The Nineteenth Byte

Alex A.

06:15

I need a Python person. @Mego, are you still around?

Anonymous

@AlexA. You know it

Alex A.

\o/

Anonymous

> implying that I sleep

Alex A.

Say I have a dict where each key and value is a string. But I have a desired mapping from key to value type. What's the most efficient way to do that?

somebody

value type?

Anonymous

06:17

Can you elaborate?

Alex A.

Say I have something like {"id": "1", "date": "2016-01-04"} and I want {"id": 1, "date": <datetime object>}

For each key in the dict, there's a specific type I want the value to have.

Anonymous

Probably iterate over the keys, and select the construction method based on the key

Anonymous

Lemme type up the awful example of how I would probably do it

Alex A.

So something like for key in dict: if key == "id": dict[key] = int(dict[key]), etc?

somebody

84

A: Fastest way to convert a dict's keys & values from `unicode` to `str`?

DATA = { u'spam': u'eggs', u'foo': frozenset([u'Gah!']), u'bar': { u'baz': 97 }, u'list': [u'list', (True, u'Maybe'), set([u'and', u'a', u'set', 1])]} def convert(data): if isinstance(data, basestring): return str(data) elif isinstance(data, collections.Mapping): ...

wait what

can't you just parse as json?

Alex A.

06:21

It's not that the entire dict is a string

Both the keys and the values are strings

But I want the values to be of different types, where the specific type depends on the key

somebody

can't you have

MAPPING = {
    'id': int
    'date': time.strptime
}

Anonymous

ideone.com/4xQLrR

somebody

then do MAPPING[key](value)?

Alex A.

Yeah, I thought about that. Wasn't sure whether that'd be more efficient.

Anonymous

It's probably the most Pythonic way

Anonymous

06:26

Other than "don't get yourself in this situation"

Alex A.

I'm parsing XML, but the raw XML has everything as strings.

Looks like the mapping thing is what you suggest as well

Anonymous

Eww gross

Alex A.

Cool beans

Anonymous

Yeah the mapping is how I'd do it

Anonymous

You could also make it a defaultdict with str as the default value

Anonymous

06:27

@somebody That doesn't work; iter(data) only iterates over keys

Anonymous

You'd need data.items() or data.iteritems()

Alex A.

^ Isn't one of those deprecated or removed in Python 3?

Anonymous

data.items() in 3 is data.iteritems() in 2

Alex A.

Oh, okay

Anonymous

In 3, basically everything returns an iterator object rather than a list

Anonymous

06:29

Which means a lot of itertools stuff got promoted to builtins - 3's map is 2's itertools.imap

Alex A.

Would it be more Pythonic and/or efficient to use get with a default of lambda x: x when I'm applying the mapping?

(I could be remembering Python's get incorrectly; I'm used to Julia's.)

Anonymous

That's how get works, and sort of

Anonymous

The defaultdict is probably what most Pythonista would say :P

Anonymous

With defaultdict (the default arg is ugly as sin though)

Anonymous

The default_factory argument expects a function, and since the values are functions, you have to provide a function that, when called, returns the identity function

Alex A.

06:40

Gross

somebody

/me needs help

Anonymous

Yeah I dunno what could be done to make it cleaner

Alex A.

Your original method seems much cleaner

somebody

i have no idea how to make codegolf.stackexchange.com/questions/tagged/tips into a github pages + jekyll repo

Anonymous

Step 1: don't

Alex A.

06:42

^

somebody

why not?

Anonymous

Why?

somebody

idk

it's a lot easier see tips

not half a page for each suggestion, instead it's one line

not 4 pages of python tips, 2 screens on a laptop/desktop

Alex A.

I don't see how GitHub Pages and Jekyll would solve that

somebody

the markdown should be the easy part

just have a table: tip | before | after

Liam

06:50

since someone was posing about radiation-hardened quines, it is impossible to write one in a language like C right?

Anonymous

Yep

Anonymous

Erase any character in main, gg

Justin

@Mego I'll believe it if you show me a proof :P

Liam

can someone give me a one-liner about how the one in ruby works

(i don't know how ruby works at all)

Justin

link?

Liam

06:52

github.com/mame/radiation-hardened-quine

Anonymous

In C, C++, Java, or any similar language that requires a certain string to be in the program to be valid, it's impossible to write a true radiation-hardened quine

Anonymous

(C# and VB are similarly impossible)

Anonymous

Simple remove a letter from main/Main and you're hosed

Justin

github.com/mame/radiation-hardened-quine/blob/master/rquine.rb

This is a strange smiley face...

