我在ruby中有一个哈希,看起来像这样:
{
"admin_milestones"=>"1",
"users_milestones"=>"0",
"admin_goals"=>"1",
"users_goals"=>"0",
"admin_tasks"=>"1",
"users_tasks"=>"0",
"admin_messages"=>"1",
"users_messages"=>"0",
"admin_meetings"=>"1",
"users_meetings"=>"0"
}
我正在寻找一种解决方案,可以将这个哈希值分成两部分,一部分值为1,另一部分哈希值为0。
答案 0 :(得分:29)
您可以按其值分组哈希:
h1 = {
"admin_milestones"=>"1",
"users_milestones"=>"0",
"admin_goals"=>"1",
"users_goals"=>"0",
"admin_tasks"=>"1",
"users_tasks"=>"0",
"admin_messages"=>"1",
"users_messages"=>"0",
"admin_meetings"=>"1",
"users_meetings"=>"0"
}
h2 = h1.group_by{|k,v| v}
它将生成一个按其值分组的哈希:
h2 = {"1"=>[["admin_milestones", "1"], ["admin_goals", "1"], ["admin_tasks", "1"], ["admin_messages", "1"], ["admin_meetings", "1"]],
"0"=>[["users_milestones", "0"], ["users_goals", "0"], ["users_tasks", "0"], ["users_messages", "0"], ["users_meetings", "0"]]}
答案 1 :(得分:6)
如果你想要一个数组作为答案,最干净的解决方案是分区方法。
zeros, ones = my_hash.partition{|key, val| val == '0'}
答案 2 :(得分:3)
只需Hash.select
:
h1.select { |key, value| value == '0' } #=> {"users_milestones"=>"0", "users_goals"=>"0", ...}
h1.select { |key, value| value == '1' } #=> {"admin_milestones"=>"1", "admin_goals"=>"1", ...}
返回值取决于您的Ruby版本。 Ruby 1.8返回一个数组数组,而Ruby 1.9返回一个哈希,如上例所示。
答案 3 :(得分:0)
您应该在group_by
数组上使用keys
并将该值用作分组元素:
h1 = {
"admin_milestones"=>"1",
"users_milestones"=>"0",
"admin_goals"=>"1",
"users_goals"=>"0",
"admin_tasks"=>"1",
"users_tasks"=>"0",
"admin_messages"=>"1",
"users_messages"=>"0",
"admin_meetings"=>"1",
"users_meetings"=>"0"
}
# group_by on the keys, then use the value from the hash as bucket
h2 = h1.keys.group_by { |k| h1[k] }
puts h2.inspect
将哈希从值返回到键数组:
{
"1" => [
[0] "admin_milestones",
[1] "admin_goals",
[2] "admin_tasks",
[3] "admin_messages",
[4] "admin_meetings"
],
"0" => [
[0] "users_milestones",
[1] "users_goals",
[2] "users_tasks",
[3] "users_messages",
[4] "users_meetings"
]
}
答案 4 :(得分:0)
与https://stackoverflow.com/a/56164608/14718545类似,您可以使用group_by
,但与then
类似,在这种情况下,您将避免实例化额外的变量。
{
"admin_milestones" => "1",
"users_milestones" => "0",
"admin_goals" => "1",
"users_goals" => "0",
"admin_tasks" => "1",
"users_tasks" => "0",
"admin_messages" => "1",
"users_messages" => "0",
"admin_meetings" => "1",
"users_meetings" => "0"
}.then { |h| h.keys.group_by { |k| h[k] } }
{"1"=>["admin_milestones", "admin_goals", "admin_tasks", "admin_messages", "admin_meetings"],
"0"=>["users_milestones", "users_goals", "users_tasks", "users_messages", "users_meetings"]}