Error in Interp1 function when searching for a corresponding matrix

1 次查看(过去 30 天)
I have eight 1024X1024 matrices corresponding to 8 thicknesses (0,5,10,15,20,25,30,35cm). My code below aims to retrieve a correct matrix for a given thickness, e.g. Th12...
I am getting this error:
??? Error using ==> interp1 at 259 The values of X should be distinct.
Error in ==> Predict_EPIDM at 122 CorrFactor = interp1(col,A,Th);
What is wrong with the code? __________________________________________________________________________
% There are 8 PT 0,5,10,15,20,25,30,35cm
PT=PhysicalThickness;
% A is matrices 1024X1024 for the 8PTs. Size(A) returns 1024X1024X8.
A = cat(3, A{:});
% define the column number nearest to the input Th
Th=12;
colindex=min(find(PT>=Th));
% return the 'in between' A matrix
col=A(:,colindex)+((A(:,colindex)-A(:,colindex-1))*(Th-PT(colindex))/(PT(colindex)-PT(colindex-1)));
% Using the RDepid value to find MUx
CorrFactor = interp1(col,A,Th);

采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-7-24
编辑:Andrei Bobrov 2012-7-24
Let A - cell array your matrices.
A = arrayfun(@(ii)randi(10,5),1:8,'un',0);
Da = cellfun(@(x)x(:),A,'un',0);
Dn = cat(2,Da{:});
Th = [0,5,10,15,20,25,30,35];
PP = arrayfun(@(x)interp1(Th,Dn(x,:),'linear','pp'),(1:size(Dn,1))','un',0);
outfun = @(th)reshape(cellfun(@(x)ppval(x,th),PP),size(A{1}));
outfun(12);
or
t = 12;
idx = find(Th > t,1,'first')-[1 0];
out = diff(cat(3,A{idx}),[],3)/diff(Th(idx))*(t - Th(idx(1)))+A{idx(1)};
  2 个评论
Yun Inn
Yun Inn 2012-8-22
编辑:Yun Inn 2012-8-22
Andrei,
I chose option 2 of the solution that u gave. If Th remains as [0,5,10,15,20,25,30,35], but 't' is now a matrix, how do I make each pixel search for the right value? Also, some pixels in 't' matrix contain value smaller than 0 and bigger than 35.

请先登录,再进行评论。

更多回答(1 个)

Walter Roberson
Walter Roberson 2012-7-29
Please review the guide to tags and retag this question; see http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags

类别

Help CenterFile Exchange 中查找有关 NaNs 的更多信息

标签

尚未输入任何标签。

Community Treasure Hunt

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

Start Hunting!

Translated by