Re: [Factor-talk] Dedupe by Slot

The *.extras vocabularies are places to incubate new words. We haven't done the best job of documenting them and promoting to the core/basis vocabularies. > On Nov 18, 2016, at 8:21 AM, Alexander Ilin wrote: > > Hello, Björn! > > 18.11.2016, 18:25, "Björn Lindqvist"

Re: [Factor-talk] Dedupe by Slot

Hello, Björn! 18.11.2016, 18:25, "Björn Lindqvist" : > USE: sequences.extras > [ id>> ] sort-with [ id>> ] group-by [ second first ] map I could not find `group-by` using the Browser. Grepping the source tree, it turned up in `grouping.extras`. > USE:

Re: [Factor-talk] Dedupe by Slot

John, thank you very much! : )Really helpful stuff! 18.11.2016, 18:13, "John Benediktsson" :P.S., Hah I should have called it unique-by, it's too early in the morning! P.P.S., I committed this word into sets.extras, with one small change besides the name which is to size the

Re: [Factor-talk] Dedupe by Slot

2016-11-18 15:36 GMT+01:00 Alexander Ilin : > Hello, all! > > I have an interesting little task for you today. > > Let's say you have a sequence of tuples, and you want to remove all tuples > with duplicate ids, so that in the new sequence there is only one tuple with >

Re: [Factor-talk] Dedupe by Slot

Maybe something like this: : duplicates-by ( seq quot: ( elt -- key ) -- seq' ) HS{ } clone '[ @ _ ?adjoin ] filter ; inline Then you can use it: IN: scratchpad { 1 2 3 4 5 } [ 2/ ] duplicates-by { 1 2 4 } IN: scratchpad sequence-of-tuples [ hash>> ] duplicates-by It would

[Factor-talk] Dedupe by Slot

Hello, all! I have an interesting little task for you today. Let's say you have a sequence of tuples, and you want to remove all tuples with duplicate ids, so that in the new sequence there is only one tuple with each id. Here's my solution: TYPED: dedupe-by-hash ( seq: sequence --