I need help with my While Loop!

조회 수: 3 (최근 30일)
Uma Dixit
Uma Dixit 2020년 2월 23일
댓글: darova 2020년 2월 26일
I have an original array (3D) called 'pixelArray' that I eventually need to crop 3 different ways. So I have 3 different sets of x1,x2,y1,y2 numbers that crop the original array in different ways.
I currently have it set up like this:
x1 = [52, 52, 52];
x2 = [480, 480, 480];
y1 = [4, 1, 205] ;
y2 = [392, 205, 392];
n = 1;
while n < 3
%crop rows
pixelArray(1:y1(n),:,:)=[];
pixelArray(y2(n):end,:,:) = [];
%crop columns
pixelArray(:,1:x1(n),:)=[];
pixelArray(:,x2(n):end,:)=[];
n =n+1;
end
But I want to save each iteration's new pixelArray as its own matrix. What is the best way to go about doing this?
  댓글 수: 3
Uma Dixit
Uma Dixit 2020년 2월 26일
Oh, very good point. How do you suggest I make sure the original pixelArray matrix stays intact in its dimensions?
darova
darova 2020년 2월 26일
Remove elements from the end?
>> a = 1:10
a =
Columns 1 through 9
1 2 3 4 5 6 7 8 9
Column 10
10
>> a(7:9) = []
a =
1 2 3 4 5 6 10
>> a(1:3) = []
a =
4 5 6 10

댓글을 달려면 로그인하십시오.

채택된 답변

Jakob B. Nielsen
Jakob B. Nielsen 2020년 2월 24일
편집: Jakob B. Nielsen 2020년 2월 24일
Index them! Either as a structure, or an array with a 4th dimension. I personally favour structures, since 4D data makes my head hurt ;) dot indexing is great for looping results.
Also note that you will only get 2 runs of your loop, as after the 2nd run n will equal 3, and your loop terminates before running a 3rd time since the condition is less than 3. Either make the condition less than or equal to 3, or you can use a for loop instead, as your while loop is essentially a for loop where you do the counting yourself - something you dont need to do in this case.
for n=1:3
%crop rows
pixelArray(1:y1(n),:,:)=[];
pixelArray(y2(n):end,:,:) = [];
%crop columns
pixelArray(:,1:x1(n),:)=[];
pixelArray(:,x2(n):end,:)=[];
Crops(n).pixelArray=pixelArray;
end

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Matrices and Arrays에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by