I have a starting time specified as a year-month character, e.g. "2020-12". From the start, for each of T consecutive months, I need to generate n different dates (year-month-day), where the day is random.
Any help will be useful!
The data I'm working on:
data <- data.frame(
data = sample(seq(as.Date('2000/01/01'), as.Date('2020/01/01'), by="day"), 500),
price = round(runif(500, min = 10, max = 20),2),
quantity = round(rnorm(500,30),0)
)
func <- function(start, months, n) {
startdate <- as.Date(paste0(start, "-01"))
enddate <- seq(startdate, by = "month", length.out = months)
months <- seq_len(months)
enddate_lt <- as.POSIXlt(enddate)
enddate_lt$mon <- enddate_lt$mon + 1
enddate_lt$mday <- enddate_lt$mday - 1
days_per_month <- as.integer(format(enddate_lt, format = "%d"))
days <- lapply(days_per_month, sample, size = n)
dates <- Map(`+`, enddate, days)
do.call(c, dates)
}
set.seed(2021)
func("2020-12", 4, 3)
# [1] "2020-12-08" "2020-12-07" "2020-12-15" "2021-01-27" "2021-01-08" "2021-01-13" "2021-02-21" "2021-02-07" "2021-02-28"
# [10] "2021-03-28" "2021-03-07" "2021-03-15"
func("2020-12", 5, 2)
# [1] "2020-12-06" "2020-12-16" "2021-01-08" "2021-01-10" "2021-02-24" "2021-02-13" "2021-03-20" "2021-03-29" "2021-04-19"
# [10] "2021-04-28"
func("2020-12", 2, 10)
# [1] "2020-12-29" "2020-12-30" "2020-12-04" "2020-12-15" "2020-12-09" "2020-12-27" "2020-12-05" "2020-12-06" "2020-12-23"
# [10] "2020-12-17" "2021-01-03" "2021-01-20" "2021-01-05" "2021-01-22" "2021-01-23" "2021-01-06" "2021-01-10" "2021-01-07"
# [19] "2021-01-19" "2021-01-12"
Most of the dancing with POSIXlt objects is because it gives us clean (base R) access to the number of days in a month, which makes sampleing the days in a month rather simple. It can also be done (code-golf shorter) using the lubridate package, but I don't know that that is any more correct than this code is.
This just dumps out a sequence of random dates, with n days per month. It does not sort within each month, though it does output the months in order. (That's not a difficult extension, there just wasn't a requirement for it.) It doesn't put out a frame, you can easily extend this to fit in a frame or call data.frame(date = do.call(c, dates)) on the last line, depending on what you need to do with the output.
You could convert the start time to a class for monthly data, zoo::yearmon. Then use as.Date.yearmon and its frac argument ("a number between 0 and 1 inclusive that indicates the fraction of the way through the period that the result represents") with random values from runif (uniform between 0 and 1) to convert to a random date within each year-month.
start = "2020-12"
T = 3
n = 2
library(zoo)
set.seed(1)
as.Date(as.yearmon(start) + rep((1:T)/12, each = n), frac = runif(T * n))
# [1] "2021-01-08" "2021-01-12" "2021-02-16" "2021-02-25" "2021-03-07" "2021-03-27"
Related
I am performing creel surveys and am attempting to construct a random date generator that weights the weekends higher than the weekdays. So far I have a simplistic random date generator that does not take into account the day type. We expect more angling pressure on the weekends (as that is when more people have time to fish) but do not have a way to select random days without including bias. I would like to select 15 days within a given month.
I've already generated a simplistic random date generator:
dates <- data.frame(seq.Date(as.Date(day.start),as.Date(day.end),by="day"))
dates
sample(dates$seq.Date.as.Date.day.start...as.Date.day.end...by....day.., size = 15, replace = FALSE)
[1] "2019-11-10" "2019-11-06" "2019-11-04" "2019-11-27" "2019-11-30" "2019-11-15"
[7] "2019-11-18" "2019-11-21" "2019-11-13" "2019-11-01" "2019-11-19" "2019-11-25"
[13] "2019-11-07" "2019-11-02" "2019-11-23"
Ideally I would have an end product that allows me to input the month start and end and outputs 15 random days.
