MaplePrimes - answers and comments on Question, external image as background in 2d plot

Background image

jzivku — Tue, 09 Mar 2010 23:54:08 Z

You can import images into Maple to view, but they can't be used as a background for a plot.
I apologize for this inconenience.

Jonny
Maplesoft Technical Support Analyst

listdensityplot

Robert Israel — Wed, 10 Mar 2010 13:14:34 Z

For a 2-d plot (of a monochrome image), you can try listdensityplot:

> plots:-listdensityplot(immy,style=patchnogrid);

(you may want to rotate that 90 degrees clockwise).

Unfortunately listdensityplot doesn't allow an m * n * 3 Array which you would get from a colour image, but you could use listdensityplot on one of the colour components.

Also this might not work very well on big images, especially in Standard GUI.

Thus:

> immy:=ImageTools:-Read(
       "c:/Program Files/Maple 13.02/data/images/tree.jpg"):
   reds:= immy(..,..,1);
   plottools[rotate](plots[listdensityplot](reds,
      style=patchnogrid), -Pi/2);

How we can change the

herclau — Wed, 10 Mar 2010 18:56:54 Z

How we can change the background plot, as shown in this picture?

Gracias

background images (2D and 3D)

pagan — Wed, 10 Mar 2010 20:44:00 Z

Here's the code to my original reply (the one that got accidentally replaced by Thomas's reply entitled "kernelopts(datadir)".

It's not fast, but I'm sure it could be made faster, and smarter. The basic idea is that an image, once read into Maple, can be put up as either a 3D or a 2D plot.

First the 3D plot. (Thanks, Thomas, for the datadir tip.)

restart:
a,b:=-10,10:
P1:=plots:-spacecurve([sin(t),t,10],t=a..b,color=black):
#immy:=ImageTools:-Read("/usr/local/maple10.03/data/images/phone.jpg"):
immy:=ImageTools:-Read(cat(kernelopts(datadir),"/images/phone.jpg")):
m,n:=op(map2(op,2,[rtable_dims(immy)])):
P2a:=ImageTools:-Preview(immy):
P2:=plottools:-translate(P2a,-floor(m/2),-floor(n/2),1):
P2:=plottools:-scale(P2,(b-a)/m,(b-a)/n,1):
plots:-display([P2,P1],orientation=[0.0,0.1]);

Next, the 2D plot, using P2a from above.

Z:=op(2,op(4,op(1,P2a))):
Zc:=Array(1..265*265,1..3,(i,j)->Z[265-max(1,iquo(i,265))+1,265-irem(i,265),j],datatype=float[8]):
M:=Matrix(265*265,2,datatype=float[8]):
V:=Vector(265,i->265-i+1,datatype=float[8]):
Vc:=Vector(265,datatype=float[8]):
for i from 1 to 265 do
  ArrayTools:-Fill(265,i,Vc,0,1);
  M[(i-1)*265+1..(i-1)*265+265,2]:=Vc:
  M[(i-1)*265+1..(i-1)*265+265,1]:=V:
end do:
P3a:=plots:-pointplot(convert(M,listlist),color=[seq(COLOR(RGB,seq(Zc[i][k],k=1..3)),i=1..265^2)]):
P3:=plottools:-translate(P3a,-floor(m/2),-floor(n/2)):
P3:=plottools:-scale(P3,(b-a)/m,(b-a)/n):
P4:=plot(sin(x),x=a..b,color=black):
plots:-display([P3,P4]);

Anyway, it can be done, is the essence.

Background image

Axel Vogt — Wed, 10 Mar 2010 00:11:44 Z

So we can not have Fat Freddy's cat looking at a Riemannian surface ?

Background colour

Robert Israel — Thu, 11 Mar 2010 02:05:42 Z

The simplest way is to display the plot together with a solid rectangle of the appropriate size. Regions in a display are rendered "front to back", so make
the background rectangle the last item in the display (this will only matter, I think, if there are filled regions in your plot). Thus:

> P1:= plot(whatever...): 
# say for this plot x goes from x1 to x2 and y from y1 to y2 
  Background:= plottools[rectangle]([x1,y1],[x2,y2],colour=cyan):
  plots:-display([P1, Background]);