how to store a double loop result into a matrix column by column?

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LEI Li
LEI Li el 29 de Sept. de 2021
Comentada: LEI Li el 29 de Sept. de 2021
I have a matrix cycle; What I am doing is to find out certain points satisfying the condition and store them into Array up and down;
When cycle1 is one column, it could work; but when cycle1 increases to 2 columns, then the double loop doesn't work.
up and down's size is not certain; I just estimate their size 100*12;
now the code has no error, but give wrong results.
please give me some suggestion about this code.
cycle1 = [butterworth_filter1,butterworth_filter2];
[m,n] = size(cycle1)
up = zeros(100,12);
down = zeros(100,12);
k=1;
for j = 1:n
for i = 1:m
if i == m
break
elseif cycle1(i,j) < 1.5e-5 && cycle1(i+1,j) > 1.5e-5
up(k,:) = i + 1;
k = k+1;
elseif cycle1(i,j) > 1.5e-5 && cycle1(i+1,j) < 1.5e-5
down(k,:) = i;
k = k+1;
end
end
end
  6 comentarios
Geoff Hayes
Geoff Hayes el 29 de Sept. de 2021
So if (i,j) is "too big" then is k good enough? If not, where is the error?
Stephen23
Stephen23 el 29 de Sept. de 2021
LEI Li's incorrectly posted "Answer" moved here:
from the test code, up has the result; now each column is the same; I need to chech the random matrix;
next thing is there should not exist 0 in the column; this is the reason I use k;
previously when I deal with one dimension array, I use
up(end+1) = i + 1; to store the location; but now I need to deal with a matrix; I stuck here.

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Respuestas (1)

Jan
Jan el 29 de Sept. de 2021
You are using one counter k for the up and the down lists.
cycle1 = [butterworth_filter1,butterworth_filter2];
[m, n] = size(cycle1)
up = zeros(100, 12);
down = zeros(100, 12);
kup = 1;
kdown = 1;
for j = 1:n
for i = 1:m - 1 % Simpler than: if i==m, break
if cycle1(i,j) < 1.5e-5 && cycle1(i+1,j) > 1.5e-5
up(kup,:) = i + 1; % Are you sure you want to set the complete row to i+1?
kup = kup + 1;
elseif cycle1(i,j) > 1.5e-5 && cycle1(i+1,j) < 1.5e-5
down(kdown, :) = i;
kdown = kdown + 1;
end
end
end
  1 comentario
LEI Li
LEI Li el 29 de Sept. de 2021
close all;
clear;
clc;
cycle1 = abs(randn(100,12));
for i = 1:12
plot(cycle1(:,i))
hold on
end
[m,n] = size(cycle1)
up = zeros(100,12);
down = zeros(100,12);
kup=1;
kdown=1;
for j = 1:n
for i = 1:m-1
% if i == m-1
% break
if cycle1(i,j) < 0.2 && cycle1(i+1,j) > 0.2
up(kup,:) = i + 1; %up only has one value now??
kup = kup+1;
elseif cycle1(i,j) > 0.2 && cycle1(i+1,j) < 0.2
down(kdown,:) = i;
kdown = kdown+1; %down only has one value now??
end
end
end
figure (4)
plot(cycle1)
hold on
plot(kup, cycle1(kup), 'go', 'MarkerSize', 8, 'MarkerFaceColor', 'g')
plot(kdown, cycle1(kdown), 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r')
xlabel('Time(s)');ylabel('EMG(V)');title('Muscle Activation & Cycles')
legend ('Subject 1')

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