I have data from several years and each record has a date value (YYYY-MM-DD). I want to label each record with the season that it fell into. For example, I want to take all the records from December 15 to March 15, across all years, and put "Winter" in a season column. Is there a way in R to specify a sequence of dates using just the month and date, regardless of year?
Lubridate quarter command doesn't work because I have custom dates to define the seasons and the seasons are not all of equal length, and I can't just do month(datevalue) %in% c(12,1,2,3) because I need to split the months in half (i.e. March 15 is winter and March 16 is spring).
I could manually enter in the date range for each year in my dataset (e.g. Dec 15 2015 to March 15 2015 or Dec 15 2016 to Mar 15 2016, etc...), but is there a better way?
You can extract the month and date out of the date column and use case_when to assign Season based on those two dates.
library(dplyr)
library(lubridate)
df %>%
mutate(day = day(Date),
month = month(Date),
Season = case_when(#15 December to 15 March as Winter
month == 12 & day >= 15 |
month %in% 1:2 | month == 3 & day <= 15 ~ "Winter",
#Add conditions for other season
)
)
We assume that when the question says that winter is "Dec 15 2015 to March 15 201 or Dec 15 2016 to Mar 15 2016" what is really meant is that winter is Dec 16, 2015 to Mar 15, 2016 or Dec 16, 2016 to Mar 15, 2017.
Also it is not clear what the precise output is supposed to be but in each case below we provide a second argument which takes a vector giving the season names or numbers. The default is that winter is reported as 1, spring is 2, summer is 3 and fall is 4 but you could pass a second argument of c("Winter", "Spring", "Summer", "Fall") instead or use other names if you wish.
1) yearmon/yearqtr Convert to Date class and subtract 15. Then convert that to yearmon class which represents dates internally as year + fraction where fraction = 0 for January, 1/12 for February, ..., 11/12 for December. Add 1/12 to get to the next month. Convert that to yearqtr class which represents dates as year + fraction where fraction is 0, 1/4, 2/4 or 3/4 for the 4 quarters and take cycle of that which gives the quarter number (1, 2, 3 or 4).
If we knew that the input x was a Date vector as opposed to a character vector then we could simplify this by replacing as.Date(x) in season.
library(zoo)
season <- function(x, s = 1:4)
s[cycle(as.yearqtr(as.yearmon(as.Date(x) - 15) + 1/12))]
# test
d <- c(as.Date("2020-12-15") + 0:1, as.Date("2021-03-15") + 0:1)
season(d)
## [1] 4 1 1 2
season(d, c("Winter", "Spring", "Summer", "Fall"))
## [1] "Fall" "Winter" "Winter" "Spring"
2) base The above could be translated to base R using POSIXlt. Subtract 15 as before and then add 1 to the month to get to the next month. Finally extract the month and ensure that is is less than or equal to the third month.
season.lt <- function(x, s = 1:4) {
lt <- as.POSIXlt(as.Date(d) - 15)
lt$mon <- lt$mon + 1
s[as.POSIXlt(format(lt))$mon %/% 3 + 1]
}
# test - d defined in (1)
is.season.lt(d)
## [1] 4 1 1 2
3) lubridate We can follow the same logic in lubridate like this:
season.lub <- function(x, s = 1:4)
s[(month((as.Date(x) - 15) %m+% months(1)) - 1) %/% 3 + 1]
# test - d defined in (1)
season.lub(d)
## [1] 4 1 1 2
Related
In R, how can I produce a list of dates of all 2nd to last Wednesdays of the month in a specified date range? I've tried a few things but have gotten inconsistent results for months with five Wednesdays.
To generate a regular sequence of dates you can use seq with dates for parameter from and to. See the seq.Date documentation for more options.
Create a data frame with the date, the month and weekday. And then obtain the second to last wednesday for each month with the help of aggregate.
day_sequence = seq(as.Date("2020/1/1"), as.Date("2020/12/31"), "day")
df = data.frame(day = day_sequence,
month = months(day_sequence),
weekday = weekdays(day_sequence))
#Filter only wednesdays
df = df[df$weekday == "Wednesday",]
result = aggregate(day ~ month, df, function(x){head(tail(x,2),1)})
tail(x,2) will return the last two rows, then head(.., 1) will give you the first of these last two.
Result:
month day
1 April 2020-04-22
2 August 2020-08-19
3 December 2020-12-23
4 February 2020-02-19
5 January 2020-01-22
6 July 2020-07-22
7 June 2020-06-17
8 March 2020-03-18
9 May 2020-05-20
10 November 2020-11-18
11 October 2020-10-21
12 September 2020-09-23
There are probably simpler ways of doing this but the function below does what the question asks for. it returns a named vector of days such that
They are between from and to.
Are weekday day, where 1 is Monday.
Are n to last of the month.
