18

Ostatnio nie było dość szereg pytań dotyczących rekreacji z xkcd -Style wykresy jak na poniższym jednego ( 1064 ) w różnych językach .

Twoim celem jest odtworzenie istniejącego komiksu xkcd tak blisko oryginału, jak to możliwe. Komiks musi mieć co najmniej jeden prosty wykres lub wykres kołowy lub podobny element.

Najwyższa głosowana odpowiedź wygrywa , nie wcześniej niż 10 dni po pierwszej ważnej odpowiedzi.

Wpis musi zawierać kod wyświetlający komiks xkcd, jak opisano powyżej, w języku graficznym, co najmniej częściowo wykorzystujący jego narzędzie do kreślenia (Mathematica, MatLab itp.)

komiks nie może być przechowywany jako obraz, musi zostać wygenerowany proceduralnie
nie może być żadnych linków do oryginalnej strony internetowej, jej kopii lustrzanej ani jej zawartości na dysku twardym itp.

Głosy powinny oceniać podobieństwo wyników do oryginału oraz, w mniejszym stopniu, piękno rozwiązania. Nie ma tu golfa kodowego, ale spodziewane jest rozwiązanie tak krótkie i proste, jak to możliwe.

popularity-contest vsz
źródło

3

Większość głosów nie jest odpowiednim obiektywnym warunkiem wygranej - a w jaki sposób głosy boolowskie mają na celu uwzględnienie wielu elementów w różnym zakresie? Myślę, że to pytanie może mieć obiektywny warunek zwycięstwa oparty na średnim błędzie na piksel podzielonym przez długość kodu lub coś podobnego, ale w obecnej formie zachęca ludzi do kopiowania i wklejania odpowiedzi, które były mocno oceniane w innych stosach.

Peter Taylor

1

@PeterTaylor: dlatego określiłem, że musi to być istniejący komiks xkcd, więc nie będzie kopiowania. Większość entuzjastów to powszechnie stosowane kryteria (wraz z „pierwszym rozwiązaniem”, które tak bardzo mi się nie podoba), w przeciwnym razie rzadko można znaleźć tutaj problemy, które nie są kodowaniem (i nie wygrane przez Golfscript)

vsz

8

Oto dźgnięcie.

Postscriptum

[ Liczba usunięta, ponieważ nie jest to pytanie dotyczące golfa. ]

xkcd1064

Dodano sekcję „styl”, aby wyglądała bardziej ręcznie.

xkcd1064, więcej „ręcznie rysowanych”

Edycja: Helvetica reaguje lepiej niż Kurier. Poprawiono i ulepszyłem tę sekcję „stylu”. I dodał komentarze.

xkcd1064, jeszcze bardziej „ręcznie rysowane” niż wcześniej

%!

%futz with a coordinate pair
/f { 2 { rand 100 mod 100 div 1.47 mul
        dup .4 mul sub add
        exch } repeat } def

/op 2 array def %original point
/cp 2 array def %current point
/setcp { 2 copy cp astore pop } def %set current point
/difcp { cp aload pop %x' y' x y   %diff between cp and point on stack
    3 2 roll exch sub %x' x y'-y
    3 1 roll sub %y'-y x'-x
    exch %dx dy
} def
/scale2 { /n exch def  %scale two values
    n mul exch n mul exch
} def
/add2 {
    3 2 roll add
    3 1 roll add exch
} def
%futz with a path
/fpath {
    2 {
    %flattenpath
    % replace lines with curves
    [ { setcp cp op copy pop /moveto cvx }
        {
            %cp aload pop 4 2 roll 2 copy
            2 copy difcp %x3 y3 x30 y30
            2 copy 1 4 div scale2 %x3 y3 x30 y30 (1/3)x30 (1/3)y30
            cp aload pop add2 %x3 y3 x30 y30 x1 y1
            4 2 roll 3 4 div scale2 %x3 y3 x1 y1 (2/3)x30 (2/3)y30
            cp aload pop add2 %x3 y3 x1 y1 x2 y3
            6 4 roll %x1 y1 x2 y2 x3 y3
            setcp
            /curveto cvx }
        { setcp /curveto cvx }
        { op cp copy pop /closepath cvx } pathforall ] cvx newpath exec
    % chop the curves
    flattenpath
    % futz all the points
    [ { f
        /moveto cvx }
        { f
            /lineto cvx }
        { %f
            f 6 2 roll
            f 6 2 roll
            f 6 2 roll
            2 copy
            /curveto cvx
            3 1 roll
            3{f}repeat /lineto cvx % extra triple-futz'd line after each curve
        }
        { /closepath cvx } pathforall ] cvx newpath exec
    } repeat
} def

