finding values from matrix

Dear All
I have a matrix which is given by A= [20 140]. I want to find the location of elements in A which are greter than 45. Here is the matlab script which I tried to work upon but it gives me the whole value rather than location with respect to each coumn.
for i = 1:20
Loc(ii,:)= find(A(i,:)>45);
end
Thank you for your help

Risposte (3)

Azzi Abdelmalek
Azzi Abdelmalek il 15 Lug 2013
Modificato: Azzi Abdelmalek il 15 Lug 2013
out=A(A>45)
If you want the location:
idx=find(A>45) % corresponding indices
out=A(idx)

3 Commenti

This gives me the total values. What I want is the position(location) of values which are greater than 45. The locationhas to be with respect to the column ex. for column 1, 5, 65 ,78 etc... for column two 4, 5,8,12,...
Ok
A=randi(200,5) % Example
[ii,jj]=find(A>45)
Azzi Abdelmalek
Azzi Abdelmalek il 15 Lug 2013
Modificato: Azzi Abdelmalek il 15 Lug 2013
A=[0 46 10 1; 47 5 48 5; 3 19 25 70]
out=arrayfun(@(x) max(find(A(x,:)>45)),1:size(A,1))

Accedi per commentare.

Matt J
Matt J il 15 Lug 2013
[i,j]=find(A>45);

7 Commenti

I think am not clear with you. Is it not possible to get the location per co;umn? this gives me the over all location? isn't that? tnx
Matt J
Matt J il 15 Lug 2013
Modificato: Matt J il 15 Lug 2013
Show us the desired output for this small example matrix
A=[0 46 10 1; 47 5 48 5; 3 19 25 70]
And explain what form the output should take (cell, matrix, structure, etc...)
The answer will be 2 for the first column 1 and 3 foe the second as well as 4 for the third column. the output should be in str. Thank you
Matt J
Matt J il 15 Lug 2013
The example matrix I gave you has 4 columns...
sorry, it has to be rows
for k=1:size(A,1)
S(k).locs = find(A(k,:)>45);
end
Thanks!!!

Accedi per commentare.

Iain
Iain il 15 Lug 2013
A = [45 46; 42 43];
logical_address = A>45;
A(logical_address) %gives you a single 46.
linear_index = find(logical_address);
A(linear_index) % gives you a single 46.
[rowno, colno ] = sub2ind(size(A),linear_index);
A(rowno,colno ) % gives you a single 46
Looking at your code, I think you're trying to do that row by row.
find(A(i,:)>45) will give a vector containing the column number of each column which is >45 in that row of A - the length of this may vary from row to row, so you can't store the result in a simple vector - you'd need to use a cell array.

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