for a particular contour level,should the pixels have the same intensity

Question

sparsh garg 2021-10-10

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1531276-for-a-particular-contour-level-should-the-pixels-have-the-same-intensity

评论： Walter Roberson 2021-12-8

I know that imcontour basically gives you a contour at a particular level.

My question is that the pixels that lie along that particular contou level r they al should have the same intensity and pixel values right.

for example in the below image(which shows the contour for an image at level 50),all pixels that lie along this contour should have their pixel intensity values somewhere close to 50 right?

l

But,when I save this as jpg and run imtools on it I get this image where the pixel values along the contour are not havint heir intensity levels close to 50.

So to conclude

1.do all pixels along the contour (obtained at level X) have the same intensity values?

2 Could it be possible that when I saved the contour map as jpg,the values were changed and that's why I am getting the result shown in the second image.Also to view pixel values can I use impixelinfo or improfile?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2021-10-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1531276-for-a-particular-contour-level-should-the-pixels-have-the-same-intensity#answer_805471

No, all pixels along a closed contour do not have to have the same intensity. What if there just simply weren't any pixels of the contour level intensity? You can either leave the contour open, or if you close it, it will have to create a path in between values above the level and below the level.

8 个评论
显示 6更早的评论隐藏 6更早的评论

Walter Roberson 2021-10-10

Contour plots work on the assumption that the data available in the array is a sampling of a continuous function. That, for example, if a pixel with value 180 is beside a pixel of value 140, that there is some point between them that was value 150.

If you call contour() or contourf() and look at the first output, the coordinates returned are floating point values rather than integers. They represent continuous locations at which the algorithm figures that if you were to interpolate at that continuous point using the known values supplied in the discrete array, that the interpolated value would match the target level.

For example if the 180 was at x = 1 and the 140 was at x = 2, then it might calculate that at x = 1 3/4 that the value would have been 150. And that 1 3/4 would be the location returned in the contour matrix.

Now, that 1 3/4 would be in data units, not in pixels. Eventually those floating point data units need to be rendered to pixels. But pixels have integer locations. In this particular case if you happen to be able to allocate a multiple of 4 physical pixels per data unit, you could position the 1 3/4 exactly in one physical pixel... but you probably do not happen to have exactly a multiple of 4 physical pixels per data unit (and remember that with different arrays and different target levels, the location might be for example 1.45302116).

What do you do when you have an array location given in data units and that location does not match exactly into an integer location of pixels for display on the screen, or an integer location of pixels for storing into a file? Should you discard all locations that do not turn out as exact integers? Should you round them to the nearest underlying integer? Should you "turn on" the pixels that are passed through? Should you "turn on" the pixels that are passed through but with different intensities depending upon mathematically "how much" of the pixel is being occupied by the line ?

The floating point calculation of the locations where the value would interpolate to the target value (under the model of continuous linear surfaces) might be quite good, but pixels to draw in or array locations to turn on are discrete units.