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comp.lang.ruby

representing a span of time

Brad Tilley

9/5/2006 9:02:00 PM

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

--
Posted via http://www.ruby-....
3 Answers

Ara.T.Howard

9/5/2006 9:27:00 PM

0

Ken Bloom

9/6/2006 3:23:00 AM

0

On Tue, 05 Sep 2006 15:26:41 -0600, ara.t.howard wrote:

> On Wed, 6 Sep 2006, Brad Tilley wrote:
>
>> 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?
>
[SNIP-CODE]
>
> harp:~ > ruby a.rb
> a: 2006-09-05 15:26:06.341147 -06:00
> b: 2006-09-05 17:28:06.341147 -06:00
> d.seconds: 7320.0
> d.minutes: 122.0
> d.hours: 2.03333333333333
> d.days: 0.0847222222222222

I think he wants something more like this, where all of the different
options collectively add up to the total time interval (rather than yours
where each different unit of time nonetheless represents the whole span)

class TimeSpan
# takes input in seconds because that's what Time#to_i returns
def initialize(seconds)
@raw_seconds=seconds
breakdown
end

attr_reader :seconds,:minutes,:hours,:days
attr_reader :raw_seconds

private
attr_writer :seconds,:minutes,:hours,:days

# [:x=,y] represents that y is the maximum number of x before
# carrying over to the next larger unit of time
FACTORS=[[:seconds=,60],[:minutes=,60],
[:hours=,24],[:days=,nil]]

def breakdown
higherintervals=@raw_seconds
FACTORS.each do |method, max|
higherintervals, thisinterval= higherintervals.divmod(max) unless max==nil
thisinterval=higherintervals if max==nil
send(method,thisinterval)
end
end
end

irb(main):020:0> a=Time.now.to_i
=> 1157512271
irb(main):021:0> b=Time.now.to_i
=> 1157512296
irb(main):025:0> TimeSpan.new(b-a)
=> #<TimeSpan:0xa7cbfc58 @days=0, @hours=0, @minutes=0,
@seconds=25, @raw_seconds=25>

Can't do years and months without more complicated math, and a definite
knowledge of what the start date and end date are. But at least this
should be closer.

--Ken Bloom



--
Ken Bloom. PhD candidate. Linguistic Cognition Laboratory.
Department of Computer Science. Illinois Institute of Technology.
http://www.iit.edu...

Daniel DeLorme

9/6/2006 9:03:00 AM

0

Brad Tilley wrote:
> 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?


This may not be exactly what you had in mind, but it's better than the other
proposed "solutions" when it comes to displaying the number of months or years
between two dates: (uses rails activesupport)

class Time
def time_span(other = nil)
other ||= self.utc? ? Time.now.utc : Time.now
other.time_spent(self)
end
def time_spent(other = nil)
other ||= self.utc? ? Time.now.utc : Time.now
n = other - self
case n.abs
when 0...60 #1.minute
"%d seconds" % n
when 0...3600 #1.hour
"%.1f minutes" % (n / 60)
when 0...86400 #1.day
"%.1f hours" % (n / 3600)
else
sign = "-" if self > other
t1,t2 = [other,self].sort
if t2.year != t1.year and t2 >= t1.advance(:years=>1)
nb_years = t2.year - t1.year
t = t1.advance(:years => nb_years)
t = t1.advance(:years => (nb_years -= 1)) if t > t2
"#{sign}%.1f years" % (nb_years + (t2 - t).to_f / (t.advance(:years=>1)
- t))
elsif t2 >= t1.advance(:months=>1)
nb_months = (t1.year==t2.year ? 0 : 12) + t2.month - t1.month
t = t1.advance(:months => nb_months)
t = t1.advance(:months => (nb_months -= 1)) if t > t2
"#{sign}%.1f months" % (nb_months + (t2 - t).to_f /
(t.advance(:months=>1) - t))
else
"%.1f days" % (n / 1.0.day)
end
end
end
end

>> t.time_span(t - 26.seconds)
=> "26 seconds"
>> t.time_span(t - 26.hours)
=> "1.1 days"
>> t.time_span(t - 26.days)
=> "26.0 days"
>> t.time_span(t - 26.months)
=> "2.1 years"