How to get ISO week numbers from date
0
I would like to get the ISO week number from a given date in string format, where Monday is the first day of the week, and the week that contains the first Thursday of the year is considered to be the first week.
From other answers, I think strftime("2017-11-18", format = "%V") suits my purpose the best.
However it doesn't work on Windows.
Any suggestions for alternatives with only the base package in R?
r
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
asked Nov 27 '17 at 11:10
noname
331113
add a comment |
0
I would like to get the ISO week number from a given date in string format, where Monday is the first day of the week, and the week that contains the first Thursday of the year is considered to be the first week.
From other answers, I think strftime("2017-11-18", format = "%V") suits my purpose the best.
However it doesn't work on Windows.
Any suggestions for alternatives with only the base package in R?
r
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
asked Nov 27 '17 at 11:10
noname
331113
add a comment |
0
0
0
I would like to get the ISO week number from a given date in string format, where Monday is the first day of the week, and the week that contains the first Thursday of the year is considered to be the first week.
From other answers, I think strftime("2017-11-18", format = "%V") suits my purpose the best.
However it doesn't work on Windows.
Any suggestions for alternatives with only the base package in R?
r
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
asked Nov 27 '17 at 11:10
noname
331113
I would like to get the ISO week number from a given date in string format, where Monday is the first day of the week, and the week that contains the first Thursday of the year is considered to be the first week.
From other answers, I think strftime("2017-11-18", format = "%V") suits my purpose the best.
However it doesn't work on Windows.
Any suggestions for alternatives with only the base package in R?
r
r
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
asked Nov 27 '17 at 11:10
noname
331113
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
asked Nov 27 '17 at 11:10
noname
331113
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
edited Nov 27 '17 at 11:29
h3rm4n
3,218820
3,218820
asked Nov 27 '17 at 11:10
noname
331113
asked Nov 27 '17 at 11:10
noname
331113
asked Nov 27 '17 at 11:10
noname
331113
331113
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
3
Use the package lubridate
it has a function isoweek()which gives you the ISOWeek of a given date
lubridate::isoweek("2017-11-18")
[1] 46
Now you just want to use the base packege. Here ist the code from lubridate for the ISO week
function (x)
{
xday <- make_datetime(year(x), month(x), day(x))
dn <- 1 + (wday(x) + 5)%%7
nth <- xday + ddays(4 - dn)
jan1 <- make_datetime(year(nth), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
we can take that an make it to something which only uses the base package.
myIsoweek <- function (x)
{
dateList <- as.list(strsplit(as.character(as.Date(x)),split = "-")[[1]])
names(dateList) <- c("year","month","day")
weekday <- as.POSIXlt(x)[["wday"]] + 1
xday <- ISOdate(dateList$year, dateList$month, dateList$day)
dn <- 1 + (weekday + 5)%%7
nth <- xday + 86400*(4 - dn)
jan1 <- ISOdate(format(nth,format = "%Y"), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
|
show 1 more comment
1
data.table has an implementation of isoweek which you might want to just port (it's easy to replicate with base functionality)
# data.table approach:
isoweek <- function(x) {
x = as.IDate(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.IDate(7L * (as.integer(x + 3L) %/% 7L))
year_start <- as.IDate(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
Ported to be strictly base:
isoweek <- function(x) {
x = as.Date(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.Date(7L * (as.integer(x + 3L) %/% 7L), origin = '1970-01-01')
year_start <- as.Date(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
add a comment |
0
I believe strftime("2017-11-18", format = "%W") should work in Windows.
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
You are correct. Can you tryformat(as.POSIXct(as.Date("2017-11-18")), "%V")instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…
– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
add a comment |
0
This should work
format(as.Date("2017-02-015"),"%W")
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
Use the package lubridate
it has a function isoweek()which gives you the ISOWeek of a given date
lubridate::isoweek("2017-11-18")
[1] 46
Now you just want to use the base packege. Here ist the code from lubridate for the ISO week
function (x)
{
xday <- make_datetime(year(x), month(x), day(x))
dn <- 1 + (wday(x) + 5)%%7
nth <- xday + ddays(4 - dn)
jan1 <- make_datetime(year(nth), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
we can take that an make it to something which only uses the base package.
