Julia

Dennis

03:57

How do I turn a nested array into multi-dimensional one?

2

Alex A.

Oh my god so much lol

(note that when I say "lol" it means I'm literally laughing out loud)

This is incredible

@Dennis That's actually a really good question to which I do not have an immediate answer.

Okay

So

Dennis

hcat(A...) seems to do the trick.

Alex A.

Oh really?

damn

That's way easier than what I just came up with

Dennis

Potentially follwed by ', which is what I was going for.

Alex A.

I was going to do a reshape reduce vcat

Yours is 800% golfier

Probably a hell of a lot more efficient too

WTF WHY HAVE I BEEN DOING reduce vcat WHEN vcat... WORKS?!

Dennis

04:13

¯\_(ツ)_/¯

Alex A.

My life is a lie

@Dennis Good use case: sum(isodd,x) is shorter than sum(i->i%2>0,x)

Dennis

sum(x%2)

Alex A.

Okay so my example isn't perfect

I actually didn't realize that % applied elementwise

Dennis

It does. sum(x&1) would also work.

Alex A.

Oh nice!

Well, if I were writing production code, I would get the number of odd elements in an array using something like count(isodd, x) because it should be both readable and reasonably efficient.

So that's a good practical use case for iseven/isodd

Dennis

04:23

I guess...

I've tried overloading the - operator, but it doesn't seem to do anything.

Alex A.

??

What do you mean it doesn't do anything?

Dennis

Oh, I need to specify the types... Bah.

Alex A.

Oh, yes.

-(a::Type1, b::Type2) = some cool stuff

Dennis

Hm, that didn't work like I wanted it to...

julia> -(x::Int,y::Int)=x+y
- (generic function with 204 methods)

julia> diff([1, 2, 3])
ERROR: BoundsError: attempt to access 3-element Array{Int64,1}:

 in diff at linalg/generic.jl:47SYSTEM: show(lasterr) caused an error

Alex A.

Good god why would you redefine subtracting integers as addition?!

Dennis

04:29

To abuse diff to do things pairwise.

Alex A.

Then you should overload diff, not -...

Dennis

More bytes.

Alex A.

Overloading a function in a code golf submission is going to be pretty long regardless because you have to specify the types

If the type has a long name you can do -{T<:YourLongTypeName}(a::T,b::T) which is a tad shorter

Have you looked at the source for diff?

There's this: github.com/JuliaLang/julia/blob/…

Looks like that's what's getting called

Since diff is literally just [a[i+1]-a[i]for i=1:endof(a)-1], it's probably shorter to just reimplement it entirely to do what you want than trying to overload things.

Dennis

Probably...

Alex A.

Are you still trying to do the Char substraction business?

Dennis

04:39

No, not anymore.

Alex A.

Oh, okay

Dennis

04:59

@AlexA. It worked. \o/

0

A: Are there More Hard Objects or Soft Objects

Julia, 99 bytes g=t->2t+[t[2:end,:];0t[1,:]] x->("Hard","Equal","Soft")[cmp(sum(g(g(x.>' ')')[:]∩(4,7)-5.5),0)+2] The anonymous function expects a two-dimensional Char array as argument. Verification julia> g=t->2t+[t[2:end,:];0t[1,:]] (anonymous function) julia> f=x->("Hard","Equal","Soft"...

Alex A.

+1

I was going to suggest g(t)= instead of g=t-> but I guess it's the same length

g(t)= will be faster though

Dennis

I'd rather define g inside f (seems cleaner), but horizontal scrolling.

Alex A.

Haha yeah

Dennis

Oh, how come?

Alex A.

Lambdas are currently slow compared to "regular" functions

I guess there's a package called FastAnonymous that has an @anon macro that makes lambdas as fast as if the function were inlined and I think they're incorporating some of the speedups into Base eventually

Dennis

05:05

Considering how often lambda are used in sum and such, that sounds like a good idea.

Alex A.

Yep

Dennis

Wait, Julia has sign, right?

Alex A.

Yes

Sp3000

o/

Alex A.

Hey!

Dennis

05:06

Hi!

Sp3000

Alex would cringe, but I wonder about "ESHqoaufratdl"[...:3:]

Alex A.

And cringed I have

How does that even what

Sp3000

Only problem, of course, is that Equal is the first one :/

Dennis

The shortest I found was x->"ESHqoaufratdl"[(sign(sum(g(g(x.>' ')')∩(4,7)-5.5))+3)%3+1:3:end], which is longer...

Alex A.

Ohhhhhhhhh I thought ... was supposed to be a splat

Dennis

05:18

s/end/13

Alex A.

That's why I was confused

Sp3000

I meant to leave out the end or 13, hence the cringe :P

Hmm I should use something else instead of ... for Julia I guess

Alex A.

That works?

Sp3000

<blah> maybe

Deprecated

Hence cringe

Alex A.

Doesn't work at all for me

> ERROR: syntax: missing last argument in "1:3:" range expression

Dennis

05:19

It's a syntax error in 0.4.5...

