How do I return an array of days and hours from a

问题:

How do I return an array of days and hours from a range? So far I have tried:

(48.hours.ago..Time.now.utc).map { |time| { :hour => time.hour } }.uniq

Returns:

[{:hour=>1}, {:hour=>2}, {:hour=>3}, {:hour=>4}, {:hour=>5}, {:hour=>6}, {:hour=>7}, {:hour=>8}, {:hour=>9}, {:hour=>10}, {:hour=>11}, {:hour=>12}, {:hour=>13}, {:hour=>14}, {:hour=>15}, {:hour=>16}, {:hour=>17}, {:hour=>18}, {:hour=>19}, {:hour=>20}, {:hour=>21}, {:hour=>22}, {:hour=>23}, {:hour=>0}] 

Not ideal, since its iterates over every second. Which takes a long time. And I get several warning messages that say:

/Users/Chris/.rvm/gems/ruby-1.9.2-p290/gems/activesupport-3.2.2/lib/active_support/time_with_zone.rb:328: warning: Time#succ is obsolete; use time + 1

I am trying to return something like:

[{:day => 25, :hour=>1}, {:day => 25, :hour=>2}, {:day => 25, :hour=>3}, {:day => 25, :hour=>4} ... {:day => 26, :hour=>1}, {:day => 26, :hour=>2}, {:day => 26, :hour=>3}, {:day => 26, :hour=>4}] 

回答1:

Use Range#step, but as a precaution, convert the dates to integers first (apparently ranges using integers have step() optimized—YMMV). As a matter of style, I also truncate to the hour first.

Here's some quick 'n' dirty code:

#!/usr/bin/env ruby

require 'active_support/all'

s=48.hours.ago
n=Time.now

st=Time.local(s.year, s.month, s.day, s.hour).to_i
en=Time.local(n.year, n.month, n.day, n.hour).to_i

result = (st..en).step(1.hour).map do |i|
  t = Time.at(i)
  { day: t.day, hour: t.hour }
end

puts result.inspect

Yields:

[{:day=>25, :hour=>11}, {:day=>25, :hour=>12}, {:day=>25, :hour=>13}, 
{:day=>25, :hour=>14}, {:day=>25, :hour=>15}, {:day=>25, :hour=>16},
...

回答2:

stime = 48.hours.ago
etime=Time.now.utc
h = []
while stime <= etime
  h.push({ :day => stime.day, :hour => stime.hour })
  stime += 1.hour
end