Can Ruby print out time difference (duration) read

问题:

Can Ruby do something like this?

irb(main):001:0> start = Time.now
=> Thu Nov 05 01:02:54 -0800 2009

irb(main):002:0> Time.now - start
=> 25.239

irb(main):003:0> (Time.now - start).duration
=> "25 seconds"

(the duration method doesn't exist now)... and similarly, report

23 minutes and 35 seconds
1 hour and 33 minutes
2 days and 3 hours

(either report the whole duration, up to how many seconds, or report up to 2 numbers and units (if day and hour is reported, then no need to tell how many minutes))

回答1:

Here's a quick and simple way to implement this. Set predefined measurements for seconds, minutes, hours and days. Then depending on the size of the number, output the appropriate string with the those units. We'll extend Numeric so that you can invoke the method on any numeric class (Fixnum, Bignum, or in your case Float).

class Numeric
  def duration
    secs  = self.to_int
    mins  = secs / 60
    hours = mins / 60
    days  = hours / 24

    if days > 0
      "#{days} days and #{hours % 24} hours"
    elsif hours > 0
      "#{hours} hours and #{mins % 60} minutes"
    elsif mins > 0
      "#{mins} minutes and #{secs % 60} seconds"
    elsif secs >= 0
      "#{secs} seconds"
    end
  end
end

回答2:

Have a look at the Rails DateHelper.distance_of_time_in_words method. It will give you a great starting place. Despite being loaded with magic numbers, the approach should work for you.

回答3:

There is a gem available https://rubygems.org/gems/time_diff

Which gives the difference in a hash

回答4:

Try a ruby gem for that https://rubygems.org/gems/time_difference - Time Difference gem for Ruby Documentation at https://github.com/tmlee/time_difference

start_time = Time.new(2013,1)
end_time = Time.new(2014,1)
TimeDifference.between(start_time, end_time).in_years
=> 1.0

回答5:

Time difference a pretty printed string:

 class Numeric
   def duration
     rest, secs = self.divmod( 60 )  # self is the time difference t2 - t1
     rest, mins = rest.divmod( 60 )
     days, hours = rest.divmod( 24 )

     # the above can be factored out as:
     # days, hours, mins, secs = self.duration_as_arr
     #
     # this is not so great, because it could include zero values:
     # self.duration_as_arr.zip ['Days','Hours','Minutes','Seconds']).flatten.join ' '

     result = []
     result << "#{days} Days" if days > 0
     result << "#{hours} Hours" if hours > 0
     result << "#{mins} Minutes" if mins > 0
     result << "#{secs} Seconds" if secs > 0
     return result.join(' ')
    end
  end

Time difference as an Array:

  class Numeric
    def duration_as_arr
      rest, secs = self.divmod( 60 )
      rest, mins = rest.divmod( 60 )
      days, hours = rest.divmod( 24 )
      [days, hours, mins, secs]
    end
  end

Example:

  x = 1209801.079257
  x.duration
   => "14 Days 3 Minutes 21.079257000004873 Seconds" 
  x.duration_as_arr
   => [14, 0, 3, 21.079257000004873] 

回答6:

Building on Michael Richard's answer, here's a replacement for the if block that gets English pluralization right, and won't say things like "14 days and 0 hours":

if days > 0
  hour_remainder = hours % 24
  if hour_remainder > 0
    hour_str = hour_remainder == 1 ? 'hour' : 'hours'
    "#{days} days and #{hour_remainder} #{hour_str}"
  elsif days == 1
    "#{days} day"
  else
    "#{days} days"
  end
elsif hours > 0
  min_remainder = mins % 60
  if min_remainder > 0
    min_str = min_remainder == 1 ? 'minute' : 'minutes'
    "#{hours} hours and #{min_remainder} #{min_str}"
  elsif hours == 1
    "#{hours} hour"
  else
    "#{hours} hours"
  end
elsif mins > 0
  sec_remainder = secs % 60
  if sec_remainder > 0
    sec_str = sec_remainder == 1 ? 'second' : 'seconds'
    "#{mins} minutes and #{sec_remainder} #{sec_str}"
  elsif minutes == 1
    "#{mins} minute"
  else
    "#{mins} minutes"
  end
elsif secs == 1
  "#{secs} second"
elsif secs >= 0
  "#{secs} seconds"
end

回答7:

I could not withstand to put a generic solution here - although: has a year 365 days?

