Replacing values in a matrix by specified values

3 ビュー (過去 30 日間)
Carolin Brueckmann
Carolin Brueckmann 2015 年 6 月 10 日
コメント済み: Carolin Brueckmann 2015 年 6 月 13 日
Hi there,
I have a three dimensional array of simulated data (dimensions are 10000,16,312 or #trials, TimeSeries, horizons). I would like to replace values above / below a pre-specified threshold with the threshold values. I have calculated the threshold values for each individual time series in MinAcceptableVal(:,i) and MaxAcceptableVal(:,i). When I run the code I do not receive an error message, but the values above the threshold are not cut off.
for i=1:nIndices
simulatedReturnsEVT1(simulatedReturnsEVT1(:,i,:)<MinAcceptableVal(:,i))=MinAcceptableVal(:,i);
simulatedReturnsEVT1(simulatedReturnsEVT1(:,i,:)>MaxAcceptableVal(:,i))=MaxAcceptableVal(:,i);
end
I have tried to use the code in a different form (see below) before and it worked perfectly. Matlab seems to be having problems with me introducing different cutoff levels for the different time series variables (i).
simulatedReturnsEVT1(simulatedReturnsEVT1<-1)=-1;
simulatedReturnsEVT1(simulatedReturnsEVT1>1)=1;
I would be very happy about any Hints!
Best, Carolin
  2 件のコメント
James Tursa
James Tursa 2015 年 6 月 10 日
What are the dimensions of MinAcceptableVal and MaxAcceptableVal?
Rong Yu
Rong Yu 2015 年 6 月 10 日
I also think it is a dimensional issue. You could try to create a one-dimensional time series of your threshold values instead of two-dimensional.
for i=1:nIndices
simulatedReturnsEVT1(simulatedReturnsEVT1(:,i,:)<MinAcceptableVal(i))=MinAcceptableVal(i);
simulatedReturnsEVT1(simulatedReturnsEVT1(:,i,:)>MaxAcceptableVal(i))=MaxAcceptableVal(i);
end

サインインしてコメントする。

採用された回答

Image Analyst
Image Analyst 2015 年 6 月 10 日
Assign simulatedReturnsEVT1(:,i,:)>MaxAcceptableVal(:,i) to a variable and see what the dimension are.
indexesToClip = simulatedReturnsEVT1(:,i,:) > MaxAcceptableVal(:,i);
theSizes = size(indexesToClip)
You're basically taking a slice (plane) out of the 3D volume and comparing it to a value. So it might be a 2D array. On the other hand, it might be a N-by-1-by-M array, which I think should be okay. But if it's N-by-M, then that would be a problem.
  1 件のコメント
Carolin Brueckmann
Carolin Brueckmann 2015 年 6 月 13 日
Hi Image Analyst,
thanks a lot for your help! I was able to implement the code now and it works perfectly :-)

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeMatrix Indexing についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by