Question

我有一个如下所示的数组：

booklist = [000A,000B,000C]

我想成对组合所有元素，以便这是输出数组：

combinedbooks = [000A|000B, 000A|000C, 000B|000C]

我所做的是以下内容：

combinedbooks= Array.new
    for i in 0..booklist.length
        for j in i+1..booklist.length                  
            combinedbooks.push(booklist(i) + "|" + booklist(j))
        end
    end

它不起作用，我不知道为什么

Answer 1

嵌套for循环的替代方法是使用#combination：

[25] pry(main)> booklist
=> ["000A", "000B", "000C"]
[26] pry(main)> booklist.combination(2).map { |i| i.join("|") }
=> ["000A|000B", "000A|000C", "000B|000C"]

Answer 2

您的代码实际上存在两个问题。第一个是数组的特定元素应该用括号[]引用，而不是括号()。第二个是，由于Ruby中的数组是零索引的，list[list.length]不存在（list[list.length-1]是数组中的最后一个元素）。如果我们解决了这两个问题，你最终得到：

booklist = ["000A", "000B", "000C"]

combinedbooks = Array.new
for i in 0..booklist.length-1
  for j in i+1..booklist.length-1
    combinedbooks.push(booklist[i] + "|" + booklist[j])
  end
end

哪个有效。

Ruby是一种非常强大的语言，所以实际上有一种更简单的方法可以使用combination，map和join方法组合Array，{{3}}和{{3}}方法。 3}}：

booklist = ["000A", "000B", "000C"]
combinedbooks = booklist.combination(2).map{|combination| combination.join("|")}

Answer 3

booklist = ['000A', '000B', '000C']
joined_booklist = []
booklist.each_with_index do |book, i|
   next if i == (booklist.length - 1)
  ((i+1)..(booklist.length-1)).each do |j|
     joined_booklist << book + "|" + booklist[j]
  end
end

Answer 4

这是另一个until loop

的解决方案

booklist = ["000A","000B","000C"]

# duplicate the array
booklist_dup = booklist

result_array = []

until booklist_dup.empty?
  # takes the first element out of the array
  first_el = booklist_dup.shift

  # create a combination with each element of the remaining array
  booklist.each{ |el| result_array << first_el + "|" + el}
end

# => ["000A|000B", "000A|000C", "000B|000C"]

Answer 5

对这种行为使用Array＃组合：

booklist = ['000A', '000B', '000C']
booklist.combination(2).map { |pair| pair.join('|') }

Answer 6

你可以尝试这样的事情。

booklist = ['000A', '000B', '000C']
pairs = booklist.combination(2).to_a
pairs.map!{|a,b|"#{a}|#{b}"}

将数组元素组合成红宝石

6 个答案: