Averaging every n elements of a vector in matlab - matlab

I would like to average every 3 values of an vector in Matlab, and then assign the average to the elements that produced it.
Examples:
x=[1:12];
y=%The averaging operation;
After the operation,
y=
[2 2 2 5 5 5 8 8 8 11 11 11]
Therefore the produced vector is the same size, and the jumping average every 3 values replaces the values that were used to produce the average (i.e. 1 2 3 are replaced by the average of the three values, 2 2 2). Is there a way of doing this without a loop?
I hope that makes sense.
Thanks.

I would go this way:
Reshape the vector so that it is a 3×x matrix:
x=[1:12];
xx=reshape(x,3,[]);
% xx is now [1 4 7 10; 2 5 8 11; 3 6 9 12]
after that
yy = sum(xx,1)./size(xx,1)
and now
y = reshape(repmat(yy, size(xx,1),1),1,[])
produces exactly your wanted result.
Your parameter 3, denoting the number of values, is only used at one place and can easily be modified if needed.

You may find the mean of each trio using:
x = 1:12;
m = mean(reshape(x, 3, []));
To duplicate the mean and reshape to match the original vector size, use:
y = m(ones(3,1), :) % duplicates row vector 3 times
y = y(:)'; % vector representation of array using linear indices

Related

MATLAB - Returning a matrix of sums of elements corresponding to the same kind

Overview
An n×m matrix A and an n×1 vector Date are the inputs of the function S = sumdate(A,Date).
The function returns an n×m vector S such that all rows in S correspond to the sum of the rows of A from the same date.
For example, if
A = [1 2 7 3 7 3 4 1 9
6 4 3 0 -1 2 8 7 5]';
Date = [161012 161223 161223 170222 160801 170222 161012 161012 161012]';
Then I would expect the returned matrix S is
S = [15 9 9 6 7 6 15 15 15;
26 7 7 2 -1 2 26 26 26]';
Because the elements Date(2) and Date(3) are the same, we have
S(2,1) and S(3,1) are both equal to the sum of A(2,1) and A(3,1)
S(2,2) and S(3,2) are both equal to the sum of A(2,2) and A(3,2).
Because the elements Date(1), Date(7), Date(8) and Date(9) are the same, we have
S(1,1), S(7,1), S(8,1), S(9,1) equal the sum of A(1,1), A(7,1), A(8,1), A(9,1)
S(1,2), S(7,2), S(8,2), S(9,2) equal the sum of A(1,2), A(7,2), A(8,2), A(9,2)
The same for S([4,6],1) and S([4,6],2)
As the element Date(5) does not repeat, so S(5,1) = A(5,1) = 7 and S(5,2) = A(5,2) = -1.
The code I have written so far
Here is my try on the code for this task.
function S = sumdate(A,Date)
S = A; %Pre-assign S as a matrix in the same size of A.
Dlist = unique(Date); %Sort out a non-repeating list from Date
for J = 1 : length(Dlist)
loc = (Date == Dlist(J)); %Compute a logical indexing vector for locating the J-th element in Dlist
S(loc,:) = repmat(sum(S(loc,:)),sum(loc),1); %Replace the located rows of S by the sum of them
end
end
I tested it on my computer using A and Date with these attributes:
size(A) = [33055 400];
size(Date) = [33055 1];
length(unique(Date)) = 2645;
It took my PC about 1.25 seconds to perform the task.
This task is performed hundreds of thousands of times in my project, therefore my code is too time-consuming. I think the performance will be boosted up if I can eliminate the for-loop above.
I have found some built-in functions which do special types of sums like accumarray or cumsum, but I still do not have any ideas on how to eliminate the for-loop.
I would appreciate your help.
You can do this with accumarray, but you'll need to generate a set of row and column subscripts into A to do it. Here's how:
[~, ~, index] = unique(Date); % Get indices of unique dates
subs = [repmat(index, size(A, 2), 1) ... % repmat to create row subscript
repelem((1:size(A, 2)).', size(A, 1))]; % repelem to create column subscript
S = accumarray(subs, A(:)); % Reshape A into column vector for accumarray
S = S(index, :); % Use index to expand S to original size of A
S =
15 26
9 7
9 7
6 2
7 -1
6 2
15 26
15 26
15 26
Note #1: This will use more memory than your for loop solution (subs will have twice the number of element as A), but may give you a significant speed-up.
Note #2: If you are using a version of MATLAB older than R2015a, you won't have repelem. Instead you can replace that line using kron (or one of the other solutions here):
kron((1:size(A, 2)).', ones(size(A, 1), 1))

How to add elements of a matrix to create a new matrix?

