I have a data set of size 11490x1. the data is recorded every 0.25 second(i.e. 4hz). So, 1 second accounts for 4 data points. The goal here is to further create sub sets every 3 seconds, meaning that I want to look at data every 3 seconds and analyze it. for example: if I had data such as [1 2 3 4 5 6 8 2 4 2 4 3 2 4 2 5 2 5 24 2 5 1 5 1], I want to have a sub set [1 2 3 4 5 6 8 2 4 2 4 3 ] and so on...
Any help would be appreciate.
It really depends on how you plan to "analyse" your data. The simplest way is to use a loop:
n = 4*3;
breaks = 0:n:numel(data)
for i = 1:numel(breaks)-1
sub = data(breaks(i)+1:breaks(i+1));
%// do analysis
%// OR sub{i} = data(breaks(i)+1:breaks(i+1));
end
A vectorized approach might use reshape(data,[],12) after padding data so that mod(numel(data),12)==0
A third way might be to break your matrix up into a cell array using mat2cell or in a for loop like above but instead of sub=... rather use sub{i}=...
Related
I can "Vectorize" the circshift command but I'm having trouble adding dimensions to it.
See code below with working FOR loop that I'm trying to vectorize using dimensions
clear all,clf reset,tic,clc , close all
function [outMat] = vectcircshift(vectToShift,shiftVector)
%This function generates a matrix where each row is a circshift of the
%original vector from the specified interval in the shiftVector;
%
%Inputs
%vectToShift: is the original vector you want to circshift multiple times
%shiftVector: is the vector of the circshift sizes;
%
%Outputs
%outMat: is a matrix were every row is circshift by the amount in the
% shiftVector
[n,m]=size(vectToShift);
if n>m
inds=(1:n)';
i=toeplitz(flipud(inds),circshift(inds,[1 0]));
outMat=vectToShift(i(shiftVector,:));
outMat=circshift(outMat,[0,-1]); %shift to include original signal first
else
inds=1:m;
i=toeplitz(fliplr(inds),circshift(inds,[0 1]));
outMat=vectToShift(i(shiftVector,:));
outMat=circshift(outMat,[0,-1]); %shift to include original signal first
end
end
%%----Working FOR LOOP below I'm trying to vectorize.
ndim=0;
ndim_tot=[1:3] %total dimensions
for ndim=1:length(ndim_tot)
ndim=ndim+0
if ndim==1
array_sort(ndim,:)=circshift(ndim_tot,[0 ndim-1]) %start at row of sort array
else
array_sort(ndim,:)=circshift(ndim_tot,[0 mod(-ndim,length(ndim_tot))+1]) %next start of row of sort array
endif
array_sort= array_sort(ndim,:)
array_dim(:,:,ndim)=vectcircshift([1:5],array_sort)
endfor
I tired the syntax below but that logic won't work.
ndim_tot=[1:3]; %number of dimensions
array_dim2(:,:,ndim_tot)=vectcircshift([1:5],[1:3])
I get an error nonconformant arguments(op1 is 0x0x1, op2 is 3x5)
My goal is to create a multidimensional array that circshifts a signal / array and also creates and shifts it in multiple dimensions.
Example: of what the multidimensional array would look like
if I start with a signal / array a1=[1 2 3 4 5]
I'm trying to have it create.
array_dim(:,:,1)=
[
1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
]
array_dim(:,:,2)=
[
5 1 2 3 4
4 5 1 2 3
1 2 3 4 5
]
array_dim(:,:,3)=
[
4 5 1 2 3
1 2 3 4 5
5 1 2 3 4
]
Please note: the the numbers won't be sequential I just used it as an example to help explain things a little easier.
PS: I'm using Octave 4.2.2
Not clear why you are shifting in mod 3, but here is a loop assignment using shift
a1=[1 2 3 4 5];
array_dim=zeros(3,5,3);
for i=0:2
array_dim(:,:,i+1)=[shift(a1,i);
shift(a1,mod(i+1,3));
shift(a1,mod(i+2,3))];
endfor
array_dim
and the output fits your example
array_dim =
ans(:,:,1) =
1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
ans(:,:,2) =
5 1 2 3 4
4 5 1 2 3
1 2 3 4 5
ans(:,:,3) =
4 5 1 2 3
1 2 3 4 5
5 1 2 3 4
I am trying to smooth the temporal history of each pixel in my matrix- in other words, trying to smooth each pixel through both 'space' (mxn) and 'time'(third dimension). I am using the function movmean to create an average of each pixel in time of a 1000x1000x8 matrix.
I am currently using the following code to take an average, using a window size of 5, operating along the third dimension:
av_matrix = movmean(my_matrix,5,3)
This is creating an average as expected, but I'm wondering if the window is just operating in the mxn direction and not taking the average along the third dimension as well.
To compute a moving average along the n dimensions of an n-dimensional array (the "window" is an n-dimensional rectangle), the simplest way is to use convolution (see convn).
