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.
Related
This question is about how to replace missing days and months in a data frame using R. Considering the data frame below, 99 denotes missing day or month and NA represents dates that are completely unknown.
df<-data.frame("id"=c(1,2,3,4,5),
"date" = c("99/10/2014","99/99/2011","23/02/2016","NA",
"99/04/2009"))
I am trying to replace the missing days and months based on the following criteria:
For dates with missing day but known month and year, the replacement date would be a random selection from the middle of the interval (first day to the last day of that month). Example, for id 1, the replacement date would be sampled from the middle of 01/10/2014 to 31/10/2014. For id 5, this would be the middle of 01/04/2009 to 30/04/2009. Of note is the varying number of days for different months, e.g. 31 days for October and 30 days for April.
As in the case of id 2, where both day and month are missing, the replacement date is a random selection from the middle of the interval (first day to last day of the year), e.g 01/01/2011 to 31/12/2011.
Please note: complete dates (e.g. the case of id 3) and NAs are not to be replaced.
I have tried by making use of the seq function together with the as.POSIXct and as.Date functions to obtain the sequence of dates from which the replacement dates are to be sampled. The difficulty I am experiencing is how to automate the R code to obtain the date intervals (it varies across distinct id) and how to make a random draw from the middle of the intervals.
The expected output would have the date of id 1, 2 and 5 replaced but those of id 3 and 4 remain unchanged. Any help on this is greatly appreciated.
This isn't the prettiest, but it seems to work and adapts to differing month and year lengths:
set.seed(999)
df$dateorig <- df$date
seld <- grepl("^99/", df$date)
selm <- grepl("^../99", df$date)
md <- seld & (!selm)
mm <- seld & selm
df$date <- as.Date(gsub("99","01",as.character(df$date)), format="%d/%m/%Y")
monrng <- sapply(df$date[md], function(x) seq(x, length.out=2, by="month")[2]) - as.numeric(df$date[md])
df$date[md] <- df$date[md] + sapply(monrng, sample, 1)
yrrng <- sapply(df$date[mm], function(x) seq(x, length.out=2, by="12 months")[2]) - as.numeric(df$date[mm])
df$date[mm] <- df$date[mm] + sapply(yrrng, sample, 1)
#df
# id date dateorig
#1 1 2014-10-14 99/10/2014
#2 2 2011-02-05 99/99/2011
#3 3 2016-02-23 23/02/2016
#4 4 <NA> NA
#5 5 2009-04-19 99/04/2009
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
I have a R time series data, where I am calculating the means for all values up to a particular date, and storing this means in the date + 4 quarters. The dates are all month ends. To achieve this, I am looking to increment 4 quarters to a date. My question is how can I add 4 quarters to an R date data-type. An illustration:
a <- as.Date("2006-01-01")
b <- as.Date("2011-01-01")
date_range <- quarter(seq.Date(a, b, by = "quarter"), with_year = TRUE)
> date_range[1] + 1
[1] 2007.1
> date_range[1] + quarter(1)
[1] 2007.1
> date_range[1] + 0.25
[1] 2006.35
One possible way I am thinking is to get year-quarter dates, and then adding 4 to it. But wasn't sure what is the best way to do this?
The problem is that quarters have different lengths. Q1 is shortest because it includes February (though it ties with Q2 in leap years). Things like this make "adding a quarter to a date" poorly defined. Even adding months to a date can be tricky at the ends months - what is 1 month after January 31?
Beginnings of months are more straightforward, and I would recommend you use the 1st day of quarters rather than the last (if you must use a specific date). lubridate provides functions like floor_date() and ceiling_date() to which you can pass unit = "quarter" and they will return the first day of the current or subsequent quarter, respectively. You can also always add months(3) to a day at the beginning of a month, though of course if your intention is to add 4 quarters you may as well just add 1 year.
Just add 12 months or a year instead?
Or if it must be quarters, define yourself a function, like so:
quarters <- function(x) {
months(3*x)
}
and then use it to add to the date sequence:
date_range <- seq.Date(a, b, by = "quarter")
date_range + quarters(4)
Lubridate has a function for quarters already included. This is a much better solution than creating your own function.
https://www.rdocumentation.org/packages/lubridate/versions/1.7.4/topics/quarter
Old answer but to those arriving here, lubridate has a function %m+%that adds months and preserves monthends.
a <- as.Date("2006-01-01")
Add future months worth of dates:
The original poster wanted 4 quarters in future so that will be 12 months.
future_date <- a %m+% months(12)
future_date
[1] "2007-01-01"
You could also do years as the period:
future_date <- a %m+% years(1)
Remove months from date:
Subtract dates with %m-%
If you wanted a date 3 months ago from 1/1/2006:
past_date <- a %m-% months(3)
past_date
[1] "2005-10-01"
Example with dates not at end of months:
mplus will preserve days in month:
as.Date("2022-10-10") %m-% months(3)
[1] "2022-07-10"
For more, see documentation on "Add and subtract months to a date without exceeding the last day of the new month"
Note that other answers that use Date class will give irregularly spaced series and so are unsuitable for time series analysis.
To do this in such a way that time series analyses can be performed and noting the zoo tag on the question, the yearmon class represents year/month as year + fraction where fraction is 0 for Jan, 1/12 for Feb, 2/12 for Mar, ..., 11/12 for Dec. Thus adding 4 quarters is just a matter of adding 1. (Adding x quarters is done by adding x/4.)
library(zoo)
ym <- yearmon(2006) + 0:11/12 # months in 2006
ym + 1 # one year later
Also this converts yearmon objects to end-of-month Date and in the second line Date to yearmon. Using frac = 0 or omitting frac in the first line would convert to beginning of month dates.
d <- as.Date(ym, frac = 1) # d is Date vector of end-of-months
as.yearmon(d) # convert Date vector to yearmon
If your input dates represent quarters then there is also the yearqtr class which represents a year/quarter as year + fraction where fraction is 0, 1/4, 2/4, 3/4 for the 4 quarters of a year. Adding 4 quarters is done by adding 1 (or to add x quarters add x/4).
yq <- as.yearqtr(2006) + 0:3/4 # all quarters in 2006
yq + 1 # one year later
Conversions work similarly to yearmon:
d <- as.Date(ym, frac = 1) # d is Date vector of end-of-quarters
as.yearqtr(d) # convert Date vector to yearqtr
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
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