Find array elements that meet a condition an put them in a secondary array

5 Ansichten (letzte 30 Tage)
Hello all, I have a 3D matrix A. For example suppose that A is as follows:
A(:,:,1) =
0.1835 0.0811 0.4359
0.3685 0.9294 0.4468
0.6256 0.7757 0.3063
0.7802 0.4868 0.5085
A(:,:,2) =
0.5108 0.3786 0.9390
0.8176 0.8116 0.8759
0.7948 0.5328 0.5502
0.6443 0.3507 0.6225
I want to find the elements in A that are >0.5 and put them in an array B. Each row of array B should be [i,j,k,value]. In above example my array B would be like:
3 1 1 0.6256
4 1 1 0.7802
2 2 1 0.9294
3 2 1 0.7757
...
After this I want to print array B as it is in a text file ('Example.txt'). Thank you for your help.

Akzeptierte Antwort

Star Strider
Star Strider am 21 Dez. 2015
The find and ind2sub will do exactly what you want:
A(:,:,1) = [0.1835 0.0811 0.4359
0.3685 0.9294 0.4468
0.6256 0.7757 0.3063
0.7802 0.4868 0.5085];
A(:,:,2) = [0.5108 0.3786 0.9390
0.8176 0.8116 0.8759
0.7948 0.5328 0.5502
0.6443 0.3507 0.6225];
idx = find(A > 0.5);
[i,j,k] = ind2sub(size(A), idx);
B = [i j k A(idx)]
B =
3 1 1 0.6256
4 1 1 0.7802
2 2 1 0.9294
3 2 1 0.7757
4 3 1 0.5085
1 1 2 0.5108
2 1 2 0.8176
3 1 2 0.7948
4 1 2 0.6443
2 2 2 0.8116
3 2 2 0.5328
1 3 2 0.939
2 3 2 0.8759
3 3 2 0.5502
4 3 2 0.6225

Weitere Antworten (4)

Image Analyst
Image Analyst am 21 Dez. 2015
Here are two different ways, a for loop way, and a vectorized way:
A(:,:,1) =[...
0.1835 0.0811 0.4359
0.3685 0.9294 0.4468
0.6256 0.7757 0.3063
0.7802 0.4868 0.5085];
A(:,:,2) =[...
0.5108 0.3786 0.9390
0.8176 0.8116 0.8759
0.7948 0.5328 0.5502
0.6443 0.3507 0.6225]
% In tuitive brute force "for" loop way:
numRows = nnz(A(:)>0.5);
out = zeros(numRows, 4);
currentRow = 1
for z = 1 : size(A, 3)
for row = 1 : size(A, 1)
for col = 1 : size(A, 2)
if A(row, col, z) > 0.5
out(currentRow, 1:4) = [row, col, z, A(row, col, z)];
currentRow = currentRow + 1;
end
end
end
end
out
% Vectorized way:
bigA = A(A(:)>0.5);
actualSubscripts = find(A(:) > 0.5)
[row, col, z] = ind2sub(size(A), actualSubscripts)
out = [row, col, z, bigA]

Azzi Abdelmalek
Azzi Abdelmalek am 21 Dez. 2015
ll=find(A>0.5)
[ii,jj,kk]=ind2sub(size(A),ll)
out=[ii,jj,kk,A(ll)]

Andrei Bobrov
Andrei Bobrov am 21 Dez. 2015
t = A > .5;
[m,n,~] = size(A);
[ii,jj] = find(t);
B = [ii,rem(jj-1,n)+1,ceil(jj/n),A(t)];

H R
H R am 21 Dez. 2015
Bearbeitet: H R am 21 Dez. 2015
Thank you for your help.
As a secondary filtering on B, I would like to only retain the rows when the first element of any row in B >2, and the second element of B in any row <3 and put them in a new array C. In other words, my array C has the following properties [i>2,j<3,k, value]
  1 Kommentar
Star Strider
Star Strider am 21 Dez. 2015
This seems to work (continuing my previous code):
Crows = (B(:,1)>2) & (B(:,2)<3); % Logical Vector: 1=Meets Criteria
C = [B(Crows,1) B(Crows,2) k(Crows) B(Crows,4)] % New Filtered Matrix
C =
3 1 1 0.6256
4 1 1 0.7802
3 2 1 0.7757
3 1 2 0.7948
4 1 2 0.6443
3 2 2 0.5328

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Tags

Noch keine Tags eingegeben.

Community Treasure Hunt

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

Start Hunting!

Translated by