Bug 5968 – std.algorithm.group by key function + groupFull

Andrej Mitrovic has asked to split the following array in three arrays/ranges, according to the splitting predicate x<32: [64, 64, 64, 32, 31, 16, 32, 33, 64] A solution using group(): import std.stdio, std.algorithm; void main() { auto arr = [64, 64, 64, 32, 31, 16, 32, 33, 64]; int last = 0; foreach (g; group!q{ (a < 32) == (b < 32) }(arr)) { writeln(arr[last .. last+g[1]]); last += g[1]; } } Output: [64, 64, 64, 32] [31, 16] [32, 33, 64] Andrei has suggested the second item of the tuples that group() yields to be a lazy range instead just of counter (untested code). This is an improvement, and it makes group() closer to the Python itertools.groupby(). With this change the code becomes simpler: import std.stdio, std.algorithm; void main() { auto arr = [64, 64, 64, 32, 31, 16, 32, 33, 64]; foreach (g; group!q{ (a < 32) == (b < 32) }(arr)) writeln(g[1]); // g[1] is lazy } In Python groupby uses a key mapping function, like D schwartzSort(), that's more handy: >>> from itertools import groupby >>> arr = [64, 64, 64, 32, 31, 16, 32, 33, 64] >>> [list(g) for h,g in groupby(arr, key = lambda x: x < 32)] [[64, 64, 64, 32], [31, 16], [32, 33, 64]] I suggest to change D group() to follow the Python groupby() design on this too. With this the D code improves further (untested): import std.stdio, std.algorithm; void main() { auto arr = [64, 64, 64, 32, 31, 16, 32, 33, 64]; foreach (g; group!q{ a < 32 }(arr)) writeln(g[1]); } Implementation note: unlike schwartzSort() there isn't a need to memorize the results of all the key mapping functions, this avoids slow memory allocations. With tuple unpacking syntax sugar: import std.stdio, std.algorithm; void main() { auto arr = [64, 64, 64, 32, 31, 16, 32, 33, 64]; foreach ((h, g); group!q{ a < 32 }(arr)) writeln(g); }

I suggest to introduce a new function std.algorithm.groupFull that yields tuples where the second field of the tuple is a lazy range of all the grouped items, as in Python groupby. This Python2 program returns all the longest words that have ordered chars: from itertools import groupby o = (w for w in map(str.strip, open("words.txt")) if sorted(w)==list(w)) print list(next(groupby(sorted(o, key=len, reverse=True), key=len))[1]) A similar program using groupFull: import std.stdio, std.algorithm, std.range, std.file, std.string; void main() { "words.txt" .readText() .splitter() .filter!isSorted() .array() .sort!q{a.length > b.length}() .groupFull!q{a.length == b.length}() .front[1] .writeln(); }

*** Issue 11097 has been marked as a duplicate of this issue. ***

There is chunkBy now.