I want to hollow out a boolean array:

1 1 1 1 0 0 0 1 1 1 → 1 0 0 1 0 0 0 1 0 1

how do I do it?

@Razetime What is the expected output for 0 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0?

0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0

basically convert 1s between other 1s to zeroes

{(2⊥⍵)∊2 3 6}⌺3 should work

Basically tests if each 3-width window has 01x or x10 pattern

wow

except that it doesn't work when the length is 1 or 0, because ⌺ doesn't

@Bubbler behold

      'Buy24Beers'(({(2⊥⍵)∊2 3 6}⌺3)⍤∊⊂⊣)⎕A,⎕D
┌───┬──┬─────┐
│Buy│24│Beers│
└───┴──┴─────┘

@Razetime f⍢⌽∨f←2</0, in dzaima/APL (the equivalent vanilla Dyalog APL is ⌽⍤f⍤⌽∨f←2</0∘,)

Works for any input length and is probably a bit faster in general:

     ∇ ⌺ 3
      M←⌽⍤f⍤⌽∨f←2</0∘, ⋄ S←{(2⊥⍵)∊2 3 6}⌺3
      s←0 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0
      l←2=?1e6⍴2
      ]runtime -c M⊢s S⊢s

  M⊢s → 7.3E¯7 |     0% ⎕
  S⊢s → 4.0E¯5 | +5414% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
      ]runtime -c M⊢l S⊢l

  M⊢l → 1.0E¯3 |       0%
  S⊢l → 3.1E0  | +308800% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

I think I've come up with another version of median, but I don't know how it differs from the one in APLcart {X←⍵[⍋⍵]⋄.5×(X+⌽X)[⌈.5×≢X]}

Aha, it's a train (for reference: (2÷⍨1⊥⊢⌷⍨∘⊂⍋⌷⍨∘⊂∘⌈2÷⍨0 1+≢))

@Sherlock9 Yours doesn't work on higher-rank arrays (though that could be fixed).

It also does a lot of unnecessary work, sorting the entire array, and adding all the values instead of only the two relevant ones.

Ah, I have not gotten the hang of higher rank arrays yet

What does the median of a higher-rank array look like? Flatten the whole array and then get the median value of all the elements?

@Adám very cool

sadly i realized my program would get too bloated with a non-regex approach

@Adám does the APLcart example work on higher rank?

Yes.

      {X←⍵[⍋⍵]⋄.5×(X+⌽X)[⌈.5×≢X]}3 3⍴⍳9
RANK ERROR
      {X←⍵[⍋⍵] ⋄ 0.5×(X+⌽X)[⌈0.5×≢X]}3 3⍴⍳9

@Sherlock9 No, that'd be f, for any f. It does the same as +⌿÷≢

am I using it wrong?

@Razetime No, it simply doesn't work on high-rank arguments.

@Adám Doesn't the one in APLCart also sort the entire array with X ← ⍵[⍋⍵]..?

@RGS @Razetime I think you're mixing up APLcart's median with Sherlock9's median.

@Adám Ah, maybe.

oh ok

      (2÷⍨1⊥⊢⌷⍨∘⊂⍋⌷⍨∘⊂∘⌈2÷⍨0 1+≢)3 3⍴9
9 9 9

i see

The best way to find out the name of the current function is by looking at the stack?

@Adám What do you mean by f?

@Sherlock9 f is a function. What Adám means is that "flatten the whole array and then do something" is the same as writing the code f,

Where f is the function that does something.

Oh I see

Thanks

So if you have a function f that computes the median, then "flatten whole array and compute median" would be f,array.

Np. As far as my understanding goes, most functions look at high-dimensional arrays like "containers" of lower-dimensional things, in the sense that it is useful to compute some operation on those lower-dimensional subarrays separately.

So what this does is calculate the median ... element/row/plane along the first axis?

Yes, (2÷⍨1⊥⊢⌷⍨∘⊂⍋⌷⍨∘⊂∘⌈2÷⍨0 1+≢) calculates the median along the first axis.

for matrices, that means along the columns.

