R Extract Hours from Time in factor Format

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I have a vector of time that is in the format of factor. For instance, Time[1] is 8:01:01, class(Time[1]) is factor.

Now I want to extract hours and minute from this vector. What is the most computational efficient way of doing it? My vector is very large. Thank you so much,

Try this:

format(strptime(Time,"%H:%M:%S"),'%H:%M')
[1] "08:01"

Extract time from factor column in R, The Time variable in your dataset is formatted "%d/%m/%Y %H:%M" but your code's format is "%m/%d/%Y" . Perhaps, that's why you're getting NAs . Humidity � R Extract Hours from Time in factor Format. Ask Question Asked 4 years ago. Active 16 days ago. Viewed 31k times 12. I have a vector of time that is in the format of

One way with the lubridate package would be:

Time <- factor("08:01:01")
lubridate::hms(as.character(Time))
Edit

To extract hours and minutes:

library(lubridate)
Time <- factor("08:01:01")

# parese date
a <- hms(as.character(Time))

# get hours
hour(a)

# get minutes
minute(a)

R: how to separate date-time data types, However, we can also use hour() , minute() , and second() to extract specific time information. today() and now() provide neatly formatted date-time information. Warning in Ops.factor(left, right): '-' not meaningful for factors If two times (using any of the date or date/time classes) are subtracted, R will return the results in the form of a time difference, which represents a difftime object. For example, New York City experienced a major blackout on July 13, 1997, and another on August 14, 2003.

You could also try a combination of converting your factor to a POSIXt object and then using the format function. I'm not sure how efficient this method is though.

Time <- factor("08:01:01")
hour = format(as.POSIXct(Time,format="%H:%M:%S"),"%H")
minute = format(as.POSIXct(Time,format="%H:%M:%S"),"%M")

paste(hour,minute,sep=":")
[1] "08:01"

Woking with Dates and Times with lubridate in R, Date-time must be a POSIXct, POSIXlt, Date, Period, chron, yearmon, yearqtr, zoo , zooreg, timeDate, xts, its, ti, jul, timeSeries, and fts objects. ## bed.time rise.time sleep.time efficiency ## 1 2013-09-01 23:05:24 2013-09-02 08:03:29 24264s (~6.74 hours) 75.2 ## 2 2013-09-02 22:51:09 2013-09-03 07:34:21 28512s (~7.92 hours) 90.8 ## 3 2013-09-04 00:09:16 2013-09-04 07:45:06 25236s (~7.01 hours) 92.3 ## 4 2013-09-04 23:43:31 2013-09-05 07:07:17 22428s (~6.23 hours) 84.2 ## 5 2013-09-06 00

What about this

 Time  <- factor("08:01:01")
 Time2 <- strptime(Time, "%H:%M:%S")
 Time3 <- format(Time2, "%H:%M")

Time3

[1] "08:01"

hour function, Thus dates in R will generally have a numeric mode, and the class To extract the components of the dates, the weekdays, months, days or The default format for times consists of the hour, minutes and seconds, separated by colons. fdate = factor(format(rdates$Date,'%Y')) > fdate [1] 2000 2000 2000� The most difficult part of dealing with time data can be converting it into the right format. Once a time or date is stored in R’s internal format then a number of basic operations are available. The thing to keep in mind, though, is that the units you get after an operation can vary depending on the magnitude of the time values.

Simpler solution: format(Sys.time(), "%H:%M")

Dates and Times in R, This tutorial explores working with date and time field in R. We will overview the from Harvard Forest # Factors=FALSE so strings, series of letters/words/ numerals, Also, we often want to quickly extract some portion of the data (e.g., months). in a format that we are used to seeing (e.g., second, min, hour, day of month,� The datetime vector is imported into r from a text file, thus I do not simply have the vector in format YYYY-dd-mm but as YYYY-dd-mm HH:MM therefore by replacing the time above with my vector it will not work. How would I convert the vector I have i.e. in the format YYYY-mm-dd HH:MM into a POSIXct vector? – user1407388 May 21 '12 at 11:31

Time Series 02: Dealing With Dates & Times in R, not hard to set the format when creating or printing date-times. In fact it's not a bad value: Factor w/ 3 levels "5","6","7": 1 1 2 2 1 1 2 3 1 1. ## $ ct. : POSIXct notice how hour is the only attribute to change print(unlist(time1)). If you want it in POSIX format, the only way would be to leave it as it is, and extract just the "time" part everytime you display it. But internally it will always be date + time anyway. If you want it in numeric, however, you can simply convert it into a number.

[PDF] Handling date-times in R, character string giving a date-time format as used by strptime . Alternatively, format can be specified for character vectors or factors: if it is not specified of a time (such as the day of the week) just convert it to class "POSIXlt" and extract the tz = "GMT") # in UTC ## SPSS dates (R-help 2006-02-16) z <- c(10485849600, � The hour in GMT is (time %% 86400) / 3600 and this can be used as the x-axis of the plot: plot((as.numeric(times) %% 86400)/3600, val) To adjust for a time zone, adjust the time before taking the modulus, by adding the number of seconds that your time zone is ahead of GMT. For example, US central daylight saving time (CDT) is 5 hours behind GMT.

as.POSIXlt: Date-time Conversion Functions, how do we add/subtract dates? how do we extract components such as year, month, day etc. character/string; number; factor (categorical/qualitative) This default format used in R is the ISO 8601 standard for date/time. ISOdate(year = 2019, month = 12, day = 12, hour = 8, min = 5, sec = 3, tz = "UTC") To format a date for pretty printing in R, you use format(), which takes a POSIXct or POSIXlt datetime as input, together with a formatting string. Using the example where the object xct is the day and time of the Apollo landing, you can format this date and time in many different ways.

Comments
  • Dirk's approach worked for me - Check it out stackoverflow.com/questions/23348992/… or stackoverflow.com/questions/16239646/…
  • Thank you so much for your suggestion. How do you get the actual hours and minutes after applying hms though? Thanks,
  • Thank you so much. I will test them to see which is fastest.
  • Thank you so much. That's the current method I'm using. I was hoping to make it faster though.
  • maybe try the chron package. require(chron) times(Time,out.format="h:m"). The only problem with this method is that the hour and minutes aren't separated by a colon. output: [1] 0801. But maybe this will speed things up for you.