Question

有没有办法循环遍历两个数组并让它们每次引用某个元素？例如：

first_array = ['a', 'b', 'c', 'd']
second_array = ['e', 'f', 'g', 'h']

first_array.each |item| puts item.get(second_array)

结果看起来像：

a would work with e
b would work with f
c would work with g    
d would work with h

我试图这样做，以便当first_array中的变量传递给second_array时，它会移动到second_array中的下一个变量，跳过之前的使用变量。

Answer 1

那是zip。

first_array.zip(second_array){|e1, e2| ...}

Answer 2

你可以这样做：

first_array.each_with_index { |v, i| puts "#{v} can work with #{second_array[i]}" }
# a would work with e
# b would work with f
# c would work with g
# d would work with h

Answer 3

假设，在您的示例中，两个数组的大小相同，您可以使用Array#transpose：

a1 = %w{a b c d}
  #=> ['a', 'b', 'c', 'd']  
a2 = %w{e f g h}
  #=> ['e', 'f', 'g', 'h']
[a1,a2].transpose { |s1,s2| puts "#{s1} and #{s2}" }
  #-> a and e
  #   b and f
  #   c and g
  #   d and h

每当您使用数量相等的数组a（此处为[a1,a2]）并希望对每个[a[0][i],a[1][i],..]操作i时，您始终可以选择使用transpose或Array#zip。从某种意义上说，它们是Yin and Yang。

您也可以使用索引：

a1.each_index { |i] puts "#{a1[i]} and #{a2[i]}" }

Answer 4

给定第二个数组始终至少与第一个数组一样长

first_array.each_with_index |item,index| do
  puts "#{item} would work with #{second_array[index]}"
end

如何循环跳过以前引用元素的数组？

4 个答案: