从Ruby数组中删除顺序重复
发布时间:2020-12-16 20:03:17 所属栏目:百科 来源:网络整理
导读:假设我有以下数组,我想摆脱连续的重复: arr = [1,1,4,3,5,1] 我想得到以下内容: = [1,1] 如果有一些比我的解决方案(或其变体)更简单和更有效率,那将是巨大的: (arr + [nil]).each_cons(2).collect { |i| i[0] != i[1] ? i[0] : nil }.compact 要么 (arr +
假设我有以下数组,我想摆脱连续的重复:
arr = [1,1,4,3,5,1] 我想得到以下内容: => [1,1] 如果有一些比我的解决方案(或其变体)更简单和更有效率,那将是巨大的: (arr + [nil]).each_cons(2).collect { |i| i[0] != i[1] ? i[0] : nil }.compact 要么 (arr + [nil]).each_cons(2).each_with_object([]) { |i,memo| memo << i[0] unless i[0] == i[1] } 编辑: 编辑: 当我们来的时候,我会对他们作出反应: require 'fruity' arr = 10000.times.collect { [rand(5)] * (rand(4) + 2) }.flatten compare do abdo { (arr + [nil]).each_cons(2).collect { |i| i[0] != i[1] ? i[0] : nil }.compact } abdo2 { (arr + [nil]).each_cons(2).each_with_object([]) { |i,memo| memo << i[0] unless i[0] == i[1] } } arup { arr.chunk(&:to_i).map(&:first) } arupv2 { arr.join.squeeze.chars.map(&:to_i) } agis { i = 1 a = [arr.first] while i < arr.size a << arr[i] if arr[i] != arr[i-1] i += 1 end a } arupv3 { arr.each_with_object([]) { |el,a| a << el if a.last != el } } end 基准测试结果: agis is faster than arupv3 by 39.99999999999999% ± 10.0% arupv3 is faster than abdo2 by 1.9x ± 0.1 abdo2 is faster than abdo by 10.000000000000009% ± 10.0% abdo is faster than arup by 30.000000000000004% ± 10.0% arup is faster than arupv2 by 30.000000000000004% ± 10.0% 如果我们使用: arr = 10000.times.collect { rand(4) + 1 } # less likelihood of repetition 我们得到: agis is faster than arupv3 by 19.999999999999996% ± 10.0% arupv3 is faster than abdo2 by 1.9x ± 0.1 abdo2 is similar to abdo abdo is faster than arupv2 by 2.1x ± 0.1 arupv2 is similar to arup 解决方法
使用
Enumerable#chunk 进行如下操作:
arr = [1,1] arr.chunk { |e| e }.map(&:first) # => [1,1] # if you have only **Fixnum**,something magic arr.chunk(&:to_i).map(&:first) # => [1,1] UPDATE 根据@abdo’s评论,这里是另一个选择: arr.join.squeeze.chars.map(&:to_i) # => [1,1] 另一个选择 arr.each_with_object([]) { |el,a| a << el if a.last != el } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |