finding values from matrix

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Ede gerlderlands
Ede gerlderlands 2013 年 7 月 15 日
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

回答 (3 件)

Azzi Abdelmalek
Azzi Abdelmalek 2013 年 7 月 15 日
編集済み: Azzi Abdelmalek 2013 年 7 月 15 日
out=A(A>45)
If you want the location:
idx=find(A>45) % corresponding indices
out=A(idx)
  3 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2013 年 7 月 15 日
Ok
A=randi(200,5) % Example
[ii,jj]=find(A>45)
Azzi Abdelmalek
Azzi Abdelmalek 2013 年 7 月 15 日
編集済み: Azzi Abdelmalek 2013 年 7 月 15 日
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))

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Matt J
Matt J 2013 年 7 月 15 日
[i,j]=find(A>45);
  7 件のコメント
Matt J
Matt J 2013 年 7 月 15 日
for k=1:size(A,1)
S(k).locs = find(A(k,:)>45);
end
Edwin Herrera Vasco
Edwin Herrera Vasco 2020 年 5 月 4 日
Thanks!!!

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Iain
Iain 2013 年 7 月 15 日
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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