Group array by nested array

Question

I have an array in this format:

[ 
  { day: 1
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 2
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 3
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 4
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 5
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 6
    intervals: [
      {
        from: 900,
        to: 1200
      },
      {
        from: 1300,
        to: 2200
      }
    ]
  },
  { day: 7
    intervals: [
      {
        from: 900,
        to: 1200
      },
      {
        from: 1300,
        to: 2200
      }
    ]
  }
]

I wan't to group them like this:

[ 
  { day: 1-5
    intervals: [
      {
        from: 900,
        to: 1200
      }
    ]
  },
  { day: 6-7
    intervals: [
      {
        from: 900,
        to: 1200
      },
      {
        from: 1300,
        to: 2200
      }
    ]
  }
]

Criteria:

Only group if intervals are the same.
Only group if matches are in chronological order, i.e 1-5 or 1-3, not 1-2-5.

How can this be achieved?

Note that your example output does not fulfill the 2nd requirement. 6-7 is only 2 days. — michael_wu
– michael_wu, Commented Apr 2, 2012 at 15:01

Michael Kohl · Accepted Answer · 2012-04-02 17:24:41Z

2

Here's a variation of @joelparkerhenderson's solution that tries to be a bit closer to your requirements re formatting of the output etc.

output = []

grouped = input.group_by do |x|
  x[:intervals]
end

grouped.each_pair do |k, v|
  days = v.map {|day| day[:day]}
  if days.each_cons(2).all? { |d1, d2| d1.next == d2 }
    output << {
      :days => days.values_at(0,-1).join('-'),
      :intervals => k
    }
  end
end

puts output

answered Apr 2, 2012 at 17:24

Michael Kohl

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3 Comments

michael_wu Over a year ago

If days 1, 2, 3, 5 have the same intervals, your code will not group 1-3, right?

Michael Kohl Over a year ago

True enough, this certainly can be improved.

Yeggeps Over a year ago

I found this to be the best starting point for my problem, thanks a lot!

michael_wu · Accepted Answer · 2012-04-02 15:16:28Z

1

This produces the required output:

by_interval = data.inject({}) do | a, e |
  i = e[:intervals]
  a[i] ||= []
  a[i] << e[:day].to_i
  a
end

result = by_interval.map do | interval, days |
  slices = days.sort.inject([]) do | a, e |
    a << [] if a == [] || a.last.last != e - 1
    a.last << e
    a
  end
  slices.map do | slice |
      {:day => "#{slice.first}-#{slice.last}", :intervals => interval }
  end
end
result.flatten!

I'm sure there are better approaches :-)

edited Apr 2, 2012 at 15:16

answered Apr 2, 2012 at 15:00

michael_wu

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Comments

michael_wu · Accepted Answer · 2012-04-02 13:59:48Z

0

You need to look into Map Method for an array. You need to remap the array and iterate over it to extract the data you want using your "grouping" logic above.

edited Apr 2, 2012 at 13:59

michael_wu

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answered Apr 2, 2012 at 13:54

Charlie Davies

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1 Comment

Yeggeps Over a year ago

Hi, thanks for your answer, I do know about map/collect, Having trouble comparing all the day hashes with each other though.

Tim Brunsmo · Accepted Answer · 2012-04-10 00:12:56Z

0

Extending @Michael Kohl's answer

output = []

grouped = schedules.as_json.group_by do |x|
  x['intervals']
end

grouped.each_pair do |k, v|
  days = v.map {|day| day['day']}
  grouped_days = days.inject([[]]) do |grouped, num|
    if grouped.last.count == 0 or grouped.last.last == num - 1
      grouped.last << num
    else
      grouped << [ num ]
    end
    grouped
  end

  grouped_days.each do |d|
    output << {
      heading: d.values_at(0, -1).uniq.join('-'),
      interval: k
    }
  end
end

output

You should probably split that up into separate methods, but you get the idea.

answered Apr 10, 2012 at 0:12

Tim Brunsmo

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