Ruby排序数组的条件相反[重复]

Possible Duplicate:
Sorting an array in descending order in Ruby

可能重复：在Ruby中按降序对数组进行排序

I want to sort an array of elements based on some condition, except in reverse order. So basically whatever it would have done and then reversed.

我想根据某些条件对元素数组进行排序，但顺序相反。所以基本上无论它做什么然后逆转。

So for example I have an array of strings and I want to sort it by decreasing string length

所以例如我有一个字符串数组，我想通过减少字符串长度对它进行排序

a = ["test", "test2", "s"]
a.sort_by!{|str| str.length}.reverse!

While this does the job...is there a way to specify the condition such that the sorting algorithm will do it in reverse?

虽然这样做了......有没有办法指定条件，以便排序算法反过来做？

2 个解决方案

#1

The length is a number so you can simply negate it to reverse the order:

长度是一个数字，所以你可以简单地否定它来反转顺序：

a.sort_by! { |s| -s.length }

If you're sorting on something that isn't easily negated then you can use sort! and manually reverse the comparison. For example, normally you'd do this:

如果你正在对不容易否定的事情进行排序，那么你可以使用排序！并手动反转比较。例如，通常你会这样做：

# shortest to longest
a.sort! { |a,b| a.length <=> b.length }

but you can swap the order to reverse the sorting:

但您可以交换订单以反转排序：

# longest to shortest
a.sort! { |a,b| b.length <=> a.length }

#2

The answer from @mu is too short is great, but only works for numbers. For the general case, you can resort to the sort method:

@mu的答案太短很好，但只适用于数字。对于一般情况，您可以使用sort方法：

 irb> a = ["foo", "foobar", "test"]
  => ["foo", "foobar", "test"]
 irb> a.sort{|a,b| a.length <=> b.length}
 => ["foo", "test", "foobar"]
 irb> a.sort{|a,b| b.length <=> a.length}
 => ["foobar", "test", "foo"]

#1