CODEWITHSUNDEEP
datetimejuliamilliseconds
Siddhesh Bhurke
Siddhesh Bhurke

Reputation: 105

How to convert Milliseconds array to MM:SS:ss format or DateTime format in Julia?

I am trying to convert milliseconds to minutes but unable to do so. I have been expecting the result in the form of MM:SS I have saved my results in a dataframe which shows the data type as milliseconds and even when I divide the entire row by 60000, the data type still remains to be milliseconds

delivery_time = delivery_timestamp - init_timestamp
783358-element Array{Millisecond,1}:
 34200000 milliseconds
 14580000 milliseconds
 27300000 milliseconds
 9180000 milliseconds 
 13560000 milliseconds
 6960000 milliseconds 
 12600000 milliseconds
 9600000 milliseconds 
 7500000 milliseconds 
 13080000 milliseconds
 9180000 milliseconds 
 9600000 milliseconds 
 27660000 milliseconds
 ⋮                    

delivery_time = (delivery_time)/60000
783358-element Array{Millisecond,1}:
 570 milliseconds
 243 milliseconds
 455 milliseconds
 153 milliseconds
 226 milliseconds
 116 milliseconds
 210 milliseconds
 160 milliseconds
 125 milliseconds
 218 milliseconds
 153 milliseconds

I am expecting the results to be

df.delivery_time
06:22
07:32
09:51
.
.
.

or even if I can get results as

df.delivery_time
570 minute
243 minute
455 minute
.
.
.

Upvotes: 1

Views: 590

Answers (2)

Siddhesh Bhurke
Siddhesh Bhurke

Reputation: 105

thank you so much for help. You solved most of my problem. I am quite naive yet in Julia started a week back. I had actually used - df.delivery_time = map((x) -> convert(Time,x),df[:15]) with this as the result 783358-element Array{Time,1}: 09:30:00 04:03:00 07:35:00 02:33:00 03:46:00 01:56:00

Upvotes: 0

Bogumił Kamiński
Bogumił Kamiński

Reputation: 69949

You can convert Millisecond to eg. Minute only with loss of precision. Here is an example that does rounding to the nearest minute:

@. Minute(round(Int, Dates.value(delivery_time) / (1000 * 60)))

If you do not need Minute values but simply floats without rounding you can do:

@. Dates.value(delivery_time) / (1000 * 60)

or e.g.

Dates.value.(delivery_time) / (1000 * 60)

if you prefer not to use @..

Finally you can write

Time(0) .+ delivery_time

which is lossless, but now the returned object is not a subtype of Period and it will "wrap around" 24 hours. If you want to avoid wrapping use DateTime instead of Time above.

Upvotes: 1

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