code logs -> 2018 -> Tue, 08 May 2018< code.20180507.log - code.20180509.log >
--- Log opened Tue May 08 00:00:38 2018
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08:32
< simon_>
I've got this probability distribution that I'm modelling as [[a,b],[c,d],[f,g]] where I want to pick a random element in the following way: x <- rand([c,d]); y <- rand([f,g]); return rand([a,b,x,y]). so the distribution can be expressed as a uniform one if I convert it into a flat list where I replicate the elements as necessary. e.g. [[a,b],[c,d],[e,f]] -> [a,a,b,b,c,d,e,f]. but when the second and third li
08:32
< simon_>
sts don't have a nice number of elements, I have to figure out the multiplication factor in a not completely trivial way.
08:34
< simon_>
I figured I'd start by counting the number of slots in total, being length([a,b]) + (has_elements([c,d]) ? 1 : 0) + (has_elements([f,g]) ? 1 : 0).
08:36
< simon_>
if [c,d] and [e,f] have the same number of elements, or if one is empty, the multiplication factor for [a,b] is length([c,d]) (or length([e,f])), e.g. [[a,b,c],[d,e,f],[]] -> [a,a,a,b,b,b,c,c,c,d,e,f]
08:38
< simon_>
(basically, a, b and c have 1/4th and d, e, f each have 1/(4*3), which I'd achieve by saying a, b and c have 3/12ths and d, e and f have 1/12ths. so I can represent the probability as a whole number of occurrences in a list.
08:40
< simon_>
in dumb cases like [[a,b],[c,d,e],[f,g]] (a and b have 1/4th, c, d and e have 1/12th, and f, g have 1/8th), I'd have to multiply the occurrence of a and b with both length([c,d,e]) and length([f,g]), but since those have common denominators, I'm thinking I can do something with GCD to reduce the total length of the list.
08:45
< simon_>
okay, so I'll repeat each element in the first list with length(second_list)*length(third_list)/gcd(length(second_list),length(third_list))
08:47
< simon_>
and I'll need to repeat the elements in the second list and third list with some factor, too.
08:53
< Vorntastic>
Lordy. Okay: first*len(second)*len(third+second*len (third)+third*len(second)
08:53
< Vorntastic>
But!
08:54
< Vorntastic>
You may prefer, because it doesn't explode with large lists...
09:01
< simon_>
I think I'll stick to having the version with three rand() calls in production, because it's quite readable. and perform the to_uniform transformation in my test only.
09:03
< simon_>
Vorntastic, there's a right-paren too little. do I place it as 'len(third+second)*len(third)+...?
09:04
< Vorntastic>
No, before the +
09:06
< Vorntastic>
http://pastebin.starforge.co.uk/12
09:06
< Vorntastic>
One rand call
09:14
< Vorntastic>
With a little work you can make this recurse down but
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10:03
<@Pi>
simon_: Is this in Haskell?
10:07
< simon_>
Pi, no, Perl.
10:07
< simon_>
Vorntastic, and thanks!
10:08
< simon_>
Pi, my pseudo-code just looks like Haskell. :P
10:08
<@Pi>
Okay, can't help you with the Perl then.
10:08
<@macdjord>
simon_: I suggest the folowing: Create the list [a, b, [c,d], [e,f]]. Select randomly from it. If you get a single item, return it; if you get a list, select a random element from /that/ and return it.
10:10
< simon_>
macdjord, I'm doing something very close to that. what I'm trying to right now is convert this to a uniform distribution so I can use the Perl module Statistics::Distributions'es chisqrprob to show that it does, with some overwhelming probability, do what I want it to.
10:10
< Vorntastic>
Which is actually what mine does though I reuse the randomness
10:10
< simon_>
Vorntastic, that's pretty neat.
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10:14
<@Pi>
simon_: In Haskell you could do this pretty generically as something like: concat . sequence $ [['c','d'], ['f','g'], ['a','b','x','y']]
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10:15
<@Pi>
So you'd want the equivalent Perl?
