Formatting dates using a condition
I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.
DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")
I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".
indicator <- !grepl("-", DateVar)
for(i in indicator == TRUE){
as.date(DateVar, origin = "1899-12-30")
It is not working for me however, so I am looking if someone can point me in the right direction.
Thanks.
r loops date
asked Nov 19 '18 at 13:11
SteveMSteveM
224
add a comment |
I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.
DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")
I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".
indicator <- !grepl("-", DateVar)
for(i in indicator == TRUE){
as.date(DateVar, origin = "1899-12-30")
It is not working for me however, so I am looking if someone can point me in the right direction.
Thanks.
r loops date
asked Nov 19 '18 at 13:11
SteveMSteveM
224
add a comment |
I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.
DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")
I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".
indicator <- !grepl("-", DateVar)
for(i in indicator == TRUE){
as.date(DateVar, origin = "1899-12-30")
It is not working for me however, so I am looking if someone can point me in the right direction.
Thanks.
r loops date
asked Nov 19 '18 at 13:11
SteveMSteveM
224
I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.
DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")
I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".
indicator <- !grepl("-", DateVar)
for(i in indicator == TRUE){
as.date(DateVar, origin = "1899-12-30")
It is not working for me however, so I am looking if someone can point me in the right direction.
Thanks.
r loops date
r loops date
asked Nov 19 '18 at 13:11
SteveMSteveM
224
asked Nov 19 '18 at 13:11
SteveMSteveM
224
asked Nov 19 '18 at 13:11
SteveMSteveM
224
asked Nov 19 '18 at 13:11
SteveMSteveM
224
asked Nov 19 '18 at 13:11
SteveMSteveM
224
224
add a comment |
add a comment |
1 Answer
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Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:
as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")
(or, without the intervening indicator assignment:
as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")
[1] "2018-05-09" "2018-02-06"
If you wish to input these dates back into DateVar, you again use the subset function:
DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")
answered Nov 19 '18 at 13:23
iodiod
3,8432722
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
add a comment |
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1 Answer
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Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:
as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")
(or, without the intervening indicator assignment:
as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")
[1] "2018-05-09" "2018-02-06"
If you wish to input these dates back into DateVar, you again use the subset function:
DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")
answered Nov 19 '18 at 13:23
iodiod
3,8432722
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
add a comment |
Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:
as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")
(or, without the intervening indicator assignment:
as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")
[1] "2018-05-09" "2018-02-06"
If you wish to input these dates back into DateVar, you again use the subset function:
DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")
answered Nov 19 '18 at 13:23
iodiod
3,8432722
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
add a comment |
Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:
as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")
(or, without the intervening indicator assignment:
as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")
[1] "2018-05-09" "2018-02-06"
If you wish to input these dates back into DateVar, you again use the subset function:
DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")
answered Nov 19 '18 at 13:23
iodiod
3,8432722
Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:
as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")
(or, without the intervening indicator assignment:
as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")
[1] "2018-05-09" "2018-02-06"
If you wish to input these dates back into DateVar, you again use the subset function:
DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")
answered Nov 19 '18 at 13:23
iodiod
3,8432722
edited Nov 19 '18 at 20:45
answered Nov 19 '18 at 13:23
iodiod
3,8432722
answered Nov 19 '18 at 13:23
iodiod
3,8432722
answered Nov 19 '18 at 13:23
iodiod
3,8432722
3,8432722
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
add a comment |
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
– SteveM
Nov 19 '18 at 14:07
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
You want to change DateVar so that the origin dates become class Date?
– iod
Nov 19 '18 at 14:10
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
– SteveM
Nov 19 '18 at 14:18
Added code above.
– iod
Nov 19 '18 at 14:19
Added code above.
– iod
Nov 19 '18 at 14:19
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
This worked beautifully. Thanks so much!
– SteveM
Nov 19 '18 at 14:40
add a comment |
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