Question

Given this hash in ruby:

h={
2010-03-01=>2
2010-03-02=>4
2010=03-03=>1
..
..
n days with working hours
}

where hash key is a date and hash value is an integer, how do I transform this hash into a new hash where keys are weeks/months/years (with aggregation)?

Let me show an example with weeks, the end result should look like:

h_weeks={7=>36, 8=>42, 9=>34 ..,..,n}

with months:

h_months={1=>170, 2=>180, 3=>146}

with years:

h_years={2009=>950, 2010=>446}

where key is a week number(month/year), and value is an aggregate of working hours within this week/month/year.

I’m writing a small working hours tracking application and would like to group this data.

Thanks.

Answer 1

h = {
  Date.parse('2010-04-02') => 3,
  Date.parse('2010-05-03') => 5,
  Date.parse('2009-04-03') => 6
}
by_year = Hash.new(0)
by_month = Hash.new(0)
by_week = Hash.new(0)
h.each{|d,v|
  by_year[d.year] += v
  by_month[d.month] += v
  by_week[d.cweek] += v
}

or, slightly more exotic:

aggregates = {}
h.each{|d,v|
  [:year, :month, :cweek].each{|period|
    aggregates[period] ||= Hash.new(0)
    aggregates[period][d.send(period)] += v
  }
}
p aggregates
#=>{:year=>{2010=>8, 2009=>6}, :month=>{4=>9, 5=>5}, :cweek=>{13=>3, 18=>5, 14=>6}}