10:18
< Vorntastic>
I think it's been misunderstood there
10:19
< simon_>
yeah, f ["ab", "cd", "ef"] == "aabbcdef"
10:19
< simon_>
and f ["abc", "def", ""] == "aaabbbcccdef"
10:20
< simon_>
the first group has one ticket each. subsequent groups have one ticket they draw for.
10:21
<@Pi>
Oh, so in "x <- rand([c,d]); y <- rand([f,g]); return rand([a,b,x,y])", a and b are static additions, and x and y are the previous selections?
10:22
< simon_>
yes
10:22
< simon_>
and f ["ab", "cde", "fg"] == "aaaaaabbbbbbccddeefffggg" (i.e. four tickets, 'a' and 'b' get as many lots by themselves as c,d,e combinened and as f,g combined.
10:22
<@Pi>
Isn't that distribution off, then?
10:23
< simon_>
off?
10:24
<@Pi>
Meaning with this example: x <- rand([c,d]); y <- rand([f,g]); return rand([a,b,x,y])
10:25
<@Pi>
You end up with a distribution of "aaaabbbbccddffgg" (or "aabbcdfg"
10:26
< simon_>
that's what I'd want here.
10:26
<@Pi>
How do you map the input list to that implementation, exactly?
10:26
<@Pi>
For say ["ab", "cde", "fg"] ?
10:27
<@Pi>
Is the input supposed to be variadic, or always taking a triple?
10:27
< simon_>
I'm not sure I understand. x <- rand([c,d,e]); y <- rand([f,g]); return rand([a,b,x,y])?
10:28
< simon_>
always taking a triple. I don't really need to make it support a variable amount of groups, even though it should extend somewhat trivially.
10:28
<@Pi>
Okay, so are they treated differently?
10:28
<@Pi>
Where are each of the three inputs intended to go in the body?
10:28
< simon_>
yeah. this is some banner selection code. the first ones are main banners. then there are fallback banners. and then there are default banners. the fallback and default banners shouldn't be picked so often (and are only added to the mix if there are too few main ones).
10:29
<@Pi>
So are three inputs intended to be main, fallback, and default?
10:29
< simon_>
yes.
10:30
<@Pi>
As it stands above, the main and fallback ones are treated with equal probability, though?
10:30
< simon_>
umm, no? a and b (main ones) have a higher probability than c and d (fallback ones) since a and x (being either c or d) have the same chance of being picked.
10:31
< simon_>
the fallback and default ones have the same probability if there are equally many of them, but if there are more fallback ones, then each individual fallback one has a lower probability.
10:35
<@Pi>
simon_: Well, unless I'm understanding the pseudo-code wrong, how are the main and fallback ones treated with different priority?
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10:36
<@Pi>
Given: x <- rand([c,d]); y <- rand([f,g]); return rand([a,b,x,y])
10:36
<@Pi>
Or equivalently (?): rand(default + rand(main) + rand(fallback)])
10:36
< simon_>
Pi, [[a,b],[c,d,e],[f,g]] means there are four slots that are picked from; a and b each get one. for the third slot, one of c,d,e is picked, and for the fourth slot, one of f,g is picked.
10:36
<@Pi>
Slots?
10:37
<@Pi>
Okay, how would you express this in terms of the original pseudocode?
10:37
<@Pi>
(Is the original pseudocode still relevant here?)
10:37
< simon_>
sub _choose ($self, $regular_slots, $fallback_slots, $default_slots) {
10:37
< simon_>
my @choices = @$regular_slots;
10:37
< simon_>
push @choices, $fallback_slots->[rand @$fallback_slots] if @$fallback_slots;
10:37
< simon_>
push @choices, $default_slots->[rand @$default_slots] if @$default_slots;
10:37
< simon_>
return $choices[rand @choices];
10:37
< simon_>
}
10:37
< simon_>
this is the Perl code I'm using.
10:39
<@Pi>
Oh, so in the original, [c,d] and [f,g] are the fallback and default, not the main and default?
10:40
< simon_>
yes.
