ~ "Oct", Month == "11" ~ "Nov", Month == "12" ~ "Dec") You can find documentation for using case_when() here: https://dplyr.tidyverse.org/reference/case_when.html
An alternative programmatic solution based on the comment left by @r2evans to do this in one line using the built-in R object month.abb:
df = mutate(df, Month = month.abb[as.numeric(Month)]) 
character and not numeric (like the column in the question)?
"12" == 12 to "12" == "12". I typically don't like relying on this (especially where floating-point numeric comparisons may be at play), and I prefer to be explicit with what I know class a variable holds. But that's just a bit paranoid/defensive programming :-)
Very simple way would be to have a vector with the months. Then you can use the numbers of the months as vector indices (after coercing them to integers).
Months <- c("Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec") df <- data.frame(Month= c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "1", "2", "3")) df$Month <- Months[as.integer(df$Month)] df #> Month #> 1 Jan #> 2 Feb #> 3 Mar #> 4 Apr #> 5 May #> 6 Jun #> 7 Jul #> 8 Aug #> 9 Sep #> 10 Oct #> 11 Nov #> 12 Dec #> 13 Jan #> 14 Feb #> 15 Mar The same can be done with the built-in R constant month.abb instead of creating your own Months vector.
Created on 2021-02-21 by the reprex package (v1.0.0)
substr(month.name,1,3)
month.abb[as.integer(df$Month)]. Themonth.abbis a base R constant. One unfortunate thing is that it is not locale-aware, so non-English months are not available via the same variable ... though it is rather easy to generate your own, since all it is is acharactervector, length 12.