How do I subset an xts object to only include weekdays (Mon-Fri, with Saturday and Sunday excluded)?
Here's what I'd do:
library(xts)
data(sample_matrix)
sample.xts <- as.xts(sample_matrix, descr='my new xts object')
x <- sample.xts['2007']
x[!weekdays(index(x)) %in% c("Saturday", "Sunday")]
EDIT:
Joshua Ulrich in comments points out a better solution using .indexwday(), one of a family of built-in accessor functions for extracting pieces of the index of xts class objects. Also, like Dirk Eddelbuettel's solution, the following should be locale-independent:
x[.indexwday(x) %in% 1:5]
By computing the day-of-the week given the date, and subsetting. In the example, I use a Date type but the cast to POSIXlt works the same way for POSIXct intra-day timestamps.
> mydates <- Sys.Date() + 0:6
> mydates
[1] "2012-01-31" "2012-02-01" "2012-02-02" "2012-02-03" "2012-02-04"
+ "2012-02-05" "2012-02-06"
> we <- sapply(mydates, function(d) { as.POSIXlt(d)$wday}) %in% c(0, 6)
> we
[1] FALSE FALSE FALSE FALSE TRUE TRUE FALSE
> mydates[ ! we ]
[1] "2012-01-31" "2012-02-01" "2012-02-02" "2012-02-03" "2012-02-06"
>
This really is not an xts question but basic date handling.
Related
I have a date in R, e.g.:
dt = as.Date('2010/03/17')
I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2008-03-17').
How would I do that?
With lubridate
library(lubridate)
ymd("2010/03/17") - years(2)
The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.
> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"
See this related question: How to subtract days in R?.
You could use seq:
R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"
If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:
> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"
# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"
Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):
R> unlist(lapply(c("2015-12-01", "2016-12-01"),
function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
[1] "2014-12-01" "2015-12-01"
This way seems to do the job as well
dt = as.Date("2010/03/17")
dt-365*2
[1] "2008-03-17"
as.Date("2008/02/29")-365*2
## [1] "2006-03-01"
cur_date <- str_split(as.character(Sys.Date()), pattern = "-")
cur_yr <- cur_date[[1]][1]
cur_month <- cur_date[[1]][2]
cur_day <- cur_date[[1]][3]
new_year <- as.integer(year) - 2
new_date <- paste(new_year, cur_month, cur_day, sep="-")
Using Base R, you can simply use the following without installing any package.
1) Transform your character string to Date format, specifying the input format in the second argument, so R can correctly interpret your date format.
dt = as.Date('2010/03/17',"%Y/%m/%d")
NOTE: If you look now at your enviroment tab you will see dt as variable with the following value "2010-03-17" (Year-month-date separated by "-" not by "/")
2) specify how many years to substract
years_substract=2
3) Use paste() combined with format () to only keep Month and Day and Just substract 2 year from your original date. Format() function will just keep the specific part of your date accordingly with format second argument.
dt_substract_2years<-
as.Date(paste(as.numeric(format(dt,"%Y"))-years_substract,format(dt,"%m"),format(dt,"%d"),sep = "-"))
NOTE1: We used paste() function to concatenate date components and specify separator as "-" (sep = "-")as is the R separator for dates by default.
NOTE2: We also used as.numeric() function to transform year from character to numeric
I have a date in R, e.g.:
dt = as.Date('2010/03/17')
I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2008-03-17').
How would I do that?
With lubridate
library(lubridate)
ymd("2010/03/17") - years(2)
The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.
> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"
See this related question: How to subtract days in R?.
You could use seq:
R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"
If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:
> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"
# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"
Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):
R> unlist(lapply(c("2015-12-01", "2016-12-01"),
function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
[1] "2014-12-01" "2015-12-01"
This way seems to do the job as well
dt = as.Date("2010/03/17")
dt-365*2
[1] "2008-03-17"
as.Date("2008/02/29")-365*2
## [1] "2006-03-01"
cur_date <- str_split(as.character(Sys.Date()), pattern = "-")
cur_yr <- cur_date[[1]][1]
cur_month <- cur_date[[1]][2]
cur_day <- cur_date[[1]][3]
new_year <- as.integer(year) - 2
new_date <- paste(new_year, cur_month, cur_day, sep="-")
Using Base R, you can simply use the following without installing any package.
1) Transform your character string to Date format, specifying the input format in the second argument, so R can correctly interpret your date format.
dt = as.Date('2010/03/17',"%Y/%m/%d")
NOTE: If you look now at your enviroment tab you will see dt as variable with the following value "2010-03-17" (Year-month-date separated by "-" not by "/")
2) specify how many years to substract
years_substract=2
3) Use paste() combined with format () to only keep Month and Day and Just substract 2 year from your original date. Format() function will just keep the specific part of your date accordingly with format second argument.
dt_substract_2years<-
as.Date(paste(as.numeric(format(dt,"%Y"))-years_substract,format(dt,"%m"),format(dt,"%d"),sep = "-"))
NOTE1: We used paste() function to concatenate date components and specify separator as "-" (sep = "-")as is the R separator for dates by default.
