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Set by lizmat on 8 June 2022.
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librasteve mcmillhj: fwiw I have tested your code (at least the bit that I can get from your link) and sort works fine for me ... 11:45
mcmillhj: and, eventually, I realise that I made the same jump that you did and when I comment out the Str method, it fails as you have already said - looks like nemokosch is spot on with this 12:54
ironically ... I then discovered this SO where the same discussion was had on 13th December 2018! 12:55
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mcmillhj What's the more idiomatic way to compare two Seqs? I know `eqv` doesn't work because they are lazy 22:56
nemokosch it doesn't? that sounds annoying 23:14
lizmat m: dd (^10).Seq.list eqv (^10).Seq.list # convert to list ? 23:15
camelia Bool::True
lizmat hmmmm
m: dd (^10).Seq.list eqv (^10).Seq.lazy.list 23:16
camelia Bool::False
lizmat hmmmm
nemokosch it's hard to say this well
there are 3 states
there is something that is just plain eager cached data 23:17
there is something where we just don't know
and there is something where the iterator is explicitly marked as lazy
mcmillhj yeah, that is what I am struggling with. I wrote a little helper function that just takes two positional arguments and compares them element by element. Sometimes one of the arguments is a List sometimes a Seq, sometimes an Array 23:18
lizmat m: dd so ((^10).Seq.lazy Zeqv (^10).Seq.lazy).all
camelia Bool::True
nemokosch a Seq can still be "don't know" rather than "explicitly lazy"
and then there is no problem
m: (1, 2, 3).Seq eqv (1, 2, 3).Seq andthen .say
Raku eval True
mcmillhj okay, so the main issue is that one of the arguments is a Seq and the other is a List then because eqv asserts against type right? 23:19
nemokosch yes, usually it does 23:20
but then I'm kinda leaning towards some of the solutions from liz 23:22
they really aren't "equivalent", just happen to have the same content when iterated
mcmillhj as long as something like this looks readable I guess I am okay. It looked weird when I wrote it this way the first time 23:28
m: my @rows = [305, 289, 460, 223, 223, 460, 289]; my $i = 3; my @left = @rows[0..$i]; my @right = @rows[$i + 1 .. * - 1]; my $closest-edge = min(@left.elems, @right.elems); say @left.reverse[0..^$closest-edge].List eqv @right[0..^$closest-edge].List;
camelia True