%futz with strokes
/oldstroke /stroke load def
/stroke { fpath oldstroke } def

%futz with fills
/oldfill /fill load def
/fill { fpath oldfill } def

%make sure rectstroke doesn't bypass being futzed with
/rectstroke {
    4 2 roll moveto
    1 index 0 rlineto
    0 exch rlineto
    neg 0 rlineto
    closepath
    stroke
} def

%futz with strings
%by making sure show uses our futz'd fill
/show { gsave currentpoint newpath moveto dup
    false charpath 1{fpath}repeat fill grestore stringwidth rmoveto } def

%fatter lines
1 setlinejoin %round joins
currentlinewidth 1.9 mul setlinewidth
%chop curves very small so there's lots of points for futzing
currentflat 6 div setflat

/Helvetica 10 selectfont

<<
/in{72 mul}
>>begin
1 in 1 in translate

0 0 6 in 4 in rectstroke
1 in 3 in moveto (WALKING BACK TO MY) show
1 in 2.8 in moveto (FRONT DOOR AT NIGHT:) show

36 3.5 in moveto
0 -2.8 in rlineto
5 in 0 rlineto
3.45 in .55 in moveto
0 20 rlineto
4.05 in .55 in moveto
0 20 rlineto
2.7 in .40 in moveto
(YARD  STEPS DOOR  INSIDE ) show
0 .05 in rmoveto
.4 in 0 rlineto
currentpoint
stroke
moveto
-10 3 rlineto
0 -6 rlineto fill

2.6 in .45 in moveto
-2 in 0 rlineto
currentpoint
stroke
moveto
10 3 rlineto
0 -6 rlineto fill

(FEAR) .9 in 2.3 in moveto show
(THAT THERE'S) .9 in 2.1 in moveto show
(SOMETHING) .9 in 1.9 in moveto show
(BEHIND ME) .9 in 1.7 in moveto show
1.8 in 1.86 in moveto
2 -26 rlineto
stroke

(FORWARD) 2.2 in 2.1 in moveto show
(SPEED) 2.2 in 1.9 in moveto show
2.75 in 2 in moveto
3 -13 rlineto
stroke

(EMBARRASSMENT) 4.5 in 2.5 in moveto show
4.6 in 2.65 in moveto
-2 20
-5 22
-27 25 rcurveto
stroke

.6 setgray %gray
40 1.6 in moveto
2.7 in 0
2.9 in 20
3.2 in 1.5 in rcurveto
10 20
20 20
30 0 rcurveto
15 -1.7 in
30 -1.8 in
1.4 in -2 in rcurveto
stroke

0 0 1 setrgbcolor %blue
40 1.2 in moveto
.9 in 27
.8 in 20
1.4 in 21 rcurveto
1 in -20
1.2 in 6
1.3 in 1.6 in rcurveto
    10 20
    20 20
    30 0 rcurveto
15 -1.8 in
30 -1.9 in
1.8 in -2.1 in rcurveto
stroke

1 0 0 setrgbcolor %red
40 .9 in moveto
3.3 in 0
3.5 in 20
3.6 in 1.5 in rcurveto
10 72
20 80
30 80 rcurveto
.7 in -10
.3 in -35
1.1 in -40 rcurveto
stroke

%thank you for scrolling all the way down to here. :)

luser droog
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czy możesz dodać do tego trochę „ręcznie rysowanej” stylizacji? :)

Sam Axe

Tak. Na początku całkowicie przeoczyłem ten aspekt. A potem, kiedy podążyłem za linkami, byłem całkowicie przytłoczony. Grube linie nie powinny być biggie. Mam od Casselmana metodę reinterpretacji ścieżki. Więc jeśli rozumiem te transformacje matematyczne, powinienem być w stanie je odtworzyć. O wiele większe zadanie niż początkowo myślałem.

luser droog

Nawet bez tych cech bardzo dobra robota.

