Finding a specific rows?

17 vues (au cours des 30 derniers jours)
JamJan
JamJan le 18 Juil 2019
Commenté : Adam Danz le 18 Juil 2019
I have this Matrix:
m = [0 0 0 0 0 0 0 0
0 3.410 0 0 0 0 0 0
10.83 11.87 7.240 3.470 1.610 1.610 0 0
20.99 18.23 17.72 6.010 3.790 3.790 1.100 0
25.26 20.40 22.70 7.590 6.990 6.990 1.550 0
22.86 19.13 24.48 8.280 9.240 9.240 2.180 0
14.92 15.73 23.19 8.280 11.61 11.61 2.690 0
10.83 11.87 22.18 8.280 12.79 12.79 3.010 0
9.050 10.69 22.70 8.490 14.91 14.91 3.500 0
9.840 11.19 23.75 8.490 16.02 16.02 3.790 0
9.840 11.54 24.48 8.840 16.43 16.43 4.100 0
8.070 10.20 25.18 9.180 17.73 17.73 4.750 0
5.230 8.350 24.48 9.180 19.28 19.28 5.800 0
3.680 6.630 25.18 8.840 22.49 22.49 6.980 2.190
0 3.190 22.70 7.840 28.06 28.06 8.280 3.320
0 0 13.03 6.010 34.68 34.68 10.88 5
0 0 0 3.470 32.87 32.87 13.86 5.300
0 0 0 0 15.14 15.14 12.27 0
0 0 0 0 1.240 1.240 1.450 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
40.53 25 0 0 0 0 0 0]
% * Edited by Adam Danz to allow for copy-pasting into command window
I want to find the first row where all the columns have a value below 10. How to do this? I have made an if statement, but this seems to specific and I was wondering if there would be another way.

Réponse acceptée

Adam Danz
Adam Danz le 18 Juil 2019
Modifié(e) : Adam Danz le 18 Juil 2019
This does not require a for-loop. In fact, the last line is the solution. The first few lines just creates an example matrix where rows 3 and 13 are all below 10.
% Create a 15x7 matrix of random integers 1:20
m = randi(20,15,7);
% Rows 3 and 13 are all below 10 (maybe others, too)
m([3,13],:) = m([3,13],:)-11;
%find the first row where all the columns have a value below 10.
rowNumber = find(all(m < 10,2),1);
% find the first row AFTER ROW N where all columns are < 10
N = 3; %start at row N
rowNumber = find(all(m(N:end,:) < 10,2),1)+N-1;
  7 commentaires
JamJan
JamJan le 18 Juil 2019
I'm looking for the transition of the positive values into the zero's, but I want to build in a certain threshold (let's say values lower than 10). So that means that I want to find the following row in matrix m.
0 0 0 0 1.240 1.240 1.450 0
Adam Danz
Adam Danz le 18 Juil 2019
I see. Here' how to compute N (the first non-zero row).
midx = all(m < 10,2);
N = find(midx == 0,1);
rowNumber = find(all(m(N:end,:) < 10,2),1)+N-1;

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Plus de réponses (1)

Mario Chiappelli
Mario Chiappelli le 18 Juil 2019
Modifié(e) : Mario Chiappelli le 18 Juil 2019
I would use a double nested For statement and a counter variable as follows:
for i = 1:31 % Number of rows
counter = 0; % Reset the counter
for j = 1:8 % Number of columns
if myMatrix(i,j) < 10
counter += 1;
end
if counter == 8 % In other words when all values are under 10
disp("Row ");
disp(i);
disp(" only contains numbers less than 10");
end
end
end
If you want it to stop after the first row is detected just add a series of break statements to break out of the for loop in the if statement that checks if the counter is equal to 8
If your matrix changes sizes over the course of the program, you are going to have to assgin the dimensions of the matrix to variables and use those as your for statement parameters.
Hope this helps :)
  1 commentaire
Adam Danz
Adam Danz le 18 Juil 2019
Have you tried to run this? There are several errors.
  • counter += 1; I think you mean counter = counter + 1;
  • if myMatrix(i,j) < 10 this might work for the example given but if the dimensions of myMatrix ever changes, this might break.
  • Same with i = 1:31 and j = 1:8 and counter == 8
Even after correction those errors, the algorithm doesn't work for this example matrix:
myMatrix =
0 0 0 10 10
10 10 0 0 0
0 0 0 0 0

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