Explanation in comments in code below:
# Generate initial data; as in question
day_start <- as.Date("2010-10-01")
day_end <- as.Date("2010-10-31")
dates <- data.frame(date = seq.Date(day_start, day_end,by="day"))
# Determine inclusion probabilities for each date; give weekend a higher
# probability.
dates$day <- as.numeric(format(dates$date, "%u"))
dates$psamp <- ifelse(dates$day >= 6, 0.2, 0.1)
# Make sure probabilites add up to requires sample size
samplesize <- 15
dates$psamp <- dates$psamp * samplesize/sum(dates$psamp)
# Do not use sample for sampling without replacement with unequal probabilities!
# The sampling package has a large number of routines for sampling without
# replacement and unequal probabilites. The following gives a fixed size sample
# (sum dates$psamp)
library(sampling)
dates$selected <- UPrandomsystematic(dates$psamp)
As for the reason why I don't use sample see, for example, https://stat.ethz.ch/pipermail/r-help/2008-February/153601.html.
Here's a somewhat general function that does what you want. It takes the start day, end day, and the weight (relative to 1) that you want to put on weekends as its own arguments, and passes on other additional arguments (size, replace, etc.) to sample. No dependencies other than base R.
However, if sampling without replacement, you may want to use the sampling package as recommended in Jan van der Laan's answer.
rday = function(
start_day = as.Date("2019-01-01"),
end_day = as.Date("2019-01-31"),
weekend_weight = 2,
...
) {
if (! "Date" %in% class(start_day)) start_day = as.Date(start_day)
if (! "Date" %in% class(end_day)) end_day = as.Date(end_day)
dates = seq(start_day, end_day, by = "1 day")
weights = rep(1, length(dates))
weights[weekdays(dates) %in% c("Saturday", "Sunday")] = 1
sample(dates, ..., prob = weights)
}
rday(size = 15)
# [1] "2019-01-24" "2019-01-07" "2019-01-21" "2019-01-15" "2019-01-27" "2019-01-04" "2019-01-30" "2019-01-12"
# [9] "2019-01-11" "2019-01-08" "2019-01-20" "2019-01-01" "2019-01-03" "2019-01-19" "2019-01-29"
I'm currently writing a script in the R Programming Language and I've hit a snag.
I have time series data organized in a way where there are 30 days in each month for 12 months in 1 year. However, I need the data organized in a proper 365 days in a year calendar, as in 30 days in a month, 31 days in a month, etc.
Is there a simple way for R to recognize there are 30 days in a month and to operate within that parameter? At the moment I have my script converting the number of days from the source in UNIX time and it counts up.
For example:
startingdate <- "20060101"
endingdate <- "20121230"
date <- seq(from = as.Date(startingdate, "%Y%m%d"), to = as.Date(endingdate, "%Y%m%d"), by = "days")
This would generate an array of dates with each month having 29 days/30 days/31 days etc. However, my data is currently organized as 30 days per month, regardless of 29 days or 31 days present.
Thanks.
The first 4 solutions are basically variations of the same theme using expand.grid. (3) uses magrittr and the others use no packages. The last two work by creating long sequence of numbers and then picking out the ones that have month and day in range.
1) apply This gives a series of yyyymmdd numbers such that there are 30 days in each month. Note that the line defining yrs in this case is the same as yrs <- 2006:2012 so if the years are handy we could shorten that line. Omit as.numeric in the line defining s if you want character string output instead. Also, s and d are the same because we have whole years so we could omit the line defining d and use s as the answer in this case and also in general if we are always dealing with whole years.
startingdate <- "20060101"
endingdate <- "20121230"
yrs <- seq(as.numeric(substr(startingdate, 1, 4)), as.numeric(substr(endingdate, 1, 4)))
g <- expand.grid(yrs, sprintf("%02d", 1:12), sprintf("%02d", 1:30))
s <- sort(as.numeric(apply(g, 1, paste, collapse = "")))
d <- s[ s >= startingdate & s <= endingdate ] # optional if whole years
Run some checks.