10:40
< simon_>
I might call _choose(..., ['a', 'b'], ['c', 'd', 'e'], ['f', 'g']) and it'll pick 'a' or 'b' with 1/4th probability each, 'c', 'd' or 'e' with 1/12th probability each, or 'f' and 'g' with 1/8th probability each.
10:40
<@Pi>
So in other words, in Pythonic pseudocode, rand(regular + rand(fallback) + rand(default))
10:41
< simon_>
yes.
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10:41
<@TheWatcher>
Idly, "pythonic" sounds like it should be a forbidden arcane script.
10:41
< simon_>
(not considering corner cases where they're empty, etc.)
10:41
<@Pi>
So fallback and default are selected with equal probability.
10:42
< simon_>
TheWatcher, it does sound arcane. tech archaeologists will wonder in 200 years.
10:42
< simon_>
Pi, they're selected with the same criterion, but the probability depends on how many of each there are.
10:42
<@Pi>
Right, I mean the choice between them.
10:42
< simon_>
ok, yeah.
10:44
< simon_>
I think I've got the part where I turn this into a flat list where elements are repeated according to their probability. next is something I've never done: test that drawing elements randomly a number of times produces something that, with some overwhelming probability, resembles the ideal distribution.
10:45
< simon_>
(the reason why I wanted to turn it into a flat list is because Perl's Statistics::Distributions doesn't support non-uniform distributions. I'm not sure a list with repeitions counts as uniform, but I figured the path there would be a lot shorter if I could just do rand([a,a,a,b,b,b,c,d,...]) and count.
10:46
< simon_>
I thought I'd use the opportunity with this assignment to try and do some solid random testing. I can probably re-use what I'm learning here to do some statistical estimations, which would be fun.
10:47
<@Pi>
So in Haskell terms, something like: foo main fallback def = do f' <- fallback; d' <- def; main ++ [f', d']
10:47
<@Pi>
That will give you the list distribution, so:
10:47
<@Pi>
> sort $ foo ['a','b'] ['c','d','e'] ['f','g']
10:47
<@Pi>
"aaaaaabbbbbbccddeefffggg"
10:48
< simon_>
oh god yes.
10:50
< simon_>
much neat.
10:52
<@Pi>
(Haskell geekery: changing the last bit to main <|> pure f' <|> pure d' would make it generic to MonadRandom or other choice / selection implementations, not just listing.)
11:07
<@Pi>
Slightly better (Haskell) implementation:
11:07
<@Pi>
foo main fallbacks = do fs' <- sequence fallbacks; main <|> fs'
11:08
<@Pi>
So that takes the main list, and then a list of fallbacks / defaults, which will be selected between with equal probability.
11:10
<@Pi>
So then:
11:10
<@Pi>
> sort $ foo ['a','b'] [['c','d','e'],['f','g']] == "aaaaaabbbbbbccddeefffggg"
11:10
<@Pi>
> sort $ foo ['a','b'] [['c','d','e']] == "aaabbbcde"
11:10
<@Pi>
> sort $ foo ['a','b'] [] == "ab"
11:10
<@Pi>
And so on.
11:10
< simon_>
yeah.
11:11
< simon_>
I should've thought about Haskell's list monad representing the non-deterministic choices that I'm looking for rand() to pick, rather than try and calculate multiplicity explicitly.
11:13
<@Pi>
The nice thing about the above is that it also works with MonadRandom. :D
11:15
< simon_>
yes, that is nice. I read this blog post on monadic testing, I think it was https://lexi-lambda.github.io/blog/2016/10/03/using-types-to-unit-test-in-haskell/ -- I liked how you can mock things easily with well-scoped type classes.
11:16
< simon_>
so instead of testing some IO () thing, you want to specifically mock filesystem stuff.
11:17
< simon_>
Pi, I think it's the mark of a really good design when you can change the type signature and your program will do something else, but somehow equivalent, in another context. :)
11:18
<@Pi>
Yup.
