R: Removing hour,min,sec from date

Question

I have a data frame contain one date column it is in POSIXCT format as shown below

"2019-02-18 00:00:31 IST"

I want date in below format

 "2019-02-18"

but when i applied

 x = as.Date( DATE ,"%Y%m%d")

it is giving below warning

Warning message:
In as.POSIXlt.POSIXct(x, tz = tz) : unknown timezone '%Y%m%d'

And i am getting below wrong( one day previous date) output

"2019-02-17"

Plz let me know whats going wrong.

No need to specify any format in this case. Just do `as.Date("2019-02-18 00:00:31 IST")` — Ronak Shah, Feb 20 '19 at 10:23
then it's your system locale issue. You are in a different timezone than the one provided in the string. — Ronak Shah, Feb 20 '19 at 10:31
You just pass the time zone within `as.Date()` this should work: `as.POSIXct("2019-02-18 00:00:31 IST") %>% as.Date(tz = 'Asia/Kolkata')` — DS_UNI, Feb 20 '19 at 10:32
@DS_UNI: it will generate problem if i get record of different time zone — learner, Feb 20 '19 at 10:38
True, I assumed that all records have the same time zone, in that case I guess your best option would be the answer by RLave, using `strftime()` or `formrat()` and then `as.Date()` — DS_UNI, Feb 20 '19 at 10:44

score 1 · Accepted Answer · answered Feb 20 '19 at 10:21

1

You could do this also with strftime():

strftime(d, format="%Y-%m-%d")
[1] "2019-02-18"

With format= you can basically choose what you want to extract. It works even with just "%Y" in order to extract the year component.

Data:

d <- as.POSIXlt("2019-02-18 00:00:31 IST")

answered Feb 20 '19 at 10:21

RLave

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it is converting date to character format – learner Feb 20 '19 at 10:28
1

`as.POSIXct()` or `as.Date()` will give you the date format back. – RLave Feb 20 '19 at 10:30

score 0 · Answer 2 · answered Feb 20 '19 at 10:20

One option would be to just cast the POSIXct datetime to a character, and then take the first 10 characters:

date <- as.POSIXct("2019-02-18 00:00:31", format="%Y-%m-%d %H:%M:%OS", tz="Asia/Kolkata")
date
date.str <- as.character(date)
substr(date.str, 1, 10)

[1] "2019-02-18 00:00:31 IST"
[1] "2019-02-18"

R: Removing hour,min,sec from date

2 Answers2