给定两个大范围数组...
A = [0..23, 30..53, 60..83, 90..113]
B = [-Float::INFINITY..13, 25..33, 45..53, 65..73, 85..93]
当我进行逻辑连接logical conjuction时...
C = A.mask(B)
然后我期望
describe "Array#mask" do
it{expect(C = A.mask(B)).to eq([0..13, 30..33, 45..53, 65..73, 90..93])}
end
感觉应该是这样的...
C = A & B
=> []
但是这是空的 因为没有一个范围是相同的。
这里有一个图示例。
.
require 'pry'
require 'rspec'
require 'benchmark'
require 'chronic'
require 'ice_cube'
require 'active_support'
require 'active_support/core_ext/numeric'
require 'active_support/core_ext/date/calculations'
A = [0..23, 30..53, 60..83, 90..113]
B = [-Float::INFINITY..13, 25..33, 45..53, 65..73, 85..93]
class Array
def mask(other)
a_down = self.map{|r| [:a, r.max]}
a_up = self.map{|r| [:a, r.min]}
b_down = other.map{|r| [:b, r.max]}
b_up = other.map{|r| [:b, r.min]}
up = a_up + b_up
down = a_down + b_down
a, b, start, result = false, false, nil, []
ticks = (up + down).sort_by{|i| i[1]}
ticks.each do |tick|
tick[0] == :a ? a = !a : b = !b
result << (start..tick[1]) if !start.nil?
start = a & b ? tick[1] : nil
end
return result
end
end
describe "Array#mask" do
context "simple integer array" do
it{expect(C = A.mask(B)).to eq([0..13, 30..33, 45..53, 65..73, 90..93])}
end
context "larger date ranges from IceCube schedule" do
it "should take less than 0.1 seconds" do
year = Time.now..(Time.now + 52.weeks)
non_premium_schedule = IceCube::Schedule.new(Time.at(0)) do |s|
s.duration = 12.hours
s.add_recurrence_rule IceCube::Rule.weekly.day(:monday, :tuesday, :wednesday, :thursday, :friday).hour_of_day(7).minute_of_hour(0)
end
rota_schedule = IceCube::Schedule.new(Time.at(0)) do |s|
s.duration = 7.hours
s.add_recurrence_rule IceCube::Rule.weekly(2).day(:tuesday).hour_of_day(15).minute_of_hour(30)
end
np = non_premium_schedule.occurrences_between(year.min, year.max).map{|d| d..d+non_premium_schedule.duration}
rt = rota_schedule.occurrences_between(year.min, year.max).map{|d| d..d+rota_schedule.duration}
expect(Benchmark.realtime{np.mask(rt)}).to be < 0.1
end
end
end
感觉用Ruby现有的核心方法无法做到这一点?我是错过了什么吗?我经常需要计算范围交集。
我也想到了,你可以使用相同的方法通过传递单个项数组来查找两个单一范围之间的交集。例如:
[(54..99)].mask[(65..120)]
我意识到我已经回答了自己的问题,但我觉得把它留在这里作为其他人的参考。