counting words

[Benjamin Thaut]

Am 28.06.2013 19:01, schrieb Brad Anderson:
> On Friday, 28 June 2013 at 16:48:08 UTC, Benjamin Thaut wrote:
>> Am 28.06.2013 18:42, schrieb Brad Anderson:
>>> On Friday, 28 June 2013 at 16:25:25 UTC, Brad Anderson wrote:
>>>> On Friday, 28 June 2013 at 16:04:35 UTC, Benjamin Thaut wrote:
>>>>> I'm currently making a few tests with std.algorithm, std.range, etc
>>>>>
>>>>> I have a arry of words. Is it possible to count how often each word
>>>>> is contained in the array and then sort the array by the count of the
>>>>> individual words by chaining ranges? (e.g. without using a foreach
>>>>> loop + hashmap)?
>>>>
>>>> If you don't mind sorting twice:
>>>>
>>>> words.sort()
>>>>     .group()
>>>>     .array()
>>>>     .sort!((a, b)=> a[1] > b[1])
>>>>     .map!(a => a[0])
>>>>     .copy(words);
>>>>
>>>> You could also do it with a hashmap to keep the count.
>>>
>>> Like so:
>>>
>>>     size_t[string] dic;
>>>     words.map!((w) { ++dic[w.idup]; return w; })
>>>          .array // eager (so dic is filled first), sortable
>>>          .sort!((a, b) { bool less = dic[a] > dic[b]; return less ||
>>> less && a < b; })
>>>          .uniq
>>>          .copy(words);
>>>
>>> It's a bit ugly and abuses side effects with the hash map. The order
>>> will differ from the other program when words have identical counts.
>>
>> I figured something like this by now too. Thank you.
>> But I don't quite understand what the copy is for at the end?
>
> Just replacing your original word list with the sorted list (which I
> just realized is wrong because it will leave a bunch of words on the
> end, oops).  You could .array it instead to get a new array or just
> store the range with auto and consume that where needed with no extra
> array allocation.

Ok, thank you very much

-- 
Kind Regards
Benjamin Thaut