CODEWITHSUNDEEP
ruby-on-railsarraysrubydatetimedictionary
DollarChills
DollarChills

Reputation: 1096

Ruby -- group items by day with shifted beginning of the day

I have a dictionary as such:

{"status": "ok", "data": [{"temp": 22, "datetime": "20160815-0330"}]}

20160815-0330 means 2016-08-15 03:30. I want to group by day and calculate min and max temp, but the day should start at 09:00 and end at 08:59. How could I do that?

Here is my code:

@results = @data['data'].group_by { |d| d['datetime'].split('-')[0] }.map do |date, values|
  [date, {
    min_temp: values.min_by { |value| value['temp'] || 0 }['temp'],
    max_temp: values.max_by { |value| value['temp'] || 0 }['temp']
  }]
end

#=> {"20160815"=>{:min_temp =>12, :max_temp =>34}}

I works, but the starting point is 00:00, not 09:00.

Upvotes: 0

Views: 44

Answers (1)

Danil Speransky
Danil Speransky

Reputation: 30473

I would suggest to use a timezone trick:

def adjust_datetime(str)
  DateTime.parse(str.sub('-', '') + '+0900').utc.strftime('%Y%m%d')
end

data.group_by { |d| adjust_datetime(d['datetime']) }

Here is an explanation:

str = '20160815-0330'
str = str.sub('-', '')   #=> '201608150330'
str = str + '+0900'      #=> '201608150330+0900'
dt = DateTime.parse(str) #=> Mon, 15 Aug 2016 03:30:00 +0900
dt = dt.utc              #=> 2016-08-14 18:30:00 UTC
dt.strftime('%Y%d%m')    #=> '20160814'

Upvotes: 1

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