How to obtain this graph???

조회 수: 5 (최근 30일)
Sabarinathan Vadivelu
Sabarinathan Vadivelu 2012년 12월 24일
댓글: Walter Roberson 2020년 4월 8일
This is my input Image.
How to get this graph??

채택된 답변

Walter Roberson
Walter Roberson 2012년 12월 24일
plot() an improfile() of the image
  댓글 수: 3
Walter Roberson
Walter Roberson 2012년 12월 24일
For which call? For the plot call, you can leave out the x, and the y value should be the value returned by improfile()
Sabarinathan Vadivelu
Sabarinathan Vadivelu 2012년 12월 24일
I found it, thanks

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추가 답변 (1개)

Winston
Winston 2020년 4월 8일
Hi, Walter, I have a question regarding the x value, it is the pixel number, but most of the time, the intensity peak is at a point which is not an interger, such as a peak (y max) at x=160.23, which means among the 161st pixel, there is a peak. so if we plot x from 1 to 200 with increasement of 1, then we will omit the peak, how to find or export the exact x value?
  댓글 수: 3
Winston
Winston 2020년 4월 8일
figure, improfile(I, x, y);
then from the profile, I can use data tips to point out the peak, giveing the corresponding x value. x starts from 0 ending with the end pixel coordinates. then how to export all the x values ?
Walter Roberson
Walter Roberson 2020년 4월 8일
It is likely that you profiled at an angle across the image, rather than on a row or column.
improfile() works by finding the longer edge of the x and y distance, and setting up integer increments along that distance, and doing linear interpolation of the other coordinates to have the same number of points. Then it does interp2() with those coordinates.
We can tell from the fact that the "x" values are non-integer, that you are looking at values belonging to the shorter distance between coordinates, the one that would have been interpolated when the other coordinate would have been exact integers.
It is questionable as to whether you can say that the peak is "really" at x = 160.23. The value you are seeing there is a weighted average of the surrounding pixels. The "real" peak is at 160 or 161.

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