我有一个包含日期的数组。
dates = [#<DateTime: 2002-07-01T00:00:00+00:00 ((2452457j,0s,0n),+0s,2299161j)>, #<DateTime: 2003-10-31T00:00:00+00:00 ((2452944j,0s,0n),+0s,2299161j)>, #<DateTime: 2003-12-01T00:00:00+00:00 ((2452975j,0s,0n),+0s,2299161j)>, #<DateTime: 2004-03-01T00:00:00+00:00 ((2453066j,0s,0n),+0s,2299161j)>, #<DateTime: 2004-03-01T00:00:00+00:00 ((2453066j,0s,0n),+0s,2299161j)>]
如果该数组的升序或降序,我如何检查该数组的排序顺序?
答案 0 :(得分:9)
如果每两个相邻元素中的第一个元素小于或等于第二个元素,则数组呈上升趋势:
def ascending? arr
arr.each_cons(2).all?{|left, right| left <= right}
end
或者(更漂亮但不幸的是更慢),您可以将数组与其自身的排序版本进行比较:
def ascending? arr
arr == arr.sort
end
答案 1 :(得分:6)
以下是处理速度的一些基准测试结果:
require 'benchmark'
require 'date'
ary = (DateTime.parse('2002-07-01T00:00:00+00:00') .. DateTime.parse('2004-03-01T00:00:00+00:00')).to_a
def ascending1? arr
arr.reduce{ |e1,e2| e1 <= e2 ? e2 : (return false) }; true
end
def ascending2? arr
arr.each_cons(2).all?{|i,j| i <= j}
end
def ascending3? arr
arr == arr.sort
end
n = 10_000
Benchmark.bm(9) do |b|
b.report('reduce') { n.times{ ascending1?(ary) } }
b.report('each_cons') { n.times{ ascending2?(ary) } }
b.report('sort') { n.times{ ascending3?(ary) } }
end
测试结果:
user system total real
reduce 1.380000 0.000000 1.380000 ( 1.381107)
each_cons 2.250000 0.000000 2.250000 ( 2.243958)
sort 0.670000 0.000000 0.670000 ( 0.675025)
答案 2 :(得分:4)
为什么不检查排序的数组是否与未排序的数组相同?
提升 array == array.sort
和array == array.sort.reverse
降序
答案 3 :(得分:0)
我认为reduce
会比each_cons
def ascending? arr
arr.reduce{ |e1,e2| e1 <= e2 ? e2 : (return false) }; true
end