删除数组中的重复项，并将其第二个值添加到数组中

I want to convert:

我想转换:

[["val1", "1"], ["val2", "2"], ["val1", "3"], ["val3", "4"], ["val2", "5"], ["val2", "6"]]

to:

[["val1", ["1", "3"]], ["val2", ["2", "5", "6"]], ["val3", ["4"]]

Duplicates of the first value ("val1") of the subarrays are removed, and their second values ("1" and "3") are put together in an array.

删除子数组的第一个值(“val1”)的重复，将它们的第二个值(“1”和“3”)放在一个数组中。

I could call uniq on the array, but that only fixes half the problem. I gave it my best crack:

我可以在数组上调用uniq，但这只能解决问题的一半。我尽了最大的努力:

Store the first value (we can call value a) of the 2d array into another 2d array and calling .uniq on the new array.
将2d数组的第一个值(我们可以调用value a)存储到另一个2d数组中，并在新的数组中调用.uniq。
The new 2d array has an empty array in value b.
新的2d数组在值b中有一个空数组。
Then loop through the new array with an if statement comparing the original arrays value a to the new arrays value a. If value a from the original array matches val a from the new array add its b value to the new arrays b value array.
然后使用if语句对新数组进行循环，该语句将原数组的值a与新数组的值a进行比较。

This was my approach but there is most likely an easier approach.

这是我的方法，但很可能有更简单的方法。

2 个解决方案

#1

[["val1", "1"], ["val2", "2"], ["val1", "3"], ["val3", "4"], ["val2", "5"], ["val2", "6"]]
.group_by(&:first).map{|k, a| [k, a.map(&:last)]}
# => [["val1", ["1", "3"]], ["val2", ["2", "5", "6"]], ["val3", ["4"]]]

#2

Here is one more way to accomplish the same:

还有一种方法可以达到同样的效果:

ary.each_with_object(Hash.new {|h, k| h[k] = []}) do |i, h|  
  h[i.first] << i.last 
end.to_a
#=> [["val1", ["1", "3"]], ["val2", ["2", "5", "6"]], ["val3", ["4"]]]

We use a Hash which will initialise the keys with empty array, and it helps in the block to push the values into the array without having to do nil check.

我们使用一个散列，它将用空数组初始化键，并且它在块中帮助将值推入数组，而无需进行nil检查。

#1

[["val1", "1"], ["val2", "2"], ["val1", "3"], ["val3", "4"], ["val2", "5"], ["val2", "6"]]
.group_by(&:first).map{|k, a| [k, a.map(&:last)]}
# => [["val1", ["1", "3"]], ["val2", ["2", "5", "6"]], ["val3", ["4"]]]

#2