Re: [Matplotlib-users] XKCD style graphs?

[Jason G. <jas...@cr...>]

On 10/4/12 4:02 AM, Pierre Haessig wrote:
> Hi Fernando,
>
> Le 04/10/2012 09:16, Fernando Perez a écrit :
>> This would make for an awesome couple of examples for the gallery, the
>> mathematica solutions look really pretty cool:
>>
>> http://mathematica.stackexchange.com/questions/11350/xkcd-style-graphs
> I've never used Mathematica so that it's pretty difficult for me to
> understand the following lines of code which I guess do the main job of
> distorting the image
>
> xkcdDistort[p_] := Module[{r, ix, iy},
>     r = ImagePad[Rasterize@p, 10, Padding -> White];
>     {ix, iy} =
>      Table[RandomImage[{-1, 1}, ImageDimensions@r]~ImageConvolve~
>        GaussianMatrix[10], {2}];
>     ImagePad[ImageTransformation[r,
>       # + 15 {ImageValue[ix, #], ImageValue[iy, #]} &, DataRange ->
> Full], -5]];
>
>
> Is there somebody there that can describe this algorithm with words
> (English or Python ;-)) ?

f@r means f(r)

a~ImageConvolve~b means ImageConvolve(a,b)  (~ treats an operator as infix)

Table[..., {2}] means [... for i in range(2)]

#+1& is a lambda function lambda x: x+1

So I think it goes something like:

def xkcdDistort(p):
     r = ImagePad(Rasterize(p), 10, Padding='White')
     (ix, iy) = [ImageConvolve(RandomImage([-1,1], ImageDimensions(r)),
                               GaussianMatrix(10))
                 for i in range(2)]
     return ImagePad(ImageTransformation(r,
                 lambda coord: (coord[0]+15*ImageValue(ix, coord),
                                coord[1]+15*ImageValue(iy, coord)),
                  DataRange='Full'),
               -5)


Thanks,

Jason