Within R, say I have a vector of some Lubridate dates:
> Date
"2012-01-01 UTC"
"2013-01-01 UTC"
Next, suppose I want to see what week number these days fall in:
> week(Date)
1
1
Lubridate is fantastic!
But wait...I'm dealing a time series with 10,000 rows of data...and the data spans 3 years.
I've been struggling with finding some way to make this happen:
> result of awesome R code here
1
54
The question: is there a succinct way to coax out a list of week numbers over multiyear periods within Lubridate? More directly, I would like the first week of the second year to be represented as the 54th week. And the first week in the third year to be represented as the 107th week, ad nauseum.
So far, I've attempted a number of hackney schemes but cannot seem to create something not fastened together with scotch tape. Any advice would be greatly appreciated. Thanks in advance.
To get the interval from a particular date to another date, you can just subtract...
If tda is your vector of dates, then
tda - min(tda)
will be the difference in seconds between them.
To get the units out in weeks:
(tda - min(tda))/eweeks(1)
To do it from a particular date:
tda - ymd(19960101)
This gives the number of days from 1996 to each value.
From there, you can divide by days per week, or seconds per week.
(tda - ymd(19960101))/eweeks(1)
To get only the integer part, and starting from January 2012:
trunc((tda - ymd(20111225))/eweeks(1))
Test data:
tda = ymd(c(20120101, 20120106, 20130101, 20130108))
Output:
1 1 53 54
Since eweeks() is now deprecated, I thought I'd add to #beroe's answer.
If tda is your date vector, you can get the week numbers with:
weeknos <- (interval(min(tda), tda) %/% weeks(1)) + 1
where %/% causes integer division. ( 5 / 3 = 1.667; 5 %/% 3 = 1)
You can do something like this :
week(dat) +53*(year(dat)-min(year(dat)))
Given you like lubridate (as do I)
year_week <- function(x,base) week(x) - week(base) + 52*(year(x) - year(base))
test <- ymd(c(20120101, 20120106, 20130101, 20130108))
year_week(test, "2012-01-01")
Giving
[1] 0 0 52 53
Related
I'm trying to get the next week day for a vector of dates in R. My approach was to create a vector of weekdays and then find the date to the weekend date I have. The problem is that for Saturday and some holidays (which are a lot in my country) i end up getting the previous week day which doesn't work.
This is an example of my problem:
vecDates = as.Date(c("2011-01-11","2011-01-12","2011-01-13","2011-01-14","2011-01-17","2011-01-18",
"2011-01-19","2011-01-20","2011-01-21","2011-01-24"))
testDates = as.Date(c("2011-01-22","2011-01-23"))
findInterval(testDates,vecDates)
for both dates the correct answer should be 10 which is "2011-01-24" but I get 9.
I though of a solution where I remove all the previous dates to the date i'm analyzing, and then use findInterval. It works but it is not vectorized and therefore kind of slow which does not work for my actual purpose.
Does this do what you want?
vecDates = as.Date(c("2011-01-11","2011-01-12",
"2011-01-13","2011-01-14",
"2011-01-17","2011-01-18",
"2011-01-19","2011-01-20",
"2011-01-21","2011-01-24"))
testDates = as.Date(c("2011-01-20","2011-01-22","2011-01-23"))
get_next_biz_day <- function(testdays, bizdays){
o <- findInterval(testdays, bizdays) + 1
bizdays[o]
}
get_next_biz_day(testDates, vecDates)
#[1] "2011-01-21" "2011-01-24" "2011-01-24"
I know how to get the week from an index, but don't know the other way around: how to create an index if I have the calendar weeks (in this case, from an SAP system with 0CALWEEK as 201501, 201502 ... 201552, 201553.
Found this:
How to Parse Year + Week Number in R?
but the day is needed and it's not clear how to set it, especially at the end of the year (Year - week - day: YEAR-53-01 does not always exist, since the first day of week 53 might be Monday, then 01 (Sunday) is not in that week.
I could try to get in the source system the first day of the corresponding week (through SQL) but thought R might do it easier...
