I am trying to subtract 3 months from February to a date using lubridate library, however, it always gives me date as 28. I know a lot of people have asked this question, but I have tried a lot of things, but still the same error. Looping through the day is an option. However, I was looking for something efficient
what my code does is this:
where test_end = "2019-02-28"
and lastnmonth = 3
as.Date(test_end) %m-% months(lastnmonth)
I always get the output as "2018-11-28" instead of "2018-11-30"
lubridate::ymd("2019-02-28") - lubridate::period(3, "months")
additional code to work-out mont-end
lubridate::ceiling_date(lubridate::ymd("2019-02-28") - lubridate::period(3, "months"), "months") - lubridate::ddays(1)
You want the last of the month -- so you can use the trick that the next is a first, move that one by the number of months -- and then substract one day.
All in one line of base R:
R> seq(as.Date("2019-02-28") + 1, length=4, by="-1 month") - 1
[1] "2019-02-28" "2019-01-31" "2018-12-31" "2018-11-30"
R>
We add one day, then use a sequence of four months (including current) backwards, and take out one day to be on the last day of the previous month.
If you just want one, simply add tail(..., 1):
R> tail(seq(as.Date("2019-02-28") + 1, length=4, by="-1 month") - 1, 1)
[1] "2018-11-30"
R>
Using Dirk's strategy we can draw-up a function to handle this type of operation:
start_date <- lubridate::ymd("2019-02-28")
mths_to_subtract <- 3
deduct_mth <- function(date_object = start_date,
mths_to_deduct = mths_to_subtract){
days_to_ceiling_date <- lubridate::ceiling_date(date_object, "months") - date_object
date_object + days_to_ceiling_date - lubridate::period(mths_to_deduct, "months") - days_to_ceiling_date
}
Gives the following answers:
> deduct_mth(ymd("2019-02-28"), 3)
[1] "2018-11-30"
> deduct_mth(ymd("2000-02-29"), 3)
[1] "1999-11-30"
Related
I want to generate a sequence of dates with one quarter interval, with a starting date and ending date. I have below code :
> seq(as.Date('1980-12-31'), as.Date('1985-06-30'), by = 'quarter')
[1] "1980-12-31" "1981-03-31" "1981-07-01" "1981-10-01" "1981-12-31"
[6] "1982-03-31" "1982-07-01" "1982-10-01" "1982-12-31" "1983-03-31"
[11] "1983-07-01" "1983-10-01" "1983-12-31" "1984-03-31" "1984-07-01"
[16] "1984-10-01" "1984-12-31" "1985-03-31"
As you can see, this is not generating right sequence, as I dont understand how the date "1981-07-01" is coming here, I would expect "1981-06-30".
Is there any way to generate such sequence correctly with quarter interval?
Thanks for your time.
The from and to dates in the question are both end-of-quarter dates so we assume that that is the general case you are interested in.
1) Create a sequence of yearqtr objects yq and then convert them to Date class. frac=1 tells it s to use the end of the month. Alternately just use yq since that directly models years with quarters.
library(zoo)
from <- as.Date('1980-12-31')
to <- as.Date('1985-06-30')
yq <- seq(as.yearqtr(from), as.yearqtr(to), by = 1/4)
as.Date(yq, frac = 1)
giving;
[1] "1980-12-31" "1981-03-31" "1981-06-30" "1981-09-30" "1981-12-31"
[6] "1982-03-31" "1982-06-30" "1982-09-30" "1982-12-31" "1983-03-31"
[11] "1983-06-30" "1983-09-30" "1983-12-31" "1984-03-31" "1984-06-30"
[16] "1984-09-30" "1984-12-31" "1985-03-31" "1985-06-30"
2) or without any packages add 1 to from and to so that they are at the beginning of the next month, create the sequence (it has no trouble with first of month sequences) and then subtract 1 from the generated sequence giving the same result as above.
seq(from + 1, to + 1, by = "quarter") - 1
Using the clock package and R >= 4.1:
library(clock)
seq(year_quarter_day(1980, 4), year_quarter_day(1985, 2), by = 1) |>
set_day("last") |>
as_date()
# [1] "1980-12-31" "1981-03-31" "1981-06-30" "1981-09-30" "1981-12-31" "1982-03-31" "1982-06-30" "1982-09-30" "1982-12-31"
# [10] "1983-03-31" "1983-06-30" "1983-09-30" "1983-12-31" "1984-03-31" "1984-06-30" "1984-09-30" "1984-12-31" "1985-03-31"
# [19] "1985-06-30"
Note that this includes the final quarter. I don't know if that was your intent.
