I have a matrix which contains values and I wish to find the index of the top n minimum values.
I use the following code for finding the minimum most value:
[r,c]=find(Result==min(min(Result)));
I cant find any other questions on stack overflow which answer the question, please help
Maybe you could do something like this:
sorted = sort(Result(:));
topten = sorted(1:10);
[~,ia,~] = intersect(Result(:),topten(:)); % // Get the indices of the top ten values
[r,c]=ind2sub(size(Result),ia); % // Convert the indices to rows and columns
Or without Intersect in the other answer
[sorted,I] = sort(Result(:));
[r,c] = ind2sub(size(Result),I(1:10)); %//Change 10 to any other required value
Use prctile (Statistics Toolbox) to find the appropriate threshold, and then use indexing to select the elements above that threshold:
x = magic(4); %// example
n = 5; %// we want the top n elements
M = numel(x);
p = prctile(x(:), (M-n)/M*100);
indices = find(x>p); %// result in the form linear indices
[row, col] = find(x>p); %// result in the form of row and column indices
In this example:
>> x
x =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> indices.'
ans =
1 8 12 13 15
>> row.'
ans =
1 4 4 1 3
>> col.'
ans =
1 2 3 4 4
>> x(indices).'
ans =
16 14 15 13 12
Example with repeated elements:
>> x = [1 1 2 5; 3 4 3 5];
>> n = 5;
gives
>> indices.'
ans =
2 4 6 7 8
>> row.'
ans =
2 2 2 1 2
>> col.'
ans =
1 2 3 4 4
>> x(indices).'
ans =
3 4 3 5 5
Related
I have a vector of 10 random numbers from 1 to 11 (median of the vector 1:11 is 6).
min = 1;
max = 11;
nVector = 10;
VectorRand = randi([min max],1,nVector);
I would like to convert the values in VectorRand to a circular array of 11 values, but with a different median. For example, median 2:
-5(8) -4(9) -3(10) -2(11) -1(1) 0(2) +1(3) +2(4) +3(5) +4(6) +5(7)
Result: 8 9 10 11 1 - 2 - 3 4 5 6 7
in the case of VectorRand = [1 3 8 4 6 8 5 2 6 8 10]
Result: -1 1 -5 2 4 -5 3 0 4 -5 -3
where the median (2 in this case) becomes zero and all the other values are translated in terms of distance from the median (e.g. 8=-5, 9=-4 10=-3 and so on).
The description is confusing, but here's my guess as to what you want:
>> % Set parameters
>> minVal = 1;
>> maxVal = 11;
>> newMedian = 2;
>> % Determine mapping
>> currMedian = median(minVal:maxVal);
>> map = circshift((minVal:maxVal) - currMedian, [0 newMedian - currMedian]);
This is shifting the range of values so that the median is 0 (this will eventually give you the distances from the target median), and then rotating it so the target median is at the center value of the ordering of values. Now, with the mapping built, you can map your vector to this new circulant ring:
>> VectorRand = [1 3 8 4 6 8 5 2 6 8 10];
>> Result = map(VectorRand)
Result =
-1 1 -5 2 4 -5 3 0 4 -5 -3
If I have matrix size(mat)= X*Y*6
let call mat(:,:,1)=A
and mat(:,:,2)=B and so on
how do i rezise mat to X*Y*12
where
mat(:,:,1)=mat(:,:,2)= A
mat(:,:,3)=mat(:,:,4)=B
and so on
You can use the following syntax:
%defines input matrix (in your case it is already defined)
m = 500;
n = 400;
z = 6;
mat = rand(m,n,z);
%initialize output matrix
newMat = zeros(m,n,z*2);
%assign old matrix values into the new matrix
newMat(:,:,1:2:end) = mat;
newMat(:,:,2:2:end) = mat;
If you have Matlab 2015a or newer you can use repelem:
N = 2; %// number of times to repeat
result = repelem(mat, 1, 1, N); %// repeat N times along 3rd dim
For older Matlab versions you can do it manually as follows:
N = 2; %// number of times to repeat
ind = ceil(1/N:1/N:size(mat,3)); %// build index with repetitions
result = mat(:,:,ind); %// apply index along desired dim
Example:
>> %// Data
>> mat = randi(9,2,4,2)
mat(:,:,1) =
5 8 9 2
7 3 1 5
mat(:,:,2) =
5 7 1 1
1 8 8 2
>> %// First approach
>> N = 2;
>> result = repelem(mat, 1, 1, N)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
>> %// Second approach
>> N = 2;
>> ind = ceil(1/N:1/N:size(mat,3));
>> result = mat(:,:,ind)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
I need to find the maximum among values with same labels, in matlab, and I am trying to avoid using for loops.
