how to filter columns
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Hello friends,
I have this script. But is very slow!
I want to filter all columns beginning with column first. After that, I want, in the place of column first, to put the second column and, in the place of column second to put the column first and filter again. And go on!
My purpose is decrease the time!
Thanks by support!
clear all
clc
q = 1;
p = 2;
r1 = 1;
r2 = 1;
tic
% Example: Filtering a matrix
load ('tum.mat'); % tum have 576 x 1000
for i1 = 1:1000
% Filter rows where the first column is greater than 15
z1 = k2(k2(:,1) == 0, :);
s1 = sum(z1, 1);
r(r1,r2) = max(s1);
temp_row = k2(:,q);
k2(:,q) = k2(:,p);
k2(:,p) = temp_row;
p = p + 1;
r1 = r1 + 1;
end
toc
Elapsed time is 337.582496 seconds.
6 个评论
Can you upload the input data? I tried running your code with a random array, but it threw an error:
clear all
clc
q = 1;
p = 2;
r1 = 1;
r2 = 1;
tic
% Example: Filtering a matrix
k2 =rand(576,1000);
% load ('tum.mat'); % tum have 576 x 1000
for i1 = 1:1000
% Filter rows where the first column is greater than 15
z1 = k2(k2(:,1) == 0, :);
s1 = sum(z1, 1);
r(r1,r2) = max(s1);
temp_row = k2(:,q);
k2(:,q) = k2(:,p);
k2(:,p) = temp_row;
p = p + 1;
r1 = r1 + 1;
end
toc
"I tried running your code with a random array, but it threw an error:"
It cannot run on data of the stated size, because p ends up being greater than the number of columns.
"Elapsed time is 337.582496 seconds."
That is incredible. It would be interesting to know which particular lines or operations are the main contributors.
"My purpose is decrease the time!"
Then why are you swapping columns around? That will be a big performance hit.
Explain what you are actually trying to achieve https://xyproblem.info/
AIRTON
2025-9-13
AIRTON
2025-9-13
Image Analyst
2025-9-13
If k2 is a matrix of random numbers, you will never have the numbers be zero so this won't work
k2(:,1) == 0 % Will never happen!
Matt J
2025-9-14
Elapsed time is 337.582496 seconds.
It does seem impossible for such a small loop to be that slow. Certainly everything excluding the load() statement is very fast, as illustrated in my timing comparison below.
The only explanation I can think of is that you have additional variables in tum.mat that are getting loaded, and which are a lot larger than k2. Or, you are loading from a super slow hard drive.
回答（1 个）
m=576;
n=1000;
k2=randi([0,1],m,n).*rand(m,n);
tic;
r=max( (k2==0)'*k2 ,[],2);
toc;
5 个评论
Just to demonstrate that the results agree:
m=576;
n=1000;
k2=randi([0,1],m,n).*rand(m,n);
tic;
Result0=max( (k2==0)'*k2 ,[],2);
toc
q = 1;
p = 1;
r1 = 1;
r2 = 1;
tic
for i1 = 1:n
% Filter rows where the first column is greater than 15
z1 = k2(k2(:,1) == 0, :);
s1 = sum(z1, 1);
r(r1,r2) = max(s1);
p = mod(p,n)+1;
temp_row = k2(:,q);
k2(:,q) = k2(:,p);
k2(:,p) = temp_row;
r1 = r1 + 1;
end
toc
Result1=r;
difference=max(abs(Result0-Result1))
@AIRTO is "my_matrix" the same as "k2" in your original code? If so, what help is required? Why isn't the answer I gave applicable to this example? I get no errors or odd behavior when I run it:
k2=[1 0 0 0 1 0 0 1 0
1 1 0 1 0 1 1 0 1
0 1 1 1 1 1 1 0 1];
r=max( (k2==0)'*k2 ,[],2)
AIRTON
2025-9-16
Stephen23
2025-9-17
@AIRTO: using numbered variables like that is a mistake. Much better to use one container array and indexing:
S = dir('./w*.mat');
for k = 1:numel(S)
F = fullfile(S(k).folder,S(k).name);
C = struct2cell(load(F));
M = C{1}; % this is your matrix.
... do whatever you want with your matrix
S(k).r1 = r1; % save whatever values you want
end
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