myIsoweek <- function (x)
{
dateList <- as.list(strsplit(as.character(as.Date(x)),split = "-")[[1]])
names(dateList) <- c("year","month","day")
weekday <- as.POSIXlt(x)[["wday"]] + 1
xday <- ISOdate(dateList$year, dateList$month, dateList$day)
dn <- 1 + (weekday + 5)%%7
nth <- xday + 86400*(4 - dn)
jan1 <- ISOdate(format(nth,format = "%Y"), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
|
show 1 more comment
3
Use the package lubridate
it has a function isoweek()which gives you the ISOWeek of a given date
lubridate::isoweek("2017-11-18")
[1] 46
Now you just want to use the base packege. Here ist the code from lubridate for the ISO week
function (x)
{
xday <- make_datetime(year(x), month(x), day(x))
dn <- 1 + (wday(x) + 5)%%7
nth <- xday + ddays(4 - dn)
jan1 <- make_datetime(year(nth), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
we can take that an make it to something which only uses the base package.
myIsoweek <- function (x)
{
dateList <- as.list(strsplit(as.character(as.Date(x)),split = "-")[[1]])
names(dateList) <- c("year","month","day")
weekday <- as.POSIXlt(x)[["wday"]] + 1
xday <- ISOdate(dateList$year, dateList$month, dateList$day)
dn <- 1 + (weekday + 5)%%7
nth <- xday + 86400*(4 - dn)
jan1 <- ISOdate(format(nth,format = "%Y"), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
|
show 1 more comment
3
3
3
Use the package lubridate
it has a function isoweek()which gives you the ISOWeek of a given date
lubridate::isoweek("2017-11-18")
[1] 46
Now you just want to use the base packege. Here ist the code from lubridate for the ISO week
function (x)
{
xday <- make_datetime(year(x), month(x), day(x))
dn <- 1 + (wday(x) + 5)%%7
nth <- xday + ddays(4 - dn)
jan1 <- make_datetime(year(nth), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
we can take that an make it to something which only uses the base package.
myIsoweek <- function (x)
{
dateList <- as.list(strsplit(as.character(as.Date(x)),split = "-")[[1]])
names(dateList) <- c("year","month","day")
weekday <- as.POSIXlt(x)[["wday"]] + 1
xday <- ISOdate(dateList$year, dateList$month, dateList$day)
dn <- 1 + (weekday + 5)%%7
nth <- xday + 86400*(4 - dn)
jan1 <- ISOdate(format(nth,format = "%Y"), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
Use the package lubridate
it has a function isoweek()which gives you the ISOWeek of a given date
lubridate::isoweek("2017-11-18")
[1] 46
Now you just want to use the base packege. Here ist the code from lubridate for the ISO week
function (x)
{
xday <- make_datetime(year(x), month(x), day(x))
dn <- 1 + (wday(x) + 5)%%7
nth <- xday + ddays(4 - dn)
jan1 <- make_datetime(year(nth), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
we can take that an make it to something which only uses the base package.
myIsoweek <- function (x)
{
dateList <- as.list(strsplit(as.character(as.Date(x)),split = "-")[[1]])
names(dateList) <- c("year","month","day")
weekday <- as.POSIXlt(x)[["wday"]] + 1
xday <- ISOdate(dateList$year, dateList$month, dateList$day)
dn <- 1 + (weekday + 5)%%7
nth <- xday + 86400*(4 - dn)
jan1 <- ISOdate(format(nth,format = "%Y"), 1, 1)
1L + as.integer(difftime(nth, jan1, units = "days"))%/%7L
}
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
edited Nov 27 '17 at 12:45
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
answered Nov 27 '17 at 11:35
Bertil Baron
2,7121716
2,7121716
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
|
show 1 more comment
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
THe OP wants to use just base R packages ...
– J_F
Nov 27 '17 at 11:41
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Due to the license I cannot use the lubridate library. That's why I'm asking for only base R packages.
– noname
Nov 27 '17 at 12:01
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Hi, I updatet my post to only use the base package
– Bertil Baron
Nov 27 '17 at 12:46
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
Great, it works perfectly!
– noname
Nov 27 '17 at 12:54
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
It has the same license as R so if you can't use that license then you can't use R either.
