Need a matlab code for calling rows in MATLAB
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I have matrix
D = [...
8.5000 0.6000 92.1000 5.3420
8.5000 0.6000 92.0000 5.4390
8.5000 0.6000 92.2000 5.3520
8.4000 0.6000 92.1000 5.4490
8.5000 0.6000 92.1000 5.3620
8.5000 0.6000 92.1000 5.4590
8.5000 0.6000 92.2000 5.3720
8.5000 0.6000 92.1000 5.4690
8.5000 0.6000 92.1000 5.3820]
Now i want to write a matlab code such that it will take first 3 rows. e.g. in this case -
8.5000 0.6000 92.1000 5.3420
8.5000 0.6000 92.0000 5.4390
8.5000 0.6000 92.2000 5.3520
then calculate the mean of each column. e.g. for 4th column - (5.342+5.439+5.350/3)= 5.377 and this will continue till n number columns. (n is multiple of 3 assumed.)
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回答 (3 件)
Image Analyst
2013 年 3 月 1 日
編集済み: Image Analyst
2013 年 3 月 1 日
Try this:
theColumnMeans = mean(D(1:3,:))
theColumnMeans =
8.5000 0.6000 92.1000 5.3777
I'm not sure what "this will continue till n number columns" means though. Please explain that part. Perhaps you meant "this will continue with every groups of 3 rows until the last row in the matrix", in which case you can try this:
theColumnMeans = mean(D(1:3,:))
theColumnMeans = mean(D(4:6,:))
theColumnMeans = mean(D(7:9,:))
% Alternate way using blockproc():
meanFilterFunction = @(theBlockStructure) mean2(theBlockStructure.data(:));
blockSize = [3 1];
blockyImage = blockproc(D, blockSize, meanFilterFunction)
In the command window:
theColumnMeans =
8.5000 0.6000 92.1000 5.3777
theColumnMeans =
8.4667 0.6000 92.1000 5.4233
theColumnMeans =
8.5000 0.6000 92.1333 5.4077
blockyImage =
8.5000 0.6000 92.1000 5.3777
8.4667 0.6000 92.1000 5.4233
8.5000 0.6000 92.1333 5.4077
0 件のコメント
Youssef Khmou
2013 年 3 月 1 日
hi,
Given your D matrix :
N=3;
D2=D(1:N,:); % select N rows
M=mean(D2) % M is vector that each elements j is the mean of column j in D2
2 件のコメント
Youssef Khmou
2013 年 3 月 1 日
編集済み: Youssef Khmou
2013 年 3 月 1 日
hi, good that it works ok you can create a function that does what you described, i tried to write one for you :
function Y=Consecu_3rMeans( X)
% Your input is matrix X
[m,n]=size(X);
Y=zeros(floor(m/3),n);
j=3;
for i=1:3:m-3
Y(floor((i+3)/3),:)=mean(X(i:j,:));
j=j+3;
end
% check if the number of rows is divisible by 3, if not then there are non
% treated rows .
N=mod(m,3);
if N~=0
Last_row=mean(X(m-3+2:end,:));
%concatenation :
Y=[Y;Last_row];
end
Note if the number of rows is not divisible by 3, then the last rows are not treated so you need to compute theirs means and add the values to the output matrix .
an example :
M=rand(20); % 20lines and 20 columns
% So we compute the mean columns wise of lines 1:3, 4:6, 7:9, %10:12,13:15,16:18 BUT 19:20 are included in the function code .
y=Consecu_3rMeans(X);
%y is of size 7x20 such that each row is mean of i:i+3 rows of input matrix , and 7 not 6 because we added a line that inculdes E[19:20] .
i hope this helps
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