Specifically, I have an array L of labels and an array V of values, same size. I need to produce an array S which contains, for each value of L, the maximum value of V. An example will explain better:
L = [1,1,1,2,2,2,3,3,3,4,4,4,1,2,3,4]
V = [5,4,3,2,1,2,3,4,5,6,7,8,9,8,7,6]
Then, the values of the output array S will be:
s(1) = 9 (the values V(i) such that L(i) == 1 are: 5,4,3,9 -> max = 9)
s(2) = 8 (the values V(i) such that L(i) == 2 are: 2,1,2,8 -> max = 8)
s(3) = 7 (the values V(i) such that L(i) == 3 are: 3,4,5,7 -> max = 7)
s(4) = 8 (the values V(i) such that L(i) == 4 are: 6,7,8,6 -> max = 8)
this can be trivially implemented by traversing the arrays L and V with a for loop, but in Matlab for loops are slow, so I was looking for a faster solution. Any idea?
This is a standard job for accumarray.
Three cases need to be considered, with increasing generality:
Integer labels.
Integer labels, specify fill value.
Remove gaps; or non-integer labels. General case.
Integer labels
You can just use
S = accumarray(L(:), V(:), [], #max).';
In your example, this gives
>> L = [1 1 1 2 2 2 3 3 3 4 4 4 1 2 3 7];
>> V = [5 4 3 2 1 2 3 4 5 6 7 8 9 8 7 6];
>> S = accumarray(L(:), V(:), [], #max).'
S =
9 8 7 8
Integer labels, specify fill value
If there are gaps between integers in L, the above will give a 0 result for the non-existing labels. If you want to change that fill value (for example to NaN), use a fifth input argument in acccumarray:
S = accumarray(L(:), V(:), [], #max, NaN).';
Example:
>> L = [1 1 1 2 2 2 3 3 3 4 4 4 1 2 3 7]; %// last element changed
>> V = [5 4 3 2 1 2 3 4 5 6 7 8 9 8 7 6]; %// same as in your example
>> S = accumarray(L(:), V(:), [], #max, NaN).'
S =
9 8 7 8 NaN NaN 6
Remove gaps; or non-integer labels. General case
When the gaps between integer labels are large, using a fill value may be inefficient. In that case you may want to get only the meaningful values in S, without fill values, i.e.skip non-existing labels. Also, it may be the case that L doesn't necessarily contain integers.
These two issues are solved by applying unique to the labels before using accumarray:
[~, ~, Li] = unique(L); %// transform L into consecutive integers
S = accumarray(Li(:), V(:), [], #max, NaN).';
Example:
>> L = [1.5 1.5 1.5 2 2 2 3 3 3 4 4 4 1 2 3 7.8]; %// note: non-integer values
>> V = [5 4 3 2 1 2 3 4 5 6 7 8 9 8 7 6 ]; %// same as in your example
>> [~, ~, Li] = unique(L); %// transform L into consecutive integers
>> S = accumarray(Li(:), V(:), [], #max, NaN).'
S =
9 5 8 7 8 6
helper=[L.', V.'];
helper=sortrows(helper,-2);
[~,idx,~]=unique(helper(:,1));
S=helper(idx,2);
What I do is: I join the two arrays as columns. Then I sort them regarding second column with biggest element first. Then I get the idx of the unique Values in L before I return the corresponding Values from V.
The solution from Luis Mendo is faster. But as far as I see his solution doesn't work if there is a zero,negative value or a noninteger inside L:
Luis solution: Elapsed time is 0.722189 seconds.
