Is there a way to convert a string of characters to a time that retains the fractional seconds, but does not add a date to the data?
Background: Data that I saved overnight (starting at 22:00 and ending at 06:00) is recorded with a time, but not a date.
I would like to be able to perform something like an "if" statement on the times (i.e. if (time < midnight) date = yesterday.... else date = today)
I don't want to use something like strptime, because strptime adds today's date to the data:
> options(digits.secs=3)
> strptime("22:59:54.807", format="%H:%M:%OS")
[1] "2019-01-20 22:59:54.807 AEDT"
When I use times from the chron package, the fractional seconds are dropped:
> options(digits.secs=3)
> times("22:59:54.807")
[1] 22:59:55
Using base::strftime(), you can obtain a character of the desired time to any divide of seconds you want. Below I have saved the current time to unit 1/1000 of a second, using the number "4" after "OS":
Using current system time:
time = Sys.time()
str(time)
POSIXct[1:1], format: "2019-01-19 15:41:28.185"
newTime = strftime(Sys.time(), format="%H:%M:%OS4")
str(newTime)
chr "15:41:28.1851"
newTime
[1] "15:41:28.1851"
Related
I have datetimes written in format "%d-%b-%Y %H:%M:%S.%OS",
so for example "25-Apr-2021 18:31:56.234",
that is to the precision of milliseconds.
And when parse that to time object I see values are not the same, sometimes it adds 1 microsecond or decreases it, or similiar things.
Why is this and what to do about this?
I want to have a time object which is exactly 56 seconds and 234 milliseconds! (and zeroes after that if it needs to add higher precision
For example some of the values it prints when I call print(as.numeric(), digits=20) command: "1615310444.7509999 1615310442.5550001",
or when I ask for difference between some 2 values, it gives: "Time difference of 0.003999949 secs" for example.
You can use the options(digits.secs) to show the milliseconds. Here is an example below. The digits.secs must be set at zero. Also note that you should change the format of the date
> dp <- options(digits.secs=0)
> dp
$digits.secs
[1] 0
> strptime("25-Apr-2021 18:31:56.234", format = "%d-%b-%Y %H:%M:%OS")
[1] "2021-04-25 18:31:56 +08"
> dp <- options(digits.secs=3)
> dp
$digits.secs
[1] 0
> strptime("25-Apr-2021 18:31:56.234", format = "%d-%b-%Y %H:%M:%OS")
[1] "2021-04-25 18:31:56.234 +08"
I have the following data. order.time is integer format. How to I convert it into time format, without adding date?
order.time
9:58:58
9:58:26
9:56:28
9:56:18
9:56:12
I have tried using parse_date_time, strptime, but all the functions add date and time zone in the final data. I only want to convert time from integer format to a format where I can draw plots with time in the x-axis
In the tidyverse, the hms package provides an S3 class for time of day:
> time <- hms::as_hms(c("9:58:58", "9:58:26", "9:56:28", "9:56:18", "9:56:12"))
> time
09:58:58
09:58:26
09:56:28
09:56:18
09:56:12
> class(time)
[1] "hms" "difftime"
> str(time)
'hms' num [1:5] 09:58:58 09:58:26 09:56:28 09:56:18 ...
- attr(*, "units")= chr "secs"
The function as_hms() also takes integer arguments, but only in seconds:
> (time_in_hours <- c(6, 9, 12))
[1] 6 9 12
> hms::as_hms(time_in_hours * 3600)
06:00:00
09:00:00
12:00:00
I would check out the anytime package. It works almost all of the time.
library(anytime)
order.time <- anytime(order.time)
Use as.POSIXct():
times <- as.POSIXct(c('9:58:58', 9:58:26), format = '%H:%M:%S')
Then you will have a time object, i.e: times[2] - times[1] will give you: Time diff of -32 secs.
You can plot these as is, also they can be formatted by using format(x, format).
Ex: format(times[1], format = '%H:%M:%S') will give you '09:58:58'.
I have a column named timings of class factor with time stamps in the following format:
1/11/07 15:15
I applied strptime on timings to generate tStamp as follows:
tStamp=strptime(timings,format="%m/%d/%Y %H:%M")
i)
The corresponding entry in tStamp looks like 0007-01-11 15:15:00 now. Why has it made 2007 or 07 into 0007? What is a correct way to generate tStamp?
ii)
After generating tStamp correctly, how do we convert it to the Unix time Seconds. (Seconds since...1970) format?
You need the lowercase %y for 2-digit years:
R> pt <- strptime("1/11/07 15:15",format="%m/%d/%y %H:%M")
R> pt
[1] "2007-01-11 15:15:00 CST"
R>
where CST is my local timezone.
And as.numeric() or as.double() converts to a double ...
R> as.numeric(pt)
[1] 1168550100
... which has fractional seconds if those are in the input:
R> options("digits.secs"=3) # show milliseconds
R> as.numeric(Sys.time()) # convert current time
[1] 1372201674.52 # now with sub0seconds.
What's the best way to get the length of time represented by an interval in lubridate, in specified units? All I can figure out is something like the following messy thing:
> ival
[1] 2011-01-01 03:00:46 -- 2011-10-21 18:33:44
> difftime(attr(ival, "start") + as.numeric(ival), attr(ival, "start"), 'days')
Time difference of 293.6479 days
(I also added this as a feature request at https://github.com/hadley/lubridate/issues/105, under the assumption that there's no better way available - but maybe someone here knows of one.)
Update - apparently the difftime function doesn't handle this either. Here's an example.