Sp3000

0.3.11, that's probably why

Dennis

ninja'd

I should get my hands on an older version.

Sp3000

Hmm I thought I had 0.4, nvm

Alex A.

You should definitely move up to 0.4

Granted, a lot of things that work fine in 0.3 are deprecated in 0.4, such as int as a synonym for parse(Int,)

And with any luck those will be dead and buried in 0.5

yesterday, by Alex A.

ಠ_ಠ

Sp3000

So I think I just had multiple Julias but PATH was set to 0.3.11

Anyway...

Alex A.

05:27

rm -rf <path/to/julia-0.3.11>

Out with the old

>:D

Sp3000

Nah, will keep for testing purposes, will just change PATH

Alex A.

One of the fun things about answering some Julia questions on SO is that the issue would be that they were using a dev version and the issue went away when they used a stable release. :P

Sp3000

Fun game idea: Let's golf Alex's Julia submissions!

Alex A.

Haha

Surely most if not all are suboptimal

Dennis

I'll start by searching for parse(Int,...) and replacing it with int(...).

Alex A.

05:35

10 mins ago, by Alex A.

yesterday, by Alex A.

ಠ_ಠ

Whenever I golf in Julia I just think of all deprecated features as already removed.

Dennis

All is fair in love and code golf.

2

Alex A.

I found an old answer of mine that uses int and I almost want to change it to use parse just to see your reaction.

Dennis

/me deletes as not a serious contender

Alex A.

eleven

Sp3000

@AlexA. (x=readline()[1]>48)?while x print(1)end:print(0)

Actually that's probably not even necessary, is it

... why did you store x again?

readline()>"0"?while 2>1 print(1)end:print(0)

Alex A.

05:56

@Sp3000 I was young and naïve

I just golfed this one

Dennis

Technically, T(x)=(print(x);x>"0"&&T(x));T(readline()) would work as well.

Sp3000

Does (a=~a) need parens?

@Dennis Hah, I was wondering about that :P nice

Alex A.

@Sp3000 I think so but I'll check

Sp3000

I think you can drop about 4 pairs of parens from that

Alex A.

o_O

@Sp3000 status-completed Thanks! :D

Sp3000

06:03

Why not Dennis'? :P

Alex A.

Oh I didn't see it

@Dennis This is nuts

Sp3000

Conclusion: Alex was indeed very young and naive

Alex A.

The ones are on newlines and it causes a stack overflow, and instead of 0 it prints false.

Nvm

It prints 0 forever too

Sp3000

You need to pipe in just "0", no trailing newline

Alex A.

Oh

Sp3000

06:08

If that bothers you, you can do x<"1"||T(x)

Dennis

x<"1" should work with and without newline.

ninja'd

Alex A.

> when the input is 1, it must continually print 1s and only stop if the program is killed or runs out of memory

Would a stack overflow constitute running out of memory?

Sp3000

Hmm... good point

Dennis

Not sure.

x=int(readline());while print(x)!=1÷x end is almost as short as the infinite recursion.

Alex A.

> int

:P

Dennis

06:16

Fine.

x=readline();while print(x)!=1÷(x>"0")end

Alex A.

XD did you have a backup solution in case I refused to use int?

Dennis

Was already working on it. :P

Alex A.

That prints infinite zeros

Dennis

No LFs plz.

Alex A.

orite

am stoop

Still does it

Dennis

06:18

How are you calling it?

Alex A.

echo "0" | julia untitled.jl

Dennis

You need echo -n.

Alex A.

What does that do?

Dennis

echo still appends a LF by default.

Alex A.

Oh, I didn't realize that. I guess that makes sense.

Dennis

06:20

printf 0 is more portable.

Alex A.

I get a division error for 0

Sp3000

That's the point :P

Alex A.

I didn't actually read the code before running it >_>

Dennis

Try this: rm -rf ...

Alex A.

~~Thank you for not sending me malware~~ ಠ_ಠ

Dennis

06:21

ninja'd?

Sp3000

Too bad you can't char mod 2

Alex A.

You mean use % with a Char argument?

Like 'a'%2?

Dennis

Oh, but % works instead of ÷.

Sp3000

I was hoping '0'%2\1 yeah

Oh wait, float div, nvm

Silly Inf

Alex A.

:P

@Sp3000 You were absolutely right. Thanks!

Maybe it was an older version where you needed parens in ternaries? Idk. For some reason I thought they were required.

Maybe they never were and I just always sucked ;-;

Sp3000

06:30

Go golf all your past ternaries :D (dunno if you have any)

Alex A.

Woo, just got a vote on Truth Machine and now I'm at exactly 19k rep. The consecutive zeros are so satisfying. :D

@Sp3000 To be clear, I thought that assignments inside of ternaries needed parentheses. That is apparently not the case or is no longer the case. ¯\_(TIL)_/¯

But I'm looking for them anyway to see if I have more stuff like that sitting around

Dennis

How about p->(a=0x0;[a=c==33?~a:c==60?a<<1:c==62?a>>1:c==64?a<<4|a>>4:a for c=p];Int(a))?

