Possible Duplicate:
ruby array element grouping可能的重复:ruby数组元素分组
Example. Given array a:
的例子。给数组:
a = [1, 2, 3]
Its length is 3 so I want to print all 2-length arrays. These are:
它的长度是3，所以我想打印所有2长度的数组。这些都是:
[1, 2]
[1, 3]
[2, 3]
I don't know if there is some method in Ruby to get subset arrays. If there is not such a method what is most efficient way to do achieve this.
我不知道Ruby中是否有什么方法可以获得子集数组。如果没有这样的方法，最有效的方法是什么?
1 个解决方案
#1
5
That's just a simple combination of 2 elements:
这只是两个元素的简单组合:
>> xs = [1, 2, 3]
>> xs.combination(xs.size - 1).to_a
=> [[1, 2], [1, 3], [2, 3]]
[EDIT] As @Joshua pointed out in a comment, the docs state that the order is not guaranteed (!). So here is a functional implementation that generates the combinations in the order you asked for. For completeness, I'll make it lazy as the original combination method:
[编辑]正如@Joshua在评论中指出的，文档声明订单没有保证(!)。这是一个功能实现，按你要求的顺序生成组合。为了完整性，我将使它作为原始的组合方法变得懒惰:
require 'enumerable/lazy'
class Array
def combinations_of(n)
if n == 0
[[]].lazy
else
0.upto(self.size - 1).lazy.flat_map do |idx|
self.drop(idx + 1).combinations_of(n - 1).map do |xs|
[self[idx]] + xs
end
end
end
end
end
#1
5
That's just a simple combination of 2 elements:
这只是两个元素的简单组合:
>> xs = [1, 2, 3]
>> xs.combination(xs.size - 1).to_a
=> [[1, 2], [1, 3], [2, 3]]
[EDIT] As @Joshua pointed out in a comment, the docs state that the order is not guaranteed (!). So here is a functional implementation that generates the combinations in the order you asked for. For completeness, I'll make it lazy as the original combination method:
[编辑]正如@Joshua在评论中指出的，文档声明订单没有保证(!)。这是一个功能实现，按你要求的顺序生成组合。为了完整性，我将使它作为原始的组合方法变得懒惰:
require 'enumerable/lazy'
class Array
def combinations_of(n)
if n == 0
[[]].lazy
else
0.upto(self.size - 1).lazy.flat_map do |idx|
self.drop(idx + 1).combinations_of(n - 1).map do |xs|
[self[idx]] + xs
end
end
end
end
end