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"}
]

我希望按时间戳（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”一起出现，即使它比未分类的条目“更旧”。我需要对这个数组进行排序，所以看起来像这样：

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

我想弄清楚如何使用group_by和sort来获得正确的结果，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

看起来有点难看，但它有效：

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中为您提供以下内容：

[
  {"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，以便稍后对这些组进行排序。

按'category'分组后，我会移动数组中nil类别的值。在这里，我使用(groups.delete(nil) || [])以防nil组为空。

现在可以再次按'timestamp'排序，数组的timestamp为其第一个哈希的timestamp。

最后flatten为我们提供了所需的数组。

Answer 3

此处需要的技巧是分配唯一组而不是nil。您只需创建一个通用的Ruby Object即可。

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

只需使用您尝试过的工具即可完成。

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

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

<强>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，你可以以可读性为代价来管理整个事情。

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

按条目属性对数组进行排序，并使用可选的分组

4 个答案: