How do I select n elements of a sequence in windows of m ? (matlab) - matlab

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...

Related

Extract elements given 2D matrix of column indices per row from a 2D matrix in MATLAB

I tried to resample my data from a block of matrix that defined its indices. Hopefully this example can make it clear:
A=rand(18400,100);
A_IDX=randi([1 100],[18400 100]);
A_IDX consist 18400 rows and 100 columns. I wanted to extract the matrix A at the A_IDX indices. Result would be something like:
A=[1 2 3; 4 5 6];
A_IDX=[1 3; 2 3];
A_Result=[1 3; 5 6];
I tried A(:,A_IDX) but that gave me 1840x184000 matrix size, which is not what I wanted to do in the first place. Anyone can help? Thanks in advance!
We could get the linear index equivalent for those indices and then simply indexing into the input array would give us the desired output. Now, to get those linear indices, we would make use of bsxfun for the math computations related to the index computations, which would basically involve scaling and offsetting.
Indexing with 2D array of column indices
For a 2D array of column indices, we would have -
function out = take_cols(a, col_idx)
n = size(a,1);
lidx = bsxfun(#plus,(col_idx-1)*n,(1:n).');
out = a(lidx);
Sample run -
>> a
a =
39 83 39 48 36
58 74 20 19 50
69 97 65 34 57
47 58 80 24 51
>> col_idx
col_idx =
2 4
3 5
1 4
2 5
>> take_cols(a, col_idx)
ans =
83 48
20 50
69 34
58 51
Indexing with 2D array of row indices
For a 2D array of row indices, it would be -
function out = take_rows(a, row_idx)
[m,n] = size(a);
lidx = bsxfun(#plus,row_idx, (0:n-1)*m);
out = a(lidx);
Sample run -
>> a
a =
39 83 39 48 36
58 74 20 19 50
69 97 65 34 57
47 58 80 24 51
>> row_idx
row_idx =
3 2 3 1 2
4 3 4 2 4
>> take_rows(a, row_idx)
ans =
69 74 65 48 50
47 97 80 19 51
This weird monster of code will give you what you want. It generates proper subscripts for each index and converts them to linear, then just indexes A linearly.
A_IDX_aux=A_IDX';
reshape(A(sub2ind(size(A),repelem(1:size(A,1),1,size(A_IDX,1)).',A_IDX_aux(:))),[size(A,1), size(A_IDX,2)]).';
I find my solution for this task too, but not so fast, as Divakar and Ander :)
Behold:
res = cell2mat(arrayfun( #(x) A(x,A_IDX(x,:)), (1:size(A,1))', 'UniformOutput',false));
It use cell2mat and I suppose it is not so fast as bsxfun, but hope is still alive and I was curios to test all the 3 solutions. And I got unobvious results!
Elapsed time is 0.000058 seconds. % Divakar
Elapsed time is 0.000077 seconds. % Andres
Elapsed time is 0.000339 seconds. % Me
This mean bsxf is fastest! But using right indexing give fast result too! And my solution was really slow. I suppose it's because of 'UniformOutput', false - I forced to convert to cells and then back, so it slow my method a lot.
Conclusion:
If you can use bsxf - use it!
Despite the fact that my method looks more visually pleasing than that of Andres, it is still slower.
So there is no any sense to post this answer :D I spend some time for current work, maybe it will help someone in future

Element-by-element max values in multidimensional matrix

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)

Extract matrix elements using a vector of column indices per row

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)

Find closest matching distances for a set of points in a distance matrix in Matlab

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.

Matlab: find mode in range

I have a matrix like:
A=
10 31 32 22
32 35 52 77
68 42 84 32
I need a function like mode but with range, for example mymode(A,10) that return 30, find most frequent number in range 0-10, 10-20, 20-30, .... and return most number in range.
You can use histc to bin your data into the ranges of your desire and then find the bin with the most members using max on the output of histc
ranges = 0:10:50; % your desired ranges
[n, bins] = histc(A(:), ranges); % bin the data
[v,i] = max(n); % find the bin with most occurrences
[ranges(i) ranges(i+1)] % edges of the most frequent bin
For your specific example this returns
ans =
30 40
which matches with your required output, as the most values in A lay between 30 and 40.
[M,F] = mode( A((A>=2) & (A<=5)) ) %//only interested in range 2 to 5
...where M will give you the mode and F will give you frequency of occurence
> A = [10 31 32 22; 32 35 52 77; 68 42 84 32]
A =
10 31 32 22
32 35 52 77
68 42 84 32
> min = 10
min = 10
> max = 40
max = 40
> mode(A(A >= min & A <= max))
ans = 32
>
I guess by the number of different answers that we may be missing your goal. Here is my interpretation.
If you want to have many ranges and you want to output most frequent number for every range, create a cell containing all desired ranges (they could overlap) and use cellfun to run mode() for every range. You can also create a cell with desired ranges using arrayfun in a similar manner:
A = [10 31 32 22; 32 35 52 77; 68 42 84 32];
% create ranges
range_step = 10;
range_start=[0:range_step:40];
range=arrayfun(#(r)([r r+range_step]), range_start, 'UniformOutput', false)
% analyze ranges
o = cellfun(#(r)(mode(A(A>=r(1) & A<=r(2)))), range, 'UniformOutput', false)
o =
[10] [10] [22] [32] [42]