Different definition of "quarter". A quarter might well be (although it is not in R) 365/4 days. Look at output of :
as.Date('1980-12-31')+(365/4)*(0:12)
#[1] "1980-12-31" "1981-04-01" "1981-07-01" "1981-09-30" "1981-12-31" "1982-04-01" "1982-07-01" "1982-09-30"
#[9] "1982-12-31" "1983-04-01" "1983-07-01" "1983-09-30" "1983-12-31"
In order to avoid the days of the month from surprising you, you need to use a starting day of the month between 1 and 28, at least in non-leap years.
seq(as.Date('1981-01-01'), as.Date('1985-06-30'), by = 'quarter')
[1] "1981-01-01" "1981-04-01" "1981-07-01" "1981-10-01" "1982-01-01" "1982-04-01" "1982-07-01" "1982-10-01"
[9] "1983-01-01" "1983-04-01" "1983-07-01" "1983-10-01" "1984-01-01" "1984-04-01" "1984-07-01" "1984-10-01"
[17] "1985-01-01" "1985-04-01"
I would like to print all the hours: minutes in a day from 00:00 to 23:59.
This part goes beyond the question, but if you want to help me, this is the whole idea:
Once that is done, I would like to calculate all the "curious" times that can be interpreted as serendipities. Patterns like: 00:00, 22:22, 01:10, 12:34, 11:44, and the like.
Later on, I would like to count all the "serendipities", and divide them to the total number of hours to know the probabilities of find a "serendipity" each time a person look at the time on his smartphone.
To be honest, I am pretty lost. There is already some months without coding. For the first part of the problem, I guess that a loop can make the task.
For the second part, an if conditional can probably make it.
For the first part of the problem I have tried loops like this
for(i in x){
for(k in y){
cat(i,":",k, ",")
}
}
For the second, something like
Assuming the digits of the time are ab:cd
if(a==b & a==c & a==d){
print(ab:cd)
TRUE
}
if(a==b & c==d){
print(ab:cd)
TRUE
}
I would like to get the whole list of numbers first. Then, the list of "serendipities", and finally the count of both to make the percentage.
I find interesting how people find patterns in numbers when they look at the time, and I would like to know how probable is to get one of these patterns out of the 24*60 = 1440
I hope I have explained myself. (I used to be better with coding and maths, but after some months, I have forgotten almost everything).
Here's a way to generate the list of all possible times.
h <- seq(from=0, to=23)
m <- seq(from=0, to=59)
h <- sprintf('%02d', h)
m <- sprintf('%02d', m)
df <- data.frame(expand.grid(h, m))
df$times <- paste0(df$Var1, ':', df$Var2)
df <- df[order(df$times), ]
df$times
Partial output
df$times[1:25]
[1] "00:00" "00:01" "00:02" "00:03" "00:04" "00:05" "00:06" "00:07" "00:08"
[10] "00:09" "00:10" "00:11" "00:12" "00:13" "00:14" "00:15" "00:16" "00:17"
[19] "00:18" "00:19" "00:20" "00:21" "00:22" "00:23" "00:24"
Length of variable
dim(df)
[1] 1440 3
We can create a sequence of 1 minute interval starting from 00:00:00 to 23:59:00 and then use format to get output in desired format.
format(seq(as.POSIXct("00:00:00", format = "%T"),
as.POSIXct("23:59:00", format = "%T"), by = "1 min"), "%H:%M")
#[1] "00:00" "00:01" "00:02" "00:03" "00:04" "00:05" "00:06" "00:07" "00:08" "00:09"
# "00:10" "00:11" "00:12" "00:13" "00:14" "00:15" "00:16" "00:17" "00:18" "00:19" ...
Yet another way of doing it:
> result <- character(1440)
> for (i in 0:1439) result[i+1L] <- sprintf("%02d:%02d",
+ i %/% 60,
+ i %% 60
+ )
> head(result)
[1] "00:00" "00:01" "00:02" "00:03" "00:04" "00:05"
> tail(result)
[1] "23:54" "23:55" "23:56" "23:57" "23:58" "23:59"
This question already has answers here:
Get Dates of a Certain Weekday from a Year in R
(3 answers)
Closed 9 years ago.
I would like to generate a dataframe that contains all the Friday dates for the whole year.
Is there a simple way to do this?
eg for December 2013: (6/12/13,13/12/13,20/12/13,27/12/13)
Thank you for your help.