Sam Axe

I świetna robota w stylizacji!

Sam Axe

7

HTML5

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  font-family: 'Humor Sans';
  font-style: normal;
  font-weight: 400;
  src: local('Humor Sans'), local('Humor Sans-Regular'), 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<title>Front Door ( from xkcd.com 1064 )</title>
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<body><h1>Front Door <div class="h1d">( from xkcd.com 1064 )</div></h1>
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    <tspan x="100.15588" y="48.485607" id="tspan7010">WALKING BACK TO MY</tspan>
    <tspan x="100.15588" y="65.285591" id="tspan7012">FRONT DOOR AT NIGHT:</tspan></text>
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    <tspan x="0.23551519" y="0.22564805">ThAT there's</tspan>
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Rozwiń fragment kodu

F. Hauri
źródło

4

Pyton

Wykorzystuje metodę nieznacznie oszukującą xkcd w matplotlib:

import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage.filters import gaussian_filter1d

plt.xkcd()

x = np.linspace(0,10,1000)
unit = np.linspace(0,1,100)

speed = np.r_[np.ones(500)*3,3+5*unit**4.,np.ones(100)*8,1+7*unit[::-1]**2,np.ones(200)]
speed = gaussian_filter1d(speed,18)

fear = np.r_[np.linspace(0.5,2,450),2+6*unit**4,np.ones(100)*8,8*unit[::-1]**2,np.zeros(250)+0.1]
fear = gaussian_filter1d(fear,18)
fear_gauss_bump = 0.8*np.exp(-np.power(x-2.5,2.)/(2*np.power(1, 2.)))
fear += fear_gauss_bump

embarrassment = np.r_[np.zeros(600)+0.1, 9.5*unit**2, 9.5-np.linspace(0,1.5,300)]
embarrassment = gaussian_filter1d(embarrassment,60)

plt.figure(figsize=(12,6))
plt.plot(x,speed,c="gray")
plt.plot(x,fear,c="#1E9FDB")
plt.plot(x,embarrassment,c="r")

plt.plot([2.5,2.7],[3.9,2.3],c='k')
plt.plot([4.3,4.5],[3.6,3.1],c='k')
plt.plot(np.linspace(7.3,7.8,15),(0.25-(np.linspace(7.3,7.8,15)-7.3)**2)**0.5+7,c='k')

ax = plt.gca()
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.get_xaxis().tick_bottom()
ax.get_yaxis().tick_left()
#plt.ylim([0,10])
ax.set_xticks([6,7])
ax.set_xticklabels(["STEPS","DOOR"])
ax.set_yticks([])

ax.text(2,4,"FEAR\nTHAT THERE'S\nSOMETHING\nBEHIND ME",horizontalalignment='center')
ax.text(3.9,3.4,"FORWARD\nSPEED",horizontalalignment='center')
ax.text(7.4,6.6,"EMBARRASSMENT")
ax.text(1,8,"WALKING BACK TO MY\nFRONT DOOR AT NIGHT:",bbox={'facecolor':'none', 'edgecolor':'black'})
ax.annotate("YARD", xy=(0.5,-0.3), xytext=(4.5,-0.4),annotation_clip=False,arrowprops={'facecolor':'black', 'shrink':0.01, 'width':0.5})
ax.annotate("INSIDE", xy=(10,-0.3), xytext=(7.5,-0.4),annotation_clip=False,arrowprops={'facecolor':'black', 'shrink':0.01, 'width':0.5})

plt.show()

Kieran Hunt
źródło

Odtwarzaj proceduralnie wykres xkcd za pomocą ulubionego narzędzia do drukowania

Odpowiedzi:

Postscriptum

HTML5

Pyton