# Polyfit on a 3d plot

조회 수: 65(최근 30일)
Fabio Carton 2022년 5월 11일
댓글: Fabio Carton 2022년 5월 12일
Hello everyone,
i'm trying to create a matrix containing the coefficient of a 3rd degree polyfit on each rows with the shape 13000x4.
For that, i have the following data shape:
Time : 13000x1 %time corresponding actually to the number of rows
Position : 13000x4 %the position is just for the sake of the plot (position of the sensor in reality). The four columns have each a constant value over the 13000 rows
Displacement: 13000x4 %measurement in global Z of four sensors.
(see image attached) My purpose is to "slice" the 3d plot at every time "t" (13000 times) and apply a polyfit on the 4 red cross (x = position, y = displacement) in order then to evaluate the curve to obtain the 2 yellow points. Ultimaly, i want to recreate a line out of the yellow points.
Up to now, i was only able to extract a few sets of points (red crosses) at a given time manually. I don't know how to reshape the matrix without creating 13000 submatrix and names associated.
I hope that my explanations are clear enough and that you can give me a little help on that one.
Fabio

댓글을 달려면 로그인하십시오.

### 채택된 답변

dpb 2022년 5월 11일
That's fairly simple, just loop over the displacement array with a set of the position data to compute that set of coefficients and evaluate them where needed.
nD=size(D,1); % the important size of D
NewD=zeros(nD,2); % for the new points -- could augment D instead but this is simpler
NewP=[300,2400]; % GUESS FOR THE NEW POINT LOCATIONS -- SET AS DESIRED
for i=1:nD % iterate over rows of D
b{i}=polyfit(P(1,:),D(i,:),3); % compute polyfit coefficients for each (hold in cell array)
NewD(i,:)=polyval(b{i},newP); % evaluate and save new positions
end
leaves you with the additional data points at the two positions -- just draw them directly or you could add to the original D array...
##### 댓글 수: 1표시숨기기 없음
Fabio Carton 2022년 5월 12일
Thank you very much!! it was pretty much what i was looking for.
I acknowledge that it can appear easy but i'm still a learner and some loops are still tricky for me.
All the best

댓글을 달려면 로그인하십시오.

R2020b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by