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.