      3 3⍴1 2 3
1 2 3
1 2 3
1 2 3

      (2÷⍨1⊥⊢⌷⍨∘⊂⍋⌷⍨∘⊂∘⌈2÷⍨0 1+≢)3 3⍴1 2 3
1 2 3

for 3D things, the first axis "pierces through" the planes:

      3 3 3⍴1 2 3 4 5 6 7 8 9
1 2 3
4 5 6
7 8 9

1 2 3
4 5 6
7 8 9

1 2 3
4 5 6
7 8 9

      (2÷⍨1⊥⊢⌷⍨∘⊂⍋⌷⍨∘⊂∘⌈2÷⍨0 1+≢)3 3 3⍴1 2 3 4 5 6 7 8 9
1 2 3
4 5 6
7 8 9

Is there an array solution to this (I have a recursive one)? Given a boolean rank-2 array, on each row, wipe all 1s before or equal to the position of the first 1 on the row before, eg given

⊢m←3 12⍴0 0 1 0 0 0 1 0 1 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0
┌→──────────────────────┐
↓0 0 1 0 0 0 1 0 1 0 0 1│
│0 1 0 1 0 0 0 1 0 0 0 0│
│1 0 0 0 1 0 0 0 0 1 0 0│
└~──────────────────────┘

f m

┌→──────────────────────┐
↓0 0 1 0 0 0 1 0 1 0 0 1│
│0 0 0 1 0 0 0 1 0 0 0 0│
│0 0 0 0 1 0 0 0 0 1 0 0│
└~──────────────────────┘

i don't see a reason that should need any recursion?

current line and line above should be enough to know the new line

      <\0⍪2⊣⌿m
0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0

@RGS that's just finding the first occurence

@dzaima I don't either :) -- hence my question -- feels like there should be a nice solution for this that I can't see.

@dzaima I know, I haven't finished typing :P

This shows the first 1 of the row above.

i think i have 25 as a dfn

      m×∨\1⍪0,1 ¯1↓<\0⍪2⊣⌿m
0 0 1 0 0 0 1 0 1 0 0 1
0 0 0 1 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0 1 0 0

Then ∨\1⍪0,1 ¯1↓ builds a mask of all the 1s you want to keep.

And then you multiply that with m.

∧

@RGS very neat

Can also do the ∨\ earlier to avoid a bit of unnecessary calculations: m×1⍪0,∨\1 ¯1↓<\0⍪2⊣⌿m

@RGS 2⊣⌿m is just an obscure way to write ¯1↓m right?

i had {⍵×(¯1↓0,⍵⍳⍤1⊢1)∘.<⍳2⊃⍴⍵}

@Adám Yes xD

@RGS Why add 0⍪ when you remove it later with 1 ¯1↓ ?

A bit shorter without adding and removing the same row for no reason at all: m×1⍪0,∨\0 ¯1↓<\¯1↓m

Ninja'd :(

@xpqz the hardest bits for me were in finding the first 1 row-wise, but RParks webinar on boolean scans reminded me of <\ , and then taking those first 1s and spreading them to the right, to create the mask, but that's just ∨\ .

aplcart.info?q=after%20first

I vaguely recall such matrices being called something, back from the days before I'd forgotten most of my maths.

What do you mean?

A matrix where there are no non-zero elements prior to the first non-zero element on the previous row -- don't they have a special name?

I may be wrong.

Kind of like triangular, but not really.

Yes. A KofT matrix. I like it.

If it doesn't have a name, then how about progressive matrix?

Works for me.

Yeah, can only think of triangular matrices, but their definition is not really the same as this one. Both KofT and progressive sound nice as names :P

Not all triangular matrices are KofT, but all KofT are upper triangular.

@xpqz this looks familiar...

@RGS try it on m←5 5⍴0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 0 0 -- I realise I may have chosen an unfortunate example/unclear explanation earlier. Having each row reflect the one before -- including earlier changes -- presumably will require recurring. I had hoped for

┌→────────┐
↓0 0 1 0 0│
│0 0 0 1 0│
│0 0 0 0 1│
│0 0 0 0 0│
│0 0 0 0 0│
└~────────┘

@rak1507 Are you still looking to avoid exam revision?

@Adám What's the problem? And is it a problem that's likely to be fixed?

well considering I have one tomorrow, I probably shouldn't...

@xpqz Row echelon form?

@Sherlock9 You win a price!

oh, I thought this was for a competition problem

@rak1507 rosalind.info/problems/sseq

that's because it is from a competition problem!

I forget which Rosalind problems double as Dyalog comp problems :/ -- good job no actual solution was posted.

@xpqz ah I see...

I've got a five-day weekend coming up and I don't plan to start Phase II of the competition until then

@RGS close enough to be a major spoiler :/

Relatedly, Ramadan Mubarak to all who celebrate it!

too late to remove now :P the entire chat is full

oh well, thanks for the nice solution guys

Eh, it's not how I'd do it and it'll continue to not be how I do it :D

Ah, nvm, doesn't quite work anyway like rgs said

@rak1507 yeah, I'm just full of empty promises

lol

I'll ... talk about other Rosalind problems instead :)

Let's talk about this one instead rosalind.info/problems/dna

I know how to do this the basic way

I keep forgetting that ⌸ is such a lovely primitive for these type of things.