head(d)
## [1] 20060101 20060102 20060103 20060104 20060105 20060106
tail(d)
## 20121225 20121226 20121227 20121228 20121229 20121230
length(d) == length(2006:2012) * 12 * 30
## [1] TRUE
2) no apply An alternative variation would be this. In this and the following solutions we are using yrs as calculated in (1) so we omit it to avoid redundancy. Also, in this and the following solutions, the corresponding line to the one setting d is omitted, again, to avoid redundancy -- if you don't have whole years then add the line defining d in (1) replacing s in that line with s2.
g2 <- expand.grid(yr = yrs, mon = sprintf("%02d", 1:12), day = sprintf("%02d", 1:30))
s2 <- with(g2, sort(as.numeric(paste0(yr, mon, day))))
3) magrittr This could also be written using magrittr like this:
library(magrittr)
expand.grid(yr = yrs, mon = sprintf("%02d", 1:12), day = sprintf("%02d", 1:30)) %>%
with(paste0(yr, mon, day)) %>%
as.numeric %>%
sort -> s3
4) do.call Another variation.
g4 <- expand.grid(yrs, 1:12, 1:30)
s4 <- sort(as.numeric(do.call("sprintf", c("%d%02d%02d", g4))))
5) subset sequence Create a sequence of numbers from the starting date to the ending date and if each number is of the form yyyymmdd pick out those for which mm and dd are in range.
seq5 <- seq(as.numeric(startingdate), as.numeric(endingdate))
d5 <- seq5[ seq5 %/% 100 %% 100 %in% 1:12 & seq5 %% 100 %in% 1:30]
6) grep Using seq5 from (5)
d6 <- as.numeric(grep("(0[1-9]|1[0-2])(0[1-9]|[12][0-9]|30)$", seq5, value = TRUE))
Here's an alternative:
date <- unclass(startingdate):unclass(endingdate) %% 30L
month <- rep(1:12, each = 30, length.out = NN <- length(date))
year <- rep(1:(NN %/% 360 + 1), each = 360, length.out = NN)
(of course, we can easily adjust by adding constants to taste if you want a specific day to be 0, or a specific month, etc.)
I would like a function that counts the number of specific days per month..
i.e.. Nov '13 -> 5 fridays.. while Dec'13 would return 4 Fridays..
Is there an elegant function that would return this?
library(lubridate)
num_days <- function(date){
x <- as.Date(date)
start = floor_date(x, "month")
count = days_in_month(x)
d = wday(start)
sol = ifelse(d > 4, 5, 4) #estimate that is the first day of the month is after Thu or Fri then the week will have 5 Fridays
sol
}
num_days("2013-08-01")
num_days(today())
What would be a better way to do this?
1) Here d is the input, a Date class object, e.g. d <- Sys.Date(). The result gives the number of Fridays in the year/month that contains d. Replace 5 with 1 to get the number of Mondays:
first <- as.Date(cut(d, "month"))
last <- as.Date(cut(first + 31, "month")) - 1
sum(format(seq(first, last, "day"), "%w") == 5)
2) Alternately replace the last line with the following line. Here, the first term is the number of Fridays from the Epoch to the next Friday on or after the first of the next month and the second term is the number of Fridays from the Epoch to the next Friday on or after the first of d's month. Again, we replace all 5's with 1's to get the count of Mondays.
ceiling(as.numeric(last + 1 - 5 + 4) / 7) - ceiling(as.numeric(first - 5 + 4) / 7)
The second solution is slightly longer (although it has the same number of lines) but it has the advantage of being vectorized, i.e. d could be a vector of dates.
UPDATE: Added second solution.
There are a number of ways to do it. Here is one:
countFridays <- function(y, m) {
fr <- as.Date(paste(y, m, "01", sep="-"))
to <- fr + 31
dt <- seq(fr, to, by="1 day")
df <- data.frame(date=dt, mon=as.POSIXlt(dt)$mon, wday=as.POSIXlt(dt)$wday)
df <- subset(df, df$wday==5 & df$mon==df[1,"mon"])
return(nrow(df))
}
It creates the first of the months, and a day in the next months.
It then creates a data frame of month index (on a 0 to 11 range, but we only use this for comparison) and weekday.
We then subset to a) be in the same month and b) on a Friday. That is your result set, and
we return the number of rows as your anwser.
Note that this only uses base R code.