NOTE2: We also used as.numeric() function to transform year from character to numeric
I have a date in R, e.g.:
dt = as.Date('2010/03/17')
I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2008-03-17').
How would I do that?
With lubridate
library(lubridate)
ymd("2010/03/17") - years(2)
The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.
> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"
See this related question: How to subtract days in R?.
You could use seq:
R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"
If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:
> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"
# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"
Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):
R> unlist(lapply(c("2015-12-01", "2016-12-01"),
function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
[1] "2014-12-01" "2015-12-01"
This way seems to do the job as well
dt = as.Date("2010/03/17")
dt-365*2
[1] "2008-03-17"
as.Date("2008/02/29")-365*2
## [1] "2006-03-01"
cur_date <- str_split(as.character(Sys.Date()), pattern = "-")
cur_yr <- cur_date[[1]][1]
cur_month <- cur_date[[1]][2]
cur_day <- cur_date[[1]][3]
new_year <- as.integer(year) - 2
new_date <- paste(new_year, cur_month, cur_day, sep="-")
Using Base R, you can simply use the following without installing any package.
1) Transform your character string to Date format, specifying the input format in the second argument, so R can correctly interpret your date format.
dt = as.Date('2010/03/17',"%Y/%m/%d")
NOTE: If you look now at your enviroment tab you will see dt as variable with the following value "2010-03-17" (Year-month-date separated by "-" not by "/")
2) specify how many years to substract
years_substract=2
3) Use paste() combined with format () to only keep Month and Day and Just substract 2 year from your original date. Format() function will just keep the specific part of your date accordingly with format second argument.
dt_substract_2years<-
as.Date(paste(as.numeric(format(dt,"%Y"))-years_substract,format(dt,"%m"),format(dt,"%d"),sep = "-"))
NOTE1: We used paste() function to concatenate date components and specify separator as "-" (sep = "-")as is the R separator for dates by default.
NOTE2: We also used as.numeric() function to transform year from character to numeric
Given a POSIXct date time, how do you extract the first day of the month for aggregation?
library(lubridate)
full.date <- ymd_hms("2013-01-01 00:00:21")
lubridate has a function called floor_date which rounds date-times down. Calling it with unit = "month" does exactly what you want:
library(lubridate)
full.date <- ymd_hms("2013-01-01 00:00:21")
floor_date(full.date, "month")
[1] "2013-01-01 UTC"
I don't see a reason to use lubridate:
full.date <- as.POSIXct("2013-01-11 00:00:21", tz="GMT")
monthStart <- function(x) {
x <- as.POSIXlt(x)
x$mday <- 1
as.Date(x)
}
monthStart(full.date)
#[1] "2013-01-01"
first.of.month <- ymd(format(full.date, "%Y-%m-01"))
first.of.month
[1] "2013-01-01 UTC"
i have another solution :
first.of.month <- full.date - mday(full.date) + 1
but it needs the library 'lubridate' or 'date.table' (aggregation with data.table)
You can simply use base R's trunc:
d <- as.POSIXct("2013-01-11 00:00:21", tz="UTC")
trunc(d, "month")
#[1] "2013-01-01 UTC"
Is there an easy way in R for me to itemize all valid days that occurred between two specified dates? For instance, I'd like the following inputs:
itemizeDates(startDate="12-30-11", endDate="1-4-12")
To produce the following dates:
"12-30-11" "12-31-11", "1-1-12", "1-2-12", "1-3-12", "1-4-12"
I'm flexible on classes and formatting of the dates, I just need an implementation of the concept.
You're looking for seq
> seq(as.Date("2011-12-30"), as.Date("2012-01-04"), by="days")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Or, you can use :
> as.Date(as.Date("2011-12-30"):as.Date("2012-01-04"), origin="1970-01-01")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Note that with : "Non-numeric arguments are coerced internally". Thus, we convert back to class Date, using as.Date method for class 'numeric' and provide origin.
Here's a function to meet your specific request
itemizeDates <- function(startDate="12-30-11", endDate="1-4-12",
format="%m-%d-%y") {
out <- seq(as.Date(startDate, format=format),
as.Date(endDate, format=format), by="days")
format(out, format)
}
> itemizeDates(startDate="12-30-11", endDate="1-4-12")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
I prefer using the lubridate package to solve datetime problems. It is more intuitive and easier to understand and use once you know it.
library(lubridate)
#mdy() in lubridate package means "month-day-year", which is used to convert
#the string to date object
>start_date <- mdy("12-30-11")
>end_date <- mdy("1-4-12")
#calculate how many days in this time interval
>n_days <- interval(start_date,end_date)/days(1)
>start_date + days(0:n_days)
[1]"2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04"
#convert to original format
format(start_date + days(0:n_days), format="%m-%d-%y")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
Reference:
Dates and Times Made Easy with lubridate
2 similar implementations in lubridate:
library(lubridate)
as_date(mdy("12-30-11"):mdy("1-4-12"))
# OR
seq(mdy("12-30-11"), mdy("1-4-12"), by = "days")
These don't format your dates in month-day-year but you can fix the formatting if you want. But year-month-day is a bit easy to work with when analyzing.