I create a custom calendar using create.calendar function from bizdays package and a vector with all 2023 year days in vector Date_A. I need to add 8 working days to every day and store it in Date_B, for that reason I used add.bizdays function but the output I get NA values for last 19 values but they should be valid dates, for example for "2023-12-13" in Date_A the output in Date_B should be "2023-12-26" but gives an NA. How can I achieve a valid result instead of that misssing values?
The code is as follows:
library(bizdays)
library(lubridate)
feriados_CR <- as.Date(c(
"2023-01-01","2023-01-02","2023-01-03","2023-01-04","2023-01-05","2023-01-06","2023-04-06",
"2023-04-07","2023-04-10","2023-05-01","2023-07-24","2023-08-02","2023-08-14","2023-09-03",
"2023-09-15","2023-12-01","2023-12-25"))
calendar <- create.calendar("Costa_Rica",
holidays = feriados_CR,
weekdays = c("saturday", "sunday"))
Date_A <- seq(ymd("2023-01-01"), ymd("2023-12-31"), by = "days")
Date_B <- add.bizdays(Date_A, 8, "Costa_Rica")
Related
I am having trouble in using the date variable in my dataset to create categories of 6 months time period. I want to create these time period categories for years between 2017-1-1 and 2020-6-30. The time period categories for each year would be from 2017-1-1 to 2017-6-30, and 2017-7-1 to 2017-12-31 until 2020-6-30.
I have used the following two types of codes to create date categories but I am getting a similar error:
#CODE1
#checking for date class
myData <- str(myData)
myData #date in factor class
#convert to date class
date_class <- as.Date(myData$date, format = "%m/%d/%Y")
myData$date_class <- as.Date(myData$date, format = "%m/%d/%Y")
myData
#creating timeperiod category 1
date_cat <- NA
myData$date_cat[which(myData$date_class >= "2017-1-1" & myData$date_class < "2017-7-1")] <- 1
#CODE2
#converting to date format
myData$date <- strptime(myData$date,format="%m/%d/%Y")
myData$date <- as.POSIXct(myData$date)
myData
#creating timeperiod category 1
date_cat <- NA
myData$date_cat[which(myData$date >= "2017-1-1" & myData$date < "2017-7-1")] <- 1
For both the codes I am getting a similar error
Error in $<-.data.frame(*tmp*, date_cat, value = numeric(0)) :
replacement has 0 rows, data has 1123
Please help me with understanding where I am going wrong.
Thanks,
Priya
Here's a function (to.interval) that returns a time interval {0, 1, 2, 3, ...}, given parameters of the event date, index date, and interval width. Probably a good idea to include error checking in the function, so if for example the event date is prior to the anchor date, it returns NA.
df <- data.frame(event.date=as.Date(c("2017-01-01", "2017-08-01", "2018-04-30")))
to.interval <- function(anchor.date, future.date, interval.days){
round(as.integer(future.date - anchor.date) / interval.days, 0)}
df$interval <- to.interval(as.Date('2017-01-01'),
df$event.date, 180 )
df
Output
event.date interval
1 2017-01-01 0
2 2017-08-01 1
3 2018-04-30 3
I am working in R, I have 20 years data and I would to check if every giving date is a UK holiday creating a categorical variable (TRUE/FALSE).
I used this code:
library(timeDate)
c <- timeDate(data$Date)
b <- isHoliday(c, holidays = GBBankHoliday(), wday = 1:6)
or
b <- isHoliday(c, holidays = HolidayLONDON(), wday = 1:6)
but it detecs only Sundays (not Christmass or other Holidays).
Does anyone have an idea what to do?