Alex A.

O_O

That's beautiful

Dennis

Also, Int(a) -> 1a.

Alex A.

Thank you both so much for all of your golfing help tonight!! I love having Julia buds (who incidentally are both way better at golfing in Julia than I am >_>).

@Dennis Whoa, I wouldn't have thought to do that

Dennis

06:46

One final touch: c==64?a<<4|a>>4:a for c=p -> c!=64?a:a<<4|a>>4for c=p

Alex A.

Ah, it eliminates the space!

Good thinking!

Dennis

@AlexA. That was an accident. I tried to shorten a<<1 to 2a.

Alex A.

¯\_(works)_/¯

Beautiful. Together we've gone from 117 down to 73. :) Granted, it was basically entirely you guys...

So I guess, together you've gone from 117 to 73. :P

I should have gone to bed about an hour or more ago but I got too caught up in the thrill of the golf.

Thanks again @Dennis and @Sp3000! <3 u both

Dennis

p->(a=0x0;[a=[a,~a,a<<1,a>>1,a<<4|a>>4][1+findlast("!<>@",c)]for c=p];1a) is just as long.

findlast and 1-based indexing make this a lot less golfy than it would be...

p->(a=0x0;[a=[~a,a<<1,a>>1,a<<4|a>>4,a][findin("!<>@$c",c)][]for c=p];1a) and p->(a=0x0;[a=[[~a,a<<1,a>>1,a<<4|a>>4][findin("!<>@",c)];a][]for c=p];1a) are also just as long...

Sp3000

11:05

Still a bit suspicious about the need for regex in subsequence, but the fact that "a"[:2] doesn't work hurts

Or "a"[1:1:2] or whatever it is that's equivalent to "a"[:1] in Python

Sp3000

11:41

Re parens for assignment: I think parens are just necessary if you have multiple statements, by the looks of it

Sp3000

12:11

This hurts :(

Dennis

15:41

@Sp3000 I think f(s,a,b,i=0)=(o=join(["$a "[i+1]!=c?c:b[i+=1]for c=s]);i<endof(a)?s:o) works.

Sp3000

I really should get used to the idea I'm not using Python and b[i+=1] works :P

(was trying to ++ before for some reason)

What's endof though?

Dennis

length minus 1 byte.

Sp3000

Ah.

Hmm getting rid of o is one byte longer because of 1-indexing :/

Dennis

Yes, I've tried that as well.

1-based indexing is painful if it isn't modular...

Sp3000

16:22

Posted, thanks :) will see if it can be golfed later, but looks hard

Dennis

18:50

@Sp3000 Martin has clarified that we can take character arrays, so you should be able to get rid of the join.

Martin Büttner

19:46

@Dennis at least Julia has mod1, but that's probably too expensive for most golfing applications where a ring buffer isn't completely necessary.

Dennis

It does? o_O

Martin Büttner

it does

and might get 3-parameter mod with offset, which I posted as a feature request at some point

github.com/JuliaLang/julia/issues/14431

github.com/JuliaLang/julia/issues/15486

Dennis

20:02

Nice.

Dennis

21:09

Argh, why doesn't Julia do some error checking before evaluating s[j]=r?

quartata

21:21

I thought no error checking was usually a good thing for golf :P

Dennis

before evaluating. If j contains a zero, it will start modifying s, then fail when it encounters the zero.

quartata

Oh.

Dennis

For whatever reason, it works like I want it to if j is a row vector...

quartata

so it goes through j element by element and only crashes when it hits the 0? I don't see why that would be an issue unless you want to catch it and keep the unmodified s

fair warning I have no idea what challenge this is for

Dennis

That's precisely what I'm doing.

0

A: Subsequence Substitution

Julia, 62 59 bytes f(s,p,r)=(i=0;try s[[i=findnext(s,c,i+1)for c=p]']=r end;s) I/O is in form of character arrays. Verification julia> f(s,p,r)=(i=0;try s[[i=findnext(s,c,i+1)for c=p]']=r end;s) f (generic function with 1 method) julia> F(s,p,r)=join(f([s...],[p...],[r...])) # string/char a...

quartata

21:34

OK so why not just copy s first?

That doesn't seem like too many extra bytes

Dennis

S=s[:] is a lot of bytes...

Anyway, transposing the column vector to yield a row vector fixed the problem.

quartata

Weird.

Dennis

Yup.

That challenge has three Julia answers now. That ought to be some kind of record...

quartata

haha

Dennis

Cool, blocks return stuff.

If I wanted to, I could rewrite my answer as f(s,p,r,_=try i=0;s[[i=findnext(s,c,i+1)for c=p]']=r end)=s. Not that it helps in any way...

Also, order of evaluation is so very, very weird.

julia> x,i,x[i]=[1,2,3],1,4
([4,2,3],1,4)

That one actually made sense. But this?

julia> x,x[i],i=[1,2,3],4,1
([4,2,3],4,1)

Ha, that weirdness actually saved a byte!

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