I'm sure there is a simpler way, but you could brute force it easy enough:
dates <- seq.Date(as.Date("2013-01-01"),as.Date("2013-12-31"),by="1 day")
dates[weekdays(dates)=="Friday"]
dates[format(dates,"%w")==5]
Building on #Frank's good work, you can find all of any specific weekday between two dates like so:
pick.wkday <- function(selday,start,end) {
fwd.7 <- start + 0:6
first.day <- fwd.7[as.numeric(format(fwd.7,"%w"))==selday]
seq.Date(first.day,end,by="week")
}
start and end need to be Date objects, and selday is the day of the week you want (0-6 representing Sunday-Saturday).
i.e. - for the current query:
pick.wkday(5,as.Date("2013-01-01"),as.Date("2013-12-31"))
Here is a way.
d <- as.Date(1:365, origin = "2013-1-1")
d[strftime(d,"%A") == "Friday"]
Alternately, this would be a more efficient approach for generating the data for an arbitrary number of Fridays:
wk1 <- as.Date(seq(1:7), origin = "2013-1-1") # choose start date & make 7 consecutive days
wk1[weekdays(wk1) == "Friday"] # find Friday in the sequence of 7 days
seq.Date(wk1[weekdays(wk1) == "Friday"], length.out=50, by=7) # use it to generate fridays
by=7 says go to the next Friday.
length.out controls the number of Fridays to generate. One could also use to to control how many Fridays are generated (e.g. use to=as.Date("2013-12-31") instead of length.out).
Takes a year as input and returns only the fridays...
getFridays <- function(year) {
dates <- seq(as.Date(paste0(year,"-01-01")),as.Date(paste0(year,"-12-31")), by = "day")
dates[weekdays(dates) == "Friday"]
}
Example:
> getFridays(2000)
[1] "2000-01-07" "2000-01-14" "2000-01-21" "2000-01-28" "2000-02-04" "2000-02-11" "2000-02-18" "2000-02-25" "2000-03-03" "2000-03-10" "2000-03-17" "2000-03-24" "2000-03-31"
[14] "2000-04-07" "2000-04-14" "2000-04-21" "2000-04-28" "2000-05-05" "2000-05-12" "2000-05-19" "2000-05-26" "2000-06-02" "2000-06-09" "2000-06-16" "2000-06-23" "2000-06-30"
[27] "2000-07-07" "2000-07-14" "2000-07-21" "2000-07-28" "2000-08-04" "2000-08-11" "2000-08-18" "2000-08-25" "2000-09-01" "2000-09-08" "2000-09-15" "2000-09-22" "2000-09-29"
[40] "2000-10-06" "2000-10-13" "2000-10-20" "2000-10-27" "2000-11-03" "2000-11-10" "2000-11-17" "2000-11-24" "2000-12-01" "2000-12-08" "2000-12-15" "2000-12-22" "2000-12-29"
There are probably more elegant ways to do this, but here's one way to generate a vector of Fridays, given any year.
year = 2007
st <- as.POSIXlt(paste0(year, "/1/01"))
en <- as.Date(paste0(year, "/12/31"))
#get to the next Friday
skip_ahead <- 5 - st$wday
if(st$wday == 6) skip_ahead <- 6 #for Saturdays, skip 6 days ahead.
first.friday <- as.Date(st) + skip_ahead
dates <- seq(first.friday, to=en, by ="7 days")
dates
#[1] "2007-01-05" "2007-01-12" "2007-01-19" "2007-01-26"
# [5] "2007-02-02" "2007-02-09" "2007-02-16" "2007-02-23"
# [9] "2007-03-02" "2007-03-09" "2007-03-16" "2007-03-23"
I think this would be the most efficient way and would also returns all the Friday in the whole of 2013.
FirstWeek <- seq(as.Date("2013/1/1"), as.Date("2013/1/7"), "days")
seq(
FirstWeek[weekdays(FirstWeek) == "Friday"],
as.Date("2013/12/31"),
by = "week"
)
I have a CSV file of 1000 daily prices
They are of this format:
1 1.6
2 2.5
3 0.2
4 ..
5 ..
6
7 ..
.
.
1700 1.3
The index is from 1:1700
But I need to specify a begin date and end date this way:
Start period is lets say, 25th january 2009
and the last 1700th value corresponds to 14th may 2013
So far Ive gotten this close to this problem:
> dseries <- ts(dseries[,1], start = ??time??, freq = 30)
How do I go about this? thanks
UPDATE:
managed to create a seperate object with dates as suggested in the answers and plotted it, but the y axis is weird, as shown in the screenshot
Something like this?
as.Date("25-01-2009",format="%d-%m-%Y") + (seq(1:1700)-1)
A better way, thanks to #AnandaMahto:
seq(as.Date("2009-01-25"), by="1 day", length.out=1700)
Plotting:
df <- data.frame(
myDate=seq(as.Date("2009-01-25"), by="1 day", length.out=1700),
myPrice=runif(1700)
)
plot(df)
R stores Date-classed objects as the integer offset from "1970-01-01" but the as.Date.numeric function needs an offset ('origin') which can be any staring date:
rDate <- as.Date.numeric(dseries[,1], origin="2009-01-24")
Testing:
> rDate <- as.Date.numeric(1:10, origin="2009-01-24")
> rDate
[1] "2009-01-25" "2009-01-26" "2009-01-27" "2009-01-28" "2009-01-29"
[6] "2009-01-30" "2009-01-31" "2009-02-01" "2009-02-02" "2009-02-03"
You didn't need to add the extension .numeric since R would automticallly seek out that function if you used the generic stem, as.Date, with an integer argument. I just put it in because as.Date.numeric has different arguments than as.Date.character.
