Well, the base function behind "ShowImage" is "Matplot".
You can use a search engine of your choice to find ShowImage.sci. "ShowColorImage" than is using "ShowImage" to create a colored Image. Hope that helps, Philipp 2016-02-09 8:04 GMT+01:00 Antoine Monmayrant <[email protected]>: > > Le Mardi 9 Février 2016 00:04 CET, Philipp Mühlmann <[email protected]> > a écrit: > > > Well, as I understand you want to plot a function over an background > image. > > > > In the case the image pixel number fits the X-Y range of your graph it's > > rather simple. > > Well, it is, unless IPD does not work on your system, which is my case! > None, absolutely none of the image processing toolboxes work on my machine. > And it's not only my machine, it's the case for most of the machines we > have at work. > So I cannot rely on an image processing toolbox to achieve what I want and > I have to rely on base functions. > > Thanks anyway for you help, > > Antoine > > > > > img = ReadImage(IPD_PATH + 'demos\teaset.png'); > > rows = size(img,'r');cols = size(img,'c'); > > img(rows,1,:) = (0); // changing left lower image corner for > > checking purpose > > // change all 3 colour > > channels to get a black pixel > > ShowColorImage(img,''); > > a = gca();a.axes_visible = ["on","on","off"];x = linspace(1,600,600);y > > = 1/2*(x);plot(x,y,'-');a.data_bounds = [0,0;cols,rows]; > > > > Two tricky things though: > > > > 1st: image coordinates (center of pixel) start at "1", hence there is a > gap > > of 0.5 between image borders and the axes > > > > 2nd: it might become tricky if you want to have negative values on the > axis. > > > > Probably there is a way to shift the image to other position within the > > coordinate system? > > > > > > So....not sure if this helps your purpose. > > > > You could also switch "ON" / "OFF" the background image by access the > > children of the figure, such as: > > > > IMGPlot = a.Children(2); // children(2) is the background in this > > caseIMGPlot.visible = 'off'; > > > > > > A complete other way would be to combine two images of same size. > > Say...background image is img1. > > > > your plot = img2. > > Plot the graph and save the figure as a temporary image. > > > > load img1 > > load temporary image > > > > If size (img1) == size(img2)....You could use SIVP module to linear > combine > > two images: > > > > img_gray = double(RGB2Gray(img));img2 = zeros(rows, cols) + > > 255;img2(rows/2,:) = 0;img3 = imlincomb( 0.5, img_gray, 0.4, img2) > > // the values here define some kind of transparency.figure(); > > ShowImage(img3,''); > > > > > > > > Best regards, > > Philipp > > > > > > > > > > 2016-02-08 17:49 GMT+01:00 Jan Åge Langeland <[email protected]>: > > > > > > > > > > > On 08.02.2016 11:47, [email protected] wrote: > > > > > > Hi everyone, > > > > > > I just failed at placing an image behind a plot. > > > I thought that would be easy: > > > - create a figure > > > - create an image uicontrol > > > - create an axis > > > - plot in the axis > > > - set axis.filled="off" > > > > > > Apparently I was wrong. > > > I did not find a way to overlap a plot with transparent background > over my > > > image. > > > It seems that the image is always above the plot no matter what order > the > > > uicontrol/axes where created. > > > I also tried to put both of them inside the same frame, but I did not > work > > > either. > > > > > > Any idea? > > > > > > > > > > > > ShowImage(im,'J2'); > > > > > > b=newaxes(); > > > b.filled = "off"; > > > plot(a) > > > > > > JÅ > > > > > > _______________________________________________ > > > users mailing list > > > [email protected] > > > http://lists.scilab.org/mailman/listinfo/users > > > > > > > > > > > > _______________________________________________ > > > users mailing list > > > [email protected] > > > http://lists.scilab.org/mailman/listinfo/users > > > > > > > > > > > > -- > > There we have the salad. > > > > > > > _______________________________________________ > users mailing list > [email protected] > http://lists.scilab.org/mailman/listinfo/users > -- There we have the salad.
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