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：
看来@ ArupRakshit的解决方案很简单.我仍然在寻找比我的解决方案更好的效率.
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
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 }