Do you have any suggestions?
(Which first day of the week would be not important , since I will create all objects the same way and then merge/cbind them, then continue the analysis. If zoo is easier, I'll go with it)
Thanks!
The problem is that all indices end in 2015-07-29:
data <- 1:4
weeks <- c('201501','201502','201552','201553')
weeks_2 <- as.Date(weeks,format='%Y%w')
xts(data, order.by = weeks_2)
[,1]
2015-07-29 1
2015-07-29 2
2015-07-29 3
2015-07-29 4
test <- xts(data, order.by = weeks_2)
index(test)
[1] "2015-07-29" "2015-07-29" "2015-07-29" "2015-07-29"
You can use as.Date() function, I think is the easiest way:
weeks <- c('201501','201502','201552','201553')
as.Date(paste0(weeks,'1'),format='%Y%W%w') # paste a dummy day
## [1] "2015-01-05" "2015-01-12" "2015-12-28" NA
Where:
%W: Week 00-53 with Monday as first day of the week
or
%U: Week 01-53 with Sunday as first day of the week
%w: Weekday 0-6 Sunday is 0
For this year, week number 53 doesn't exist. And If you want to start with 2015-01-01, just set the right week day:
weeks <- c('201500','201501','201502','201551','201552')
as.Date(paste0(weeks,'4'),format='%Y%W%w')
## [1] "2015-01-01" "2015-01-08" "2015-01-15" "2015-12-24" "2015-12-31"
You may try with substr() and lubridate
library(lubridate)
# a number from your list: 201502
# set the year
x <- ymd("2015-01-1")
# retrieve second week
week(x) <- 2
x
[1] "2015-01-08"
you can use the result for your Index or rownames().
zoo and xts are great for time series once you have set the names,
be sure to remove any column with dates from your data frame
The %tw format in Stata has the form: 1960w1 which has no equivalent in R.
Therefore %tw dates must be post-processed.
Importing a .dta file into R, the date is an integer like 1304 (instead of 1985w5) or 1426 (instead of 1987w23). If it was a simple time series you could set a starting date as follows:
ts(df, start= c(1985,5), frequency=52)
Another possibility would be:
as.Date(Camp$date, format= "%Yw%W" , origin = "1985w5")
But if each row is not a single date, then you must convert it.
The package ISOweek is based on ISO-8601 with the form "1985-W05" and does not process the Stata %tw.
The Lubridate package does not work with this format. The week() returns the number of complete seven day periods that have occurred between the date and January 1st, plus one. week function
In Stata week 1 of any year starts on 1 January, whatever day of the week that is. Stata Documentation on Dates
In the format %W of Date in R the week starts as Monday as first day of the week.
From strptime %V is
the Week of the year as decimal number (00--53) as defined in ISO
8601. If the week (starting on Monday) containing 1 January has four or more days in the new year, then it is considered week 1. Otherwise,
it is the last week of the previous year, and the next week is week 1.
(Accepted but ignored on input.) Strptime
Larmarange noted on Github that Haven doesn't interpret dates properly:
months, week, quarter and halfyear are specific format from Stata,
respectively %tm, %tw, %tq and %th. I'm not sure that there are
corresponding formats available in R. So far they are imported as
integers.
Is there a way to convert Stata %tw to a date format R understands?
Here is an Stata file with dates
This won't be an answer in terms of R code, but it is commentary on Stata weeks that can't be fitted into a comment.
Strictly, dates in Stata are not defined by the display formats that make them intelligible to people. A date in Stata is always a numeric variable or scalar or macro defined with origin the first instance in 1960. Thus it is at best a shorthand to talk about %tw dates, etc. We can use display to see the effects of different date display formats:
. di %td 0
01jan1960
. di %tw 0
1960w1
. di %tq 0
1960q1
. di %td 42
12feb1960
. di %tw 42
1960w43
. di %tq 42
1970q3
A subtle point made explicit above is that changing the display format will not change what is stored, i.e. the numeric value.
Otherwise put, dates in Stata are not distinct data types; they are just integers made intelligible as dates by a pertinent display format.
The question presupposes that it was correct to describe some weekly dates in terms of Stata weeks. This seems unlikely, as I know no instance in which a body outside StataCorp uses the week rules of Stata, not only that week 1 always starts on 1 January, but also that week 52 always includes either 8 or 9 days and hence that there is never a week 53 in a calendar year.
So, you need to go upstream and find out what the data should have been. Failing some explanation, my best advice is to map the 52 weeks of each year to the days that start them, namely days 1(7)358 of each calendar year.
Stata weeks won't map one-to-one to any other scheme for defining weeks.
More in this article on Stata weeks
It's not completely clear what the question is but the year and week corresponding to 1304 are:
wk <- 1304
1960 + wk %/% 52
## [1] 1985
wk %% 52 + 1
## [1] 5
so assuming that the first week of the year is week 1 and starts on Jan 1st, the beginning of the above week is this date:
as.Date(paste(1960 + wk %/% 52, 1, 1, sep = "-")) + 7 * (wk %% 52)
## [1] "1985-01-29"
I have 2 columns with ~ 2000 rows of dates in them. One is a variable with a visit date (df$visitdate), and the other is a birth date of the individual (df$birthday).
Wondering if there is any simple way to subtract the visit date - birth date to create the variable "age at the time of the visit", accounting for leap years, etc.
I tried to use the following code (from an answer in a similar question) but it didn't work in my case.
find number of seconds in one year:
seconds_in_a_year <- as.integer((seconds(ymd("2010-01-01")) - seconds(ymd("2009-01-01"))))
now obtain number of seconds between the 2 dates you desire
seconds_between_dates <- as.integer(seconds(date1) - seconds(date2))
your final answer for number of years in floating points will be
years_between_dates <- seconds_between_dates / seconds_in_a_year
When I tried to apply this to my data frame (note: using variables rather than specific dates, so this may be the cause) I got the following:
seconds_in_a_year <- as.integer((seconds(ymd(df$visitdate)) - seconds(ymd(df$birthday))))
Warning message:
NAs introduced by coercion
Following the code along I got a final output of:
years_between_dates
[1] 1.157407e-05 [2] 1.157407e-05
Any help is greatly appreciated!
Subtracting from a Date object another Date object gives you the time difference in days, e.g.
> dates = as.Date(c("2007-03-01", "2004-05-23"))
>
> dates[1] - dates[2]
Time difference of 1012 days
So, assuming 365 days in a year
> age_time_visit = as.numeric(dates[1] - dates[2]) / 365
> age_time_visit
[1] 2.772603
There are various answers for this scattered around the internet.
I think the one I've typically used was inspired by Professor Ripley:
http://r.789695.n4.nabble.com/Calculate-difference-between-dates-in-years-td835196.html
age_years <- function(first, second)
{
lt <- data.frame(first, second)
age <- as.numeric(format(lt[,2],format="%Y")) - as.numeric(format(lt[,1],format="%Y"))
first <- as.Date(paste(format(lt[,2],format="%Y"),"-",format(lt[,1],format="%m-%d"),sep=""))
age[which(first > lt[,2])] <- age[which(first > lt[,2])] - 1
age
}
There's another approach at https://gist.github.com/mmparker/7254445
Or you you just want to raw, decimal value of years, you can get the number of days and divide by 365.2425
Here is an approach that accounts for leap years (don't know if this has been done before, but suspect it has...).
get.age <- function(from, to) {
require(lubridate) # for leap_year(...)
n <- as.integer(to-from)
n.l <- sum(leap_year(seq(from,to,by=1)))
n.l/366 + (n+1-n.l)/365
}
get.age(as.Date("2009-01-01"),as.Date("2012-12-31"))
# [1] 4
get.age(as.Date("2012-01-01"),as.Date("2012-01-31")) # 2012 was a leap year
# [1] 0.08469945
get.age(as.Date("2011-01-01"),as.Date("2011-01-31")) # 2011 was not
# [1] 0.08493151
So the basic idea is to create a vector with one element for every day between from and to (inclusive), then for each day account for whether that day is part of a leap year or not. The we add up the leap year days and the non-leap year days separately and calculate the number of years as:
leap-year-days/366 + non-leap-year-days/365
This works for single dates (vectors of length 1). To enable this for columns of dates, as you asked, we use Vectorize(...).
