找到所有的组合中的Ruby

Question

特定标准匹配的数组中的

我有以下的数组：

a = ["melon | apple", "kiwi | melon", "apple | orange", "pineapple | kiwi"] 

我想的，得到所有字符串由字符串元素和前部的连接后半部分（"|"后） （在"|"之前）以下字符串元素。 a.combination_with_criteria(3).to_a应该输出：

["kiwi | melon", "melon | apple", "apple | orange"] 
["pineapple | kiwi", "kiwi | melon", "melon | apple"] 

a.combination(3).to_a给出了所有可能的组合，但以随机顺序。

也许为了这个目的使用散列更好。

Answer 1

a.combination(3).to_a给出了所有可能的组合，但是以随机顺序。

让我们来看看：

a.combination(3).to_a 
#=> [ 
#  ["melon | apple", "kiwi | melon", "apple | orange"], 
#  ["melon | apple", "kiwi | melon", "pineapple | kiwi"], 
#  ["melon | apple", "apple | orange", "pineapple | kiwi"], 
#  ["kiwi | melon", "apple | orange", "pineapple | kiwi"] 
# ] 

显然，这既不包含["kiwi | melon", "melon | apple", "apple | orange"]也不["pineapple | kiwi", "kiwi | melon", "melon | apple"]。

让那些为好，你必须使用permutation代替：

a.permutation(3).to_a 
#=> [ 
#  ["melon | apple", "kiwi | melon", "apple | orange"], 
#  ["melon | apple", "kiwi | melon", "pineapple | kiwi"], 
#  ["melon | apple", "apple | orange", "kiwi | melon"], 
#  ["melon | apple", "apple | orange", "pineapple | kiwi"], 
#  ["melon | apple", "pineapple | kiwi", "kiwi | melon"], 
#  ["melon | apple", "pineapple | kiwi", "apple | orange"], 
#  ["kiwi | melon", "melon | apple", "apple | orange"],  <--- here 
#  ["kiwi | melon", "melon | apple", "pineapple | kiwi"], 
#  ["kiwi | melon", "apple | orange", "melon | apple"], 
#  ["kiwi | melon", "apple | orange", "pineapple | kiwi"], 
#  ["kiwi | melon", "pineapple | kiwi", "melon | apple"], 
#  ["kiwi | melon", "pineapple | kiwi", "apple | orange"], 
#  ["apple | orange", "melon | apple", "kiwi | melon"], 
#  ["apple | orange", "melon | apple", "pineapple | kiwi"], 
#  ["apple | orange", "kiwi | melon", "melon | apple"], 
#  ["apple | orange", "kiwi | melon", "pineapple | kiwi"], 
#  ["apple | orange", "pineapple | kiwi", "melon | apple"], 
#  ["apple | orange", "pineapple | kiwi", "kiwi | melon"], 
#  ["pineapple | kiwi", "melon | apple", "kiwi | melon"], 
#  ["pineapple | kiwi", "melon | apple", "apple | orange"], 
#  ["pineapple | kiwi", "kiwi | melon", "melon | apple"], <--- here 
#  ["pineapple | kiwi", "kiwi | melon", "apple | orange"], 
#  ["pineapple | kiwi", "apple | orange", "melon | apple"], 
#  ["pineapple | kiwi", "apple | orange", "kiwi | melon"] 
# ] 

你可能已经知道select可以用来筛选出正确的元素，但如何的条件是什么样子？

让我们来匹配对：

a = 'kiwi | melon' 
b = 'melon | apple' 

我们可以split那些' | '得到部分：

a.split(' | ') #=> ["kiwi", "melon"] 
b.split(' | ') #=> ["melon", "apple"] 

这是一个比赛，如果a第一个字的最后一个字b匹配“ ：

a.split(' | ').last == b.split(' | ').first 
#=> true 

要检查这对于在阵列中的每个连续对字符串的，我们可以使用each_cons：

['kiwi | melon', 'melon | apple', 'apple | orange'].each_cons(2) do |a, b| 
    p a.split(' | ').last == b.split(' | ').first 
end 

它首先经过'kiwi | melon'和'melon | apple'到块，然后'melon | apple'和'apple | orange'。

对于这个数组，输出是：

true 
true 

要确定块11返回true将所有对，我们可以追加all?到each_cons：

['kiwi | melon', 'melon | apple', 'apple | orange'].each_cons(2).all? do |a, b| 
    a.split(' | ').last == b.split(' | ').first 
end 

而这正是什么我们可以通过select：

a.permutation(3).select do |sub_array| 
    sub_array.each_cons(2).all? do |a, b| 
    a.split(' | ').last == b.split(' | ').first 
    end 
end 
#=> [ 
#  ["kiwi | melon", "melon | apple", "apple | orange"], 
#  ["pineapple | kiwi", "kiwi | melon", "melon | apple"] 
# ] 

请注意，这仍会创建一个包含所有排列的巨大临时数组，并且会为每个比较分割字符串，因此您可能需要寻找更优化的解决方案。但是这应该让你开始。

Answer 2

这似乎工作：

def find_chains(input) 
    # Split input into usable value pairs. 
    pairs = input.map { |s| s.split(" | ") } 

    pairs.permutation(3).select do |ar| 
    ar[0][1] == ar[1][0] && ar[1][1] == ar[2][0] 
    end 
end 

input = ["melon | apple", "kiwi | melon", "apple | orange", "pineapple | kiwi"] 

find_chains(input).each do |match| 
    puts "match: " + match.map { |ar| ar.join(" | ")}.join(", ") 
end 

# Output: 
# 
# match: kiwi | melon, melon | apple, apple | orange 
# match: pineapple | kiwi, kiwi | melon, melon | apple