I want to find the minimum of a matrix say A. I could do it in this way:
NM = find(A==min(A));
but I need the minimum array of A which for it, c(NM) is not zero. How can I put this condition on finding minimum value?
Example:
c=[0,18,9,0,100,0];
A=[1,189,125,25,7,1];
I expext it returns 5.
You can use logical indexing to find the desired elements.
find(A==min(A(c~=0)))
Explanation:
Using logical indexing you first find indices of elements of c that are nonzero.
idx1 = c~=0;
then elements of A that correspond to indices of nonzeros elemets of c are extracted.
A1 = A(idx1);
then we find minimum of the extracted elements:
mn = min(A1);
again we use logical indexing to find elements of A that are equal to its minimum.
idx2 = A == mn;
finally using find the logical index idx2 converted to linear index.
result = find (idx2);
First convert your Matrix to a vector (otherwise the minimum of a Matrix is confusing, it might mean the minimum of each column).
Then you can sort the Vector and get the indexes
[val idx] = sort(A(:));
You can then use the indexes to sort your vector c
c = c(idx);
And finally get the first non-zero element of c:
c = c(c~=0);
result = c(1);
Related
I have sum of 3 cell arrays
A=72x1
B=72x720
C=72x90
resultant=A+B+C
size of resultant=72x64800
now when I find the minimum value with row and column indices I can locate the row element easily but how can I locate the column element in variables?
for example
after dong calculations for A,B,C I added them all and got a resultant in from of <72x(720x90)> or can say a matrix of integers of size <72x64800> then I found the minimum value of resultant with row and column index using the code below.
[minimumValue,ind]=min(resultant(:));
[row,col]=find(result== minimumValue);
then row got 14 and column got 6840 value..
now I can trace row 14 of all A,B,C variables easily but how can I know that the resultant column 6480 belongs to which combination of A,B,C?
Instead of using find, use the ind output from the min function. This is the linear index for minimumValue. To do that you can use ind2sub:
[r,c] = ind2sub(size(resultant),ind);
It is not quite clear what do you mean by resultant = A+B+C since you clearly don't sum them if you get a bigger array (72x64800), on the other hand, this is not a simple concatenation ([A B C]) since this would result in a 72x811 array.
However, assuming this is a concatenation you can do the following:
% get the 2nd dimension size of all matrices:
cols = cellfun(#(x) size(x,2),{A,B,C})
% create a vector with reapiting matrices names for all their columns:
mats = repelem(['A' 'B' 'C'],cols);
% get the relevant matrix for the c column:
mats(c)
so mats(c) will be the matrix with the minimum value.
EDIT:
From your comment I understand that your code looks something like this:
% arbitrary data:
A = rand(72,1);
B = rand(72,720);
C = rand(72,90);
% initializing:
K = size(B,2);
N = size(C,2);
counter = 1;
resultant = zeros(72,K*N);
% summing:
for k = 1:K
for n = 1:N
resultant(:,counter) = A + B(:,k) + C(:,n);
counter = counter+1;
end
end
% finding the minimum value:
[minimumValue,ind] = min(resultant(:))
and from the start of the answer you know that you can do this:
[r,c] = ind2sub(size(resultant),ind)
to get the row and column of minimumValue in resultant. So, in the same way you can do:
[Ccol,Bcol] = ind2sub([N,K],c)
where Bcol and Ccol is the column in B and C, respectively, so that:
minimumValue == A(r) + B(r,Bcol) + C(r,Ccol)
To see how it's working imagine that the loop above fills a matrix M with the value of counter, and M has a size of N-by-K. Because we fill M with a linear index, it will be filled in a column-major way, so the row will correspond to the n iterator, and the column will correspond to the k iterator. Now c corresponds to the counter where we got the minimum value, and the row and column of counter in M tells us the columns in B and C, so we can use ind2sub again to get the subscripts of the position of counter. Off course, we don't really need to create M, because the values within it are just the linear indices themselves.
I have a huge matrix MxN matrix, say, A=rand([M,N]); and an index vector with N integer values between 1 and M, say, RandomIndex = randi(M,[1,N]);.
Now I would like to generate a row vector with entries
result = [A(RandomIndex(1),1), A(RandomIndex(2),2), ..., A(RandomIndex(N),N)]
What would be an efficient way to do this? It should be a very cheap operation but all my implementations are slow. I don't think there is a notation in Matlab to do this directly, is there?
The fastest option so far is
indexFunction = #(r,c) A(r,c);
result = cell2mat(arrayfun(indexFunction,RandomIndex,1:N,'UniformOutput',false));
Is there a more efficient way?
Use sub2ind
A(sub2ind(size(A), RandomIndex, 1:N))
sub2ind will convert the row and column indices given by RandomIndex and 1:N to linear indices based on size(A) which you can then use to index A directly.