By n to last I mean the nth counting from the end of the month.
whichWeekday <- function(from, to, day, n, format = "%Y-%m-%d"){
from <- as.Date(from, format = format)
to <- as.Date(to, format = format)
day <- as.character(day)
d <- seq(from, to, by = "days")
m <- format(d, "%Y-%m")
f <- c(TRUE, m[-1] != m[-length(m)])
f <- cumsum(f)
wed <- tapply(d, f, function(x){
i <- which(format(x, "%u") == day)
x[ tail(i, n)[1] ]
})
y <- as.Date(wed, origin = "1970-01-01")
setNames(y, format(y, "%Y-%m"))
}
whichWeekday("2019-01-01", "2020-03-31", 4, 2)
# 2019-01 2019-02 2019-03 2019-04 2019-05
#"2019-01-23" "2019-02-20" "2019-03-20" "2019-04-17" "2019-05-22"
# 2019-06 2019-07 2019-08 2019-09 2019-10
#"2019-06-19" "2019-07-24" "2019-08-21" "2019-09-18" "2019-10-23"
# 2019-11 2019-12 2020-01 2020-02 2020-03
#"2019-11-20" "2019-12-18" "2020-01-22" "2020-02-19" "2020-03-18"
Here's my data which has 10 years in one column and 365 day of another year in second column
dat <- data.frame(year = rep(1980:1989, each = 365), doy= rep(1:365, times = 10))
I am assuming all years are non-leap years i.e. they have 365 days.
I want to create another column month which is basically month of the year the day belongs to.
library(dplyr)
dat %>%
mutate(month = as.integer(ceiling(day/31)))
However, this solution is wrong since it assigns wrong months to days. I am looking for a dplyr
solution possibly.
We can convert it to to datetime class by using the appropriate format (i.e. %Y %j) and then extract the month with format
dat$month <- with(dat, format(strptime(paste(year, doy), format = "%Y %j"), '%m'))
Or use $mon to extract the month and add 1
dat$month <- with(dat, strptime(paste(year, doy), format = "%Y %j")$mon + 1)
tail(dat$month)
#[1] 12 12 12 12 12 12
This should give you an integer value for the months:
dat$month.num <- month(as.Date(paste(dat$year, dat$doy), '%Y %j'))
If you want the month names:
dat$month.names <- month.name[month(as.Date(paste(dat$year, dat$doy), '%Y %j'))]
The result (only showing a few rows):
> dat[29:33,]
year doy month.num month.names
29 1980 29 1 January
30 1980 30 1 January
31 1980 31 1 January
32 1980 32 2 February
33 1980 33 2 February
If I have a date, say "2014-05-13" and I want to calculate the month in decimal, I would do this:
5 + 13/31 = 5.419355
How would it be possible in R to take a vector of dates and turn in it into a "month decimal" vector?
For example:
dates = c("2010-01-24", "2013-04-08", "2014-03-05", "2013-03-08", "2014-02-14",
"2004-01-28", "2006-02-21", "2013-03-28", "2013-04-01", "2006-02-14",
"2006-01-28", "2014-01-19", "2012-03-12", "2014-01-30", "2005-04-17")
library(lubridate)
month(dates) + day(dates)/31
As you can see, it would be wrong to put "31" as the diviser since the number of days differ depending on the month, and sometimes year (leap years).
So what would be the best solution?
You can use monthDaysfunction from Hmisc package
> require(Hmisc)
> library(lubridate)
> month(dates) + day(dates)/monthDays(dates)
[1] 1.774194 4.266667 3.161290 3.258065 2.500000 1.903226 2.750000 3.903226 4.033333
[10] 2.500000 1.903226 1.612903 3.387097 1.967742 4.566667
With magrittr,
library(magrittr)
library(lubridate)
dates %>% ymd() %>% { month(.) + day(.) / days_in_month(.) }
## Jan Apr Mar Mar Feb Jan Feb Mar Apr Feb Jan
## 1.774194 4.266667 3.161290 3.258065 2.500000 1.903226 2.750000 3.903226 4.033333 2.500000 1.903226
## Jan Mar Jan Apr
## 1.612903 3.387097 1.967742 4.566667
For some reason the vector gets named, so add %>% unname() if you like.
Here is a base R hack that uses a trick I've seen on SO to get the first day of the next month and subtract 1 to return the last day of the month of interest.