You can try creating wrapper functions for various holidays in the package, and extracting the dates for the holidays, and cross-referencing those dates for your analysis:
library(timeDate)
c <- timeDate(data$Date)
b <- isHoliday(c, holidays = GBBankHoliday(), wday = 1:6)
years <- list(Year = c(2019,2018,2017,2016))
year_fun <- function(year){timeDate::.easter(year)}
purrr::map(years, year_fun)
$Year
GMT
[1] [2019-04-21] [2018-04-01] [2017-04-16] [2016-03-27]
I created a new binary variable in my data which is called "holiday". If the date of my data is a UK holiday the value is 1 (TRUE) if the date is not a holiday the value is 2 (FALSE). The code is very simple:
library(timeDate)
data$holidays<-as.factor(data$Date %in% (as.Date(holidayLONDON(1990:2010))))
Let's say we have this:
ex <- c('2012-41')
This represent the week 41 from the year 2012. How would I get the month from this?
Since a week can be between two months, I will be interested to get the month when that week started (here October).
Not duplicate to How to extract Month from date in R (do not have a standard date format like %Y-%m-%d).
you could try:
ex <- c('2019-10')
splitDate <- strsplit(ex, "-")
dateNew <- as.Date(paste(splitDate[[1]][1], splitDate[[1]][2], 1, sep="-"), "%Y-%U-%u")
monthSelected <- lubridate::month(dateNew)
3
I hope this helps!
This depends on the definition of week. See the discussion of %V and %W in ?strptime for two possible definitions of week. We use %V below but the function allows one to specify the other if desired. The function performs a sapply over the elements of x and for each such element it extracts the year into yr and forms a sequence of all dates for that year in sq. It then converts those dates to year-month and finds the first occurrence of the current component of x in that sequence, finally extracting the match's month.
yw2m <- function(x, fmt = "%Y-%V") {
sapply(x, function(x) {
yr <- as.numeric(substr(x, 1, 4))
sq <- seq(as.Date(paste0(yr, "-01-01")), as.Date(paste0(yr, "-12-31")), "day")
as.numeric(format(sq[which.max(format(sq, fmt) == x)], "%m"))
})
}
yw2m('2012-41')
## [1] 10
The following will add the week-of-year to an input of year-week formatted strings and return a vector of dates as character. The lubridate package weeks() function will add the dates corresponding to the end of the relevant week. Note for example I've added an additional case in your 'ex' variable to the 52nd week, and it returns Dec-31st
library(lubridate)
ex <- c('2012-41','2016-4','2018-52')
dates <- strsplit(ex,"-")
dates <- sapply(dates,function(x) {
year_week <- unlist(x)
year <- year_week[1]
week <- year_week[2]
start_date <- as.Date(paste0(year,'-01-01'))
date <- start_date+weeks(week)
#note here: OP asked for beginning of week.
#There's some ambiguity here, the above is end-of-week;
#uncommment here for beginning of week, just subtracted 6 days.
#I think this might yield inconsistent results, especially year-boundaries
#hence suggestion to use end of week. See below for possible solution
#date <- start_date+weeks(week)-days(6)
return (as.character(date))
})
Yields:
> dates
[1] "2012-10-14" "2016-01-29" "2018-12-31"
And to simply get the month from these full dates:
month(dates)
Yields:
> month(dates)
[1] 10 1 12
I have hourly time series data for the year 2015. This data corresponds to power consumption of a big commercial building. I want to use this data to predict the usage for the year 2016. To develop a forecasting model, I need to format this data in a suitable format.
I am planning to use following features to predict the 2016 usage: (1) day of week, (2) time of the day (3) temperature, (4) year 2015 usage.
I am able to create the first 3 features but the fourth one seems tricky.
How should I arrange the 2015 data so that for a particular day of 2016 I can use the corresponding day data of year 2015. My concern is :
I should not use the weekend day data of 2015 to predict the usage of working day
There are some days in 2015, where data is missing for entire day data. For the corresponding day in 2016, how should I account for these missing readings
Here, I have created dummy data corresponding to the year 2015 and 2016.
library(xts)
set.seed(123)
seq1 <- seq(as.POSIXct("2015-01-01"),as.POSIXct("2015-12-31"), by = "hour")
data1 <- xts(rnorm(length(seq1),150,5),seq1)
seq2 <- seq(as.POSIXct("2016-01-01"),as.POSIXct("2016-09-30"), by = "hour")
data2 <- xts(rnorm(length(seq2),140,5),seq2)
Let me give an example to clarify my problem:
Suppose model is: lm( output ~ dayofweek + timeofday + temperature + lastyearusage, data = xxx)
Now suppose I want to predict the usage on 2 oct 2016(dayY), using the lastyearusage onm2 oct 2015(dayX). In this step, issue is 1) How should I ensure thatdayX is not a weekend day if dayY is a working day. I am sure that in this case if I use dayX to predict dayY without keeping a check on day type output will get messy.