Without using lubridate -
#arguments to pass to function:
whichweekday <- 5
whichmonth <- 11
whichyear <- 2013
#function code:
firstday <- as.Date(paste('01',whichmonth,whichyear,sep="-"),'%d-%m-%Y')
lastday <- if(whichmonth == 12) { '31-12-2013' } else {seq(as.Date(firstday,'%d-%m-%Y'), length=2, by="1 month")[2]-1}
sum(
strftime(
seq.Date(
from = firstday,
to = lastday,
by = "day"),
'%w'
) == whichweekday)
This question already has answers here:
How to subtract months from a date in R?
(6 answers)
Closed 4 years ago.
I am trying to add a month to a date i have. But then its not possible in a straight manner so far. Following is what i tried.
d <- as.Date("2004-01-31")
d + 60
# [1] "2004-03-31"
Adding wont help as the month wont be overlapped.
seq(as.Date("2004-01-31"), by = "month", length = 2)
# [1] "2004-01-31" "2004-03-02"
Above might work , but again its not straight forward.
Also its also adding 30 days or something to the date which has issues like the below
seq(as.Date("2004-01-31"), by = "month", length = 10)
# [1] "2004-01-31" "2004-03-02" "2004-03-31" "2004-05-01" "2004-05-31" "2004-07-01" "2004-07-31" "2004-08-31" "2004-10-01" "2004-10-31"
In the above , for the first 2 dates , month haven’t changed.
Also the following approach also failed for month but was success for year
d <- as.POSIXlt(as.Date("2010-01-01"))
d$year <- d$year +1
d
# [1] "2011-01-01 UTC"
d <- as.POSIXlt(as.Date("2010-01-01"))
d$month <- d$month +1
d
Error in format.POSIXlt(x, usetz = TRUE) : invalid 'x' argument
What is the right method to do this ?
Function %m+% from lubridate adds one month without exceeding last day of the new month.
library(lubridate)
(d <- ymd("2012-01-31"))
1 parsed with %Y-%m-%d
[1] "2012-01-31 UTC"
d %m+% months(1)
[1] "2012-02-29 UTC"
It is ambiguous when you say "add a month to a date".
Do you mean
add 30 days?
increase the month part of the date by 1?
In both cases a whole package for a simple addition seems a bit exaggerated.
For the first point, of course, the simple + operator will do:
d=as.Date('2010-01-01')
d + 30
#[1] "2010-01-31"
As for the second I would just create a one line function as simple as that (and with a more general scope):
add.months= function(date,n) seq(date, by = paste (n, "months"), length = 2)[2]
You can use it with arbitrary months, including negative:
add.months(d, 3)
#[1] "2010-04-01"
add.months(d, -3)
#[1] "2009-10-01"
Of course, if you want to add only and often a single month:
add.month=function(date) add.months(date,1)
add.month(d)
#[1] "2010-02-01"
If you add one month to 31 of January, since 31th February is meaningless, the best to get the job done is to add the missing 3 days to the following month, March. So correctly:
add.month(as.Date("2010-01-31"))
#[1] "2010-03-03"
In case, for some very special reason, you need to put a ceiling to the last available day of the month, it's a bit longer:
add.months.ceil=function (date, n){
#no ceiling
nC=add.months(date, n)
#ceiling
day(date)=01
C=add.months(date, n+1)-1
#use ceiling in case of overlapping
if(nC>C) return(C)
return(nC)
}
As usual you could add a single month version:
add.month.ceil=function(date) add.months.ceil(date,1)
So:
d=as.Date('2010-01-31')
add.month.ceil(d)
#[1] "2010-02-28"
d=as.Date('2010-01-21')
add.month.ceil(d)
#[1] "2010-02-21"
And with decrements:
d=as.Date('2010-03-31')
add.months.ceil(d, -1)
#[1] "2010-02-28"
d=as.Date('2010-03-21')
add.months.ceil(d, -1)
#[1] "2010-02-21"
Besides you didn't tell if you were interested to a scalar or vector solution. As for the latter:
add.months.v= function(date,n) as.Date(sapply(date, add.months, n), origin="1970-01-01")
Note: *apply family destroys the class data, that's why it has to be rebuilt.
