I want to piecewise-average a vector in Matlab. Vector x looks like this:
x = 1:15;
Respectively:
x = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15]
I want to find the mean value over n = 5 elements; therefore, the result-vector y should look like:
y = [1 1.5 2.5 3 4 5 6 7 8 9 10 11 12 13]
The code for generating the vector y should somehow work like this:
y = [
mean ([1])
mean ([1,2])
mean ([1,2,3])
mean ([1,2,3,4])
mean ([1,2,3,4,5])
mean ([2,3,4,5,6])
mean ([3,4,5,6,7])
mean ([4,5,6,7,8])
mean ([5,6,7,8,9])
mean ([6,7,8,9,10])
mean ([7,8,9,10,11])
mean ([8,9,10,11,12])
mean ([9,10,11,12,13])
mean ([10,11,12,13,14])
mean ([11,12,13,14,15])
]
For n < 5 elements, the program should average over n elements. For example, if there are only 3 elements available, the code should average the first 3 elements. For n > 5 elements, the program should average over the last 5 elements.
Any help is appreciated!
For such sliding summing or averaging operations, a very efficient vectorized approach would be with 1D convolution conv, like so -
n = 5
sums = conv(x,ones(1,n))
out = sums(1:numel(x))./[1:n n*ones(1,numel(x)-n)]
Try this:
x = 1:15;
for n = 1:length(x)
if n <= 5
y(n) = mean(x(1:n))
else
y(n) = mean(x(n-4:n))
end
end
Below is a kind of a brute force method.
for j=1:length(x)
A=j-4;
if A<1
A=1;
end;
y(j)=mean(x(A:j))
end;
or in a more compact form:
for j=1:length(x)
y(j)=mean(x(max(j-4,1):j));
end;
Here is an alternative method
make a matrix of all the numbers which you have to take average of in each row, here in bsxfun() function a vector four new rows are creating for each element of 1:15 row by row of previous 4 digits of the current number and then all the non zero elements are ommitted and amde 0
n =
5
A = bsxfun(#plus ,[1:15].',-(n -1):0)
A(A<0) = 0
A =
0 0 0 0 1
0 0 0 1 2
0 0 1 2 3
0 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
7 8 9 10 11
8 9 10 11 12
9 10 11 12 13
10 11 12 13 14
11 12 13 14 15
and then divide the sum of each row by number of non-zero elements in each row
>> sum(A,2)./sum(~ismember(A,0),2)
ans =
1.0000
1.5000
2.0000
2.5000
3.0000
4.0000
5.0000
6.0000
7.0000
8.0000
9.0000
10.0000
11.0000
12.0000
13.0000
Related
I want to take the union of some of the rows of a matrix x. The row numbers of the rows whose union has to be taken are given by vector r. Is there any built-in function in MATLAB that can do it?
x = [1 2 4 0 0;
3 6 5 0 0;
7 8 10 12 9;
2 4 6 7 0;
3 4 5 8 12];
r = [1, 3, 5];
I think this will work for you - first, take the submatrix x(r,:) with the rows you want, and then find all the unique values in it:
unique(x(r,:))
ans =
0
1
2
3
4
5
7
8
9
10
12
You could do it like this
>>> union(union(x(r(1),:),x(r(2),:)),x(r(3),:))
ans =
0 1 2 3 4 5 7 8 9 10 12
or set up a for loop that iterates over the vector r to compute all the unions
I have a vector of 13 entities in Matlab.
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]
I want to append values [1 2 3 4 5] at regular intervals at position 1 5 9 13 & 17.
The final value of a looks like this.
a=[1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2].
The values with italics show the appended values.
How can I do it?
Since you are looking for regular intervals, you can take advantage of the reshape and cat function:
a = [3 4 6 8 1 5 8 9 3 7 3 6 2];
v = [1 2 3 4 5];
l = [1 5 9 13 17];
interval = l(2)-l(1)-1; %computes the interval between inserts
amax = ceil(size(a,2)/interval) * interval; %calculating maximum size for zero padding
a(amax) = 0; %zero padding to allow `reshape`
b = reshape (a,[interval,size(v,2)]); %reshape into matrix
result = reshape(vertcat (v,b), [1,(size(b,1)+1)*size(b,2)]); %insert the values into the right position and convert back into vector
%remove padded zeros
final = result(result ~= 0) %remove the zero padding.
>>final =
Columns 1 through 16
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6
Columns 17 through 18
5 2
Here's an approach using boolean-indexing -
% Inputs
a = [3 4 6 8 1 5 8 9 3 7 3 6 2]
append_vals = [1 2 3 4 5]
append_interval = 4 % Starting at 1st index
% Find out indices of regular intervals where new elements are to be inserted.
