按属性分选对象的数组红宝石可能是零 [英] sorting a ruby array of objects by an attribute that could be nil

问题描述

我有我需要通过这可能是整数或零位置属性进行排序对象的数组，我需要具有零的位置是在数组的末尾的对象。现在，我可以强制返回一些值，而不是零，使得中的Array.sort不会失败的位置，但如果我使用0作为此默认值，然后将其排序的前所说的那些对象。什么是做这样的最佳方式？我应该只设置零值是'几乎'总是保证在年底有些高的离谱号码？或者是有一些其他的方式，我可能会导致Array.sort方法把nil属性对象的数组的末尾？在code是这样的：

I have an array of objects that I need to sort by a position attribute that could be an integer or nil, and I need the objects that have the nil position to be at the end of the array. Now, I can force the position to return some value rather than nil so that the array.sort doesn't fail, but if I use 0 as this default, then it puts those objects at the front of the sort. What's the best way to to do this sort? should I just set the nil values to some ridiculously high number that is 'almost' always guaranteed to be at the end? or is there some other way i could cause the array.sort method to put the nil attribute objects at the end of the array? the code looks like this:

class Parent
  def sorted_children
     children.sort{|a, b| a.position <=> b.position}
  end
end

class Child
  def position
    category ? category.position : #what should the else be??
  end
end

现在，如果我做了其他类似10亿，那么它很有可能会帮他们在数组的结束，而是因为它是任意的，我不喜欢这个解决方案

now, if i make the 'else' something like 1000000000, then it's most likely gonna put them at the end of the array, but I don't like this solution as it's arbitrary

推荐答案

怎么样在儿童定义＆LT; =＆GT; 要依据 category.position 如果类别存在，和排序的项目没有一个类别的总是比那些具有类更大？

How about in Child defining <=> to be based on category.position if category exists, and sorting items without a category as always greater than those with a category?

class Child
  # Not strictly necessary, but will define other comparisons based on <=>
  include Comparable   
  def <=> other
    return 0 if !category && !other.category
    return 1 if !category
    return -1 if !other.category
    category.position <=> other.category.position
  end
end

然后在父，你可以叫 children.sort 。

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