I've got a column with dates (df$Date) and I would like to create a new column (df$NewColumn) that adds a variable number of days depending on the day of the week (using Lubridate's wday) that is in df$Date, e.g.:
df$NewColumn <- ifelse(wday(df$Date) == 6, df$Date + 3,
ifelse(wday(df$Date) == 7, df$Date + 3,
df$Date + 2))
It is seemingly working, only problem is my date format is in a 16XXX format and I can't seem to convert it back to a proper-looking date. Thanks for your help!
One way to deal with this is to use the class() function:
R> testdate <- 16716
R> class(testdate) <- "Date"
R> testdate
[1] "2015-10-08"
As stated in the comments, ifelse changes formats. If you can avoid using it, then the format should stay the same. The following code creates the output you're looking for:
library(lubridate)
df <- data.frame(Date=as.Date(c("2011/01/01",
"2011/01/02")))
df$NewColumn <- df$Date + 2 + as.numeric(wday(df$Date)>=6)
Related
I have a date in R, e.g.:
dt = as.Date('2010/03/17')
I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2008-03-17').
How would I do that?
With lubridate
library(lubridate)
ymd("2010/03/17") - years(2)
The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.
> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"
See this related question: How to subtract days in R?.
You could use seq:
R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"
If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:
> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"
# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"
Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):
R> unlist(lapply(c("2015-12-01", "2016-12-01"),
function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
[1] "2014-12-01" "2015-12-01"
This way seems to do the job as well
dt = as.Date("2010/03/17")
dt-365*2
[1] "2008-03-17"
as.Date("2008/02/29")-365*2
## [1] "2006-03-01"
cur_date <- str_split(as.character(Sys.Date()), pattern = "-")
cur_yr <- cur_date[[1]][1]
cur_month <- cur_date[[1]][2]
cur_day <- cur_date[[1]][3]
new_year <- as.integer(year) - 2
new_date <- paste(new_year, cur_month, cur_day, sep="-")
Using Base R, you can simply use the following without installing any package.
1) Transform your character string to Date format, specifying the input format in the second argument, so R can correctly interpret your date format.
dt = as.Date('2010/03/17',"%Y/%m/%d")
NOTE: If you look now at your enviroment tab you will see dt as variable with the following value "2010-03-17" (Year-month-date separated by "-" not by "/")
2) specify how many years to substract
years_substract=2
3) Use paste() combined with format () to only keep Month and Day and Just substract 2 year from your original date. Format() function will just keep the specific part of your date accordingly with format second argument.
dt_substract_2years<-
as.Date(paste(as.numeric(format(dt,"%Y"))-years_substract,format(dt,"%m"),format(dt,"%d"),sep = "-"))
NOTE1: We used paste() function to concatenate date components and specify separator as "-" (sep = "-")as is the R separator for dates by default.
NOTE2: We also used as.numeric() function to transform year from character to numeric
I have a date in R, e.g.:
dt = as.Date('2010/03/17')
I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2008-03-17').
How would I do that?
With lubridate
library(lubridate)
ymd("2010/03/17") - years(2)
The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.
> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"
See this related question: How to subtract days in R?.
You could use seq:
R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"
If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:
> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"
# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"
Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):
R> unlist(lapply(c("2015-12-01", "2016-12-01"),
function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
[1] "2014-12-01" "2015-12-01"
This way seems to do the job as well
dt = as.Date("2010/03/17")
dt-365*2
[1] "2008-03-17"
as.Date("2008/02/29")-365*2
## [1] "2006-03-01"
cur_date <- str_split(as.character(Sys.Date()), pattern = "-")
cur_yr <- cur_date[[1]][1]
cur_month <- cur_date[[1]][2]
cur_day <- cur_date[[1]][3]
new_year <- as.integer(year) - 2
new_date <- paste(new_year, cur_month, cur_day, sep="-")
Using Base R, you can simply use the following without installing any package.
1) Transform your character string to Date format, specifying the input format in the second argument, so R can correctly interpret your date format.
dt = as.Date('2010/03/17',"%Y/%m/%d")
NOTE: If you look now at your enviroment tab you will see dt as variable with the following value "2010-03-17" (Year-month-date separated by "-" not by "/")
2) specify how many years to substract
years_substract=2
3) Use paste() combined with format () to only keep Month and Day and Just substract 2 year from your original date. Format() function will just keep the specific part of your date accordingly with format second argument.
dt_substract_2years<-
as.Date(paste(as.numeric(format(dt,"%Y"))-years_substract,format(dt,"%m"),format(dt,"%d"),sep = "-"))
NOTE1: We used paste() function to concatenate date components and specify separator as "-" (sep = "-")as is the R separator for dates by default.
