CODEWITHSUNDEEP
r
Tilsies
Tilsies

Reputation: 1

Converting Month abbreviations to numeric

I'm working with a dataset that includes dates in the following format:

[1] "19-Mar-10" "8-Feb-11" "19-Feb-11" "20-Feb-11" "20-Feb-11" "20-Feb-11" "20-Feb-11" "21-Feb-11"

I'm trying to use difftime to find the difference between two such columns. How would I go about converting the months to a numeric value to achieve this? Thanks in advance for any responses.

Upvotes: 0

Views: 401

Answers (3)

Brandon Rose MD MPH
Brandon Rose MD MPH

Reputation: 732

the strptime function in base R will work for any character to date format including yours... the abbreviations are in the help section of strptime. I agree with gss using the lubridate function if you work regularly with dates.

?strptime
dates<-c("19-Mar-10","8-Feb-11","19-Feb-11", "20-Feb-11", "20-Feb-11", 
"20-Feb-11", "20-Feb-11", "21-Feb-11")
newdates<-strptime(dates, format = "%y-%b-%d",tz="UTC")
newdates<-as.numeric(newdates)
newdates

Upvotes: 0

Skaqqs
Skaqqs

Reputation: 4140

in base R:

vec <- c("19-Mar-10", "8-Feb-11", "19-Feb-11", "20-Feb-11",
         "20-Feb-11", "20-Feb-11", "20-Feb-11", "21-Feb-11")

d <- as.Date(vec, format="%d-%b-%y")
d
[1] "2010-03-19" "2011-02-08" "2011-02-19" "2011-02-20"   
[5] "2011-02-20" "2011-02-20" "2011-02-20" "2011-02-21"

as.numeric(format(d, "%m"))
[1] 3 2 2 2 2 2 2 2

# Note that once data are in date format you can do arithmetic on them
# directly:
d[2]
[1] "2011-02-08"
d[1]
[1] "2010-03-19"
d[2]-d[1]
Time difference of 326 days

Upvotes: 3

gss
gss

Reputation: 1453

This is one solution:

library(lubridate)

vec <- "19-Mar-10"

dmy(vec)
#> [1] "2010-03-19"

Upvotes: 1

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