Another way to do this is to use RandomIndex and 1:N to return an NxN matrix and then take the diagonal of this with diag
diag(A(RandomIndex, 1:N)).'
Note: .' is used to convert the row vector returned by diag to a column vector.
M=10;N=50;
A=rand([M,N]);
RandomIndex = randi(M,[1,N]);
out = zeros(1,numel(RandomIndex));
for ii = 1:numel(RandomIndex)
out(ii)=A(RandomIndex(ii),ii);
end
Another approach would be to use sparse and logical indexing:
M = sparse(RandomIndex, 1:N, 1) == 1;
out = A(M);
The first line of code generates a logical matrix where there is only 1 true value set in each column. This is defined by each value of RandomIndex. We convert this to a logical matrix, then index into your matrix to obtain the final random vector.
Use your index directly.
M = 100;N=100;
A = rand(M,N);
% get a random index that can be as large as your matrix
% 10 rows by 1 column
idx = randi(numel(A), 10,1);
t = A(idx);
How to find the non zero negative/ positive element for each column.
size A= 3*100 and for each column we have one non zero negative/positive element. How to find the indices of those elements without using a for loop?
B = A>0;
C = A<0;
D = B-C;
D now contains a -1 if the entry is negative and a 1 if the entry is positive. Columns can be extracted by using the indices:
D(:,1) %first column
D(:,2) %second column
etc.
[row,column] = find(A>0)
would also do the trick and gives you indices as well.
Suppose in MATLAB I have a real matrix A which is n x m and a binary matrix B of the same size. The latter matrix defines the optimization set (all indices for which the element of B equals one): over this set I would like to find the maximal element of A. How can I do this?
The first idea I had is that I consider C = A.*B and look for the maximal element of C. This works fine for all matrices A which have at least one positive element, however it does not work for matrices with all negative elements.
You can do
C = A(B==1);
to give you an array of just the values of A corresponding to a value of 1 in B. And
max( C )
will give you the maximum value of A where B is 1
With this method you don't run into a problem when all values of A are negative as the zeros don't appear in C.
Obviously you can condense this to
desiredValue = max(A(B(:)==1));
I am using the colon operator to make sure that the result of A(B(:)==1) is a column vector - if B is all ones I am not sure if Matlab would return a vector or a nxm matrix (and I can't confirm right now).
update to get the index of the value, you can do:
f = find(B==1);
[m mi] = max(A(f));
maxIndex = f(mi);
And to get that back to the 2D elements:
[i j] = ind2sub(size(A), maxIndex);
very basic question: How can I find the highest or lowest value in a random matrix.
I know there is a possibility to say:
a = find(A>0.5)
but what I'm looking for would be more like this:
A = rand(5,5)
A =
0.9388 0.9498 0.6059 0.7447 0.2835
0.6338 0.0104 0.5179 0.8738 0.0586
0.9297 0.1678 0.9429 0.9641 0.8210
0.0629 0.7553 0.7412 0.9819 0.1795
0.3069 0.8338 0.7011 0.9186 0.0349
% find highest (or lowest) value
ans = A(19)%for the highest or A(7) %for the lowest value in this case
Have a look at the min() and max() functions. They can return both the highest/lowest value, and its index:
[B,I]=min(A(:)); %# note I fixed a bug on this line!
returns I=7 and B=A(7)=A(2,2). The expression A(:) tells MATLAB to treat A as a 1D array for now, so even though A is 5x5, it returns the linear index 7.
If you need the 2D coordinates, i.e. the "2,2" in B=A(7)=A(2,2), you can use [I,J] = ind2sub(size(A),I) which returns I=2,J=2, see here.
Update
If you need all the entries' indices which reach the minimum value, you can use find:
I = find(A==min(A(:));
I is now a vector of all of them.
For matrices you need to run the MIN and MAX functions twice since they operate column-wise, i.e. max(A) returns a vector with each element being the maximum element in the corresponding column of A.
>> A = rand(4)
A =
0.421761282626275 0.655740699156587 0.678735154857773 0.655477890177557
0.915735525189067 0.0357116785741896 0.757740130578333 0.171186687811562
0.792207329559554 0.849129305868777 0.743132468124916 0.706046088019609
0.959492426392903 0.933993247757551 0.392227019534168 0.0318328463774207
>> max(max(A))
ans =
0.959492426392903
>> min(min(A))
ans =
0.0318328463774207
Note that this only works for matrices. Higher dimensional arrays would require running MIN and MAX as many times as there are dimensions which you can get using NDIMS.
Try this out
A=magic(5)
[x,y]=find(A==max(max(A))) %index maximum of the matrix A
A_max=A(x,y)
[x1,y1]=find(A==min(max(A))) %index minimum of the matrix A
A_min=A(x1,y1)