11:18
<@Pi>
Or not change type signature, but just the concrete types. :)
11:18
< simon_>
for example, if this were Haskell, I might not even really care to run an actual rand() call when testing, since if it works under [], it should work under MonadRandom.
11:18
< simon_>
(i.e. I'm hoping those libraries are tested.)
11:18
<@Pi>
Right.
11:19
< simon_>
Pi, have you ever coded Haskell for money?
11:19
<@Pi>
Nope.
11:19
<@Pi>
(I wish! :D)
11:19
< simon_>
I have a friend who joined this Ethereum startup where he continued on the declarative contract DSL that he wrote a compiler in Haskell for in his BSc project.
11:20
< simon_>
he wants me to join, too, because I have a bit more experience with Haskell. but I think working for a cryptocurrency startup is a bit crazy.
11:20
<@Pi>
Neat.
11:20
<@Pi>
Well, work is work, at the end of the day.
11:21
< simon_>
right.
11:21
< simon_>
I don't mind the Perl.
11:21
< simon_>
or the C#. or the whatever-they-want.
11:21
<@Pi>
Perl is one of the few languages I'll draw a line at.
11:21
< simon_>
and TBH, I'm a little afraid that Haskell will make me tinker on unimportant details too much.
11:21
< simon_>
hehe.
11:22
<@Pi>
I read the Camel book cover to cover as a youth (my first O'Reilly programming book!), and found it so distasteful that I never actually wrote any Perl.
11:24
<@TheWatcher>
The Camel book's approach to perl is... not very good.
11:25
<@TheWatcher>
As a reference, it's useful. As a way to learn the language in a way that doesn't result in godsawful code, it's a mess.
11:26
<@Pi>
I read some other things too, but that didn't help. :)
11:26
<@Pi>
I discovered Python right after that.
11:28
<@TheWatcher>
Each to their own, if we all liked the same stuff, life'd be boring.
11:31
< simon_>
I weirdly liked Perl. it helped at the job interview. :P
11:32
< simon_>
also, I skipped most of the training. but for some reason, some number of new coders who apply here do that.
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14:12 * abudhabi does something he should've done ages ago: Removing all non-essential printers from the system.
14:14
<@TheWatcher>
Remember, when you remove a printer, it must be crushed, burnt, and buried under a crossroad to prevent the unholy, demonic revenants within from escaping
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16:34
< Emmy>
amen
16:35
< Emmy>
it should be mentioned, that according to the good book scanners fall in the same category of evil
16:35
< Emmy>
often one can find them together performing their devilry and satanism
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18:51
<&[R]>
$ ls /usr/local/etc/php.ini | awk '{print $1}' | xargs grep memory_limit # WTF
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18:53
<@TheWatcher>
"ls: cannot access '/usr/local/etc/php.ini': No such file or directory" ~
19:00
<&[R]>
It's the same as `grep memory_limit /usr/local/etc/php.ini /dev/null`
19:00
<&[R]>
(Note the /dev/null is to make it prepend the filename)
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19:13
<@gnolam>
https://twitter.com/p01arst0rm/status/992061430995324934
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22:40 macdjord|slep [macdjord@Nightstar-a1fj2k.mc.videotron.ca] has joined #code
22:40 mode/#code [+o macdjord|slep] by ChanServ
22:42 macdjord [macdjord@Nightstar-a1fj2k.mc.videotron.ca] has joined #code
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22:44 mode/#code [+o mac] by ChanServ
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22:46 macdjord [macdjord@Nightstar-a1fj2k.mc.videotron.ca] has quit [Ping timeout: 121 seconds]
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23:15 Vorntastic [uid293981@Nightstar-6br85t.irccloud.com] has quit [[NS] Quit: Connection closed for inactivity]
23:30 Emmy [Emmy@Nightstar-9p7hb1.direct-adsl.nl] has quit [Ping timeout: 121 seconds]
23:41 Kindamoody|afk is now known as Kindamoody
--- Log closed Wed May 09 00:00:39 2018
code logs -> 2018 -> Tue, 08 May 2018< code.20180507.log - code.20180509.log >

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