vget.age <- Vectorize(get.age) # vectorized version
And then a demo:
# example data set
set.seed(1) # for reproducible example
today <- as.Date("2015-09-09")
df <- data.frame(birth.date=today-sample(1000:10000,2000)) # 2000 birthdays
result <- vget.age(df$birth.date,today) # how old are they?
head(result)
# [1] 9.282192 11.909589 16.854795 25.115068 7.706849 24.865753
I have a dataset with locations and dates. I would like to calculate week of the year as number (00–53) but using Thursday as the first day of the week. The data looks like this:
location <- c(a,b,a,b,a,b)
date <- c("04-01-2013","26-01-2013","03-02-2013","09-02-2013","20-02-2013","03-03-2013")
mydf <- data.frame(location, date)
mydf
I know that there is strftime function for calculating week of year but it is only possible to use Monday or Sunday as the first day of the week.
Any help would be highly appreciated.
Just add 4 to the Date-formatted values:
> mydf$Dt <- as.Date(mydf$date, format="%d-%m-%Y")
> weeknum <- as.numeric( format(mydf$Dt+3, "%U"))
> weeknum
[1] 1 4 5 6 7 9
This uses a 0 based counting convention since that is what strftime provides and we are just piggybacking off that code base, so the first Friday in a year that begins on Tuesday as was the case in 2013 would be a 1-week result. Add 1 to the value if you want a 1 based convention. (Fundamentally, Date-formated values are in an integer sequence from the "origin" so they don't really recognize years or weeks. Adding 4 just shifts the reference frame of the underlying Date-integer.)
Edit note. Changed to an add three strategy per Gabor's advice. .... which still does not address the question of how to deal with the last week of the prior year.
Since the question stated that week goes from 00-53 we assume that the week number is the number of Thursdays in the year on or before the date in question. Thus, the first Thursday in the year begins week 1 and week 0 is assigned to any days prior to that.
(There were comments that if the first day of the year were Tuesday then that would be week 1 but if that were the case there could never be a week 0 as seems to be required in the subject so some clarification on precisely what the definition of week number is may be required. Here we are going to use the definition in the preceding paragraph but it would not be hard to change it if we knew what the definition was. For example, if we always wanted the first week in the year to be 1 even if it were a short week then we could add !is.thu(jan1(d)) to the result.)
Both of the solutions below are short enough that they could be expressed in one statement; however, we have factored them into several short functions each for clarity. The first is particularly straight forward but the second is automatically vectorized without the need for a sapply and would likely be more efficient.
1. sum Thursdays in year This solution assumes the input d is of class "Date" and just sums the number of Thursdays in the year before or on it:
is.thu <- function(x) weekdays(x) == "Thursday"
jan1 <- function(x) as.Date(cut(x, "year"))
week4 <- function(d) {
sapply(d, function(d) sum(is.thu(seq(jan1(d), d, by = "day"))))
}
We can test it like this:
d <- as.Date(c("2013-01-04", "2013-01-26", "2013-02-03", "2013-02-09",
"2013-02-20", "2013-03-03"))
week4(d) # 1 4 5 6 7 9
2. nextthu
Based on the nextfri function in the zoo quickref vignette we see that the number of days since the Epoch (1970-01-01) of the next Thursday (or the day in question if its already a Thursday) is as given by nextthu in the first line below. Applying this to the first day of the year we derive the result where d is as before:
nextthu <- function(d) 7 * ceiling(as.numeric(d) / 7)
week4a <- function(d) (as.numeric(d) - nextthu(jan1(d))) %/% 7 + 1
and here is a test
week4a(d) # 1 4 5 6 7 9
ADDED: fixed bug in second solution.