This question already has answers here:
Implementing matching pursuit algorithm
(3 answers)
Closed 6 years ago.
I'm trying to implement Matching Pursuit algorithm in Matlab.I have found out the maximum inner product value ,i m stuck with how to find out the coeffients.
help me out.
Here is the algorithm
D=[1 6 11 16 21 26 31 36 41 46
2 7 12 17 22 27 32 37 42 47
3 8 13 18 23 28 33 38 43 48
4 9 14 19 24 29 34 39 44 49
5 10 15 20 25 30 35 40 45 50];
b=[16;17;18;19;20];
n=size(D);
A1=zeros(n);
R=b;
x=[];
H=10;
if(H <= 0)
error('The number of iterations needs to be greater then 0')
end;
[c,d] = max(abs(D'*R));
Here i have used a prefined dictionary.
Thanks in advance
You can use this function based on "S. Mallat, Z. Zhang, 1993. Matching pursuit in a time frequency dictionary. IEEE Transactions Signal Processing, Vol. 41, No. 12, pp. 3397-3415."
x = MP(b,D,10);
function S = MP(y,Dictionary,iteration)
n = size(Dictionary,2);
S = zeros(n,1);
% Normalize the dictionary atoms (coloumns) to have unit norm
% It's better to implement this part out of function,
% to normalize the dictionary just one time!
%**************************************
for j = 1:n;
Dictionary(:,j) = Dictionary(:,j)/norm(Dictionary(:,j));
end
% *************************************
for i = 1:iteration
gn = Dictionary' * y / norm(y);
[MAX,index] = max(abs(gn));
y = y - MAX * Dictionary(:,index);
S(index) = MAX + S(index);
end
Related
for a=1:50; %numbers 1 through 50
for b=1:50;
c=sqrt(a^2+b^2);
if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0
pyth=[a,b,c];%pythagorean matrix
disp(pyth)
else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output
end
end
end
answer=
3 4 5
4 3 5
5 12 13
6 8 10
7 24 25
8 6 10
8 15 17
9 12 15
9 40 41
10 24 26
12 5 13
12 9 15
12 16 20
12 35 37
14 48 50
15 8 17
15 20 25
15 36 39
16 12 20
16 30 34
18 24 30
20 15 25
20 21 29
21 20 29
21 28 35
24 7 25
24 10 26
24 18 30
24 32 40
27 36 45
28 21 35
30 16 34
30 40 50
32 24 40
35 12 37
36 15 39
36 27 45
40 9 41
40 30 50
48 14 50
This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!
Thank you
welcome to StackOverflow.
The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:
pyth = zeros(0,3);
for a=1:50
for b=1:50
c = sqrt(a^2 + b^2);
if c<=50 && rem(c,1)==0
abc_sorted = sort([a,b,c]);
pyth = [pyth; abc_sorted];
end
end
end
% do final sorting outside of the loop
[~,idx] = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)
a few other tips for writing MATLAB code:
You do not need to end for or if/else stements with a semicolon
else statements cover any other case not included before, so they do not need a condition.
Some performance reommendations:
Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).
Regards,
Chris
You can also linearize your algorithm (but we're still using bruteforce):
[X,Y] = meshgrid(1:50,1:50); %generate all the combination
C = (X(:).^2+Y(:).^2).^0.5; %sums of two square for every combination
ind = find(rem(C,1)==0 & C<=50); %get the index
res = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness
Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.
I have a list of coordinates I would like to sample from a Matrix.
Is there any elegant way to do it?
Ideally, something that looks like:
A = magic(5)
A =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
r = 1:5; % row coordinates
c = 5:-1:1; % column coordinates
A(r,c)
ans =
15 14 13 12 11
Which is equivalent to
for k=1:length(r)
A(r(k), c(k))
end
I am sure someone has asked that, but I couldn't find it anywhere.
Applying #excaza comment I was able to solve this with:
rc_ids = sub2ind(size(A), r,c);
A(rc_ids)
I have a matrix of measured angles between M planes
0 52 77 79
52 0 10 14
77 10 0 3
79 14 3 0
I have a list of known angles between planes, which is an N-by-N matrix which I name rho. Here's is a subset of it (it's too large to display):
0 51 68 75 78 81 82
51 0 17 24 28 30 32
68 17 0 7 11 13 15
75 24 7 0 4 6 8
78 28 11 4 0 2 4
81 30 13 6 2 0 2
82 32 15 8 4 2 0
My mission is to find the set of M planes whose angles in rho are nearest to the measured angles.
For example, the measured angles for the planes shown above are relatively close to the known angles between planes 1, 2, 4 and 6.
Put differently, I need to find a set of points in a distance matrix (which uses cosine-related distances) which matches a set of distances I measured. This can also be thought of as matching a pattern to a mold.
In my problem, I have M=5 and N=415.
