I have an eps file looking like this:

enter image description here

I have no access to the original dataset, so replotting it is not an option. I would like to replace the ugly 80s style coordinate system (Both the captions and the lines) with a coordinate system generated by pgfplots. I have seen somewhere that this is possible, but I don't remember where and how. I can remove the captions and lines with Illustrator so that I only have the plot.

Could someone please point me to a source where it's explained how to achieve this?

I got it to run with example data from the PGFPLOTS manual. However, when I enter my coordinates:

(0,0,0) => (36.5,64.6-64.6)
(0,10,0) => (0,64.6-36.5)
(10,0,0) => (93.4,64.6-47.8)
(5,5,1) => (44.3,64.6-40.6)

I get the following error message:

Error: Sorry, 'scale uniformly' failed bec
ause its actual implementation works only if y_x = 0 (for 2d axes) or if z_x = 
0 (for 3d axes). The result will not fill the prescribed dimensions. For 2d axe
s, you may also want to consider the 'unit vector ratio' key.

What is it trying to tell me? I don't understand this error message.







    grid=both,minor tick num=1,
    xlabel=$x$,ylabel=$y$,zlabel=Local Dose
\addplot3 graphics[
    points={%  important
        (0,1,0) => (0,207-112)
        (1,0,0) => (446,207-133)
        (0.5546,0.5042,1.825) => (236,207)
        (0,0,0) => (194,207-202)
    }] {data.png};


enter image description here

  • 1
    It's the syntax \addplot3 graphics..... An example : tex.stackexchange.com/questions/78855/… See the manual for that usage. – percusse Mar 23 '13 at 10:52
  • Please add a minimal working example (MWE), otherwise this is hard to guess which options you used. In principle your complete use case is in detail described in the pgfplots manual from p. 46 on. – Alexander Mar 23 '13 at 11:46
  • Have you can checked the inside of the eps-file? There might be a small chance that the elevation matrix is still available in readable form inside it, if the plot was built as a postscript program. – g.kov Mar 23 '13 at 11:57
  • @g.kov Not the case... :( – Eekhoorn Mar 23 '13 at 12:42
  • If you open the .log file of your example, you will see the axis parameters that fitted your points argument. In your case, z={(-2.4000000000,1.54999000000000)} which appears to be wrong. And that is what the error message tries to tell you: the x component of the z vector is non-zero. I fear that somehow, the input points are not what you wanted. – Christian Feuersänger Mar 23 '13 at 19:54

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