convert Single digit day in R
I have dates in the format Apr42016, Aug12017, Apr112018. I am trying to convert in Y/m/d using R. I have tried the codes below but when I have a single digit for the day it returned NA. Anyone could help me, please?
strptime(data$date, "%b%e%Y")
as.Date (data$date, format="%b%d%Y")
as.POSIXct(data$date, format="%b%e%Y")
Thank you
r date format
edited Nov 13 '18 at 19:15
phiver
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
add a comment |
I have dates in the format Apr42016, Aug12017, Apr112018. I am trying to convert in Y/m/d using R. I have tried the codes below but when I have a single digit for the day it returned NA. Anyone could help me, please?
strptime(data$date, "%b%e%Y")
as.Date (data$date, format="%b%d%Y")
as.POSIXct(data$date, format="%b%e%Y")
Thank you
r date format
edited Nov 13 '18 at 19:15
phiver
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
add a comment |
I have dates in the format Apr42016, Aug12017, Apr112018. I am trying to convert in Y/m/d using R. I have tried the codes below but when I have a single digit for the day it returned NA. Anyone could help me, please?
strptime(data$date, "%b%e%Y")
as.Date (data$date, format="%b%d%Y")
as.POSIXct(data$date, format="%b%e%Y")
Thank you
r date format
edited Nov 13 '18 at 19:15
phiver
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
I have dates in the format Apr42016, Aug12017, Apr112018. I am trying to convert in Y/m/d using R. I have tried the codes below but when I have a single digit for the day it returned NA. Anyone could help me, please?
strptime(data$date, "%b%e%Y")
as.Date (data$date, format="%b%d%Y")
as.POSIXct(data$date, format="%b%e%Y")
Thank you
r date format
r date format
edited Nov 13 '18 at 19:15
phiver
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
edited Nov 13 '18 at 19:15
phiver
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
edited Nov 13 '18 at 19:15
phiver
13.2k92734
edited Nov 13 '18 at 19:15
phiver
13.2k92734
edited Nov 13 '18 at 19:15
phiver
13.2k92734
13.2k92734
asked Nov 13 '18 at 19:09
user2386222user2386222
345
asked Nov 13 '18 at 19:09
user2386222user2386222
345
asked Nov 13 '18 at 19:09
user2386222user2386222
345
345
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can modify the strings with sub (and add a 0 if necessary) before using as.Date:
myvec <- c("Apr42016", "Aug12017", "Apr112018") # the data
myvec2 <- sub("(?<=[^0])(?=[0-9]{5})", "0", myvec, perl = TRUE)
# [1] "Apr042016" "Aug012017" "Apr112018"
as.Date(myvec2, format = "%b%d%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
add a comment |
If you can break up the numbers before as.Date, it will make things much easier. (Borrowing Sven's look-behind.)
sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2",
c("Apr42016", "Aug12017", "Apr112018"), perl=TRUE)
# [1] "Apr-4-2016" "Aug-1-2017" "Apr-11-2018"
From here, the format should be rather straight-forward:
as.Date(sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2", c("Apr42016", "Aug12017", "Apr112018"), perl = TRUE),
format="%b-%d-%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can modify the strings with sub (and add a 0 if necessary) before using as.Date:
myvec <- c("Apr42016", "Aug12017", "Apr112018") # the data
myvec2 <- sub("(?<=[^0])(?=[0-9]{5})", "0", myvec, perl = TRUE)
# [1] "Apr042016" "Aug012017" "Apr112018"
as.Date(myvec2, format = "%b%d%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
add a comment |
You can modify the strings with sub (and add a 0 if necessary) before using as.Date:
myvec <- c("Apr42016", "Aug12017", "Apr112018") # the data
myvec2 <- sub("(?<=[^0])(?=[0-9]{5})", "0", myvec, perl = TRUE)
# [1] "Apr042016" "Aug012017" "Apr112018"
as.Date(myvec2, format = "%b%d%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
add a comment |
You can modify the strings with sub (and add a 0 if necessary) before using as.Date:
myvec <- c("Apr42016", "Aug12017", "Apr112018") # the data
myvec2 <- sub("(?<=[^0])(?=[0-9]{5})", "0", myvec, perl = TRUE)
# [1] "Apr042016" "Aug012017" "Apr112018"
as.Date(myvec2, format = "%b%d%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
You can modify the strings with sub (and add a 0 if necessary) before using as.Date:
myvec <- c("Apr42016", "Aug12017", "Apr112018") # the data
myvec2 <- sub("(?<=[^0])(?=[0-9]{5})", "0", myvec, perl = TRUE)
# [1] "Apr042016" "Aug012017" "Apr112018"
as.Date(myvec2, format = "%b%d%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
answered Nov 13 '18 at 19:20
Sven HohensteinSven Hohenstein
65.5k12100131
65.5k12100131
add a comment |
add a comment |
If you can break up the numbers before as.Date, it will make things much easier. (Borrowing Sven's look-behind.)
sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2",
c("Apr42016", "Aug12017", "Apr112018"), perl=TRUE)
# [1] "Apr-4-2016" "Aug-1-2017" "Apr-11-2018"
From here, the format should be rather straight-forward:
as.Date(sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2", c("Apr42016", "Aug12017", "Apr112018"), perl = TRUE),
format="%b-%d-%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
add a comment |
If you can break up the numbers before as.Date, it will make things much easier. (Borrowing Sven's look-behind.)
sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2",
c("Apr42016", "Aug12017", "Apr112018"), perl=TRUE)
# [1] "Apr-4-2016" "Aug-1-2017" "Apr-11-2018"
From here, the format should be rather straight-forward:
as.Date(sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2", c("Apr42016", "Aug12017", "Apr112018"), perl = TRUE),
format="%b-%d-%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
add a comment |
If you can break up the numbers before as.Date, it will make things much easier. (Borrowing Sven's look-behind.)
sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2",
c("Apr42016", "Aug12017", "Apr112018"), perl=TRUE)
# [1] "Apr-4-2016" "Aug-1-2017" "Apr-11-2018"
From here, the format should be rather straight-forward:
as.Date(sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2", c("Apr42016", "Aug12017", "Apr112018"), perl = TRUE),
format="%b-%d-%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
If you can break up the numbers before as.Date, it will make things much easier. (Borrowing Sven's look-behind.)
sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2",
c("Apr42016", "Aug12017", "Apr112018"), perl=TRUE)
# [1] "Apr-4-2016" "Aug-1-2017" "Apr-11-2018"
From here, the format should be rather straight-forward:
as.Date(sub("(?<=\D)(\d+)(\d{4})$", "-\1-\2", c("Apr42016", "Aug12017", "Apr112018"), perl = TRUE),
format="%b-%d-%Y")
# [1] "2016-04-04" "2017-08-01" "2018-04-11"
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
edited Nov 13 '18 at 19:24
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
answered Nov 13 '18 at 19:17
r2evansr2evans
26.3k33058
26.3k33058
add a comment |
add a comment |
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