My solution: Elapsed time is 2.575943 seconds.
I used:
L= ceil(rand(1,500)*10);
V= ceil(rand(1,500)*250);
and ran the code 10000 times.
Referring to Reshape row wise w/ different starting/ending elements number #Divakar came with a nice solution but, what if the number of columns is not always the same?
Sample run -
>> A'
ans =
4 9 8 9 6 1 8 9 7 7 7 4 6 2 7 1
>> out
out =
4 9 8 9 0 0
6 1 8 9 7 7
7 4 6 2 7 1
I took only the first 4 terms of A and put them in out, then fill the rest 2 empty cell with 0's. So the ncols = [4 6 6]. Unfortunately vet2mat doesn't allow vector as columns number.
Any suggestions?
You can employ bsxfun's masking capability here -
%// Random inputs
A = randi(9,1,15)
ncols = [4 6 5]
%// Initialize output arary of transposed size as compared to the desired
%// output arary size, as we need to insert values into it row-wise and MATLAB
%// follows column-major indexing
out = zeros(max(ncols),numel(ncols));
mask = bsxfun(#le,[1:max(ncols)]',ncols); %//'# valid positions mask for output
out(mask) = A; %// insert input array elements
out = out.' %//'# transpose output back to the desired output array size
Code run -
A =
5 3 7 2 7 2 4 6 8 1 9 7 5 4 5
ncols =
4 6 5
out =
5 3 7 2 0 0
7 2 4 6 8 1
9 7 5 4 5 0
You could use accumarray for that:
A = [4 9 8 9 6 1 8 9 7 7 7 4 6 2 7 1].'; %'// data
ncols = [4 6 6]; %// columns
n = max(ncols);
cs = cumsum(ncols);
ind = 1;
ind(cs+1) = 1;
ind = cumsum(ind(1:end-1)); %// `ind` tells the row for each element of A
result = accumarray(ind(:), A(:), [], #(x) {[x; zeros(n-numel(x),1)]}); %// split `A` as
%// dictated by `ind`, and fill with zeros. Each group is put into a cell.
result = [result{:}].'; %'// concatenate all cells
just lets make it simple, assume that I have a 10x3 matrix in matlab. The numbers in the first two columns in each row represent the x and y (position) and the number in 3rd columns show the corresponding value. For instance, [1 4 12] shows that the value of function in x=1 and y=4 is equal to 12. I also have same x, and y in different rows, and I want to average the values with same x,y. and replace all of them with averaged one.
For example :
A = [1 4 12
1 4 14
1 4 10
1 5 5
1 5 7];
I want to have
B = [1 4 12
1 5 6]
I really appreciate your help
Thanks
Ali
Like this?
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7];
[x,y] = consolidator(A(:,1:2),A(:,3),#mean);
B = [x,y]
B =
1 4 12
1 5 6
Consolidator is on the File Exchange.
Using built-in functions:
sparsemean = accumarray(A(:,1:2), A(:,3).', [], #mean, 0, true);
[i,j,v] = find(sparsemean);
B = [i.' j.' v.'];
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7]; %your example data
B = unique(A(:, 1:2), 'rows'); %find the unique xy pairs
C = nan(length(B), 1);
% calculate means
for ii = 1:length(B)
C(ii) = mean(A(A(:, 1) == B(ii, 1) & A(:, 2) == B(ii, 2), 3));
end
C =
12
6
The step inside the for loop uses logical indexing to find the mean of rows that match the current xy pair in the loop.
Use unique to get the unique rows and use the returned indexing array to find the ones that should be averaged and ask accumarray to do the averaging part:
[C,~,J]=unique(A(:,1:2), 'rows');
B=[C, accumarray(J,A(:,3),[],#mean)];
For your example
>> [C,~,J]=unique(A(:,1:2), 'rows')
C =
1 4
1 5
J =
1
1
1
2
2
C contains the unique rows and J shows which rows in the original matrix correspond to the rows in C then
>> accumarray(J,A(:,3),[],#mean)
ans =
12
6
returns the desired averages and
>> B=[C, accumarray(J,A(:,3),[],#mean)]
B =
1 4 12
1 5 6
is the answer.