How to convert period in milisecond to minutes / seconds , hour etc

programista · March 3, 2017, 12:40pm

How to convert period in milisecond to minutes / seconds , hour etc
julia 0.5.0
julia> delta=events[1,2].-events[1:1]
75631000 (miliseconds)

julia> convert(Dates.Minute,delta)
75631000 minutes ??? WRONG
Paul

Daneel · March 3, 2017, 1:22pm

Do you need the result to be in a time type? If it’s of some numeric type but represents milliseconds then just do the conversion. seconds = delta/1000

If you really need it in a time type then you can extend your other question.

julia> t1 = map((x) -> convert(DateTime,x),events[:,1])
julia> t2 = map((x) -> convert(DateTime,x),events[:,2])
julia> delta = t2 - t1
10-element Array{Base.Dates.Millisecond,1}:
 75631000 milliseconds 
 6031000 milliseconds  
 13531000 milliseconds 
 12685000 milliseconds 
 632000 milliseconds   
 21631000 milliseconds 
 212430000 milliseconds
 30000 milliseconds    
 30000 milliseconds    
 30000 milliseconds

or

julia> time = map((x) -> convert(DateTime,x),events)
10×2 Array{DateTime,2}:
 2016-04-01T22:00:44  2016-04-02T19:01:15
 2016-04-01T12:00:27  2016-04-01T13:40:58
 2016-04-01T14:43:50  2016-04-01T18:29:21
 2016-04-01T08:29:33  2016-04-01T12:00:58
 2016-04-01T22:53:45  2016-04-01T23:04:17
 2016-04-01T22:53:44  2016-04-02T04:54:15
 2016-04-01T02:59:52  2016-04-03T14:00:22
 2016-04-01T17:26:38  2016-04-01T17:27:08
 2016-04-01T17:32:34  2016-04-01T17:33:04
 2016-04-01T17:24:58  2016-04-01T17:25:28

julia> time[:,2] - time[:,1]
10-element Array{Base.Dates.Millisecond,1}:
 75631000 milliseconds 
 6031000 milliseconds  
 13531000 milliseconds 
 12685000 milliseconds 
 632000 milliseconds   
 21631000 milliseconds 
 212430000 milliseconds
 30000 milliseconds    
 30000 milliseconds    
 30000 milliseconds

quinnj · March 3, 2017, 4:08pm

I’m not sure what code you’re running or if it’s indeed 0.5, because I get

julia> convert(Dates.Minute, Dates.Millisecond(75631000))
ERROR: InexactError()
 in divexact(::Int64, ::Int64) at ./dates/periods.jl:376
 in convert(::Type{Base.Dates.Minute}, ::Base.Dates.Millisecond) at ./dates/periods.jl:398

julia> convert(Dates.Millisecond, Dates.Minute(2))
120000 milliseconds

julia> convert(Dates.Minute, convert(Dates.Millisecond, Dates.Minute(2)))
2 minutes

programista · March 3, 2017, 7:58pm

Nice, :
julia> convert(Dates.Minute, Dates.Millisecond(12000*60))
12 minutes

bat is unposible :
julia> convert(Dates.Minute, Dates.Millisecond(12001*60))
ERROR: InexactError()
in divexact(::Int64, ::Int64) at .\dates\periods.jl:376
in convert(::Type{Base.Dates.Minute}, ::Base.Dates.Millisecond) at .\dates\periods.jl:398

What with 12001 milliscond ?

stevengj · March 3, 2017, 8:13pm

You can also do:

julia> Dates.CompoundPeriod(Dates.Millisecond(756312342))
1 week, 1 day, 18 hours, 5 minutes, 12 seconds, 342 milliseconds

cormullion · April 15, 2017, 12:17pm

Sadly this appears to have disappeared in v0.6:

julia> Dates.CompoundPeriod(Dates.Millisecond(756312342))
756312342 milliseconds

stevengj · April 15, 2017, 1:51pm

In 0.6 you can do

julia> Dates.canonicalize(Dates.CompoundPeriod(Dates.Millisecond(756312342)))
1 week, 1 day, 18 hours, 5 minutes, 12 seconds, 342 milliseconds

cormullion · April 15, 2017, 2:04pm

Cool. That’s a nice function. With a cool name that provides some typing practice…

Karolina7 · October 2, 2018, 1:37pm

#milliseconds to seconds:
delta_sec = Dates.Second(delta)

#milliseconds to minutes:
delta_min = Dates.Minute(delta)

oxinabox · December 4, 2018, 2:22pm

For my own future reference (and anyone else who might stuble across this when googling).

