如何在Ruby中匹配條件時在數組內創建數組？

Question

例如我有一個數組

g = [0,0,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1] 

而作爲一個結果，我需要

f = [0,0,0,[1,1,1],0,[1,1],0,0,1,0,0,[1,1,1,1],0,0,0,[1,1,1]] 

需要加入的元素（1）僅當它們出現在序列中的兩個或更多次。

Answer 1

實施塞爾吉奧Tulentsev

g = [0,0,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1] 
f = [] 
ones = [] 
g.each { |x| 
    if x == 1 
    ones << x 
    else 
    if ones.size > 1 
     f << ones 
    elsif ones.size == 1 
     f += ones 
    end 
    ones = [] 
    f << x 
    end 
} 
if ones.size > 1 
    f << ones 
elsif ones.size == 1 
    f += ones 
end 

UPDATE的想法

替代

g = [0,0,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1] 
f = [] 
g.each { |x| 
    if x == 1 
    f << [] if ! f.last.is_a? Array 
    f.last << x 
    else 
    f << x 
    end 
} 
f.each_with_index { |x,i| f[i] = 1 if x == [1] } 

Answer 2

g.inject([]) do |ary, i| 
    if i == 0 or ary.last == 0 
    ary << i 
    else 
    ary[-1] = Array(ary.last) << i 
    ary 
    end 
end 

=> [0, 0, 0, [1, 1, 1], 0, [1, 1], 0, 0, 1, 0, 0, [1, 1, 1, 1], 0, 0, 0, [1, 1, 1]] 

這應該是O（n）的運行時間。

FYI：Array(thing)在數組（[thing]）中包裝thing，除非它已經是一個數組。

Answer 3

Enumerable#chunk將是一個很好的選擇。

g = [0,0,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1] 
p g.chunk{|i| i == 0}.to_a 
# >> [[true, [0, 0, 0]], [false, [1, 1, 1]], [true, [0]], 
# [false, [1, 1]], [true, [0, 0]], [false, [1]], 
# [true, [0, 0]], [false, [1, 1, 1, 1]], 
# [true, [0, 0, 0]], [false, [1, 1, 1]]] 
ar = g.chunk{|i| i == 0}.each_with_object([]) do |(e1,e2),a| 
    (!e1 && e2.size > 1) ? a.concat([e2]) : a.concat(e2) 
end 
p ar 
# >> [0, 0, 0, [1, 1, 1], 0, [1, 1], 0, 0, 1, 0, 0, 
# [1, 1, 1, 1], 0, 0, 0, [1, 1, 1] 

基準

require 'benchmark' 

g = [0,0,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1] 
n = 1000000 
Benchmark.bm(15) do |x| 
    x.report('stefan') { n.times { g.chunk{ |e| e }.flat_map { |a, b| a == 0 || b.length == 1 ? b : [b] } } } 
    x.report('priti') { n.times { g.chunk{|i| i == 0}.each_with_object([]){|(e1,e2),a| (!e1 && e2.size > 1) ? a.concat([e2]) : a.concat(e2)} } } 
    x.report('tessi') { n.times { g.inject([]) { |ary, i| (i == 0 or ary.last == 0) ? ary << i : (ary[-1] = Array(ary.last) << i); ary }}} 
end 

         user  system  total  real 
stefan   11.860000 0.000000 11.860000 (11.890553) 
priti   13.360000 0.000000 13.360000 (13.406585) 
tessi    5.650000 0.000000 5.650000 ( 5.666377) 

Answer 4

這會工作，太：

g.chunk{ |e| e }.flat_map { |a, b| a == 0 || b.length == 1 ? b : [b] } 
#=> [0, 0, 0, [1, 1, 1], 0, [1, 1], 0, 0, 1, 0, 0, [1, 1, 1, 1], 0, 0, 0, [1, 1, 1]]