There might be already a function in a package to do this, but post here a custom function to add all these kinds of calendar variables (including the week-end info) to a data.frame containing a date/hour column. Fake data:
df <- data.frame(datetime=seq(as.POSIXlt("2013/01/01 00:00:00"), as.POSIXlt("2013/12/31 23:00:00"), by="hour"), variable=rnorm(8760))
#### datetime variable
#### 1 2013-01-01 00:00:00 1.68959052
#### 2 2013-01-01 01:00:00 0.02023722
#### 3 2013-01-01 02:00:00 -0.42080942
The code for the function:
CreateCalendarVariables = function(df, id_column=NULL) {
df <- data.frame(df)
if (is.null(id_column)) stop("Id column for the datetime variable is a mandatory argument")
temp <- df[, id_column]
if ( !(class(temp)[1] %in% c("Date", "POSIXct", "POSIXt", "POSIXlt")) ){
stop("the indicated datetime variable doesn't have the suitable format")
}
require(lubridate)
df['year'] <- year(temp)
df['.quarter'] <- quarter(temp)
df['.month'] <- month(temp)
df['.week'] <- week(temp)
df['.DMY'] <- as.Date(temp)
df['.dayinyear'] <- yday(temp)
df['.dayinmonth'] <- mday(temp)
df['.weekday'] <- wday(temp, label=T, abbr=FALSE) %>% factor(., levels=levels(.)[c(2,3,4,5,6,7,1)])
df['.is_we'] <- df$.weekday %in% c("Saturday", "Sunday")
if(class(temp)[1] != "Date"){
df['.hour'] <- factor(hour(temp))
}
return(df)
}
Then you just have to specify the N° of column containing the date format. If you need for your model these variables in factor format, feel free to adapt the code:
CreateCalendarVariables(df, 2)
#### Error in CreateCalendarVariables(df, 2) :
#### the indicated datetime variable doesn't have the suitable format
CreateCalendarVariables(df, 1)
#### datetime variable year .quarter .month .week .DMY .dayinyear .dayinmonth .weekday .is_we .hour
#### 1 2013-01-01 00:00:00 1.68959052 2013 1 1 1 2012-12-31 1 1 Tuesday FALSE 0
#### 2 2013-01-01 01:00:00 0.02023722 2013 1 1 1 2013-01-01 1 1 Tuesday FALSE 1
To answer your last question, If an entire level is missing from the calibration dataset (i.e. one whole weed and you're using .Week as a predictor), you 'll need to impute the data first.
I have a data frame
> df
Age year sex
12 80210 F
13 9123 M
I want to convert the year 80210 as 26june1982. How can I do this that the new data frame contains year in day month year formate from Julian days.
You can convert Julian dates to dates using as.Date and specifying the appropriate origin:
as.Date(8210, origin=as.Date("1960-01-01"))
#[1] "1982-06-24"
However, 80210 needs an origin pretty long ago.
You should substract the origin from the year column.
as.Date(c(80210,9123)-80210,origin='1982-06-26')
[1] "1982-06-26" "1787-11-08"
There are some options for doing this job in the R package date.
See for example on page 4, the function date.mmddyy, which says:
Given a vector of Julian dates, this returns them in the form “10/11/89”, “28/7/54”, etc.
Try this code:
age = c(12,13)
year = c(8210,9123)
sex = c("F","M")
df = data.frame(cbind(age,year,sex))
library(date)
date = date.mmddyy(year, sep = "/")
df2 = transform(df,year=date) #hint provided by jilber
df2
age year sex
1 12 6/24/82 F
2 13 12/23/84 M