按条目属性对数组进行排序，可选分组

Question

我有一个如下所示的数组：

[
    {"timestamp" => 1347119549, "category" => nil},
    {"timestamp" => 1347119547, "category" => "Monkeys"},
    {"timestamp" => 1347119543, "category" => nil},
    {"timestamp" => 1347119542, "category" => "Monkeys"}
]

我想按时间戳（降序）对其进行排序，除非它的类别不是 nil，在这种情况下，它应该与它的“兄弟姐妹”一起出现，即使它比未分类的条目“旧”。我需要对这个数组进行排序，所以它看起来像这样：

[
    {"timestamp" => 1347119549, "category" => nil},
    {"timestamp" => 1347119547, "category" => "Monkeys"},
    {"timestamp" => 1347119542, "category" => "Monkeys"},
    {"timestamp" => 1347119543, "category" => nil}
]

我试图弄清楚如何使用group_by 和sort 获得正确的结果，但没有成功。

Answer 1

require 'pp'

ar = [
    {"timestamp" => 1347119549, "category" => nil},
    {"timestamp" => 1347119547, "category" => "Monkeys"},
    {"timestamp" => 1347119543, "category" => nil},
    {"timestamp" => 1347119542, "category" => "Monkeys"}
]

pp ar.group_by{|h| h['category'] ? h['category'] : h['timestamp']}.
   map{|k,v| v.sort_by{|h| -h['timestamp']}}.
   sort_by{|a| -a[0]['timestamp']}.flatten
# >> [{"timestamp"=>1347119549, "category"=>nil},
# >>  {"timestamp"=>1347119547, "category"=>"Monkeys"},
# >>  {"timestamp"=>1347119542, "category"=>"Monkeys"},
# >>  {"timestamp"=>1347119543, "category"=>nil}]

---

require 'pp'

a = [
  {"timestamp"=>1347119549, "category"=>nil},
  {"timestamp"=>1347119547, "category"=>"Monkeys"},
  {"timestamp"=>1347119543, "category"=>nil},
  {"timestamp"=>1347119542, "category"=>"Monkeys"},
  {"timestamp"=>1347119548, "category"=>"Dog"},
  {"timestamp"=>1347119544, "category"=>"Dog"}
]

pp a.group_by{|h| h['category'] ? h['category'] : h['timestamp']}.
   map{|k,v| v.sort_by{|h| -h['timestamp']}}.
   sort_by{|a| -a[0]['timestamp']}.flatten 
# >> [{"timestamp"=>1347119549, "category"=>nil},
# >>  {"timestamp"=>1347119548, "category"=>"Dog"},
# >>  {"timestamp"=>1347119544, "category"=>"Dog"},
# >>  {"timestamp"=>1347119547, "category"=>"Monkeys"},
# >>  {"timestamp"=>1347119542, "category"=>"Monkeys"},
# >>  {"timestamp"=>1347119543, "category"=>nil}]

Answer 2

只需使用您尝试过的工具即可完成。

首先sort通过tiemstamp整个数组，然后使用group_by按类别分配它们组：

arr = [
    {'timestamp' => 1347119549, 'category' => nil},
    {'timestamp' => 1347119547, 'category' => 'Monkeys'},
    {'timestamp' => 1347119543, 'category' => nil},
    {'timestamp' => 1347119542, 'category' => 'Monkeys'},
    {'timestamp' => 1347119541, 'category' => nil},
    {'timestamp' => 1347119548, 'category' => nil},
    {'timestamp' => 1347119545, 'category' => nil},
]

sorted = arr.sort_by { |elem| 0 - elem['timestamp'] }
groups = sorted.group_by { |elem| elem['category'] or Object.new }
sorted = groups.values.flatten

puts sorted

输出

{"timestamp"=>1347119549, "category"=>nil}
{"timestamp"=>1347119548, "category"=>nil}
{"timestamp"=>1347119547, "category"=>"Monkeys"}
{"timestamp"=>1347119542, "category"=>"Monkeys"}
{"timestamp"=>1347119545, "category"=>nil}
{"timestamp"=>1347119543, "category"=>nil}
{"timestamp"=>1347119541, "category"=>nil}

当然，您可以以可读性为代价将整个事情流水线化。

sorted = arr.sort_by { |elem| 0 - elem['timestamp'] }.group_by { |elem| elem['category'] or Object.new }.values.flatten

Answer 3

这里需要的技巧是分配一个 unique 组而不是 nil。你可以通过创建一个通用的 Ruby Object 来做到这一点。

orig = [
  {"timestamp"=>1347119549, "category"=>nil}, 
  {"timestamp"=>1347119547, "category"=>"Monkeys"}, 
  {"timestamp"=>1347119543, "category"=>nil}, 
  {"timestamp"=>1347119542, "category"=>"Monkeys"}]

# The "tricky bit"
grouped = orig.group_by { |x| x["category"] ?  x["category"] : Object.new  }

# Sort the siblings within the groups (note negation causes reverse order)
grouped.values.each { |siblings| siblings.sort_by! { |a| -a["timestamp"] } }

# Sort the list by first (i.e. "best" sort order) timestamp in each group 
sorted_groups = grouped.sort_by { |group_id,siblings| -siblings.first["timestamp"] }

# Remove group ids and flatten the list:
result = sorted_groups.map { |group_id,siblings| siblings }.flatten
=>  [
 {"timestamp"=>1347119549, "category"=>nil}, 
 {"timestamp"=>1347119547, "category"=>"Monkeys"}, 
 {"timestamp"=>1347119542, "category"=>"Monkeys"}, 
 {"timestamp"=>1347119543, "category"=>nil}
]

Answer 4

看起来有点难看，但确实有效：

a = [
  {"timestamp"=>1347119549, "category"=>nil},
  {"timestamp"=>1347119547, "category"=>"Monkeys"},
  {"timestamp"=>1347119543, "category"=>nil},
  {"timestamp"=>1347119542, "category"=>"Monkeys"},
  {"timestamp"=>1347119548, "category"=>"Dog"},
  {"timestamp"=>1347119544, "category"=>"Dog"}
]
groups = a.sort_by {|h| -h['timestamp']}.group_by {|h| h['category']}
sorted = (groups.delete(nil) || []) + groups.values
sorted = sorted.sort_by{|i| i.is_a?(Hash) ? -i['timestamp'] : -i.first['timestamp']}.flatten

这会在sorted 中为您提供以下信息：

[
  {"timestamp"=>1347119549, "category"=>nil},
  {"timestamp"=>1347119548, "category"=>"Dog"},
  {"timestamp"=>1347119544, "category"=>"Dog"},
  {"timestamp"=>1347119547, "category"=>"Monkeys"},
  {"timestamp"=>1347119542, "category"=>"Monkeys"},
  {"timestamp"=>1347119543, "category"=>nil}
]

我先按'timestamp'排序，这样分组再排序。

在按'category' 分组后，我将nil 类别的值移动到一个数组中。在这里，我使用(groups.delete(nil) || []) 以防nil 组为空。

现在它可以再次按'timestamp' 排序，数组的timestamp 是其第一个哈希的timestamp。

最后flatten 给了我们想要的数组。