i'm trying to find local maxima of a vector of numbers using MATLAB. The built-in findpeaks function will work for a vector such as:
[0 1 2 3 2 1 1 2 3 2 1 0]
where the peaks (each of the 3's) only occupy one position in the vector, but if I have a vector like:
[0 1 2 3 3 2 1 1 2 3 2 1 0]
the first 'peak' occupies two positions in the vector and the findpeaks function won't pick it up.
Is there a nice way to write a maxima-finding function which will detect these sort of peaks?
You can use the REGIONALMAX function from the Image Processing Toolbox:
>> x = [0 1 2 3 3 2 1 1 2 3 2 1 0]
x =
0 1 2 3 3 2 1 1 2 3 2 1 0
>> idx = imregionalmax(x)
idx =
0 0 0 1 1 0 0 0 0 1 0 0 0
Something much easier:
a = [1 2 4 5 5 3 2];
b = find(a == max(a(:)));
output:
b = [4,5]
a = [ 0 1 2 3 3 2 1 2 3 2 1 ];
sizeA = length(a);
result = max(a);
for i=1:sizeA,
if a(i) == result(1)
result(length(result) + 1) = i;
end
end
result contains the max, followed by all the values locations that are equal to max.
Related
I have a matrix with some zero values I want to erase.
a=[ 1 2 3 0 0; 1 0 1 3 2; 0 1 2 5 0]
>>a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
However, I want to erase only the ones after the last non-zero value of each line.
This means that I want to retain 1 2 3 from the first line, 1 0 1 3 2 from the second and 0 1 2 5 from the third.
I want to then store the remaining values in a vector. In the case of the example this would result in the vector
b=[1 2 3 1 0 1 3 2 0 1 2 5]
The only way I figured out involves a for loop that I would like to avoid:
b=[];
for ii=1:size(a,1)
l=max(find(a(ii,:)));
b=[b a(ii,1:l)];
end
Is there a way to vectorize this code?
There are many possible ways to do this, here is my approach:
arotate = a' %//rotate the matrix a by 90 degrees
b=flipud(arotate) %//flips the matrix up and down
c= flipud(cumsum(b,1)) %//cumulative sum the matrix rows -and then flip it back.
arotate(c==0)=[]
arotate =
1 2 3 1 0 1 3 2 0 1 2 5
=========================EDIT=====================
just realized cumsum can have direction parameter so this should do:
arotate = a'
b = cumsum(arotate,1,'reverse')
arotate(b==0)=[]
This direction parameter was not available on my 2010b version, but should be there for you if you are using 2013a or above.
Here's an approach using bsxfun's masking capability -
M = size(a,2); %// Save size parameter
at = a.'; %// Transpose input array, to be used for masked extraction
%// Index IDs of last non-zero for each row when looking from right side
[~,idx] = max(fliplr(a~=0),[],2);
%// Create a mask of elements that are to be picked up in a
%// transposed version of the input array using BSXFUN's broadcasting
out = at(bsxfun(#le,(1:M)',M+1-idx'))
Sample run (to showcase mask usage) -
>> a
a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
>> M = size(a,2);
>> at = a.';
>> [~,idx] = max(fliplr(a~=0),[],2);
>> bsxfun(#le,(1:M)',M+1-idx') %// mask to be used on transposed version
ans =
1 1 1
1 1 1
1 1 1
0 1 1
0 1 0
>> at(bsxfun(#le,(1:M)',M+1-idx')).'
ans =
1 2 3 1 0 1 3 2 0 1 2 5
For example:
>> tmp = ones(5,5)
tmp =
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
I want a command like:
tmp(colNum - 2*rowNum > 0) = 0
that modifies entries of tmp when the column number is more than twice the row number e.g. it should produce:
tmp =
1 1 0 0 0
1 1 1 1 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
As a second example, tmp(colNum - rowNum == 0) = 0 should set the diagonal elements of tmp to be zero.
A possibly more efficient solution is to use bsxfun like so
nRows = 5;
nCols = 5;
bsxfun(#(col,row)~(col - 2*row > 0), 1:nCols, (1:nRows)')
You can generalize this to just accept a function so it becomes
bsxfun(#(col,row)~f(col,row), 1:nCols, (1:nRows)')
And now just replace f with exactly the way you specify the equation in your question i.e.
f = #(colNum, rowNum)(colNum - 2*rowNum > 0)
or
f = #(colNum, rowNum)(colNum - rowNum == 0)
of course it might make more sense to specify your function to accept (row,col) instead of (col,row) as that's how MATLAB indexes
You can use meshgrid to generate a grid of 2D coordinates, then use this to impose any condition you wish. The variant you seek outputs 2 2D matrices where the first matrix gives you the column locations and the second matrix outputs the row locations.
For example, given your situation above:
>> [X,Y] = meshgrid(1:5, 1:5)
X =
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
Y =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
You can see that each unique spatial location shared between X and Y give you the desired 2D location as if you were envisioning a 2D grid.
Therefore, you would do something like this for your first situation:
[X,Y] = meshgrid(1:5,1:5); % Generate 2D coordinates
tmp = ones(5); % Generate desired matrix
tmp(X > 2*Y) = 0; % Set desired locations to 0
We get:
>> tmp
tmp =
1 1 0 0 0
1 1 1 1 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Finally for your second example:
[X,Y] = meshgrid(1:5,1:5); % Generate 2D coordinates
tmp = ones(5); % Generate desired matrix
tmp(X == Y) = 0; % Set desired locations to 0
We get:
>> tmp
tmp =
0 1 1 1 1
1 0 1 1 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
Simply put, generate a grid of 2D coordinates, then use those directly to index into your desired matrix using logical / Boolean conditions to set the desired locations to 0.