I don't understand...

Anonymous

It's the radiation symbol

Anonymous

07:00

Anonymous

See: Beta Decay's old avatar

Justin

..... Now I just feel dumb

Anonymous

It's ok, we all feel that way sometimes

Justin

I don't usually though :-)

El'endia Starman

@Justin Somehow, I think many in this room would think the same of themselves. :P

It's certainly a rarity for me.

Justin

07:10

Even if I ask a dumb question, I don't feel dumb :)

Alex A.

07:22

I feel dumb 24/7. #dumblife

Anonymous

Aww but we <3 you

Alex A.

:3

phase

07:48

Has anyone else known about this site? I love it lobste.rs

Quill

My Pyth install is still borked ;-;

Optimizer

08:19

> My borked is still Pyth install

somebody

@Quill Still borked, my Pyth install is

Quill

Yoda, I am

phase

yoda u is

New Main Posts

09:11

0

Q: A program that prints pi

You have to create a program that prints pi, with 100 decimal places. This is code golf, so the smallest program wins!

New Main Posts

09:28

0

Q: Find median of grouped continuous data

You will be given grouped continuous data, such as the following: Class Frequency 0-10 3 10-20 6 20-30 4 30-40 7 Your job is to find the median using the following formula: l+(N/2-cf)*(h/f) where l = lower limit of median class N = sum of all frequencies cf = cumulative frequency ...

code-golf math statistics

Martin Büttner

10:24

@somebody what's that github organisation?

somebody

to collect the tips from all the 'tips for <language>' threads and displaying them in an easy-to-read format

nobody likes them though, so probably gonna delete it

@Calvin'sHobbies how does env appear? the js sends and empty object to execute()

wait, nvm

Quill

10:47

⁽⁽(੭ꐦ •̀Д•́ )੭*⁾⁾ ᑦᵒᔿᵉ ᵒᐢᵎᵎ

Pyth still is bork

Glad there's a web interpreter

Neil

11:35

when a challenge requires an integer argument, are you required to use a numeric type if you have one, or can you accept it in string form?

Martin Büttner

@somebody ah well, I agree that it can be a bit annoying to read all the tips for a language, but some of them aren't as simple as "long form -> short form", and I don't think there's generally much point in duplicating all the content, especially since it's bound to get outdated.

Neil

11:54

sigh, I had just written an answer for a question when it gets closed

New Main Posts

0

Q: Weight balancing problem

Have you ever wondered, when doing shopping, how to pack your products into 2 bags you're carrying so that a total weight would be distributed among them as evenly as possible? Your task is to write a function, that takes a list of positive numbers (lets say integers for simplicity) as an input....

code-golf

somebody

12:13

@Mego finished porting Shtriped

@Calvin'sHobbies this isn't wotking for me

not printing last row

@Mego whoops, maybe not

@Mego now it's exactly the same as the JS

idk if i should port to anything other than python 2/pypy

@Dennis maybe you could add Shtriped to TIO?

LegionMammal978

12:52

pastebin.com/NC6aizRt

THE KERNEL HAS EVOLVED

RUN FOR YOUR LIVES

New Main Posts

13:31

0

Q: How to fix linker error LNK2019: unresolved external symbol and error LNK2001: unresolved external symbol

I am new to visual studio. I am going to study libxml2 library. So my first example is parsing a xml file. I follow libxml2 official tutorial. http://xmlsoft.org/tutorial/ I set the additional include directory path suitably. But there are build errors still remaining. my cpp file is in http://xm...

c++

Martin Büttner

13:41

> help me.

somebody

@MartinBüttner ?

Martin Büttner

from that off-topic question

somebody

move to SO?

Martin Büttner

I'm not going to migrate it. they can repost it there if they want

somebody

oh

also, help me. = best question ever

somebody

13:58

@Dennis is the bubblegum source available online?

Martin Büttner

@somebody esolangs.org/wiki/Bubblegum

somebody

oh

grc

14:14

ಠ_ಠ

orlp

14:29

( ͡°╭͜ʖ╮͡° )

VoteToClose

Lot of stuff getting put on hold today...

somebody

@VoteToClose ?

LegionMammal978

@VoteToClose Just created a pull request, could you check it?

Optimizer

14:45

@orlp wow french beard

quartata

@Doorknob And sure enough he got it down to 26 just now :P

Doorknob

haha, 8, not bad

VoteToClose

@LegionMammal978 Accepted.