# format dates to Date class
dates <- as.Date(dates)
# get the next month
nextMonths <- as.integer(substr(dates, 6, 7)) + 1L
# replace next month with 1 if it is equal to 13
nextMonths[nextMonths == 13] <- 1L
# extract the number of days using date formatting (%d), paste, and subtraction
dayCount <- as.integer(format(as.Date(paste(substr(dates, 1, 4),
nextMonths, "01", sep="-"))-1L, format="%d"))
dayCount
[1] 31 30 31 31 28 31 28 31 30 28 31 31 31 31 30
# get month with fraction using date formatting (%m)
as.integer(format(dates, format="%m")) + (as.integer(format(dates, format="%d")) / dayCount)
[1] 1.774194 4.266667 3.161290 3.258065 2.500000 1.903226 2.750000 3.903226 4.033333 2.500000
[11] 1.903226 1.612903 3.387097 1.967742 4.566667
Fisheries data is often collected by statistical weeks that start January 1st every year. The second week starts on the following Sunday each year.
So in 2013 Jan. 1st to Jan. 5 was week 1 and Jan. 6 to Jan.12 was week two. I am trying to calculate the statical week given a date for a number of years. My data is just dates in d-m-y format (i.e 16-6-1990) and I want a statistical week output in R code.
An example would be:
> d <- as.Date(c("01-01-2013","06-01-2013","01-01-2006","08-01-2006"),"%d-%m-%Y")
And the desired result would be:
> statweek(d)
[1] 1 2 1 2
Try this:
> d <- as.Date("01-01-2013", "%d-%m-%Y") + 0:7 # first 8 days of 2013
> d
[1] "2013-01-01" "2013-01-02" "2013-01-03" "2013-01-04" "2013-01-05"
[6] "2013-01-06" "2013-01-07" "2013-01-08"
>
> ufmt <- function(x) as.numeric(format(as.Date(x), "%U"))
> ufmt(d) - ufmt(cut(d, "year")) + 1
[1] 1 1 1 1 1 2 2 2
Note: The first Sunday in the year is defined as the start of week 1 by %U which means that if the year does not start on Sunday then we must add 1 to the week so that the first week is week 1 rather than week 0. ufmt(cut(d, "year")) equals one if d's year starts on Sunday and zero otherwise so the formula above reduces to ufmt(d) if d's year starts on Sunday and ufmt(d)+1 if not.
UPDATE: corrections so Jan starts at week 1 even if year starts on a Sunday, e.g. 2006.
Here is the statweek function. The main argument can be a character vector of dates (the default after reading a data.frame, for example). You can specify the format of the dates (has a default: format="%d-%m-%Y")
d1 <- c("01-01-2013","06-01-2013","01-01-2006","08-01-2006") # format="%d-%m-%Y"
d2 <- c("01/01/2013","06/01/2013","01/01/2006","08/01/2006") # format="%d/%m/%Y"
statweek = function(dates, format="%d-%m-%Y", ...) {
# convert to Date
dates = as.Date(dates, format=format, ...)
# get correction for the first week of the year (0 if 1-Jan not a Sunday)
firstweek = 1 - as.numeric(format(as.Date(cut(dates, "year")), "%U"))
output = as.numeric(format(dates, "%U")) + firstweek
return(output)
}
And the examples:
statweek(d1)
[1] 1 2 1 2
statweek(d1, format="%d-%m-%Y")
[1] 1 2 1 2
statweek(d2, format="%d/%m/%Y")
[1] 1 2 1 2
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Find the day of a week in R
I have a data for days like 11-01-2011 etc. But I want to add the data corresponding
the date as Monday, Tuesday etc. Is there any R package that contains the information of the dates with days?
weekdays(as.Date('16-08-2012','%d-%m-%Y'))
[1] "Thursday"
The lubridate package is great for this sort of stuff.
> wday(as.Date('16-08-2012','%d-%m-%Y'))
[1] 5
> wday(as.Date('16-08-2012','%d-%m-%Y'), label=TRUE)
[1] Thurs
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat
> wday(as.Date('16-08-2012','%d-%m-%Y'), label=TRUE, abbr = FALSE)
[1] Thursday
Levels: Sunday < Monday < Tuesday < Wednesday < Thursday < Friday < Saturday
Here is some information to create your own library or routine
Constants:
day_of_month
the day of the month
e.g. if input mm-dd-yyy then dd
month:
march = 1
april = 2
may = 3
...
year
yy[yy] (last to digits from yyyy)
*subtract 1 if month jan or feb
e.g. if input date is 02-01-2012 (mm-dd-yyyy)
year = (12-1) = 11
century
[yy]yy (first two digits from yyyy)
e.g. if input year is 2012 then 20 = century
* year 2000, 1900, ... are 20-1, 19-1 respectively
ALGORITHM
step1: floor(century / 4)
step2: year
step3: floor(year/4)
step4: floor(month*2.6 -0.2) #this is the leap year correction
step5: day_of_month
step6: add step1...step5
step7: divide by 7 # modulo 7 in codespeak
step8: the remainder is the day of the week
To Interpret Results:
Sun = 0, Mon = 1, Tues = 3, etc..
Not a library, but as the public service jingle goes...
"Read: The More you Know"
Ref: http://www.faqs.org/faqs/sci-math-faq/dayWeek/