Given A is symmetry matrix with size n and
A =
1 2 3 4 5 % The Position
1 [0 5 2 4 1
2 5 0 3 0 2
3 2 3 0 0 0
4 4 0 0 0 5
5 1 2 0 5 0]
B is a row vector that permute the matrix A row and column
B = [2 4 1 5 3]
The output that I want is
C =
2 4 1 5 3 % The New Position given by Matrix B
2 [0 0 5 2 3
4 0 0 4 5 0
1 5 4 0 1 2
5 2 5 1 0 0
3 3 0 2 0 0]
I can get the output by using simple for loop
index = [2,4,1,5,3];
C = zeros(5,5);
for i = 1:5
for j = 1:5
% Position of in square matrix n
% (i,j) = (i-1)*n + j
C(i,j) = A((index(i)-1)*5+index(j));
end
end
However, if I want to permute a matrix with size 80x80, then I need to run 1600 times in order to get the output.
Is there any simple trick to do it instead of using for loop?
You should be able to rearrange your matrices as follows:
C = A(index,index);
This rearranges each dimension according to the index variable independently.
This question already has answers here:
General method to find submatrix in matlab matrix
(3 answers)
Closed 8 years ago.
I have the vector:
1 2 3
and the matrix:
4 1 2 3 5 5
9 8 7 6 3 1
1 4 7 8 2 3
I am trying to find a simple way of locating the vector [1 2 3] in my matrix.
A function returning either coordinates (Ie: (1,2:4)) or a matrix of 1s where there is a match a 0s where there isn't, Ie:
0 1 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
So far, the only function I've found is is 'ismember', which however only tells me if the individual components of the vector appear in the matrix. Suggestions?
Use strfind with a linearized version of the matrix, and then convert linear indices to subindices. Care should be taken to remove matches of the vector spanning different rows.
mat = [1 2 3 1 2 3 1 2;
3 0 1 2 3 5 4 4]; %// data
vec = [1 2 3]; %// data
ind = strfind(reshape(mat.',[],1).', vec);
[col row] = ind2sub(fliplr(size(mat)), ind);
keep = col<=size(mat,2)-length(vec)+1; %// remove result split across rows
row = row(keep);
col = col(keep);
Result for this example:
>> row, col
row =
1 1 2
col =
1 4 3
meaning the vector appears three times: row 1, col 1; row 1, col 4; row 2, col 3.
The result can be expressed in zero-one form as follows:
result = zeros(fliplr(size(mat)));
ind_ones = bsxfun(#plus, ind(keep).', 0:numel(vec)-1);
result(ind_ones) = 1;
result = result.';
which gives
>> result
result =
1 1 1 1 1 1 0 0
0 0 1 1 1 0 0 0
One way to get the starting location of the vector in the matrix is using colfilt:
>> A = [1 2 3 1 2 3 1 2; ...
3 0 1 2 3 5 4 4]; % matrix from Luis Mendo
>> T = [1 2 3];
>> colFun = #(x,t) all(x==repmat(t,1,size(x,2)),1);
>> B = colfilt(A,size(T),'sliding',colFun,T(:))
B =
0 1 0 0 1 0 0 0
0 0 0 1 0 0 0 0
That gives a mask of the center points, which translate to (row,col) coordinates:
>> [ii,jj]=find(B);
>> locs = bsxfun(#minus,[ii jj],floor((size(T)-1)/2))
locs =
1 1
2 3
1 4
I want to create a matrix like
A = [0 0 0 0 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 1 1 1;
0 1 1 1 1]
based on a vector indicating how many '0's should precede '1's on each row:
B = [4 3 3 3 2 1]
Is there a loopless way to do this ?
You don't mention in your question how the horizontal size of the array should be defined (the number of ones).
For predefined width you can use this code:
width = 5;
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,width-x) ], B, 'UniformOutput', false)');
If you want that A has minimal width, but still at least one 1 in every row:
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,max(B)+1-x) ], B, 'UniformOutput', false)');
A shorter “old-school” way to achieve this without a loop would be as follows:
A = repmat(B',1,max(B)+1)<repmat([1:max(B)+1],size(B,2),1)
If you want to have a minimum number of ones
min_ones=1; %or whatever
A = repmat(B',1,max(B)+min_ones)<repmat([1:max(B)+min_ones],size(B,2),1)
I don’t know how this compares speedwise to #nrz’s approach (I’ve only got Octave to hand right now), but to me it's more intuitive as it’s simply comparing a max(B) + min_ones * column tiling of B:
4 4 4 4 4
3 3 3 3 3
3 3 3 3 3
3 3 3 3 3
2 2 2 2 2
1 1 1 1 1
with a row tiling of [1 : max(B) + min_ones]
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
To generate:
A =
0 0 0 0 1
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
0 0 1 1 1
0 1 1 1 1
This requires only one line, and seems to be faster than previous solutions based on repmat or arrayfun:
%// Example data
ncols = 5;
B = [4 3 3 3 2 1];
%// Generate A
A = bsxfun(#gt, 1:ncols, B(:));