– G. Grothendieck
Nov 27 '17 at 13:25
|
show 1 more comment
1
data.table has an implementation of isoweek which you might want to just port (it's easy to replicate with base functionality)
# data.table approach:
isoweek <- function(x) {
x = as.IDate(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.IDate(7L * (as.integer(x + 3L) %/% 7L))
year_start <- as.IDate(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
Ported to be strictly base:
isoweek <- function(x) {
x = as.Date(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.Date(7L * (as.integer(x + 3L) %/% 7L), origin = '1970-01-01')
year_start <- as.Date(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
add a comment |
1
data.table has an implementation of isoweek which you might want to just port (it's easy to replicate with base functionality)
# data.table approach:
isoweek <- function(x) {
x = as.IDate(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.IDate(7L * (as.integer(x + 3L) %/% 7L))
year_start <- as.IDate(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
Ported to be strictly base:
isoweek <- function(x) {
x = as.Date(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.Date(7L * (as.integer(x + 3L) %/% 7L), origin = '1970-01-01')
year_start <- as.Date(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
add a comment |
1
1
1
data.table has an implementation of isoweek which you might want to just port (it's easy to replicate with base functionality)
# data.table approach:
isoweek <- function(x) {
x = as.IDate(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.IDate(7L * (as.integer(x + 3L) %/% 7L))
year_start <- as.IDate(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
Ported to be strictly base:
isoweek <- function(x) {
x = as.Date(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.Date(7L * (as.integer(x + 3L) %/% 7L), origin = '1970-01-01')
year_start <- as.Date(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
data.table has an implementation of isoweek which you might want to just port (it's easy to replicate with base functionality)
# data.table approach:
isoweek <- function(x) {
x = as.IDate(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.IDate(7L * (as.integer(x + 3L) %/% 7L))
year_start <- as.IDate(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
Ported to be strictly base:
isoweek <- function(x) {
x = as.Date(x) # number of days since 1 Jan 1970 (a Thurs)
nearest_thurs = as.Date(7L * (as.integer(x + 3L) %/% 7L), origin = '1970-01-01')
year_start <- as.Date(format(nearest_thurs, '%Y-01-01'))
1L + (nearest_thurs - year_start) %/% 7L
}
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
edited Nov 21 at 9:16
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
answered Nov 20 at 6:05
MichaelChirico
19.7k859109
19.7k859109
add a comment |
add a comment |
0
I believe strftime("2017-11-18", format = "%W") should work in Windows.
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
You are correct. Can you tryformat(as.POSIXct(as.Date("2017-11-18")), "%V")instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…
– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
add a comment |
0
I believe strftime("2017-11-18", format = "%W") should work in Windows.
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
You are correct. Can you tryformat(as.POSIXct(as.Date("2017-11-18")), "%V")instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…
– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
add a comment |
0
0
0
I believe strftime("2017-11-18", format = "%W") should work in Windows.
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
I believe strftime("2017-11-18", format = "%W") should work in Windows.
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
answered Nov 27 '17 at 11:20
Cihan Ceyhan
681310
681310
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
You are correct. Can you tryformat(as.POSIXct(as.Date("2017-11-18")), "%V")instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…
– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
add a comment |
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
You are correct. Can you tryformat(as.POSIXct(as.Date("2017-11-18")), "%V")instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…
– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
Hi Cihan, I've tried %W, it doesn't give the ISO week. I saw in it's documentation it mentions "first Monday of the year as day 1 of week 1"
– noname
Nov 27 '17 at 11:28
1
1
You are correct. Can you try
format(as.POSIXct(as.Date("2017-11-18")), "%V") instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…– Cihan Ceyhan
Nov 27 '17 at 11:35
You are correct. Can you try
format(as.POSIXct(as.Date("2017-11-18")), "%V") instead? I've found a discussion that's on this topic here: r.789695.n4.nabble.com/…– Cihan Ceyhan
Nov 27 '17 at 11:35
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
It returns the same empty string for me.
– noname
Nov 27 '17 at 12:01
add a comment |
0
This should work
format(as.Date("2017-02-015"),"%W")
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
add a comment |
0
This should work
format(as.Date("2017-02-015"),"%W")
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
add a comment |
0
0
0
This should work
format(as.Date("2017-02-015"),"%W")
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
This should work
format(as.Date("2017-02-015"),"%W")
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
edited Nov 27 '17 at 11:41
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
answered Nov 27 '17 at 11:26
Smiley Bcc
1149
1149
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
add a comment |
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi Smiley, this only give the day of the week instead of the week number. Also it use Sunday as first day of the week.
– noname
Nov 27 '17 at 11:32
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
Hi noname, sorry i meant the capital "W" but still it wouldn't help you answer with the need of monday to be first day of the week.
– Smiley Bcc
Nov 27 '17 at 11:40
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
If there is really no way, you can consider using cut() function to manually define your own customized week
– Smiley Bcc
Nov 27 '17 at 11:42
add a comment |
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