I really tried to get my head around it but have run out of time. So currently I'm using the simplest method: iterating over every possible combination of 3 planes but this is slow and currently written only for M=3. I then return a list of matching planes sorted by a matching score:
function [scores] = which_zones(rho, angles)
N = size(rho,1);
scores = zeros(N^3, 4);
index = 1;
for i=1:N-2
for j=(i+1):N-1
for k=(j+1):N
found_angles = [rho(i,j) rho(i,k) rho(j,k)];
score = sqrt(sum((found_angles-angles).^2));
scores(index,:)=[score i j k];
index = index + 1;
end
end;
end
scores=scores(1:(index-1),:); % was too lazy to pre-calculate #
scores=sortrows(scores, 1);
end
I have a feeling pdist2 might help but not sure how. I would appreciate any help in figuring this out.
There is http://www.mathworks.nl/help/matlab/ref/dsearchn.html for closest point search, but that requires same dimensionality. I think you have to bruteforce find it anyway because it's just a special problem.
Here's a way to bruteforce iterate over all unique combinations of the second matrix and calculate the score, after that you can find the one with the minimum score.
A=[ 0 52 77 79;
52 0 10 14;
77 10 0 3;
79 14 3 0];
B=[ 0 51 68 75 78 81 82;
51 0 17 24 28 30 32;
68 17 0 7 11 13 15;
75 24 7 0 4 6 8;
78 28 11 4 0 2 4;
81 30 13 6 2 0 2;
82 32 15 8 4 2 0];
M = size(A,1);
N = size(B,1);
% find all unique permutations of `1:M`
idx = nchoosek(1:N,M);
K = size(idx,1); % number of combinations = valid candidates for matching A
score = NaN(K,1);
idx_triu = triu(true(M,M),1);
Atriu = A(idx_triu);
for ii=1:K
partB = B(idx(ii,:),idx(ii,:));
partB_triu = partB(idx_triu);
score = norm(Atriu-partB_triu,2);
end
[~, best_match_idx] = min(score);
best_match = idx(best_match_idx,:);
The solution of your example actually is [1 2 3 4], so the upperleft part of B and not [1 2 4 6].
This would theoretically solve your problem, and I don't know how to make this algorithm any faster. But it will still be slow for large numbers. For example for your case of M=5 and N=415, there are 100 128 170 583 combinations of B which are a possible solution; just generating the selector indices is impossible in 32-bit because you can't address them all.
I think the real optimization here lies in cutting away some of the planes in the NxN matrix in a preceding filtering part.
I have a vector called time, which contains time values. I'd like obtain a vector of indexes of time in which the value is between threshold x and threshold y.
This is undoubltedly trivial to do, but I'm struggling with Matlab syntax a bit, here. Any help would be greatly appreciated.
Blz
time=5:20
idx = find(time > 10 & time < 15) % indices
values=time(time(:)>10 & time(:)<15) % values
which give
time =
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
idx =
7 8 9 10
values =
11 12 13 14
I have a routine that returns a list of integers as a vector.
Those integers come from groups of sequential numbers; for example, it may look like this:
vector = 6 7 8 12 13 14 15 26 27 28 29 30 55 56
Note that above, there are four 'runs' of numbers (6-8, 12-15, 26-30 & 55-56). What I'd like to do is forward the longest 'run' of numbers to a new vector. In this case, that would be the 26-30 run, so I'd like to produce:
newVector = 26 27 28 29 30
This calculation has to be performed many, many times on various vectors, so the more efficiently I can do this the better! Any wisdom would be gratefully received.
You can try this:
v = [ 6 7 8 12 13 14 15 26 27 28 29 30 55 56];
x = [0 cumsum(diff(v)~=1)];
v(x==mode(x))
This results in
ans =
26 27 28 29 30
Here is a solution to get the ball rolling . . .
vector = [6 7 8 12 13 14 15 26 27 28 29 30 55 56]
d = [diff(vector) 0]
maxSequence = 0;
maxSequenceIdx = 0;
lastIdx = 1;
while lastIdx~=find(d~=1, 1, 'last')
idx = find(d~=1, 1);
if idx-lastIdx > maxSequence
maxSequence = idx-lastIdx;
maxSequenceIdx = lastIdx;
end
d(idx) = 1;
lastIdx=idx;
end
output = vector(1+maxSequenceIdx:maxSequenceIdx+maxSequence)
In this example, the diff command is used to find consecutive numbers. When numbers are consecutive, the difference is 1. A while loop is then used to find the longest group of ones, and the index of this consecutive group is stored. However, I'm confident that this could be optimised further.
Without loops using diff:
vector = [6 7 8 12 13 14 15 26 27 28 29 30 55 56];
seqGroups = [1 find([1 diff(vector)]~=1) numel(vector)+1]; % beginning of group
[~, groupIdx] = max( diff(seqGroups)); % bigger group index
output = vector( seqGroups(groupIdx):seqGroups(groupIdx+1)-1)
output vector is
ans =
26 27 28 29 30
Without loops - should be faster
temp = find ( ([(vector(2:end) - vector(1:end-1))==1 0])==0);
[len,ind]=max(temp(2:end)-temp(1:end-1));
vec_out = vector(temp(ind)+1:temp(ind)+len)