You need to be careful with edge effects, that is, when the convolution kernel (or n-dimensional window) partially slides out of the data. What movmean does is average over the actual data points only. To achieve that behaviour you can
compute the sum over the kernel via convolution with the 'same' option; and then
divide each entry by the number of actual data points from which it was computed. This number can also be obtaind via convolution, namely, applying the kernel to an array of ones.
So, all you need is:
my_matrix = randi(9,5,5,3); % example 3D array
sz = [3 3 2]; % 3D window size
av_matrix = convn(my_matrix, ones(sz), 'same') ... % step 1
./convn(ones(size(my_matrix)), ones(sz), 'same'); % step 2
Check:
The following examples use
>> my_matrix
my_matrix(:,:,1) =
6 8 2 1 8
4 6 7 9 8
4 5 1 4 3
5 5 8 7 9
3 6 6 4 9
my_matrix(:,:,2) =
8 8 5 3 6
8 9 6 9 1
9 5 6 2 2
1 7 4 1 2
5 4 7 4 9
my_matrix(:,:,3) =
6 5 8 6 6
1 6 8 6 1
5 5 1 6 7
1 1 2 9 8
1 2 6 1 2
With edge effects:
>> mean(mean(mean(my_matrix(1:2,1:2,1:2))))
ans =
7.125000000000000
>> av_matrix(1,1,1)
ans =
7.125000000000000
Without edge effects:
>> mean(mean(mean(my_matrix(1:3,1:3,1:2))))
ans =
5.944444444444445
>> av_matrix(2,2,1)
ans =
5.944444444444445
I'm trying to reshape an array, perform an operation and then reshape it back to the original. See example of the output I'm trying to get. I can get a and b but I'm having trouble getting c to look like a again.
Step 1) (the original array)
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
Step 2) (reshape and perform some operation)
1,1,1,2,2,2,3,3,3,4,4,4,5,5,5
Step 3) (array is reshaped back to the original size to look like step 1) this is what I want
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
I can get the variables a and b but I'm not sure how to reshape c from b to look like a again see example code and output below
a=[repmat(1,[1,3]);repmat(2,[1,3]);repmat(3,[1,3]);repmat(4,[1,3]);repmat(5,[1,3])]
[rw,col]=size(a)
b=reshape(a',1,rw*col)
c=reshape(b,rw,col)
a=
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
b=1,1,1,2,2,2,3,3,3,4,4,4,5,5,5
c =
1 2 4
1 3 4
1 3 5
2 3 5
2 4 5
Ps: I'm using Octave 4.0 which is like Matlab.
MATLAB and Octave use column-major ordering so you'll need to reshape the result with that in mind. The data will be filled down the columns first but you want it to fill the columns first. To achieve this, you can specify the number of columns as the number of rows provided to reshape and then transpose the result
c = reshape(b, 3, []).'
Or more flexibly
c = reshape(b, flip(size(a))).'
This question pops up quite often in one form or another (see for example here or here). So I thought I'd present it in a general form, and provide an answer which might serve for future reference.
Given an arbitrary number n of vectors of possibly different sizes, generate an n-column matrix whose rows describe all combinations of elements taken from those vectors (Cartesian product) .
For example,
vectors = { [1 2], [3 6 9], [10 20] }
should give
combs = [ 1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20 ]
The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired n-column matrix:
vectors = { [1 2], [3 6 9], [10 20] }; %// input data: cell array of vectors
n = numel(vectors); %// number of vectors
combs = cell(1,n); %// pre-define to generate comma-separated list
[combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two
%// comma-separated lists is needed to produce the rows of the result matrix in
%// lexicographical order
combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1
combs = reshape(combs,[],n); %// reshape to obtain desired matrix
A little bit simpler ... if you have the Neural Network toolbox you can simply use combvec:
vectors = {[1 2], [3 6 9], [10 20]};
combs = combvec(vectors{:}).' % Use cells as arguments
which returns a matrix in a slightly different order:
combs =
1 3 10
2 3 10
1 6 10
2 6 10
1 9 10
2 9 10
1 3 20
2 3 20
1 6 20
2 6 20
1 9 20
2 9 20
If you want the matrix that is in the question, you can use sortrows:
combs = sortrows(combvec(vectors{:}).')
% Or equivalently as per #LuisMendo in the comments:
% combs = fliplr(combvec(vectors{end:-1:1}).')
which gives
combs =
1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20
If you look at the internals of combvec (type edit combvec in the command window), you'll see that it uses different code than #LuisMendo's answer. I can't say which is more efficient overall.
If you happen to have a matrix whose rows are akin to the earlier cell array you can use:
vectors = [1 2;3 6;10 20];
vectors = num2cell(vectors,2);
combs = sortrows(combvec(vectors{:}).')
I've done some benchmarking on the two proposed solutions. The benchmarking code is based on the timeit function, and is included at the end of this post.
