You may do so by moving the Stroke>Thickness slider to the right. Make sure to increase the width of the stroke so that it is easy to close the contour when drawing. Img = use the Front End Graphic Tools to draw the polygon over the region that interests you (right click mouse button):įollow by getting the mask that corresponds to your polygon (again, right click button):įinally, make a binary image out of the mask and use it to recover the sum of the pixel inside the polygon: Infinity]ĮDIT: If you'd like to define your area of interest using a free-hand contour, use the Free-hand line tool instead of polygon. The data can be produced by a matrix S: S = Table - Sin, ] To explain my problem better, here an easy example. It is possible to plot the matrix like in the image. I have measurement data in a matrix, I want to select some of them and then add them up. If there's anything unclear, I really recommand you to watch this video made by 3Blue1Brown, it shows this in a visual way.I am having a (mathematica 8.0.1.0-) problem which I cannot solve by myself. The visual effect of this blurring technique is a smooth. We are simplely getting linearly independent basis vectors after this linear transformation. It is a widely used effect in graphics software, typically to reduce image noise and reduce detail. This is the reason why we need to get rref of the transpose of A. Therefore, the image of A is just the span of the basis vectors after this linear transformation in this case, span ((a1, a2), (b1, b2)). However, we could consider this transformation as it transforms the basis vectors to all the columns A has. If we apply A as a linear transformation to the standard base, aka the identity matrix, we get A itself. Image of a matrix is basically all the vectors you can obtain after this linear transformation. I could give an explanation for the most appreciated answer why image is calculated in this way. Use it like this: Export 'image-file-name.png', image, ImageResolution -> 500 Share. It Increases the resolution of exported image but does not change the scale. The previous answers on transformation were all good, but I have the outlined steps on how to find $\mathrm$$Īgain, there is a familiar subspace of the matrix $A$ associated with the kernel, look carefully at the definition and you should be able to figure out what it is. The best way I find is to use ImageResolution property. After a long night of studying I finally figured out the answer to these.
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