splitting dates in R

Question

The time coloumn of my dataframe looks like the following

# time
# %Y-%m-%d
# 2000-01-01
# 2000-01-02
# ...

I would like to split the date coloumn in one for the year and one for the month (I dont need the days) So the output should look like the following:

# year  month
# %Y    %m
# 2000  01
# 2000  02
# ...

score 1 · Accepted Answer · answered Apr 28 '20 at 12:58

In the lubridate package, use the year() and month() functions to extract them, respectively, from a date object.

library(data.table)
library(lubridate)

DT <- as.data.table(yourDataFrame)

DT[, Date := ymd(time)]
DT[, year := year(Date)]
DT[, month := month(Date)]

ThomasIsCoding · Answer 2 · 2020-04-28T20:00:43.723

1

Yo can try separate, e.g.,

library(dplyr)
library(tidyr)

df %>% separate(d,c("year","month"),extra = "drop")

which gives

> df %>% separate(d,c("year","month"),extra = "drop")
  year month
1 2020    01
2 2020    02

Data

df <- data.frame(d = c("2020-01-01","2020-02-01"))

edited Apr 28 '20 at 20:00

answered Apr 28 '20 at 13:02

ThomasIsCoding

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Or use `extract`, and ini `separate`, there is an option to `extra = 'drop'` if you speciify only two columns – akrun Apr 28 '20 at 18:42
1

@akrun Thanks a lot! Now I learn a bit more about `separate` (I didn't use `dplyr`) before :P – ThomasIsCoding Apr 28 '20 at 19:59

score 1 · Answer 3 · answered Apr 28 '20 at 13:05

1

Base R:

# data:
time<-seq.Date(from = as.Date("2000/1/1"),to=as.Date("2000/5/1"),by="days")
df<-data.frame(ID=1:length(time), time= time)

# months
df$month<-month(df$time)
# or:
df$month.name<-months(df$time)


df

answered Apr 28 '20 at 13:05

user12256545

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splitting dates in R

3 Answers3