LegionMammal978

It's interesting how a packet can almost instantaneously cross 5000 miles under the Atlantic Ocean and back just to ping a server

Dennis

@somebody Sure. The JavaScript or Python version?

quartata

15:01

There's a JS version? I thought the only implementation was the reference Python one on esolangs

VoteToClose

@LegionMammal978 Internet is awesome. c:

orlp

guys

I have a problem

LegionMammal978

Yes?

VoteToClose

^

orlp

I have a histogram of occurences

e.g. A = 25, B = 3, C = 0, D = 1, E = 10

blablabla

LegionMammal978

15:09

Okay...

orlp

I want to turn this into cumulative probabilities

this is not so hard

you sum

(25 + 3 + 0 + 1 + 10) = 39

LegionMammal978

Then take the ratios

orlp

Indeed, so A = 25/39, B = 28/39, C = 28/39, D = 29/39, E = 39/39

but now comes the hard part

I want the ratios as 15-bit fixed point

where the cumulative probability doesn't go from 0 to 1 as a real number, but from 0 to 0x7fff as an integer

LegionMammal978

*0x8000?

orlp

and the second requirement is that if a probability has non-zero chance, it must not get round down to 0

@LegionMammal978 I should've said from 0 up to and including 0x7fff

the third requirement is that the cumulative fixed point probability represents the histogram as accurately as possible

LegionMammal978

15:12

*0x7fff, then round up?

orlp

and the last requirement is that the cumulative probability must be exactly 0 to 0x7fff

@LegionMammal978 then you will go over 0x7fff

because rounding up makes the total larger

LegionMammal978

No

The maximum is 1

1 * 0x7fff is 0x7fff

Rounded up is still 0x7fff

orlp

oh, then you don't fulfill the second requirement

LegionMammal978

Why?

orlp

since 0.9999 * 0x7fff would round up to 0x7fff

and the last 0.0001 just got rounded to zero

LegionMammal978

15:14

The requirement only talks about a non-zero chance

orlp

yes

if an event has a cumulative probability of 0.9999

then there is an event after it with a non-zero probability less or equal than 0.0001

LegionMammal978

*0x7ffd+1

0x0000 represents 0

0x7ffff represents 1

orlp

... let me illustrate by example

@LegionMammal978 try your approach on probabilities A = 99999/100000 and B = 1/100000

LegionMammal978

f(A) is 0x7fffe and f(B) is 0x0001

orlp

@LegionMammal978 how did you get 0x7fffe?

you rounded up, didn't you?

LegionMammal978

15:17

I'm rounding normally this time

Except that 0x0000 and 0x7fff are reserved for 0 and 1

orlp

@LegionMammal978 then please exactly explain your algorithm

because right now it's just handwaving

LegionMammal978

2 mins ago, by LegionMammal978

*0x7ffd+1

2 mins ago, by LegionMammal978

0x0000 represents 0

orlp

with what rounding?

LegionMammal978

2 mins ago, by LegionMammal978

0x7ffff represents 1

Rounding normally otherwise

orlp

'normal' is not a rounding mode

what rounding?

nearest integer?

LegionMammal978

15:19

yes

orlp

banker's rounding?

ok, here is where your scheme goes wrong

A = 99998/100000, B = 1/100000 and C = 1/100000

that becomes 0x7ffe, 0x1 and 0x1

LegionMammal978

Okay

orlp

total = 0x8000 > 1

starting to see the issue?

LegionMammal978

Kinda

Is there a definite range to the amount of data

orlp

Well, for the problem to be possible the number of symbols must be < 0x7fff

but other than that a division is always possible

the problem is to make the division as representative of the histogram as possible

LegionMammal978

15:22

Can you just use the numerator and then store the denominator separately?

orlp

Nope, the cumulative probability must exactly become a 15-bit integer.

LegionMammal978

What is this for?

orlp

arithmetic coding

LegionMammal978

Very tight requirements

orlp

well, a solution is always simple

assign 1 to every symbol with nonzero probability except the last, which gets whatever is left

the problem is to make the solution represent the histogram as accurately as possible

LegionMammal978

15:26

It's tricky with just 15 bits

But wait

You have 15 bits per symbol!

orlp

@LegionMammal978 nope

the cumulative probability goes from 0 to 0x7fff

LegionMammal978

7 mins ago, by orlp

that becomes 0x7ffe, 0x1 and 0x1

orlp

e.g. to give an example if it were to go from 0 to 10

LegionMammal978

15 * 3 = 45 bits total

orlp

nope

you don't understand

the output has to be

such that the sum of the probabilities is exactly 0x7fff

LegionMammal978

15:28

I know

Just saying that there doesn't have to be a 1-to-1 mapping between words and symbols

orlp

@LegionMammal978 what do you mean with a word?

