I would like to generate a random date in an interval, say a random date in 2000-2010. Naturally it would need to take care of leap years.
How would I do this in Julia?
Assuming that by date you mean a day you can do:
julia> using Dates
julia> rand(Date(2000, 1, 1):Day(1):Date(2010, 12, 31))
2004-03-13
This constructs a range of Dates which is efficient as it doesn't have to allocate an array with all Dates, and then picks a random one in that range.
Related
I want to create a time series with date and quantity as variables. However I always have zero observation on Sundays. Therefore I want to define the week as 6-days in length in R. Any suggestions?
I have this UTC date in a Google spreadsheet: 2018-10-18T08:55:13Z and would like to convert it to Unix timestamp (1539852913). I tried this formula, but it's unable to recognize the timevalue:
=DATEVALUE(MID(A1;1;10)) + TIMEVALUE(MID(A1;12;8))
If I can get a valid date and time, I can use this formula to convert to Unix timestamp:
=(A1-$C$1)*86400
Does anyone have a solution for this?
Simpler:
=86400*(left(substitute(A1,"T"," "),19))-2209161600
Replaces T with space and cuts off Z, leaving what's left recognisable as date and time in arithmetical calculations. Convert day and time index into seconds and adjust for the offset.
Assuming your date has proceeding zeros for single digit days and month, pull each date string part and drop it into the DATE formula as follows:
Year
=LEFT(A1,4)
Month
=MID(A1,6,2)
Day
=MID(A1,9,2)
Use the date formula
=DATE(year,month,day)
=DATE(LEFT(A1,4),MID(A1,6,2),MID(A1,9,2))
A similar process can be used for TIME
Hour
=MID(A1,12,2)
Minutes
=MID(A1,15,2)
Seconds
=MID(A1,18,2)
Time
=TIME(Hour,Minutes,Seconds)
=TIME(MID(A1,12,2),MID(A1,15,2),MID(A1,18,2))
1) There are other methods
2) The formulas will need to be adapted if you do not have leading 0 for each unit. In that case you would need to use FIND to identify the position of key characters and measure the distance between them to determine if there was a single digit unit or double digit unit.
Since the date is the integer part (left of the decimal) represents the number of days since 1900/01/01 (with that date being 1) and decimal portion represents time in terms of fraction of a day, to get a full date and time, you would add the date formula to the time formula as follows:
=DATE(LEFT(A1,4),MID(A1,6,2),MID(A1,9,2))+TIME(MID(A1,12,2),MID(A1,15,2),MID(A1,18,2))
I have a dataset in .csv, and I have added in a column on my own in the csv that takes the total time taken for a task to be completed. There are two other columns that consists of the start time and the end time, and that is where I calculated the total time taken column from. The format of the start time and end time columns are in the datetime format 5/7/2018 16:13 while the format of the total time taken column is 0:08:20(H:MM:SS).
I understand that for datetime, it is possible to use the functions as.Date or as.POSIXlt to change the variable type from a factor to that of date. Is there a function that I can convert my total time taken column to (from that of factor) so that I can use it to plot scatterplots/plots in general? I tried as.numeric but the numbers that come out are gibberish and do not correspond to the original time.
If you want to plot the total time taken for each row, then I would suggest just plotting that difference as seconds. Here is a code snippet which shows how you can convert your start or end date into a numerical value:
start <- "5/7/2018 16:13"
start_date <- as.POSIXct(start, format="%d/%m/%Y %H:%M")
as.numeric(start_date)
[1] 1530799980
The above is a UNIX timestamp, which is number of seconds since the epoch (January 1, 1970). But, since you want a difference between start and end times, this detail does not really matter for you, and the difference you get should be valid.
If you want to use minutes, hours, or some other time unit, then you can easily convert.
I'm using R to analyze 365 days of data collected on over 40,000 events. The events occur at various times of the day. I wish to aggregate the events and calculate means at various intervals such as 2, 8, 12 hour or daily. I've seen CUT and AGGREGATE used but it does not appear to provide the intervals as required.
Any suggestions would be greatly appreciated.
To use the CUT function one must first define the break points. In order to do that, use the seq function.
mydateseq<-seq(as.POSIXct("2016-01-01"), by="2 hour", length.out = 20)
There are options to set the start/stop points or the number of elements. In this example the breaks are set every 2 hours but this is adjustable. See ?seq.POSIXt for more help. Be sure to set the start/stop to completely capture the date range of interest.
Once the date sequence is defined this can be passed to cut function to aggregate or use the group_by function in the dplyr package.
I have some numbers that represent dates in milliseconds since epoch, 00:00:00 Coordinated Universal Time (UTC), Thursday, 1 January 1970
1365368400000,
1365973200000,
1366578000000
I'm converting them to date format:
as.Date(as.POSIXct(my_dates/1000, origin="1970-01-01", tz="GMT"))
answer:
[1] "2013-04-07" "2013-04-14" "2013-04-21"
How to convert these strings back to milliseconds since epoch?
Here are your javascript dates
x <- c(1365368400000, 1365973200000, 1366578000000)
You can convert them to R dates more easily by dividing by the number of milliseconds in one day.
y <- as.Date(x / 86400000, origin = "1970-01-01")
To convert back, just convert to numeric and multiply by this number.
z <- as.numeric(y) * 86400000
Finally, check that the answer is what you started with.
stopifnot(identical(x, z))
As per the comment, you may sometimes get numerical rounding errors leading to x and z not being identical. For numerical comparisons like this, use:
library(testthat)
expect_equal(x, z)
I will provide a simple framework to handle various kinds of dates encoding and how to go back an forth. Using the R package ‘lubridate’ this is made very easy using the period and interval classes.
When dealing with days, it can be easy as one can use the as.numeric(Date) to get the number of dates since the epoch. To get any unit of time smaller than a day one can convert using the various factors (24 for hours, 24 * 60 for minutes, etc.) However, for months, the math can get a bit more tricky and thus I prefer in many instances to use this method.
library(lubridate)
as.period(interval(start = epoch, end = Date), unit = 'month')#month
This can be used for year, month, day, hour, minute, and smaller units through apply the factors.
Going the other way such as being given months since epoch:
library(lubridate)
epoch %m+% as.period(Date, unit = 'months')
I presented this approach with months as it might be the more complicated one. An advantage to using period and intervals is that it can be adjusted to any epoch and unit very easily.