Convert Date difference to Float64

New to Julia
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[Edit: Sorry, folks, I’ve realized that I simplified my actual problem too much. The solutions you kindly offered are solutions to my example, but they don’t work for my actual problem! To avoid confusion, I post a more realistic problem above my original posting.]

I first calculate the center of a month and then need to calculate how many days there are between a time origin and the center of the month.

using Dates
const t0 = DateTime(1950,1,1) # origin
const t1 = DateTime(1986,1,1)
const t2 = t1 + Dates.Month(1)
const tm = t1 + (t2 - t1)/2 # center of the month
display(tm)
# LoadError: InexactError: divexact(Int64, 13164.5)
tm_in_days = Day(tm - t0)

The problem is the conversion of the interval tm - t0 to Day fails.

Of course, we can utilize the ideas posted below to convert both t1 - t0 and t2 - t1 to integer days and then calculate t1 - t0 + (t2 - t1)/2 all in days.

But that’d be so inelegant a solution that I suspect there are better ways.

[Original posting]
I’m wondering how to convert the difference between two DateTimes in days to a Float64:

using Dates
const t1 = DateTime(1986,1,1)
const t2 = DateTime(1986,2,1)

#s = Second(t2 - t1) / 2
d = Day(t2 - t1) / 2 # I want 15.5::Float64

The error is ERROR: LoadError: InexactError: Int64(15.5). How does one get 15.5 as Float64 ?

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julia> d = Day(t2 - t1)
31 days

julia> d.value/2
15.5
1 Like
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julia> Day(t2 - t1) / Day(2)
15.5
6 Likes
4

Julia avoids auto-rounding for arithmetic operations, so you need to explicitly round the result:

julia> round(tm - t0, Day)
13165 days

As an aside, it’s recommended to post a reply if you edit your original post, otherwise no notifications are sent and your edit might go unseen.

2 Likes
5

Thank you for your input, but unfortunately, that’s not what I need. I need a Float64 representation of tm - t0 in days. That is, the value should be 13164.5.

it’s recommended to post a reply if you edit your original post

Thank you. That’s definitely important advice.

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By the way, how do you find details of a library module like “Dates”. I spent quite a lot of time on the official documentation Dates · The Julia Language , but I haven’t found a description of the value attribute of the Period type or of the division of Period by another Period, which is used in the excellent answers to my original posting.

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Use the source, Luke.

https://github.com/JuliaLang/julia/tree/master/stdlib/Dates

https://github.com/JuliaLang/julia/blob/master/stdlib/Dates/src/periods.jl

2 Likes
8

You could do Hour(tm-t0)/Hour(24).

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9

Thank you all for your help. So, it seems that the Dates module isn’t designed to make it straightforward to obtain an approximate floating-point representation of a Period.

So, I guess the most general solution is

using Dates
const t0 = DateTime(1950,1,1) # origin
const t1 = DateTime(1986,1,1)
const t2 = t1 + Dates.Month(1)
const tm = t1 + (t2 - t1) ÷ 7 # 1/7 of the month
println(tm)
tm_in_days  = Microsecond(tm-t0)/Microsecond(24*60*60*1e6)
tm_in_days2 = round(tm-t0, Second)/Second(24*60*60) # approximate
println(tm_in_days, ", ", tm_in_days2)

I’ve noticed that the floating point division / doesn’t always work on a Period because it’s ultimately a kind of integer representing the time interval down to microseconds.

So, to get a Float64 number which is as accurate as possible, we convert the Period to an integer microsecond and then divid it by a day in microsecond.