@Sherlock9 What is "the basic way"?

nested dfns and an each

+/'ACGT'∘.=⊢

@RGS ⌸ isn't particularly nice here as it requires an order and stuff

      {⍵∘{+/⍺=⍵}¨'ACGT'}'AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC'
┌→──────────┐
│20 12 17 21│
└~──────────┘

@rak1507 You can prepend 'ACGT' and then subtract 1, but the ∘.= makes much more sense.

yeah

Oh ∘.=

Nuts

(≢⍤⊢-≢)⌸'ACTG',⊢ fun

Ok I'm so confused

(≢⍤⊢-≢)..?

I get ≢⍤⊢, I just don't understand why the -≢ works

To subtract 1, that is.

bc it subtracts ⍺≢⍵ which is always 1

Yeah that's nasty

:)

lol

((¯1+≢⍤⊢)⌸'ACTG',⊢)

Much better.

an extra byte

But the readability increased by a factor of 24352623452345.6

This is my favourite so far: rosalind.info/problems/orf -- good opportunity for dyadic transposing

might as well use {1-⍨≢⍵}

@xpqz I've not done many of them but rosalind.info/problems/grph is quite fun

@rak1507 Graphs are fun, in general.

@rak1507 I think I did that one. Let's se

(key data) ← FASTA 'rosalind_grph-2.txt'
m ← ∘.{(¯3↑⍺)≡(3↑⍵)} ⍨data
(0 0⍉m) ← 0
r ← (2÷⍨≢pairs) 2⍴pairs ← key[∊⍸m]

looks similar to mine

although a lot simpler for some reason haha

rosalind.info/problems/hamm is just +/≠⌿ :D

@rak1507 or +/≠ if the two strings are the left and right args :P

true

1⊥≠ for extra obscurity

≢9/⍨≠ for extra⍣2⊢ obscurity

Although longer.

⍷⍨∘⊃⍨⍤0 1 lol

frustrating that the rank operator encloses the left argument or something

so ⍷⍤0 1 doesn't do what I expected

@rak1507 Say, what‽

@Adám i assume that's about this (which of course gets much more annoying dyadically). ⍤ asks for ⍤-left and ⍤-right more than ¨ really

Oh, that it doesn't disclose then :-)

well, that's equivalent to effectively enclosing it

Except (2 3⍴⍳6)(≡⍤1) 4 5 6 surely doesn't enclose anything.

well yeah, it doesn't have 0 in ⍵⍵

(⍪(1 2)(3 4)) (∊⍤0 2) (⍪(2 1)(3 4)) surely doesn't enclose anything.

So, I was hacking on the KAP UI, and figured it was time to render arrays in a nicer way. No need to use plain text-based rendering when I have an entire UI framework at my fingertips. Now, has anyone given any thought to a nice way to do this? I mean, I've experimented with using a JavaFX grid widget, which is nice enough (looks a bit like a HTML table). I've also tried using a regular table widget, which makes it look a lot more like a real table (think spreadhseet).

Surely someone has experimented with this? I've been thinking about using a grid for small tables, and moving to a scrollable table widget for larger ones. However, there is then the issue of a small table that contains a large one. That'll look weird.

What about rank>2?

@EliasMårtenson Yes, there is the array editor included with Dyalog.

@Adám I have some ideas, although none of them particularly great.

I do intend to implement a 3D array viewer using OpenGL. And yes, it will be just as useless as you imagine. I haven't written OpenGL code in a long time though, so I will just do it as a joke. :-)

@Adám How do I get to it?

@Adám Is it Windows only?

It is.

I've been wanting to write an array viewer/editor in HTML/JS for a while.

@Adám I was thinking of implementing something like that, but I felt it wouldn't be very useful.

I'm thinking about having a button next to a printed object in the output REPL. If you click on it, a spreadsheet-like editor can be opened up for browsing/editing the content.

Does ⎕ not work for writing to stdout?

I am trying to use it with 'dyalog load=..' and no output is done

maybe i should try using ⎕sh

@rcabaco Are you sure you shouldn't be using LX=?

@Adám didn't know about LX=. I am trying to use ANSI escape control sequences in the terminal.

but ⎕SH'tput clear' does not work either.

@Adám maybe i didn't explain myself correctly. I was thinking that it was possible to load a .aplf with load= and have it print stuff to the terminal. But now I see that it does not work that way.

maybe when scripting is available.