I have a series of decimal numbers (marathon race split times): 64.90, etc., and I want to convert it into HH:MM:SS format using R so that I can us the result to do time math. The answer I am looking for is: 1:04:54.
chron doesn't seem to be doing what I'm expecting it to do.
chron::times(64.90)
Time in days:
[1] 64.9
First time on this site, so be kind. Thanks.
chron times are measured in days and since you apparently have minutes divide the input by the number of minutes in a day:
library(chron)
times(64.90 / (24 * 60))
## [1] 01:04:54
You could try lubridate::seconds_to_period
library(lubridate)
seconds_to_period(64.90)
[1] "1M 4.90000000000001S"
library(hms)
as.hms(64.90*60)
output
01:04:54
Related
I have decimal hours in format 245.85 equalling to 245:51:00 in [hh]:mm:ss format.
I want to transform the decimal hours to hh:mm format, but how do I do it?
the original calculation that renders 245.85 is:
library(lubridate)
time_length(hm("7 27")*33,unit = "hours")
what I want is 245:51 or 245:51:00
If I use as.period I get days too - like in:
as.period(dhours(time_length(hm("7 27")*33,"hours")))
[1] "10d 5H 51M 0S"
and for background - my aim is to multiply hours and minutes (e.g. 7:27) by an arbitrary integer (e.g. 33) and get result back in hh:mm format - avoiding days (as in as.period example above). Say if a piece of work takes 7 hours and 27 minutes and we give me 33 pieces of such work to do per year, it should take me about this many work hours (and minutes) to do.
If it's really only the H:M:S format that gives you trouble, try
library(hms)
hms(hours=245.85)
which yields 245:51:00
I have a vector of decimal numbers, which represent 'decimal day', the fraction of a day. I want to convert it into HH:MM format using R.
For example, the number 0.8541667 would correspond to 20:30. How can I convert the numbers to HH:MM format?
Using chron:
chron::times(0.8541667)
#[1] 20:30:00
Try this:
R> format(as.POSIXct(Sys.Date() + 0.8541667), "%H:%M", tz="UTC")
[1] "20:30"
R>
We start with a date--which can be any date, so we use today--and add your desired fractional day.
We then convert the Date type into a Datetime object.
Finally, we format the hour and minute part of the Datetime object, ensuring that UTC is used for the timezone.
One option with data.table:
> library(data.table)
> structure(as.integer(0.4305556*60*60*24), class="ITime")
[1] "10:20:00"
We convert from day fraction to seconds since midnight; coerce to integer; and apply ITime class. (ITime works with integer-stored seconds since midnight.)
Other resources:
#GaborGrothendieck re chron package and link to his R News article with Thomas Petzoldt about converting from Excel in particular Converting a time decimal/fraction representing days to its actual time in R?
#JorisChau re RStudio's hms package how to convert excel internal coding for hours to hours in R?
I read data from an xls file. Apparently, the time is not in the right format. It is as follows (for example)
0.3840277777777778
0.3847222222222222
0.3854166666666667
Indeed, they should be
09:12
09:13
09:13
I don't know how to convert it to the right format. I searched several threads and all of them are about converting the date (with/without time) to the right format.
Can somebody give me any clues?
You can use as.POSIXct after having multiplied your number by the number of seconds in a day (60 * 60 * 24)
nTime <- c(0.3840277777777778, 0.3847222222222222, 0.3854166666666667)
format(as.POSIXct((nTime) * 86400, origin = "1970-01-01", tz = "UTC"), "%H:%M")
## [1] "09:13" "09:14" "09:15"
Another option is times from chron
library(chron)
times(nTime)
#[1] 09:13:00 09:14:00 09:15:00
To strip off the seconds,
substr(times(nTime),1,5)
#[1] "09:13" "09:14" "09:15"
data
nTime <- c(0.3840277777777778, 0.3847222222222222, 0.3854166666666667)
For people who want the opposite way: given the 09:13:00, get 0.3840278
as.numeric(chron::times("09:13:00"))
Essentially, the idea is that one whole day is 1,so noon (12pm) is 0.5.
Lots of people ask how to strip the time and keep the date, but what about the other way around? Given:
myDateTime <- "11/02/2014 14:22:45"
I would like to see:
myTime
[1] "14:22:45"
Time zone not necessary.
I've already tried (from other answers)
as.POSIXct(substr(myDateTime, 12,19),format="%H:%M:%S")
[1] "2013-04-13 14:22:45 NZST"
The purpose is to analyse events recorded over several days by time of day only.
Thanks
Edit:
It turns out there's no pure "time" object, so every time must also have a date.
In the end I used
as.POSIXct(as.numeric(as.POSIXct(myDateTime)) %% 86400, origin = "2000-01-01")
rather than the character solution, because I need to do arithmetic on the results. This solution is similar to my original one, except that the date can be controlled consistently - "2000-01-01" in this case, whereas my attempt just used the current date at runtime.
I think you're looking for the format function.
(x <- strptime(myDateTime, format="%d/%m/%Y %H:%M:%S"))
#[1] "2014-02-11 14:22:45"
format(x, "%H:%M:%S")
#[1] "14:22:45"
That's character, not "time", but would work with something like aggregate if that's what you mean by "analyse events recorded over several days by time of day only."
If the time within a GMT day is useful for your problem, you can get this with %%, the remainder operator, taking the remainder modulo 86400 (the number of seconds in a day).
stamps <- c("2013-04-12 19:00:00", "2010-04-01 19:00:01", "2018-06-18 19:00:02")
as.numeric(as.POSIXct(stamps)) %% 86400
## [1] 0 1 2
simple question in lubridate--I want to convert an hms object into its appropriate number of seconds since the start of the day.
For instance
library(lubridate)
hms("12:34:45")
then I want to know exactly how long 12 hours, 34 minutes, and 45 seconds is, in seconds
something obvious like
seconds(hms("12:34:45"))
just returns
45s
which is not what I want. How do I convert these hms values into seconds? I'd like to use lubridate
R>lubridate::period_to_seconds(hms("01:00:00"))
gives expected 3600 seconds as numeric counting from 00:00:00
or in the case above:
R>period_to_seconds(hms("12:34:45"))
It doesn't matter which package you use -- it will have convert a date / datetime object into a POSIXct representation of seconds since the epoch. So you may as well do it in base R -- so here deploy ISOdatetime() with an arbitrary day, using today:
R> difftime(ISOdatetime(2012,7,2,12,34,45), ISOdatetime(2012,7,2,0,0,0))
Time difference of 12.5792 hours
So we want seconds:
R> difftime(ISOdatetime(2012,7,2,12,34,45), ISOdatetime(2012,7,2,0,0,0),
+ unit="secs")
Time difference of 45285 secs
And we can cast to numbers:
R> as.numeric(difftime(ISOdatetime(2012,7,2,12,34,45), +
ISOdatetime(2012,7,2,0,0,0), unit="secs"))
[1] 45285
Edit: And getting back to lubridate, this is arguably a bug:
> hms("12:34:45") - hms("00:00:00")
[1] 12 hours, 34 minutes and 45 seconds
R> as.numeric(hms("12:34:45") - hms("00:00:00"))
[1] 45
R>