LegionMammal978

15-bit block

For example, in my old attempted solution, I had 0x7ffe00010001 for A,B,C combined

orlp

no

you had 0x7ffe + 0x1 + 0x1 = 0x8000 combined

LegionMammal978

No, I'm talking about the amount of bits we have

For storage

orlp

why is that relevant?

I never mentioned storage requirements?

LegionMammal978

15:33

True, but better storage utilization means better accuracy

orlp

?

what?

LegionMammal978

Let me think

Wait, do you know if there is a solution?

orlp

Of course

There always is

Assign 0x1 to every nonzero symbol except the last, which gets whatever is left

LegionMammal978

But an accurate one, that is

orlp

it's a bad solution (not very representative of the histogram), but it is one

@LegionMammal978 define 'accurate'

LegionMammal978

15:37

One that can represent differences between values

orlp

That's too vague

LegionMammal978

How?

orlp

I really shouldn't need to explain why that's too vague

LegionMammal978

Your solution makes it seem as if every symbol (except the last) has the same value

mınxomaτ

Feb 12 at 20:06, by mınxomaτ

When you sign up for the wrong exam timeslot and have to study a whole semester worth of stuff in like 50h...

Had this exam today...

orlp

15:40

My solution just shows that there is always a valid solution, then you can start to shuffle probabilities around to better match the histogram

LegionMammal978

I see

Another possible one is to just multiply by 0x7fff, rounding towards 50%

Leave the last symbol out

VoteToClose

@mınxomaτ Feel good about it?

LegionMammal978

as it is just 1- the sum of the others

Actually, there's an idea

mınxomaτ

@VoteToClose I have no idea....

VoteToClose

Well, good luck. :D

LegionMammal978

15:43

Take a set of possible formats, and choose one based off of the last symbol's value

mınxomaτ

10 weeks minimum to grade. Plenty of time to forget :D

orlp

@LegionMammal978 formats?

LegionMammal978

Come up with several solutions

and choose the one that best fits the data

And there's my solution

orlp

Again, too vague man.

Be specific

LegionMammal978

So, say you have your solution, right?

orlp

15:44

The terrible one?

Sure.

LegionMammal978

Let's assume it best fits the data somehow

Now set the first n-1 symbols according to that (so set them to 1)

Say this solution is solution #0x4567

So set the last symbol to that

Because we can calculate its value

orlp

How can I set the 'last' symbol to a solution?

where does the 0x4567 come from?

LegionMammal978

It's your 0x4567'th solution to this problem

Out of your set of solutions

For example,

orlp

that impossible

if you have set the first n-1 symbols

the last symbol is ALWAYS fixed

since the sum must be 0x7fff

LegionMammal978

take A=1/1000, B=1/1000, C=1/1000, and D=9997/1000

And your terrible solution works best somehow

Then you have 0x1, 0x1, 0x1, and 0x4567

orlp

15:49

no

that doesn't sum to 0x7fff

LegionMammal978

We can calculate D from A, B, and C

orlp

so it's not a valid solution

LegionMammal978

But it does

orlp

0x1 + 0x1 + 0x1 + 0x4567 != 0x7fff

LegionMammal978

When decoding, we choose D so that it does sum to 0x7fff

The 0x4567 is metadata

orlp

15:49

.....

there is no 'decoding'

there is no 'metadata'

LegionMammal978

Then when reading the representation

orlp

what 'representation'?

what the hell are you going on about

you simply have to output the cumulative probability using an array of fixed-point 15-bit integers

LegionMammal978

The representation of the data

orlp

there is no 'representation'

you are severely confused

LegionMammal978

Then what have we been doing to the histogram, if not representing it?

orlp

15:52

1 min ago, by orlp

you simply have to output the cumulative probability using an array of fixed-point 15-bit integers

we've not been doing anything to the histogram

it's not a representation

at best it's a model

LegionMammal978

You've been converting it to an array

I'm proposing a different conversion process

orlp

I'm not interested in that

LegionMammal978

Which requirement does it break?

My proposed solution works at least as well as yours

orlp

You still haven't given a solution that is sufficiently precise to verify

LegionMammal978

It's more of a meta-solution

Its effectiveness depends on the 0x8000 regular solutions you choose

If you want a regular one, here:

Set n-1 symbols to 0x1

Set the last one to 0x2345, as it doesn't really matter because we can compute its value from the others anyway

(Of course, that's just being wasteful)

Transcript for