Polar coordinates of image.

If your f is an array rather than a function, then f(x,y) would not be typical. The first dimension corresponds to rows and the second dimension correspond to columns, and normally rows corresponds to Y and columns corresponds to x. If your f were an array then normally you would index f(row,column) which would correspond most closely to f(y,x) .

However, rows and columns do not typically correspond to y and x, because your y and x are not typically restricted to small consecutive positive integers. You would normally have some kind of correspondance index or formula, such as

column = floor(x*10) + 51

which might be used if you were willing to approximate x by 1/10 increment for an x range starting from -5.0

If you had an array fpol then fpol(r,theta) would have similar issues to above, namely that your r and theta would be restricted to small positive integers. You would be more likely to have an fpol(rows, columns) with translation formulas such as row = 1 + r * 100 and column = floor(theta*100) + 501 to support r increments of 1/100 starting from 0 and theta increments of 1/100 starting from -5.00 .

But you really need to be clear on whether you are working with an array fpol or a polar function

When you start talking about multipling by matrices in this context, you should start suspecting that lurking somewhere around is a transformation matrix such as a rotation matrix . In such a case what you would probably be working with is a matrix with two columns of coordinates, and the matrix multiplication would transform the coordinates to new coordinates.

Stewart Tan 2019-9-7

编辑：Stewart Tan 2019-9-7

@Walter Roberson, spot on!! the image function

is multiplied by a rotation matrix. I've been trying to figure it out for weeks now.

Walter Roberson 2019-9-7

在 MATLAB Online 中打开

You will probably find that

is a function that produces

and

coordinate pairs, and that the rotation matrix operates on those to produce

and

coordinate pairs

Rotation matrices can also be created in terms of polar coordinates

[r(:), theta(:), zeros(numel(r),1)] * [1 0 0; 0 1 dtheta; 0 0 1]

which would increate theta -> theta+dtheta ... though for that simple case it would typically be easier to just do theta = theta + dtheta

请先登录，再进行评论。

Answer 2

Walter Roberson 2019-9-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/479165-polar-coordinates-of-image#answer_390754

在 MATLAB Online 中打开

If what you have is a formula in terms of x and y, and you need to convert it to polar, and you have the symbolic toolbox, then use

syms r theta
polar_formula = simplify( subs(YourFormula, {x, y}, {r*cos(theta), r*sin(theta)}) )

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Walter Roberson 2019-9-7

See also https://www.mathworks.com/matlabcentral/fileexchange/17933-polar-to-from-rectangular-transform-of-images?focused=5095463&tab=function

请先登录，再进行评论。

Polar coordinates of image.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（2 个）

4 个评论
显示 2更早的评论隐藏 2更早的评论

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

另请参阅

类别

标签

产品

Community Treasure Hunt

Polar coordinates of image.

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（2 个）

4 个评论 显示 2更早的评论隐藏 2更早的评论

1 个评论 显示 -1更早的评论隐藏 -1更早的评论

另请参阅

类别

标签

产品

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

4 个评论
显示 2更早的评论隐藏 2更早的评论

1 个评论
显示 -1更早的评论隐藏 -1更早的评论