% This should create that array [1,5,9,13,17]
N_total = numel(a) + numel(append_vals)
append_idx = find(rem(0:N_total-1,append_interval)==0)
% Get boolean array with 1s at inserting indices, 0s elsewhere
append_mask = ismember(1:N_total,append_idx)
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(~append_mask) = a
out(append_mask) = append_vals
Alternatively, we can also use linear-indexing and avoid creating append_mask, like so -
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(append_idx) = append_vals
out(setdiff(1:numel(out),append_idx)) = a
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]; % // Your original values
pos = [1 5 9 13 17]; % // The position of the values you want to insert
b=[1 2 3 4 5]; % // The values you want to insert
% // Pre-allocate a vector with the total size to hold the resulting values
r = zeros(size(a,2)+size(pos,2),1);
r(pos) = b % // Insert the appended values into the resulting vector first
r3 = r.' <1 % // Find the indices of the original values. These will be zero in the variable r but 1 in r3
ans =
0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1
ind= find(r3==1) % // Find the indices of the original values
ind =
2 3 4 6 7 8 10 11 12 14 15 16 18
r(ind) = a; % // Insert those into the resulting vector.
r.'
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
You can use this function to append a bunch of values to an existing vector, given their positions in the new vector:
function r=append_interval(a,v,p)
% a - vector with initial values
% v - vector containing values to be inserted
% p - positions for values in v
lv=numel(v); % number of elements in v vector
la=numel(a); % number of elements in a vector
column_a=iscolumn(a); % check if a is a column- or row- wise vector
tot_elements=la+lv;
% size of r is tha max between the total number of elements in the two vectors and the higher positin in vector p (in this case missing positions in a are filled with zeros)
lr=max([max(p) tot_elements]);
% initialize r as nan vector
r=zeros(column_a*(lr-1)+1,~column_a*(lr-1)+1)/0;
% set elements in p position to the corresponding values in v
r(p)=v;
% copy values in a in the remaining positions and fill with zeros missing entries (if any)
tot_missing_values=lr-tot_elements;
if(tot_missing_values)
remaining_values=cat(2-iscolumn(a),a,zeros(column_a*(tot_missing_values-1)+1,~column_a*(tot_missing_values-1)+1));
else
remaining_values=a;
end
% insert values
r(isnan(r))=remaining_values;
You can use row-wise or column-wise vectors; the orientation of r will be the same of that of a.
Input:
a =
3 4 6 8 1 5 8 9 3 7 3 6 2
v =
1 2 3 4 5
p =
1 5 9 13 17
Output:
>> append_interval(a,v,p)
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
Every sequence of positive positions is allowed and the function will pad for you with zeros the final vector, in case you indicate a position exceding the sum of the original vector and added items.
For example, if:
v3 =
1 2 3 4 5 6 90
p3 =
1 5 9 13 17 30 33
you get:
append_interval(a,v3,p3)
ans =
Columns 1 through 19
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2 0
Columns 20 through 33
0 0 0 0 0 0 0 0 0 0 6 0 0 90
Hope this will help.
As I am trying to multiply a m x n Matrix with a p-dimensional vector, I am stumbling across some difficulties.
Trying to avoid for loops, here is what I am looking to achieve
enter code here
M = [1 2 3; p = [1;2;3]
4 5 6;
7 8 9]
I want to obtain a 3x3x3 matrix, where the slices in third dimension are simply the entries of M multiplied by the respective entry in p.
Help is much appreciated
You can use bsxfun with permute for a vectorized (no-loop) approach like so -
out = bsxfun(#times,M,permute(p(:),[3 2 1]))
You would end up with -
out(:,:,1) =
1 2 3
4 5 6
7 8 9
out(:,:,2) =
2 4 6
8 10 12
14 16 18
out(:,:,3) =
3 6 9
12 15 18
21 24 27
With matrix-multiplication -
out = permute(reshape(reshape(M.',[],1)*p(:).',[size(M) numel(p)]),[2 1 3])
Having the values of time sequence, I would like to reshape it into a nx4 matrix [X y], for the purpose of using these values as input and output values for machine learning algorithm.
X(i) is a 1x3 input vector and y is output scalar value.
The algorithm takes as an input every 2nd sequence value (3 values) in order to predict the 4th value.
To give a practical example, let's say we have a sequence
[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16]
The [X y] matrix should be the following:
[1 3 5 7; 2 4 6 8; 9 11 13 15; 10 12 14 16]
To get every second row I wrote the following code:
vec1 = timeSeries(1:2:end);
XyVec1 = reshape(vec1,4,[])'
similarly it could be written to get even numbers:
vec2 = timeSeries(2:2:end);
XyVec2 = reshape(vec2,5,[])'
The thing that I don't know how to do is to interleave matrix vec1 and vec2 rows to get
[vec(1,:); vec2(1,:);vec1(2,:), vec2(2,:)...]
Does anyone know how to interleave the rows of two (or more) matrices?