> (d1 <- as.POSIXct("2011-03-12 12:00:00", 'America/Chicago'))
[1] "2011-03-12 12:00:00 CST"
> (d2 <- d1 + days(1)) # Gives desired result
[1] "2011-03-13 12:00:00 CDT"
> (i2 <- d2 - d1)
[1] 2011-03-12 12:00:00 -- 2011-03-13 12:00:00
> difftime(attr(i2, "start") + as.numeric(i2), attr(i2, "start"), 'days')
Time difference of 23 hours
As I mention below, I think one nice way to handle this would be to implement a /.interval function that doesn't first cast its input to a period.
The as.duration function is what lubridate provides. The interval class is represented internally as the number of seconds from the start, so if you wanted the number of hours you could simply divide as.numeric(ival) by 3600, or by (3600*24) for days.
If you want worked examples of functions applied to your object, you should provide the output of dput(ival). I did my testing on the objects created on the help(duration) page which is where ?interval sent me.
date <- as.POSIXct("2009-03-08 01:59:59") # DST boundary
date2 <- as.POSIXct("2000-02-29 12:00:00")
span <- date2 - date #creates interval
span
#[1] 2000-02-29 12:00:00 -- 2009-03-08 01:59:59
str(span)
#Classes 'interval', 'numeric' atomic [1:1] 2.85e+08
# ..- attr(*, "start")= POSIXct[1:1], format: "2000-02-29 12:00:00"
as.duration(span)
#[1] 284651999s (9.02y)
as.numeric(span)/(3600*24)
#[1] 3294.583
# A check against the messy method:
difftime(attr(span, "start") + as.numeric(span), attr(span, "start"), 'days')
# Time difference of 3294.583 days
This question is really old, but I'm adding an update because this question has been viewed many times and when I needed to do something like this today, I found this page. In lubridate you can now do the following:
d1 <- ymd_hms("2011-03-12 12:00:00", tz = 'America/Chicago')
d2 <- ymd_hms("2011-03-13 12:00:00", tz = 'America/Chicago')
(d1 %--% d2)/dminutes(1)
(d1 %--% d2)/dhours(1)
(d1 %--% d2)/ddays(1)
(d1 %--% d2)/dweeks(1)
Ken, Dividing by days(1) will give you what you want. Lubridate doesn't coerce periods to durations when you divide intervals by periods. (Although the algorithm for finding the exact number of whole periods in the interval does begin with an estimate that uses the interval divided by the analagous number of durations, which might be what you are noticing).
The end result is the number of whole periods that fit in the interval. The warning message alerts the user that it is an estimate because there will be some fraction of a period that is dropped from the answer. Its not sensible to do math with a fraction of a period since we can't modify a clock time with it unless we convert it to multiples of a shorter period - but there won't be a consistent way to make the conversion. For example, the day you mention would be equal to 23 hours, but other days would be equal to 24 hours. You are thinking the right way - periods are an attempt to respect the variations caused by DST, leap years, etc. but they only do this as whole units.
I can't reproduce the error in subtraction that you mention above. It seems to work for me.
three <- force_tz(ymd_hms("2011-03-12 12:00:00"), "")
# note: here in TX, "" *is* CST
(four <- three + days(1))
> [1] "2011-03-13 12:00:00 CDT"
four - days(1)
> [1] "2011-03-12 12:00:00 CST"
Be careful when divinding time in seconds to obtain days as then you are no longer working with abstract representations of time but in bare numbers, which can lead to the following:
> date_f <- now()
> date_i <- now() - days(23)
> as.duration(date_f - date_i)/ddays(1)
[1] 22.95833
> interval(date_i,date_f)/ddays(1)
[1] 22.95833
> int_length(interval(date_i,date_f))/as.numeric(ddays(1))
[1] 22.95833
Which leads to consider that days or months are events in a calendar, not time amounts that can be measured in seconds, miliseconds, etc.
The best way to calculate differences in days is avoiding the transformation into seconds and work with days as a unit:
> e <- now()
> s <- now() - days(23)
> as.numeric(as.Date(s))
[1] 18709
> as.numeric(as.Date(e) - as.Date(s))
[1] 23
However, if you are considering a day as a pure 86400 seconds time span, as ddays() does, the previous approach can lead to the following:
> e <- ymd_hms("2021-03-13 00:00:10", tz = 'UTC')
> s <- ymd_hms("2021-03-12 23:59:50", tz = 'UTC')
> as.duration(e - s)
[1] "20s"
> as.duration(e - s)/ddays(1)
[1] 0.0002314815
> as.numeric(as.Date(e) - as.Date(s))
[1] 1
Hence, it depends on what you are looking for: time difference or calendar difference.
How can I add one hour to all the elements of the index of a zoo series?
I've tried
newseries <- myzooseries
index(newseries) <- index(myzooseries)+times("1:00:00")
but I get the message
Incompatible methods ("Ops.dates", "Ops.times") for "+"
thanks
My index is a chron object with date and time but I've tried with simpler examples and I can't get it
This is easily solved by adding the time you want in a numerical fashion :
newseries <- myzooseries
index(newseries) <- index(myzooseries) + 1/24
chron objects are represented as decimal numbers, so you can use that to calculate. A day is 1, so an hour is 1/24, a minute 1/1440 and so on. You can see this easily if you use the function times. This gives you the times of the object tested, eg :
> A <- chron(c("01/01/97","01/02/97","01/03/97"))
> B <- A + 1/24
> B
[1] (01/01/97 01:00:00) (01/02/97 01:00:00) (01/03/97 01:00:00)
> times(A)
Time in days:
[1] 9862 9863 9864
> times(B)
Time in days:
[1] 9862.042 9863.042 9864.042
> times(B-A)
[1] 01:00:00 01:00:00 01:00:00
> times(A[3]-B[1])
Time in days:
[1] 1.958333
Convert to POSIXct, add 60*60 (1h in s) and then convert back.