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How to write a for loop to look at the rows of a M X N matrix?

조회 수: 1 (최근 30일)
i Have a M X N matrix, and I need to divide the rows to n =130 data points and find the max value in that, and then move down the row by (n+1):2n,(2n+1):3n.. until the end, and I need to repeat this iteration for all the columns in the data, and save the data alongside the row number. could you please help me with this. the code which I have written is below:
*newdepth corresponds to length of my data in mm, and the input function asks me to scan the data for a desired depth.
section = input('Size of section (mm): ');
sectionrows = (section*(rsize))/(newdepth);
sectionrows=round(sectionrows);
Row= [];
nsections = (rsize/sectionrows);
nsections =round(nsections);
for k = 1:sectionrows
for i = 1:nsections
Vmax = max(max(hil1(k+(i-1)*sectionrows,:))); % Reflected maximum voltage amplitude from normal tissue
Row=[Row Vmax];
end
end

채택된 답변

Honglei Chen
Honglei Chen 2012년 8월 23일
편집: Honglei Chen 2012년 8월 23일
It looks like below is a simplified version of what you want to do
a is a matrix with 6 rows and the max is found for every 3 rows
a = rand(6,10);
sz_a = size(a);
n = 3;
max(reshape(max(reshape(a,n,[])),[],sz_a(2)),[],2)
  댓글 수: 2
Srikanta Sharma
Srikanta Sharma 2012년 8월 24일
Hi Honglei,
Thank you for your answer, but I seem to have a problem with the code you suggested, I get the following error when I use your code.
my Matrix has 10019 rows and 824 columns. and I want to crawl down 200 rows every time, until the end and find a max value in that and save it as a vector for all the columns. Many thanks in advance.
Error using ==> reshape Product of known dimensions, 300, not divisible into total number of elements, 8255656.
Jan
Jan 2012년 8월 24일
편집: Jan 2012년 8월 24일
@Srikanta: Yes, Honglei's version is just a simplified methods, as mentioned. You have to adjust the maximum possible size by your own.

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Jan
Jan 2012년 8월 24일
편집: Jan 2012년 8월 24일
data = rand(10019, 824);
[row, col] = size(data);
len = floor(row / 200);
block = reshape(data(1:len * 200, :), 200, len, col);
result = squeeze(max(block, [], 1));

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