Try
result = zeros(size(vec1,1)+size(vec2,1),size(vec1,2));
result(1:2:end,:) = vec1;
result(2:2:end,:) = vec2;
Reuse matlab indexing facilities ot insert elements in correct rows
Sample octave mock-up: http://ideone.com/RVgmYA
There is this one-liner option
result = kron(vec1, [1;0]) + kron(vec2, [0;1]);
However, #Joel Falcou is faster. Having set the input vectors as
vec1 = rand(1000,1000);
vec2 = -rand(1000,1000);
it gives
Elapsed time is 0.007620 seconds. (indexing)
Elapsed time is 0.054607 seconds. (kron)
Good luck :) figuring out what's going on with those reshape(), permutes():
a = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16];
reshape(permute(reshape(a,2,4,[]),[2,1,3]),4,[])'
the result
ans =
1 3 5 7
2 4 6 8
9 11 13 15
10 12 14 16
To interleave the vectors as mentioned in the end of your question you can use
reshape([vec1, vec2]', 4, [])'
for
vec1 =
1 3 5 7
9 11 13 15
vec2 =
2 4 6 8
10 12 14 16
it returns
>> reshape([vec1, vec2]', 4, [])'
ans =
1 3 5 7
2 4 6 8
9 11 13 15
10 12 14 16
I have the following two matrices
c=[1 0 1.05
1 3 2.05
1 6 2.52
1 9 0.88
2 0 2.58
2 3 0.53
2 6 3.69
2 9 0.18
3 0 3.22
3 3 1.88
3 6 3.98]
f=[1 6 3.9
1 9 9.1
1 12 9
2 0 0.3
2 3 0.9
2 6 1.2
2 9 2.5
3 0 2.7]
And the final matrix should be
n=[1 6 2.52 3.9
1 9 0.88 9.1
2 0 2.58 0.3
2 3 0.53 0.9
2 6 3.69 1.2
2 9 0.18 2.5
3 0 3.22 2.7]
The code I used gives as a result only the last row of the previous matrix [n].
for j=1
for i=1:rs1
for k=1
for l=1:rs2
if f(i,j)==c(l,k) && f(i,j+1)==c(l,k+1)
n=[f(i,j),f(i,j+1),f(i,j+2), c(l,k+2)];
end
end
end
end
end
Can anyone help me on this?
Is there something more simple?
Thanks in advance
You should learn to use set operations and avoid loops wherever possible. Here intersect could be extremely useful:
[u, idx_c, idx_f] = intersect(c(:, 1:2) , f(:, 1:2), 'rows');
n = [c(idx_c, :), f(idx_f, end)];
Explanation: by specifying the 'rows' flag, intersect finds the common rows in c and f, and their indices are given in idx_c and idx_f respectively. Use vector subscripting to extract matrix n.
Example
Let's use the example from your question:
c = [1 0 1.05;
1 3 2.05
1 6 2.52
1 9 0.88
2 0 2.58
2 3 0.53
2 6 3.69
2 9 0.18
3 0 3.22
3 3 1.88
3 6 3.98];
f = [1 6 3.9
1 9 9.1
1 12 9
2 0 0.3
2 3 0.9
2 6 1.2
2 9 2.5
3 0 2.7];
[u, idx_c, idx_f] = intersect(c(:, 1:2) , f(:, 1:2), 'rows');
n = [c(idx_c, :), f(idx_f, end)];
This should yield the desired result:
n =
1.0000 6.0000 2.5200 3.9000
1.0000 9.0000 0.8800 9.1000
2.0000 0 2.5800 0.3000
2.0000 3.0000 0.5300 0.9000
2.0000 6.0000 3.6900 1.2000
2.0000 9.0000 0.1800 2.5000
3.0000 0 3.2200 2.7000
According to this answer on Mathworks support you can use join from the statistics toolbox, specifically in your case, an inner join.
Unfortunately I don't have access to my computer with matlab on it, but give it a try and let us know how/if it works.
You can reduce the number of loops by comparing both the first and second columns of at once, then using the "all" function to only collapse the values if they both match. The following snippet replicates the "n" array you had provided.
n = [];
for r1 = 1:size(c, 1)
for r2 = 1:size(f,1)
if all(c(r1, [1 2]) == f(r2, [1 2]))
n(end+1, 1:4) = [c(r1,:) f(r2,3)];
end
end
end
If you insist on doing this in a loop you need to give n the proper dimension according
to the loop counter you are using, or concatenate it to itself of each iteration (this can be very slow for big matrices). For example, writing:
for j=1
for i=1:rs1
for k=1
for l=1:rs2
m=m+1;
if f(i,j)==c(l,k) && f(i,j+1)==c(l,k+1)
n(m,:)=[f(i,j),f(i,j+1),f(i,j+2), c(l,k+2)];
end
end
end
end
end
will save into the m-th row the for numbers when the loop reaches a counter value of m.
However, just be aware that this can be done also without a nested loop and an if condition, in a vectorized way. For example, instead of the condition if f(i,j)==c(l,k)... you can use ismember etc...
How about without any for loops at all (besides in native code)
mf = size(f,1);
mc = size(c,1);
a = repmat(c(:,1:2),1,mf);
b = repmat(reshape((f(:,1:2))',1,[]),mc,1);
match = a == b;
match = match(:, 1 : 2 : 2*mf) & match(:, 2 : 2 : 2*mf);
crows = nonzeros(diag(1:mc) * match);
frows = nonzeros(match * diag(1:mf));
n = [c(crows,:),f(frows,3)]