如何按两个条件对红宝石数组进行排序 [英] How to sort a ruby array by two conditions

问题描述

我想按两个不同的条件对该数组进行排序.

I want to sort this array by two different conditionals.

首先，我想按类型对数组进行排序:类型可以是(1,2,3,4)，我想按4-1-2-3的顺序对其进行排序.

First I want to sort the array by type: A type can be either (1,2,3,4) and I want to sort them in this order 4 - 1 - 2 - 3.

然后在每种不同的类型中，我想按降序对它们进行排序.

Then within each different type I want to sort them by a percentage descending.

所以排序后的数组看起来像这样:

So a sorted array would look like this:

[
  <OpenStruct percent=70, type=4>,
  <OpenStruct percent=60, type=4>,
  <OpenStruct percent=50, type=4>,
  <OpenStruct percent=73, type=1>,
  <OpenStruct percent=64, type=1>,
  <OpenStruct percent=74, type=2>
]ect

如何完成这种排序?目前，我只能按降序排序.

How can I accomplish this sort? Currently I can only sort by type descending.

array = array.sort_by {|r| r.type }

推荐答案

这应该做到:

require 'ostruct'
arr = [
  OpenStruct.new(percent: 73, type: 1),
  OpenStruct.new(percent: 70, type: 4),
  OpenStruct.new(percent: 60, type: 4),
  OpenStruct.new(percent: 50, type: 4),
  OpenStruct.new(percent: 64, type: 1),
  OpenStruct.new(percent: 74, type: 2)
]


puts arr.sort_by { |a| [a.type % 4, -a.percent] }

输出:

#<OpenStruct percent=70, type=4>
#<OpenStruct percent=60, type=4>
#<OpenStruct percent=50, type=4>
#<OpenStruct percent=73, type=1>
#<OpenStruct percent=64, type=1>
#<OpenStruct percent=74, type=2>

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