Methods for smoothing contour lines

Question

Andrew Sol 2022-10-20

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1831563-methods-for-smoothing-contour-lines

评论： Star Strider 2022-10-21

采纳的回答： Star Strider

P.mat

在 MATLAB Online 中打开

I have a database P with columns X, Y and Z (file above):

N = 150; % Number Of Points Desired
xv = linspace(min(P(:,1)), max(P(:,1)), N);
yv = linspace(min(P(:,2)), max(P(:,2)), N);
[X,Y] = ndgrid(xv, yv);
Z = griddata(P(:,1), P(:,2), P(:,3), X, Y);
contourf(X, Y, Z, 35)
Zi = interp2(P(:,1), P(:,2), P(:,3), X, Y) ;

With the code above, I get the following plot:

How to make contour lines neater and smoother?

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Answer 1

Star Strider 2022-10-20

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1831563-methods-for-smoothing-contour-lines#answer_1079768

编辑：Star Strider 2022-10-20

在 MATLAB Online 中打开

I dislike going to external sites.

Increasing ‘N’ would be the option I would use, initially experimenting with 250 and perhaps 500 or greater, depending on the available memory and the result provided. That should increase the resolution of the vectors, and therefore the resolution of the matrices derived from them.

EDIT — (20 Oct 2022 at 15:45)

Using the supplied data —

LD = load(websave('P','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1163343/P.mat'));

P = LD.P;

N = 500; % Number Of Points Desired

xv = linspace(min(P(:,1)), max(P(:,1)), N);

yv = linspace(min(P(:,2)), max(P(:,2)), N);

[X,Y] = ndgrid(xv, yv);

Z = griddata(P(:,1), P(:,2), P(:,3), X, Y);

Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.

figure

contourf(X, Y, Z, 35)

I looked at the first column of ‘P’ to see if reshaping would work. It will not, because the first indices of the different unique values of ‘P(:,1)’ do not have constant distances between them.

The data simply do not appear to be smooth (i.e. may have limited precision), and that is reflected in the contours even with a relatively high precision in the interpolating matrices.

This is likely as good as it gets.

.

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Star Strider 2022-10-20

在 MATLAB Online 中打开

I doubt that any sort of smoothing would help, specifically because the data are not consistently sampled, that is they are neither regularly sampled nor are they sampled in any consistent manner. It would be difficult to use any sort of smoothing function (such as the Savitzky-Golay filter) with them for those reasons.

If they are data, sampling them with greater precision might be worthwhile in a subsequent experiment. I have no idea what the data represent, so I have no idea how to create them with greater precision.

The default interpolation method is 'linear' so experimenting with the 'natural' and 'cubic' methods —

LD = load(websave('P','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1163343/P.mat'));

P = LD.P;

N = 750; % Number Of Points Desired

xv = linspace(min(P(:,1)), max(P(:,1)), N);

yv = linspace(min(P(:,2)), max(P(:,2)), N);

[X,Y] = ndgrid(xv, yv);

Z = griddata(P(:,1), P(:,2), P(:,3), X, Y, 'natural');

Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.

figure

contourf(X, Y, Z, 35)

title('''natural'' Method')

Z = griddata(P(:,1), P(:,2), P(:,3), X, Y, 'cubic');

Warning: Duplicate x-y data points detected: using average values for duplicate points.

figure

contourf(X, Y, Z, 35)

title('''cubic'' Method')

Even increasing ‘N’ further and using differnt ‘method’ options fails to make any meaningful changes in the result.

.

Andrew Sol 2022-10-21

I seem to have found the reason why the contour lines are "noisy". Once again, I'll bring it up. There is some space in which the robot moves. The range of position change along the x-coordinate = [-4;4], and the range of position change along the y-coordinate = [-5;5]. We generate a huge number of uniformly distributed points from these ranges and see if the robot is in or out of the valid area. The invalid position is marked on the contour plot with a "white" area. So that's where the "noise" comes from. When the value of the acceptable location criterion was calculated, the /ceil/ command was applied to this value. That's it. I don't know how I missed it, but the fact remains. When I removed it (and increased the number of contour lines), the graph began to look like this:

But some problems still remain. They stem from the fact that you need a lot of points to make a nice graph. And doing it by hand is extremely difficult and difficult. Here is a graph with a much smaller (by several orders) number of measurements.

It would be nice to somehow "smooth" it. Eliminate "creases" on the borders and fill in the "voids" within it. Perhaps surface approximation will help here?

Star Strider 2022-10-21

在 MATLAB Online 中打开

I doubt that using ceil would change the memory requirements —

v1 = pi; 
v2 = ceil(pi); 
whos v1 v2 
  Name      Size            Bytes  Class     Attributes

  v1        1x1                 8  double              
  v2        1x1                 8  double              

If you want to use less memory, the single precision option (or even integers, such as int8) is possible —

v1 = single(pi);  
v2 = int16(pi);
v3 = int8(pi);
whos v1 v2 v3
  Name      Size            Bytes  Class     Attributes

  v1        1x1                 4  single              
  v2        1x1                 2  int16               
  v3        1x1                 1  int8                

however many MATLAB functions require double precision, so this could involve converting from single (or integer) to double and back again as necessary, likely also multiplying and dividing by powers-of-10 to retain at least some decimal precision in the integers, considerably slowing the code.

Since I have no idea what your code does, I can’t suggest specific methods to make it more efficient or less msmory intensive.

In any event, keeping as much precision as possible in the interim calculations will create a better final result.

.

Andrew Sol 2022-10-21

And thank you for not passing by!

Star Strider 2022-10-21

As always, my pleasure!

This was an interesting problem!

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Methods for smoothing contour lines

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