I would like to generate a sequence of dates from 10,000 B.C.E. to the present. This is easy for 0 C.E. (or A.D.):
ADtoNow <- seq.Date(from = as.Date("0/1/1"), to = Sys.Date(), by = "day")
But I am stumped as to how to generate dates before 0 AD. Obviously, I could do years before present but it would be nice to be able to graph something as BCE and AD.
To expand on Ricardo's suggestion, here is some testing of how things work. Or don't work for that matter.
I will repeat Joshua's warning taken from ?as.Date for future searchers in big bold letters:
"Note: Years before 1CE (aka 1AD) will probably not be handled correctly."
as.integer(as.Date("0/1/1"))
[1] -719528
as.integer(seq(as.Date("0/1/1"),length=2,by="-10000 years"))
[1] -719528 -4371953
seq(as.Date(-4371953,origin="1970-01-01"),Sys.Date(),by="1000 years")
# nonsense
[1] "0000-01-01" "'000-01-01" "(000-01-01" ")000-01-01" "*000-01-01"
[6] "+000-01-01" ",000-01-01" "-000-01-01" ".000-01-01" "/000-01-01"
[11] "0000-01-01" "1000-01-01" "2000-01-01"
> as.integer(seq(as.Date(-4371953,origin="1970-01-01"),Sys.Date(),by="1000 years"))
# also possibly nonsense
[1] -4371953 -4006710 -3641468 -3276225 -2910983 -2545740 -2180498 -1815255
[9] -1450013 -1084770 -719528 -354285 10957
Though this does seem to work for graphing somewhat:
yrs1000 <- seq(as.Date(-4371953,origin="1970-01-01"),Sys.Date(),by="1000 years")
plot(yrs1000,rep(1,length(yrs1000)),axes=FALSE,ann=FALSE)
box()
axis(2)
axis(1,at=yrs1000,labels=c(paste(seq(10000,1000,by=-1000),"BC",sep=""),"0AD","1000AD","2000AD"))
title(xlab="Year",ylab="Value")
Quite some time has gone by since this question was asked. With that time came a new R package, gregorian which can handle BCE time values in the as_gregorian method.
Here's an example of piecewise constructing a list of dates that range from -10000 BCE to the current year.
library(lubridate)
library(gregorian)
# Container for the dates
dates <- c()
starting_year <- year(now())
# Add the CE dates to the list
for (year in starting_year:0){
date <- sprintf("%s-%s-%s", year, "1", "1")
dates <- c(dates, gregorian::as_gregorian(date))
}
starting_year <- "-10000"
# Add the BCE dates to the list
for (year in starting_year:0){
start_date <- gregorian::as_gregorian("-10000-1-1")
date <- sprintf("%s-%s-%s", year, "1", "1")
dates <- c(dates, gregorian::as_gregorian(date))
}
How you use the list is up to you, just know that the relevant properties of the date objects are year and bce. For example, you can loop over list of dates, parse the year, and determine if it's BCE or not.
> gregorian_date <- gregorian::as_gregorian("-10000-1-1")
> gregorian_date$bce
[1] TRUE
> gregorian_date$year
[1] 10001
Notes on 0AD
The gregorian package assumes that when you mean Year 0, you're really talking about year 1 (shown below). I personally think an exception should be thrown, but that's the mapping users needs to keep in mind.
> gregorian::as_gregorian("0-1-1")
[1] "Monday January 1, 1 CE"
This is also the case with BCE
> gregorian::as_gregorian("-0-1-1")
[1] "Saturday January 1, 1 BCE"
As #JoshuaUlrich commented, the short answer is no.
However, you can splice out the year into a separate column and then convert to integer. Would this work for you?
The package lubridate seems to handle "negative" years ok, although it does create a year 0, which from the above comments seems to be inaccurate. Try:
library(lubridate)
start <- -10000
stop <- 2013
myrange <- NULL
for (x in start:stop) {
myrange <- c(myrange,ymd(paste0(x,'-01-01')))
}