In 1.0 it is written type last.

julia> round(Millisecond(12_001*60), Minute)
12 minutes

The logic behind this is that your are also allowed to specify things like

julia> round(Millisecond(12_001*60), Minute(5))
10 minutes

See discussion on

github.com/JuliaLang/julia

Add rounding support for `Periods`

JuliaLang:master ← invenia:gn/period-rounding

opened 04:11PM - 17 Oct 17 UTC

spurll

+196 -9

Allows (most) subtypes of `Period` to be rounded with `floor`, `ceil`, and `roun…d`, also allowing for easy conversion between `Period` subtypes where we would otherwise encounter an `InexactError`. ```julia julia> x = Dates.Millisecond(29394943) 29394943 milliseconds julia> convert(Dates.Hour, x) ERROR: InexactError: divexact(Int64, 8.165261944444444) Stacktrace: [1] divexact(::Int64, ::Int64) at ./dates/periods.jl:400 [2] convert(::Type{Base.Dates.Hour}, ::Base.Dates.Millisecond) at ./dates/periods.jl:422 julia> round(x, Dates.Hour) 8 hours ``` As with the date rounding functions, this approach also supports rounding to values (e.g., the nearest 15 minutes) as well as types: ```julia julia> x = Dates.Hour(910) 910 hours julia> ceil(x, Dates.Hour(12)) 912 hours ``` I can see at least two potential concerns being raised about this approach: 1. As discussed in https://github.com/JuliaLang/julia/issues/18574, there may be some confusion/contention surrounding the typical rounding signature (`round(value, precision)`) and the convert signature (`convert(type, value)`). 2. ~For certain very large values, the rounding may still result in an `InexactError`. This is because all values are converted to `Nanosecond` for rounding, so we can't easily accommodate values that exceed `typemax(Int64)` nanoseconds (which is a little over 292 years). This could be mitigated by using a lower-precision `Period` type when nanosecond precision is not required.~ **Edit:** Point 2 above has been mitigated via `promote`.

bernhard · October 8, 2019, 11:48am

Do you know a canonical way of converting a compound period to (e.g.) minutes (with digits)

julia> t=Minute(3)+Second(26)
3 minutes, 26 seconds

julia> want=3+26/60
3.433333333333333

I suppose I can always (??) convert to something small (say Nanosecond or Millisecond) and then divide, but there must be a more elegant way (or rather, an existing function) for this.

yakir12 · October 8, 2019, 11:56am

I’d do:

DateTime(0) + t - DateTime(0)

but I’m also curious to see if there’s a better way…

bernhard · October 8, 2019, 12:00pm

ok, I guess this works (as for elegance…)

julia> t=Minute(3)+Second(26)
3 minutes, 26 seconds

julia> (DateTime(0) + t - DateTime(0))/convert(Millisecond,Minute(1))
3.433333333333333

stevengj · October 8, 2019, 12:49pm

You can do

Dates.toms(t) / Dates.toms(Minute(1))

using the undocumented Dates.toms function to convert time periods to floating-point milliseconds.

TheLateKronos · February 17, 2022, 12:56pm

This seems to be undocumented:

Should it perhaps be documented under " Conversion Functions"?

aplavin · February 17, 2022, 1:25pm

Also see (my) package DateFormats.jl. It converts between different date/time formats, including regular decimal numbers.

That is, instead of

do period_decimal(Minute, t).
For example,

using DateFormats

julia> period_decimal(Minute, Second(123))
2.05

# and inverse conversion
julia> period_decimal(Minute, 2.05) == Second(123)
true

Note that, unlike Dates.toms, it doesn’t truncate everything below 1 ms to zero.