Say I have a matrix
x = [1 2 3 4 ]
and I want to create a matrix that adds two elements and outputs a third?
y = [1+2 3+4]
y = [3 7]
For four values, I can just do y=[x(1)+x(2) x(3)+x(4)].
How would I go about doing that if x was bigger and variable (say a thousand elements)? How would I program y?
I've been struggling with this for two days now... thanks in advance...
A simple way would be to reshape the original array x into a matrix containing k rows, k being the number of elements you are summing together (here 2, sorry I don't know how to explain this), and then simply call sum on this matrix which will calculate the sum of each column and output it into a new vector y.
Example:
x = [1:10]
block = 2
r = reshape(x,block,[])
So here r looks like this:
r =
1 3 5 7 9
2 4 6 8 10
Therefore calling sum with the 1st dimension on r yields the following:
y = sum(reshape(x,block,[]),1)
y =
3 7 11 15 19

Matlab: How to create a CDF array

I am fairly new to Matlab and am trying to learn for school. I have created a vector of values with fixed differences between consecutive values. E.g. A = [1 2.5 4 5.5 7 8.5 10 ...].
I also have another vector of random values, e.g. B = [3 7 1 2 3 4 8 0 ...].
I want to create a new vector of the same size of A, which has numbers indicating the number of values in B which are less than or equal to each value in A.
In this example, C = [2 3 6 6 7 8 ...]
Thanks in advance!
Context: I am working on a CDF function
You can use bsxfun() to implement element-wise comparisons between arrays:
C = sum( bsxfun(#le, B', A) )
Here we're passing bsxfun() the "less than or equal to" function handle, #le . This produces a length(B) by length(A) logical array. We simply sum down the rows to get the total number of TRUE's.

matrix get min values of a matrix before max values occurred

I was trying to get the min values of a matrix before the max values of the matrix occurred. I have two matrices: matrix data and matrix a. Matrix a is a subset of matrix data and is composed of the max values of matrix data. I have the following code but obviously doing something wrong.
edit:
Matrix a are the max values of matrix data. I derived it from:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
matrix b code:
for x=1:size(data,1)
b(x)=min(data(x,(x<data==a)));
end
b=b'
clear x
matrix data matrix a matrix b
1 2 3 4 4 1
6 5 4 7 7 4
9 6 12 5 12 6
I need all the min values that occurred before to matrix a occurred in matrix data
Short and simple:
[a,idxmax] = max(data,[],2);
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
which is the same as
b=NaN(1,size(data,1)); % preallocation!
for ii=1:size(data,1)
b(ii) = min(data(ii,1:idxmax(ii)));
end
Ignore maximum itself
If you want minimum of everything really before (and not including the maximum), it's possible that the maximum is the first number, and you try taking minimum of an empty matrix. Solution then is to use cell output, which can be empty:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii)-1)), 1:size(data,1),'uni',false);
Replace empty cells with NaN
If you want to replace empty cells to Nan and then back to a matrix use this:
b(cellfun(#isempty,b))={NaN};
b=cell2mat(b);
or simply use the earlier version and replace b(ii) with NaN when it is equal to a(ii) same outcome:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
b(b'==a) = NaN
Example:
data=magic(4)
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
outputs:
a' = 16 11 12 15
b =
16 5 6 4
and
b =[1x0 double] [5] [6] [4]
for the 2nd solution using cell output and ignoring the maximum itself also.
And btw:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
can be replaced with
a=max(data,[],2);
It's not pretty but this is the only way I found so far of doing this kind of thing without a loop.
If loops are ok I would recommend Gunther Struyf answer as the most compact use of matlab's in-built array looping function, arrayfun.
Some of the transposition etc may be superfluous if you're wanting column mins instead of row...
[mx, imx] = max(data');
inds = repmat(1:size(data,2), [size(data,1),1]);
imx2 = repmat(imx', [1, size(data,2)]);
data2 = data;
data2(inds >= imx2) = inf;
min(data2');
NOTE: if data is not needed we can remove the additional data2 variable, and reduce the line count.
So to demonstrate what this does, (and see if I understood the question correctly):
for input
>> data = [1,3,-1; 5,2,1]
I get minima:
>> min(data2')
ans = [1, inf]
I.e. it only found the min values before the max values for each row, and anything else was set to inf.
In words:
For each row get index of maximum
Generate matrix of column indices
Use repmat to generate a matrix, same size as data where each row is index of maximum
Set data to infinity where column index > max_index matrix
find min as usual.

What kind of matrix is in Matlab: "W = ones(1:numEntries); "?

One simple question..
Let's suppose in Matlab we have a variable: numEntries = 8
I know "ones" is used to create "1's matrixes" but what's the result of the following?
W = ones(1:numEntries);
And,if possible,where I can find a good reference about matrixes in Matlab? I looked in official documentation but it is not so clear to me.
Thank you very much
ones creates a matrix of ones. ones works for any number of dimensions, however if you specify only one argument, you'll get a square matrix of ones of that size. Otherwise, you'll get a N-dimensional matrix of ones, N being the number of arguments you give to ones.
If numEntries is 8, 1:numEntries creates a 1x8 vector 1 2 3 4 5 6 7 8. This vector being given as input to ones, you'll get a 8-dimensional matrix of ones, in which the size of 1st dimension (vertical) is 1, the size of 2nd dimension (horizontal) is 2, the size of third dimension is 3, and so on, until the 8th dimension, that has size of 8.
numEntries = 8;
W = ones(1:numEntries);
% To check sizes of each dimension of W:
size(W)
ans =
1 2 3 4 5 6 7 8
% To calculate the total number of elements in W:
prod(size(W))
ans =
40320
% Edit: to get the number of elements in W, numel(W) is faster than prod(size(W)) :
numel(W)
ans =
40320
It's an N-D (N-dimensional) array where N = numEntries.
More specifically, it's a N-D array of size 1 x 2 x ... x N filled with all ones.