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.
Related
I'd like to obtain all unique products for a given vector.
For example, given a:
a = [4,10,12,3,6]
I want to obtain a matrix that contains the results of:
4*10
4*12
4*3
4*6
10*12
10*3
10*6
12*3
12*6
3*6
Is there a short and/or quick way of doing this in MATLAB?
EDIT: a may contain duplicate numbers, giving duplicate products - and these must be kept.
Given:
a =
4 10 12 3 6
Construct the matrix of all pairwise products:
>> all_products = a .* a.'
all_products =
16 40 48 12 24
40 100 120 30 60
48 120 144 36 72
12 30 36 9 18
24 60 72 18 36
Now, construct a mask to keep only those values below the main diagonal:
>> mask = tril(true(size(all_products)), -1)
mask =
0 0 0 0 0
1 0 0 0 0
1 1 0 0 0
1 1 1 0 0
1 1 1 1 0
and apply the mask to the product matrix:
>> unique_products = all_products(mask)
unique_products =
40
48
12
24
120
30
60
36
72
18
If you have the Statistics Toolbox, you can abuse pdist, which considers only one of the two possible orders for each pair:
result = pdist(a(:), #times);
One option involves nchoosek, which returns all combinations of k elements out of a vector, each row is one combination. prod computes the product of rows or columns:
a = [4,10,12,3,6];
b = nchoosek(a,2);
b = prod(b,2); % 2 indicates rows
Try starting with this. Have the unique function filter out the result of multiplying a by itself.
b = unique(a*a')
I have a few multidimensional matrices of dimensions mxnxt, where each element in mxn is an individual sensor input, and t is time. What I want to do is analyse only the peak values for each element in mxn over t, so I would end up with a single 2D matrix of mxn containing only max values.
I know there are are ways to get a single overall max value, but is there a way to combine this with element-by-element operations like bsxfun so that it examines each individual element over t?
I'd be grateful for any help you can give because I'm really stuck at the moment. Thanks in advance!
Is this what you want?
out = max(A,[],3); %// checking maximum values in 3rd dimension
Example:
A = randi(50,3,3,3); %// Random 3x3x3 dim matrix
out = max(A,[],3);
Results:
A(:,:,1) =
35 5 8
38 12 42
23 46 27
A(:,:,2) =
50 6 39
4 49 41
23 1 44
A(:,:,3) =
5 41 10
20 22 14
13 46 8
>> out
out =
50 41 39
38 49 42
23 46 44
You can call max() with the matrix and select the dimension (look the documentation) on which the operation will be calculated, e.g
M = max(A,[],3)
I have an array which looks similar to:
0 2 3 4 0 0 7 8 0 10
0 32 44 47 0 0 37 54 0 36
I wish to remove all
0
0
from this to get:
2 3 4 7 8 10
32 44 47 37 54 36
I've tried x(x == 0) = []
but I get:
x =
2 32 3 44 4 47 7 37 8 54 10 36
How can I remove all zero columns?
Here is a possible solution:
x(:,all(x==0))=[]
You had the right approach with x(x == 0) = [];. By doing this, you would remove the right amount of elements that can still form a 2D matrix and this actually gives you a vector of values that are non-zero. All you have to do is reshape the matrix back to its original form with 2 rows:
x(x == 0) = [];
y = reshape(x, 2, [])
y =
2 3 4 7 8 10
32 44 47 37 54 36
Another way is with any:
y = x(:,any(x,1));
In this case, we look for any columns that are non-zero and use these locations to index into x and extract out those corresponding columns.
Result:
y =
2 3 4 7 8 10
32 44 47 37 54 36
Another way which is more for academic purposes is to use unique. Assuming that your matrix has all positive values:
[~,~,id] = unique(x.', 'rows');
y = x(:, id ~= 1)
y =
2 3 4 7 8 10
32 44 47 37 54 36
We transpose x so that each column becomes a row, and we look for all unique rows. The reason why the matrix needs to have all positive values is because the third output of unique assigns unique ID to each unique row in sorted order. Therefore, if we have all positive values, then a row of all zeroes would be assigned an ID of 1. Using this array, we search for IDs that were not assigned a value of 1, and use those to index into x to extract out the necessary columns.
You could also use sum.
Sum over the columns and any column with zeros only will be zeros after the summation as well.
sum(x,1)
ans =
0 34 47 51 0 0 44 62 0 46
x(:,sum(x,1)>0)
ans =
2 3 4 7 8 10
32 44 47 37 54 36
Also by reshaping nonzeros(x) as follows:
reshape(nonzeros(x), size(x,1), [])
ans =
2 3 4 7 8 10
32 44 47 37 54 36
I have an MxN matrix and I want a column vector v, using the vector s that tells me for each row in the matrix what column I will take.
Here's an example:
Matrix =
[ 4 13 93 20 42;
31 18 94 64 02;
7 44 24 91 15;
11 20 43 38 31;
21 42 72 60 99;
13 81 31 87 50;
32 22 83 24 04]
s = [4 4 5 4 4 4 3].'
And the desired output is:
v = [20 64 15 38 60 87 83].'
I thought using the expression
Matrix(:,s)
would've work but it doesn't. Is there a solution without using for loops to access the rows separately?
It's not pretty, and there might be better solutions, but you can use the function sub2ind like this:
M(sub2ind(size(M),1:numel(s),s'))
You can also do it with linear indexing, here is an example:
M=M'; s=s';
M([0:size(M,1):numel(M)-1]+s)
Quick MATLAB question.
What would be the best/most efficient way to select a certain number of elements, 'n' in windows of 'm'. In other words, I want to select the first 50 elements of a sequence, then elements 10-60, then elements 20-70 ect.
Right now, my sequence is in vector format(but this can easily be changed).
EDIT:
The sequences that I am dealing with are too long to be stored in my RAM. I need to be able to create the windows, and then call upon the window that I want to analyze/preform another command on.
Do you have enough RAM to store a 50-by-nWindow array in memory? In that case, you can generate your windows in one go, and then apply your processing on each column
%# idxMatrix has 1:50 in first col, 11:60 in second col etc
idxMatrix = bsxfun(#plus,(1:50)',0:10:length(yourVector)-50); %'#
%# reshapedData is a 50-by-numberOfWindows array
reshapedData = yourVector(idxMatrix);
%# now you can do processing on each column, e.g.
maximumOfEachWindow = max(reshapedData,[],1);
To complement Kerrek's answer: if you want to do it in a loop, you can use something like
n = 50
m = 10;
for i=1:m:length(v)
w = v(i:i+n);
% Do something with w
end
There's a slight issue with the description of your problem. You say that you want "to select the first 50 elements of a sequence, then elements 10-60..."; however, this would translate to selecting elements:
1-50
10-60
20-70
etc.
That first sequence should be 0-10 to fit the pattern which of course in MATLAB would not make sense since arrays use one-indexing. To address this, the algorithm below uses a variable called startIndex to indicate which element to start the sequence sampling from.
You could accomplish this in a vectorized way by constructing an index array. Create a vector consisting of the starting indices of each sequence. For reuse sake, I put the length of the sequence, the step size between sequence starts, and the start of the last sequence as variables. In the example you describe, the length of the sequence should be 50, the step size should be 10 and the start of the last sequence depends on the size of the input data and your needs.
>> startIndex = 10;
>> sequenceSize = 5;
>> finalSequenceStart = 20;
Create some sample data:
>> sampleData = randi(100, 1, 28)
sampleData =
Columns 1 through 18
8 53 10 82 82 73 15 66 52 98 65 81 46 44 83 9 14 18
Columns 19 through 28
40 84 81 7 40 53 42 66 63 30
Create a vector of the start indices of the sequences:
>> sequenceStart = startIndex:sequenceSize:finalSequenceStart
sequenceStart =
10 15 20
Create an array of indices to index into the data array:
>> index = cumsum(ones(sequenceSize, length(sequenceStart)))
index =
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
>> index = index + repmat(sequenceStart, sequenceSize, 1) - 1
index =
10 15 20
11 16 21
12 17 22
13 18 23
14 19 24
Finally, use this index array to reference the data array:
>> sampleData(index)
ans =
98 83 84
65 9 81
81 14 7
46 18 40
44 40 53
Use (start : step : end) indexing: v(1:1:50), v(10:1:60), etc. If the step is 1, you can omit it: v(1:50).
Consider the following vectorized code:
x = 1:100; %# an example sequence of numbers
nwind = 50; %# window size
noverlap = 40; %# number of overlapping elements
nx = length(x); %# length of sequence
ncol = fix((nx-noverlap)/(nwind-noverlap)); %# number of sliding windows
colindex = 1 + (0:(ncol-1))*(nwind-noverlap); %# starting index of each
%# indices to put sequence into columns with the proper offset
idx = bsxfun(#plus, (1:nwind)', colindex)-1; %'
%# apply the indices on the sequence
slidingWindows = x(idx)
The result (truncated for brevity):
slidingWindows =
1 11 21 31 41 51
2 12 22 32 42 52
3 13 23 33 43 53
...
48 58 68 78 88 98
49 59 69 79 89 99
50 60 70 80 90 100
In fact, the code was adapted from the now deprecated SPECGRAM function from the Signal Processing Toolbox (just do edit specgram.m to see the code).
I omitted parts that zero-pad the sequence in case the sliding windows do not evenly divide the entire sequence (for example x=1:105), but you can easily add them again if you need that functionality...