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minimum of an array

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Rakshmy  C S
Rakshmy C S on 10 Dec 2011
Hi, I have a 10*10 array.I need to find the minium of all the rows of the array except selected ones.For eg. to find the minimum of the array except 2nd and fifthrow.

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 13 Dec 2011
r = [1 3 5];
A1 = A;
A1(r,:) = nan;
[c,idx] = min(A1(:));
[i1,j1] = ind2sub(size(A1),idx);
  1 Comment
Rakshmy  C S
Rakshmy C S on 13 Dec 2011
thanks a lot.its working.

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More Answers (3)

Sean de Wolski
Sean de Wolski on 12 Dec 2011
A = [999, 25.019, 1.414, 25.806, 1.414, 3.316, 4.472, 29.782, 26.248, 28.248
25.019, 999, 25.059, 5.291, 25.495, 25.278, 25.573, 8.774, 3.316, 7.348
1.414, 25.059, 999, 25.612, 2.449, 4.358, 5.477, 29.546, 26.134, 28.035
25.806, 5.291, 25.612, 999, 26.495, 26.514, 27.092, 5.385, 4.795, 8.124
1.414, 25.495, 2.449, 26.495, 999, 2.236, 3.162, 30.577, 26.739, 28.635
3.316, 25.278, 4.358, 26.514, 2.236, 999, 1.732, 30.822, 26.645, 28.513
4.472, 25.573, 5.477, 27.092, 3.162, 1.732, 999, 31.416, 26.925, 28.670
29.782, 8.774, 29.546, 0, 30.577, 30.822, 31.416, 999, 7.874, 10.535
26.248, 3.316, 26.134, 4.795, 26.739, 26.645, 26.925, 7.874, 999, 4.358
28.248, 7.348, 28.035, 8.124, 28.635, 28.513, 28.670, 10.535, 4.358, 999];
exc = [1 3 5];
[Amin idc] = min(A(setdiff(1:size(A,1), exc),:),[],2);
[Amin idr] = min(Amin);
idc = idc(idr);
idr = idr+sum(exc<=idr);
fprintf('Minimum %f at row %i col %i\n',Amin,idr,idc);
Minimum 0.000000 at row 8 col 4
I added a zero to test it.
  5 Comments
Andrei Bobrov
Andrei Bobrov on 13 Dec 2011
small corrected
r1 = setdiff(1:size(A,1),exc)
[Amin,idr] = min(A(r1,:))
[Amin,idc] = min(Amin)
idr = r1(idr(idc))
or
r1 = setdiff(1:size(A,1),exc)
[Amin,c1] = min(A(r1,:),[],2)
[Amin,r2] = min(Amin)
idc = c1(r2)
idr = r1(r2)
Rakshmy  C S
Rakshmy C S on 13 Dec 2011
Thank you. this code is also working

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Walter Roberson
Walter Roberson on 10 Dec 2011
Create a new matrix that includes only the desired information, and apply the minimum to that.
This can possibly be done without any assignment to variables, by using indexing, but whether you will be able to handle things that way depends on your intention.
B = A; %work on a copy, not the original
B([2 5],:) = []; %remove the 2nd and 5th rows
%now apply the appropriate min() function to B.
or
apply min() to A(setdiff(1:size(A,1), [2 5]),:)
I am being vague about the min because I cannot tell whether you mean minimum across each row with rows 2 and 5 happening not to be wanted; or if you minimum down each column after row 2 and 5 have been ignored; or if you want the minimum over the entire array but excluding the content of rows 2 and 5.
  2 Comments
Rakshmy  C S
Rakshmy C S on 10 Dec 2011
i want to find the minimum over the entire array but excluding the content of row2 and 5
Walter Roberson
Walter Roberson on 10 Dec 2011
min(min(A(setdiff(1:size(A,1), [2 5]),:)))

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Mohsen  Davarynejad
Mohsen Davarynejad on 10 Dec 2011
My guess is the the following scales better:
B = A; %work on a copy, not the original
B([2 5],:) = []; %remove the 2nd and 5th rows
[Min, MinIndex] = min(B(:));
  9 Comments
Rakshmy  C S
Rakshmy C S on 12 Dec 2011
Sorry its not working, i am having the values as given below.It is a 10*10 array. I want to find the minimum value of the elements of array excluding [1,3,5].So the value is 1.7321. and the index is 6,7 or 7,6. But after executing the above code i get 6,9 as the index.
999, 25.019, 1.414, 25.806, 1.414, 3.316, 4.472, 29.782, 26.248, 28.248
25.019, 999, 25.059, 5.291, 25.495, 25.278, 25.573, 8.774, 3.316, 7.348
1.414, 25.059, 999, 25.612, 2.449, 4.358, 5.477, 29.546 26.134, 28.035
25.806, 5.291, 25.612, 999 26.495, 26.514, 27.092, 5.385, 4.795, 8.124
1.414, 25.495, 2.449, 26.495, 999, 2.236, 3.162, 30.577, 26.739, 28.635
3.316, 25.278, 4.358, 26.514, 2.236, 999 1.732, 30.822, 26.645, 28.513
4.472, 25.573, 5.477, 27.092, 3.162, 1.732, 999 31.416 26.925, 28.670
29.782, 8.774, 29.546, 5.385, 30.577, 30.822, 31.416, 999 7.874, 10.535
26.248, 3.316, 26.134, 4.795, 26.739, 26.645, 26.925, 7.874 999 4.358
28.248, 7.348, 28.035, 8.124, 28.635, 28.513, 28.670, 10.535, 4.358 999
Walter Roberson
Walter Roberson on 12 Dec 2011
I see what you mean, and I have an idea of how to correct for it, but I need to work on other things now.

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