Additional I put an abs when converting self.to_int

class Numeric
    def duration
        steps=[60, 60, 24, 365,0]
        names=[:seconds, :minutes, :hours, :days, :years]
        results=[]
        stepper = self.to_int.abs
        steps.each { |div|
            if stepper>0
                if div>0
                    results<<stepper % div
                    stepper/=div
                else
                    results << stepper
                end
            end
        }
        e= results.empty? ? 0 : results.count-1
        mt= e>0 ? results[e-1] : 0
        et=results[e] || 0

        et.to_s+" "+names[e].to_s + (mt>0 ? " "+mt.to_s+" "+names[e-1].to_s : '')
    end
end

and with translation

class Numeric
    def i18n_duration
        steps=[60, 60, 24, 365,0]
        names=[:seconds, :minutes, :hours, :days, :years]
        results=[]
        stepper = self.to_int.abs
        steps.each { |div|
            if stepper>0
                if div>0
                    results<<stepper % div
                    stepper/=div
                else
                    results << stepper
                end
            end
        }
        e= results.empty? ? 0 : results.count-1
        mt= e>0 ? results[e-1] : 0
        et=results[e] || 0

        I18n.t("datetime.distance_in_words.x_#{names[e]}", count: et) +
             (mt>0 ? " "+I18n.t("datetime.distance_in_words.x_#{names[e-1]}", count: mt):'')
    end
end

回答8:

As an alternative you can do this:

start = DateTime.now.strftime('%Q').to_i / 1000.0
#=> 1409163050.088
sleep 3.5
#=> 3
now_ms = DateTime.now.strftime('%Q').to_i / 1000.0
#=> 1409163053.606
'%.3f' % (now_ms - start)
#=> "3.518"

回答9:

I've given an alternative implementation for this when used in script logs here [copied from there]:

How to generate a human readable time range using ruby on rails

If you want to show significant durations in the seconds to days range, an alternative would be (as it doesn't have to perform the best):

def human_duration(secs, significant_only = true)
  n = secs.round
  parts = [60, 60, 24, 0].map{|d| next n if d.zero?; n, r = n.divmod d; r}.
    reverse.zip(%w(d h m s)).drop_while{|n, u| n.zero? }
  if significant_only
    parts = parts[0..1] # no rounding, sorry
    parts << '0' if parts.empty?
  end
  parts.flatten.join
end
start = Time.now
# perform job
puts "Elapsed time: #{human_duration(Time.now - start)}"

human_duration(0.3) == '0'
human_duration(0.5) == '1s'
human_duration(60) == '1m0s'
human_duration(4200) == '1h10m'
human_duration(3600*24) == '1d0h'
human_duration(3600*24 + 3*60*60) == '1d3h'
human_duration(3600*24 + 3*60*60 + 59*60) == '1d3h' # simple code, doesn't round
human_duration(3600*24 + 3*60*60 + 59*60, false) == '1d3h59m0s'

Alternatively you may be only interested in stripping the seconds part when it doesn't matter (also demonstrating another approach):

def human_duration(duration_in_seconds)
  n = duration_in_seconds.round
  parts = []
  [60, 60, 24].each{|d| n, r = n.divmod d; parts << r; break if n.zero?}
  parts << n unless n.zero?
  pairs = parts.reverse.zip(%w(d h m s)[-parts.size..-1])
  pairs.pop if pairs.size > 2 # do not report seconds when irrelevant
  pairs.flatten.join
end

Hope that helps.

回答10:

I came with this solution

def formate_duration(started, ended)
  differences = ["secs", "mins", "hours", "days"].reduce([[nil, [(ended - started).round, nil]]]) do |acc, unit|
    mod = unit == "hours" ? 24 : 60
    # will return [ over, unit_value ]
    # over is than used for calculation in next step
    # accumulator is in format [["unit" [ over, value]]] and we are interesting in latest over in each step
    # => This is why main diff is in this place in accumulator
    entry = acc[acc.size-1][1][0].divmod(mod)
    acc << [ unit, entry ]
  end

  # this will do string conversion and reverse in one step
  str = differences.drop(1).reduce("") do |acc, current|
    if (current[1][1] > 0)
      "#{current[1][1]} #{current[0]} #{acc}"
    else acc end
  end

  str.empty? ? "now" : str
end

Be aware that this won't work for differences bigger than 60 days (In that case you will need to add condition)

回答11:

time_difference = current_time - old_time



 def seconds_fraction_to_time(time_difference)
    days = hours = mins = 0
      mins = (seconds / 60).to_i
      seconds = (seconds % 60 ).to_i
      hours = (mins / 60).to_i
      mins = (mins % 60).to_i
      days = (hours / 24).to_i
      hours = (hours % 24).to_i
    return [days,hours,mins,seconds]
  end

then you can print it out what ever way you wish,

i.e

if(days > 0)
return "#{days} Days #{hours} Hours"
else
return "#{hours} Hours #{mins} Minutes"
end