Representing a span of time

ruby-talk
1

Are these values seconds since epoch? If so:

irb(main):002:0> Time.at(1157489583.2798 - 64299600.0)
=> Sun Aug 22 09:53:03 MDT 2004

Regards,

Dan

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-----Original Message-----
From: list-bounce@example.com
[mailto:list-bounce@example.com] On Behalf Of Brad Tilley
Sent: Tuesday, September 05, 2006 3:02 PM
To: ruby-talk ML
Subject: representing a span of time

Say I have two time objects represented as floats like so:

x = 64299600.0
y = 1157489583.2798

I want to subtract x from y and then represent the difference
as years,
months, days, hours, minutes and seconds.

I don't see how the Time library would do this. Has anyone done
something like this? If so, how?

Thanks,
Brad

2

Berger, Daniel wrote:

x = 64299600.0
Brad

Are these values seconds since epoch? If so:

irb(main):002:0> Time.at(1157489583.2798 - 64299600.0)
=> Sun Aug 22 09:53:03 MDT 2004

They are dates in time from Time.mktime(1972, 1, 15, 0, 0, 0) and
Time.now() represented as floats. Basically, I would like to show the
difference in a human significant way (years, months, days, etc.)

···

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3

harp:~ > cat a.rb
require 'rubygems'
require 'timeunits'

a = Time.mktime(1972, 1, 15, 0, 0, 0).utc
b = Time.now.utc

delta = b - a

puts delta.decades
puts delta.years
puts delta.months
puts delta.days
puts delta.minutes
puts delta.seconds

harp:~ > ruby a.rb
3.76567099435856
37.6567099435856
451.880519323027
12652.6545410448
18219822.5391045
1093189352.34627

-a

···

On Wed, 6 Sep 2006, Brad Tilley wrote:

Berger, Daniel wrote:

x = 64299600.0
Brad

Are these values seconds since epoch? If so:

irb(main):002:0> Time.at(1157489583.2798 - 64299600.0)
=> Sun Aug 22 09:53:03 MDT 2004

They are dates in time from Time.mktime(1972, 1, 15, 0, 0, 0) and
Time.now() represented as floats. Basically, I would like to show the
difference in a human significant way (years, months, days, etc.)

They are dates in time from Time.mktime(1972, 1, 15, 0, 0, 0) and
Time.now() represented as floats. Basically, I would like to show the

--
what science finds to be nonexistent, we must accept as nonexistent; but what
science merely does not find is a completely different matter... it is quite
clear that there are many, many mysterious things.
- h.h. the 14th dalai lama

4

a = Time.mktime(1972, 1, 15, 0, 0, 0).utc
b = Time.now.utc

delta = b - a

puts delta.decades
puts delta.years

harp:~ > ruby a.rb
3.76567099435856
37.6567099435856 #This should be 34 years... what is it 37? (2006 - 1972 = 34)

···

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5

basically because it's using 28 days months since it's impossible to tell when
one says

   2.months

whether that should be 28 + 31 vs. 29 + 31 days, etc. so 1.month is simply
equal to 4.weeks and 1.year is 12.months. however, i that is not good
behaviour so the latest release will give:

     harp:~ > cat a.rb
     require 'yaml'
     require 'rubygems'
     require 'timeunits'

     a = Time.mktime(1972, 1, 15, 0, 0, 0).utc
     b = Time.now.utc

     delta = b - a
     y 'delta.years' => delta.years

     seconds_per_day = 1.day
     y 'seconds_per_day' => seconds_per_day

     n_days = 0 and ((a += seconds_per_day and n_days += 1) while a < b)
     y 'n_days' => n_days

     days_per_year = 365
     n_years = Float(n_days)/Float(days_per_year)
     y 'n_years' => n_years

     harp:~ > ruby a.rb
     delta.years: 34.6651137623302
     seconds_per_day: 86400
     n_days: 12653
     n_years: 34.6657534246575

i may re-consider the method used to set the delta fields to another method.

thanks for the bug report.

-a

···

On Wed, 6 Sep 2006, Brad Tilley wrote:

a = Time.mktime(1972, 1, 15, 0, 0, 0).utc
b = Time.now.utc

delta = b - a

puts delta.decades
puts delta.years

harp:~ > ruby a.rb
3.76567099435856
37.6567099435856 #This should be 34 years... what is it 37? (2006 - 1972 = 34)

--
what science finds to be nonexistent, we must accept as nonexistent; but what
science merely does not find is a completely different matter... it is quite
clear that there are many, many mysterious things.
- h.h. the 14th dalai lama