The vector version brings:
d=c(as.Date('2010/01/01'), as.Date('2010/01/31'))
add.months.v(d,1)
[1] "2010-02-01" "2010-03-03"
Hope you liked it))
Vanilla R has a naive difftime class, but the Lubridate CRAN package lets you do what you ask:
require(lubridate)
d <- ymd(as.Date('2004-01-01')) %m+% months(1)
d
[1] "2004-02-01"
Hope that helps.
The simplest way is to convert Date to POSIXlt format.
Then perform the arithmetic operation as follows:
date_1m_fwd <- as.POSIXlt("2010-01-01")
date_1m_fwd$mon <- date_1m_fwd$mon +1
Moreover, incase you want to deal with Date columns in data.table, unfortunately, POSIXlt format is not supported.
Still you can perform the add month using basic R codes as follows:
library(data.table)
dt <- as.data.table(seq(as.Date("2010-01-01"), length.out=5, by="month"))
dt[,shifted_month:=tail(seq(V1[1], length.out=length(V1)+3, by="month"),length(V1))]
Hope it helps.
"mondate" is somewhat similar to "Date" except that adding n adds n months rather than n days:
> library(mondate)
> d <- as.Date("2004-01-31")
> as.mondate(d) + 1
mondate: timeunits="months"
[1] 2004-02-29
Here's a function that doesn't require any packages to be installed. You give it a Date object (or a character that it can convert into a Date), and it adds n months to that date without changing the day of the month (unless the month you land on doesn't have enough days in it, in which case it defaults to the last day of the returned month). Just in case it doesn't make sense reading it, there are some examples below.
Function definition
addMonth <- function(date, n = 1){
if (n == 0){return(date)}
if (n %% 1 != 0){stop("Input Error: argument 'n' must be an integer.")}
# Check to make sure we have a standard Date format
if (class(date) == "character"){date = as.Date(date)}
# Turn the year, month, and day into numbers so we can play with them
y = as.numeric(substr(as.character(date),1,4))
m = as.numeric(substr(as.character(date),6,7))
d = as.numeric(substr(as.character(date),9,10))
# Run through the computation
i = 0
# Adding months
if (n > 0){
while (i < n){
m = m + 1
if (m == 13){
m = 1
y = y + 1
}
i = i + 1
}
}
# Subtracting months
else if (n < 0){
while (i > n){
m = m - 1
if (m == 0){
m = 12
y = y - 1
}
i = i - 1
}
}
# If past 28th day in base month, make adjustments for February
if (d > 28 & m == 2){
# If it's a leap year, return the 29th day
if ((y %% 4 == 0 & y %% 100 != 0) | y %% 400 == 0){d = 29}
# Otherwise, return the 28th day
else{d = 28}
}
# If 31st day in base month but only 30 days in end month, return 30th day
else if (d == 31){if (m %in% c(1, 3, 5, 7, 8, 10, 12) == FALSE){d = 30}}
# Turn year, month, and day into strings and put them together to make a Date
y = as.character(y)
# If month is single digit, add a leading 0, otherwise leave it alone
if (m < 10){m = paste('0', as.character(m), sep = '')}
else{m = as.character(m)}
# If day is single digit, add a leading 0, otherwise leave it alone
if (d < 10){d = paste('0', as.character(d), sep = '')}
else{d = as.character(d)}
# Put them together and convert return the result as a Date
return(as.Date(paste(y,'-',m,'-',d, sep = '')))
}
Some examples
Adding months
> addMonth('2014-01-31', n = 1)
[1] "2014-02-28" # February, non-leap year
> addMonth('2014-01-31', n = 5)
[1] "2014-06-30" # June only has 30 days, so day of month dropped to 30
> addMonth('2014-01-31', n = 24)
[1] "2016-01-31" # Increments years when n is a multiple of 12
> addMonth('2014-01-31', n = 25)
[1] "2016-02-29" # February, leap year
Subtracting months
> addMonth('2014-01-31', n = -1)
[1] "2013-12-31"
> addMonth('2014-01-31', n = -7)
[1] "2013-06-30"
> addMonth('2014-01-31', n = -12)
[1] "2013-01-31"
> addMonth('2014-01-31', n = -23)
[1] "2012-02-29"
addedMonth <- seq(as.Date('2004-01-01'), length=2, by='1 month')[2]
addedQuarter <- seq(as.Date('2004-01-01'), length=2, by='1 quarter')[2]
I turned antonio's thoughts into a specific function:
library(DescTools)
> AddMonths(as.Date('2004-01-01'), 1)
[1] "2004-02-01"
> AddMonths(as.Date('2004-01-31'), 1)
[1] "2004-02-29"
> AddMonths(as.Date('2004-03-30'), -1)
[1] "2004-02-29"
I'm working with some time data and I'm having problems converting a time difference to years and months.
My data looks more or less like this,
dfn <- data.frame(
Today = Sys.time(),
DOB = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
Patient = factor(1:26, labels = LETTERS))
First I subtract the data of birth (DOB) form today's data (Today).
dfn$ageToday <- dfn$Today - dfn$DOB
This gives me the Time difference in days.
dfn$ageToday
Time differences in days
[1] 1875.866 1872.866 1869.866 1866.866 1863.866
[6] 1860.866 1857.866 1854.866 1851.866 1848.866
[11] 1845.866 1842.866 1839.866 1836.866 1833.866
[16] 1830.866 1827.866 1824.866 1821.866 1818.866
[21] 1815.866 1812.866 1809.866 1806.866 1803.866
[26] 1800.866
attr(,"tzone")
[1] ""
This is where first part of my question comes in; how do I convert this difference to years and months (rounded to months)? (i.e. 4.7, 4.11, etc.)
I read the ?difftime man page and the ?format, but I did not figure it out.
Any help would be appreciated.
Furthermore, I would like to melt my final object and if I try using melt on the data frame above using this command,
require(plyr)
require(reshape)
mdfn <- melt(dfn, id=c('Patient'))
I get this strange warning I haven't see before
Error in as.POSIXct.default(value) :
do not know how to convert 'value' to class "POSIXct"
So, my second question is; how do I create a time diffrence I can melt alongside my POSIXct variables? If I melt without dfn$ageToday everything works like a charm.
Thanks, Eric
The lubridatepackage makes working with dates and times, including finding time differences, really easy.
library("lubridate")
library("reshape2")
dfn <- data.frame(
Today = Sys.time(),
DOB = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
Patient = factor(1:26, labels = LETTERS))
dfn$diff <- new_interval(dfn$DOB, dfn$Today) / duration(num = 1, units = "years")
mdfn <- melt(dfn, id=c('Patient'))
class(mdfn$value) # all values are coerced into numeric
The new_interval() function calculates the time difference between two dates. Note that there is a function today() that could substitute for your use of Sys.time. Finally note the duration() function that creates a standard, ehm, duration that you can use to divide the interval by a length of standard units, in this case, a unit of one year.
In case you want to preserve the contents of Today and DOB, then you may want to convert everything to character first and reconvert later...
library("lubridate")
library("reshape2")
dfn <- data.frame(
Today = Sys.time(),
DOB = seq(as.POSIXct('2007-03-27 00:00:01'), len= 26, by="3 day"),
Patient = factor(1:26, labels = LETTERS))
# Create standard durations for a year and a month
one.year <- duration(num = 1, units = "years")
one.month <- duration(num = 1, units = "months")
# Calculate the difference in years as float and integer
dfn$diff.years <- new_interval(dfn$DOB, dfn$Today) / one.year
dfn$years <- floor( new_interval(dfn$DOB, dfn$Today) / one.year )
# Calculate the modulo for number of months
dfn$diff.months <- round( new_interval(dfn$DOB, dfn$Today) / one.month )
dfn$months <- dfn$diff.months %% 12
# Paste the years and months together
# I am not using the decimal point so as not to imply this is
# a numeric representation of the diference
dfn$y.m <- paste(dfn$years, dfn$months, sep = '|')
# convert Today and DOB to character so as to preserve them in melting
dfn$Today <- as.character(dfn$Today)
dfn$DOB <- as.character(dfn$DOB)
# melt using string representation of difference between the two dates
dfn2 <- dfn[,c("Today", "DOB", "Patient", "y.m")]
mdfn2 <- melt(dfn2, id=c('Patient'))
# alternative melt using numeric representation of difference in years
dfn3 <- dfn[,c("Today", "DOB", "Patient", "diff.years")]
mdfn3 <- melt(dfn3, id=c('Patient'))