NOTE2: We also used as.numeric() function to transform year from character to numeric
I can't figure out how to turn Sys.Date() into a number in the format YYYYDDD. Where DDD is the day of the year, i.e. Jan 1 would be 2016001 Dec 31 would be 2016365
Date <- Sys.Date() ## The Variable Date is created as 2016-01-01
SomeFunction(Date) ## Returns 2016001
You can just use the format function as follows:
format(Date, '%Y%j')
which gives:
[1] "2016161" "2016162" "2016163"
If you want to format it in other ways, see ?strptime for all the possible options.
Alternatively, you could use the year and yday functions from the data.table or lubridate packages and paste them together with paste0:
library(data.table) # or: library(lubridate)
paste0(year(Date), yday(Date))
which will give you the same result.
The values that are returned by both options are of class character. Wrap the above solutions in as.numeric() to get real numbers.
Used data:
> Date <- Sys.Date() + 1:3
> Date
[1] "2016-06-09" "2016-06-10" "2016-06-11"
> class(Date)
[1] "Date"
Here's one option with lubridate:
library(lubridate)
x <- Sys.Date()
#[1] "2016-06-08"
paste0(year(x),yday(x))
#[1] "2016160"
This should work for creating a new column with the specified date format:
Date <- Sys.Date
df$Month_Yr <- format(as.Date(df$Date), "%Y%d")
But, especially when working with larger data sets, it is easier to do the following:
library(data.table)
setDT(df)[,NewDate := format(as.Date(Date), "%Y%d"
Hope this helps. May have to tinker if you only want one value and are not working with a data set.
Help me to find difference between times.For eg: these are the date and time
2015-11-24 16:49:14
2014-12-02 16:52:43
Need the result in HH:MM:SS format using r.
As you need difference between only the time, ignoring the dates you can first extract the time using strptime
x <- strptime(substr(a, 12, 19), format="%H:%M:%S")
y <- strptime(substr(b, 12, 19), format="%H:%M:%S")
Then using the seconds_to_period function of lubridate package you can get the time difference and then format the output using sprintf
library(lubridate)
temp <- seconds_to_period(as.numeric(difftime(y, x, units = "secs")))
sprintf('%02d:%02d:%02d', hour(temp), minute(temp), second(temp))
# [1] "00:03:29"
data
a <- as.POSIXct("2015-11-24 16:49:14")
b <- as.POSIXct("2014-12-02 16:52:43")
Following code to get the difference
library(lubridate)
interval(ymd_hms("2015-11-2416:17:38"),ymd_hms("2015-11-24 14:19:44"))
span<-interval(as.POSIXct("2015-11-24 16:17:38"),
as.POSIXct("2015-11-24 14:19:44"))
as.period(span)
Format of answer
> -1H -57M -54S
Also display the difference in year, month & date
Currently my dataframe has dates displayed in the 'Date' column as 01/01/2007 etc I would like to convert these into a week/year value i.e. 01/2007. Any ideas?
I have been trying things like this and getting no where...
enviro$Week <- strptime(enviro$Date, format= "%W/%Y")
You have to first convert to date, then you can convert back to the week of the year using format, for example:
### Converts character to date
test.date <- as.Date("10/10/2014", format="%m/%d/%Y")
### Extracts only Week of the year and year
format(test.date, format="Week number %W of %Y")
[1] "Week number 40 of 2014"
### Or if you prefer
format(date, format="%W/%Y")
[1] "40/2014"
So, in your case, you would do something like this:
enviro$Week <- format(as.Date(enviro$Date, format="%m/%d/%Y"), format= "%W/%Y")
But remember that the part as.Date(enviro$Date, format="%m/%d/%Y") is only necessary if your data is not in Date format, and you also should put the right format parameter to convert your character to Date, if that is the case.
What is the class of enviro$Date? If it is of class Date there is probably a better way of doing this, otherwise you can try
v <- strsplit(as.character(enviro$Date), split = "/")
weeks <- sapply(v, "[", 2)
years <- sapply(v, "[", 3)
enviro$Week <- paste(weeks, years, sep = "/")