Order Hash and delete first key-value pair

Question

I have a Hash with timestamp as keys.

hash = {
  "2016-05-31T22:30:58+02:00" => {
          "path" => "/",
        "method" => "GET"
  },
  "2016-05-31T22:31:23+02:00" => {
          "path" => "/tour",
        "method" => "GET"
  },
  "2016-05-31T22:31:05+02:00" => {
          "path" => "/contact_us",
        "method" => "GET"
  }
}

I order the collection and get the first pair like this:

hash.sort_by {|k, _| k}.first.first

But how do I remove it?

The delete method requires you to know the exakt spelling of the key. Of course I could return the key and then use it in the delete method, but I was thinking if there was any more straight forward way?

Note that `sort_by` returns an array. – Stefan May 31 '16 at 21:07 — Stefan, May 31 '16 at 21:07
I have now clarified the question. – Fellow Stranger Jun 01 '16 at 08:14 — Fellow Stranger, Jun 01 '16 at 08:14

Kelsey Hannan · Answer 1 · 2016-05-31T21:12:30.160

10

Also note that shift can be used on the Hash values.

hash.shift

Removes the first key-value pair from the hash. Works on arrays too.

edited May 31 '16 at 21:12

answered May 31 '16 at 20:58

Kelsey Hannan

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Is not an array of hashes, it's simply a hash. `.shift` still works though. – Leonel Galán May 31 '16 at 21:02
2

You need to add the line `hash` after `hash.shift`, or possibly replace your line with `hash.tap { |h| hash.shift }`. – Cary Swoveland May 31 '16 at 21:34
You only need to add `hash` after if you are returning the value, but the statement alone modifies hash, what happens later is out of the scope of the question. – Leonel Galán Jun 01 '16 at 14:09
1

I wasn't aware of [Hash#shift](http://ruby-doc.org/core-2.2.0/Hash.html#method-i-shift) (why no `Hash#pop`?). Good to know. – Cary Swoveland Jun 01 '16 at 16:18

Leonel Galán · Accepted Answer · 2016-06-02T15:07:33.357

9

The method you are looking for is hash.keys it returns an array of the keys:

hash.delete(hash.keys.min)

EDIT: I've updated the answer to reflect that keys must be sorted first, this has been added in the original question and brought up by @Shadwell in comments to this post.

I replaced hash.keys.sort.first for hash.keys.min as suggested by @Cary Swoveland, it is not only more performant but better semantically.

edited Jun 02 '16 at 15:07

answered May 31 '16 at 20:52

Leonel Galán

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You'd need to sort the keys too I think to get the first one in order : `hash.delete(hash.keys.sort.first)` (I appreciate the title says an ordered hash but the question then sorts the keys explicitly) – Shadwell May 31 '16 at 20:57
I guess it depends if the keys are sorted or not, but that's not clear in the question. But yeah, you could sort them out too. – Leonel Galán May 31 '16 at 21:03
yeah, agreed, it's slightly unclear. Think you have the right approach though in both cases. – Shadwell May 31 '16 at 21:04
@Shadwell, why do you have to sort to find the smallest key? – Cary Swoveland May 31 '16 at 23:41
I suggest `hash.keys.min` or `hash.min_by(&:first).first` rather than `hash.keys.sort.first`. Sorting should be avoided when only the smallest or largest value is needed. – Cary Swoveland Jun 01 '16 at 16:17
Thanks for the suggestion, it definitely looks/reads better and I'll be using it. But why do you say sort should be avoided? As far as I can tell both `.min` and `.sort` depend on `<=>`. – Leonel Galán Jun 01 '16 at 16:33
1

Finding the minimum is much faster than sorting. Think about it, finding the minimum requires just a single pass through the array. – Cary Swoveland Jun 01 '16 at 20:29
I don't know how I missed that! Thanks again for keeping up with me and explaining, you are right, a sort is likely O(n log n), while finding the min is just O(n). – Leonel Galán Jun 02 '16 at 15:05
You don't have to state that you have updated your answer. Consider putting attributions in comments, to keep your answers "clean". If (in a comment) you thank the author of a comment for a suggestion and say that you've edited your answer to incorporate it), that's sufficient. – Cary Swoveland Jun 04 '16 at 14:18

Cary Swoveland · Answer 3 · 2016-06-01T15:08:29.773

2

As the OP has not stated that the hash's keys are in sorted order, we must assume that there is no guarantee that they are.

hash = { "2016-05-31T22:31:05+02:00"=>{ "path"=>"/tour", "method"=>"GET" },
         "2016-05-31T22:30:58+02:00"=>{ "path"=>"/", "method"=>"GET" },
         "2016-05-31T22:31:23+02:00"=>{ "path"=>"/contact_us", "method"=>"GET" } }

First, find the smallest key (the second one):

smallest_key = hash.keys.min
  #=> "2016-05-31T22:30:58+02:00"

This is obviously more efficient than sorting the keys then taking the smallest.

Because the date-time strings are in iso8601 format, they can be sorted as strings, without having to first convert them to time objects.

Then use Hash#reject to obtain the desired hash:

hash.reject { |k,_| k == smallest_key }
  #=> {"2016-05-31T22:31:05+02:00"=>{"path"=>"/tour", "method"=>"GET"},
  #    "2016-05-31T22:31:23+02:00"=>{"path"=>"/contact_us", "method"=>"GET"}}

To change hash in place, write

hash.delete(smallest_key }
hash

edited Jun 01 '16 at 15:08

answered May 31 '16 at 21:19

Cary Swoveland

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Don't loop to find the entry matching the key, that's O(n), if you have the key, simply use it, access to hashes is O(1). – Leonel Galán Jun 01 '16 at 14:11
Thanks, @Leito. That's a good suggestion for the case where the hash is to be mutated. (I did an edit.) If the hash is not to be mutated (the working assumption), one could write `h = hash.dup; h.delete(smallest_key); h`, but I prefer `reject`, which reads better (imo), does not require a non-block variable, is one line versus three and is probably not much worse performance-wise. – Cary Swoveland Jun 01 '16 at 15:13
You are right, it depends on the desired result, mutateted hash or not. Performance depends on n, for smaller n's probably not, for bigger n's yes. – Leonel Galán Jun 01 '16 at 15:28
Downvoter: note that the OP does not reorder the hash's keys. – Cary Swoveland Jun 01 '16 at 20:04

Order Hash and delete first key-value pair

3 Answers3