I consider two cases: three vectors of size n, and three vectors of sizes n/10, n and n*10 respectively (both cases give the same number of combinations). n is varied up to a maximum of 240 (I choose this value to avoid the use of virtual memory in my laptop computer).
The results are given in the following figure. The ndgrid-based solution is seen to consistently take less time than combvec. It's also interesting to note that the time taken by combvec varies a little less regularly in the different-size case.
Benchmarking code
Function for ndgrid-based solution:
function combs = f1(vectors)
n = numel(vectors); %// number of vectors
combs = cell(1,n); %// pre-define to generate comma-separated list
[combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two
%// comma-separated lists is needed to produce the rows of the result matrix in
%// lexicographical order
combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1
combs = reshape(combs,[],n);
Function for combvec solution:
function combs = f2(vectors)
combs = combvec(vectors{:}).';
Script to measure time by calling timeit on these functions:
nn = 20:20:240;
t1 = [];
t2 = [];
for n = nn;
%//vectors = {1:n, 1:n, 1:n};
vectors = {1:n/10, 1:n, 1:n*10};
t = timeit(#() f1(vectors));
t1 = [t1; t];
t = timeit(#() f2(vectors));
t2 = [t2; t];
end
Here's a do-it-yourself method that made me giggle with delight, using nchoosek, although it's not better than #Luis Mendo's accepted solution.
For the example given, after 1,000 runs this solution took my machine on average 0.00065935 s, versus the accepted solution 0.00012877 s. For larger vectors, following #Luis Mendo's benchmarking post, this solution is consistently slower than the accepted answer. Nevertheless, I decided to post it in hopes that maybe you'll find something useful about it:
Code:
tic;
v = {[1 2], [3 6 9], [10 20]};
L = [0 cumsum(cellfun(#length,v))];
V = cell2mat(v);
J = nchoosek(1:L(end),length(v));
J(any(J>repmat(L(2:end),[size(J,1) 1]),2) | ...
any(J<=repmat(L(1:end-1),[size(J,1) 1]),2),:) = [];
V(J)
toc
gives
ans =
1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20
Elapsed time is 0.018434 seconds.
Explanation:
L gets the lengths of each vector using cellfun. Although cellfun is basically a loop, it's efficient here considering your number of vectors will have to be relatively low for this problem to even be practical.
V concatenates all the vectors for easy access later (this assumes you entered all your vectors as rows. v' would work for column vectors.)
nchoosek gets all the ways to pick n=length(v) elements from the total number of elements L(end). There will be more combinations here than what we need.
J =
1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 3 4
1 3 5
1 3 6
1 3 7
1 4 5
1 4 6
1 4 7
1 5 6
1 5 7
1 6 7
2 3 4
2 3 5
2 3 6
2 3 7
2 4 5
2 4 6
2 4 7
2 5 6
2 5 7
2 6 7
3 4 5
3 4 6
3 4 7
3 5 6
3 5 7
3 6 7
4 5 6
4 5 7
4 6 7
5 6 7
Since there are only two elements in v(1), we need to throw out any rows where J(:,1)>2. Similarly, where J(:,2)<3, J(:,2)>5, etc... Using L and repmat we can determine whether each element of J is in its appropriate range, and then use any to discard rows that have any bad element.
Finally, these aren't the actual values from v, just indices. V(J) will return the desired matrix.
how to Compare Values in Array on Specific Indexes in Matlab?
Suppose:
A= [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8]
so i want to Find that
on index 2,3,4,5,6 values or same or not ?
Note: Index can be Dynamically Entered.
Number(length) of Values in Array also can be changed..
To check if they are all equal: use diff to subtract pairs of values, and then check if all those differences are 0.
A = [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8];
ind = [2 3 4 5 6];
result = ~any(diff(A(ind)));
This is faster than using unique. With A and ind as in your example,
>> tic
for cont = 1:1e5
result = ~any(diff(A(ind)));
end
toc
tic
for cont = 1:1e5
result=numel(unique(A(ind)))==1;
end
toc
Elapsed time is 0.371142 seconds.
Elapsed time is 4.754007 seconds.
Hey this should do the trick:
A= [2 2 3 3 3 4 4 4 5 5 6 6 7 8 8];
B= [1,3,5];
C=A(B);
result=numel(unique(C))==1;
Here A is your data.
B is the index vector.
C contains the elements corresponding to the index vector.
result is 1 if all values were the same and 0 otherwise.
You can even "shorten" the code further by joining the two line:
result=numel(unique(A(B)))==1;
There are some ways, it depends on your taste.
For example, if the variable indexing contain the corresponding indexes:
unique(A(indexing));
will give you a vector with the unique elements in the sub-vector A(indexing). Then you just need to check the length:
length(unique(A(indexing))) == 1
I would avoid the use of numel when the function length is available (it is much more clearer what you are trying to achieve).
Other option is to compare the first element to the rest of the element in the sub-vector:
sub_vector = A(indexing);
all(sub_vector == sub_vector(1));
The second option assumes that the sub-vector will never be empty!