Consider the matrix
c =
1 2
3 4
m = 2;
n = 2;
% mean
% sum1 = uint32(0);
b4 = sum(c);
b5 = sum(b4');
c5 = b5 / ( m * n )
% standard deviation
sum2 = uint32(0);
for i = 1 : m
for j = 1 : n
b = ( double(c(i,j)) - c5 ) ^ 2 ;
sum2 = sum2 + b ;
end
end
sum3 = sum2 / ( m * n );
std_dev = sqrt(double(sum3))
% skewness
sum9 = 0;
for i = 1 : m
for j = 1 : n
skewness_old = ( ( double(c(i,j)) - c5 ) / ( std_dev) )^ 3 ;
sum9 = sum9 + skewness_old ;
end
end
skewness_new = sum9 / ( m * n )
The skewness result is 0
If I use the matlab function skewness,
skewness(c)
c =
1 2
3 4
skewness(c)
ans =
0 0
Why is the function skewness returning two 0's, While the formula returns only one 0
MATLAB function SKEWNESS by default calculates skewness for each column separately. For the whole matrix do skewness(c(:)).
Related
Suppose there is a value n input from a user and it goes in to the following for loop code. Is there a way to vectorize the following code?
A = 1:n
B = [1 1;1 1]
for i = 1:n
B = B + A(i)*B;
end
Let's have a look at a specific example:
n = 5;
A = 1:n;
B = [1 1; 1 1];
for i = 1:n
B = B + A(i) * B;
end
B
The result is:
B =
720 720
720 720
First of all, I would re-write the loop:
n = 5;
A = 1:n;
B = [1 1; 1 1];
for i = 1:length(A)
B = B * (A(i) + 1);
end
B
That way, it's more obvious, that your loop variable i simply iterates all elements in A.
Also: B + A(i) * B is the same as B * (A(i) + 1).
Now, we see, that inside the loop, you're basically calculating:
B = B * (A(1) + 1) * (A(2) + 1) * (A(3) + 1) ...
The product over all elements in A (or here: A + 1) can be simplified by using MATLAB's prod function:
n = 5;
A = 1:n;
B = [1 1; 1 1];
B = B * prod(A + 1)
Let's check the result:
B =
720 720
720 720
In that very special case for A = 1:n, the product prod(A + 1) is simply the factorial of n + 1, such that we could also use MATLAB's factorial function:
n = 5;
B = [1 1; 1 1];
B = B * factorial(n + 1)
i want to declare that x will be variable for binary 0,1
i and k represent facilities in flow matrix, j and q represent location in distance matrix ..
x(i,j) mean that x will be equal to 1 if i (facility) is assign in j (location) ..
x(i, j) = 1 if facility i is assigned to location j and if otherwise, xij = 0,
so otherwise mean that if x(k,q) =1 , x(i,J) will be 0...
example of the manual calculation
Min =(f i1,k1 * d j1,q1 * x i1,j1 * x k1,q1) + (f i1,k1 * d j1,q2 * x i1,j1 * x k1,q2) + (f i1,k1 * d j1,q3 * x i1,j1 * x k1,q3)....
( 0*0* 1*1 ) + ( 0* 6* 1*0 ) + ( 0*8 * 1 *0 ).....
I want to *xi1,j1 * xk1,q1 to be 0 or 1.. if i choose i1, j1=1 the other will be 0.. for example i2,j1 will be equal to 0
below is the coding
clc;
clear;
%sum sum sum sum(fik*djq*xij*xkq)
%i,k= facilities
%j,q= location
%f(i,k)= flow between facilities i and k
%d(j,q)= distance between locations j and q
%xij = 1 if facility i is assigned to location j and if otherwise, xij = 0
% Flow matrix: flow assigning facility i (column) to facility k (row)
f = [0 5 7 9;
5 0 4 6;
7 4 0 3;
9 6 3 0];
%Distance matrix: distance assigning location j (column) to location q (row)
d = [0 6 8 9;
6 0 5 1;
8 5 0 2;
9 1 2 0];
z= 0;
nf= 4;
nd= 4;
for i=1:nf
for j=1:nf
for k=1:nd
for q=1:nd
z = min('z','f(i,k)*d(j,q)*x(i,j)*x(k,q)');
end
end
end
end
%Constraints
%The first set of constraints requires that each facility gets exactly one
%location, that is for each facility, the sum of the location values
%corresponding to that facility is exactly one
Constraints.constr1 = sum(x,2) == 1;
%The second set of constraints are inequalities. These constraints specify
%that each office has no more than one facility in it.
Constraints.constr2 = sum(x,1) == 1;
disp (z);
by using recperm
this recperm is used to fine permuation 0,1 in the coding
i have a vector and a scalar as input in a problem , the problem is to compute the largest product of n consecutive number of the vector and output the product and the index of the element of the vector that is the first term of the product.
E.g vector =[ 1 2 3 4 5 6] , n=3
I'm supposed to get '3' (i e n) consecutive number of vector whose product is the largest .
in this case it will be 4*5*6
so output will be 120 and 4 as the index.
now if vector has fewer than 'n' elements, the function outputon returns 0 and -1 as the output.
please i need ideas on how to achieve this
You can make a simple implementation with loops:
a = [1 2 3 4 5 6]
w = 3
n = length(a)
maximum = -1
for i = 1:n-w
p = prod(a(i:i+w))
if (p > maximum)
maximum = p
end
end
maximum
Or, you can use the nlfilter from the image processing palette.
a = []
w = 3
if (length(a) >= w)
products = nlfilter(a, [1 w], #(x) prod(x))
res = max(products)
else
res = -1
end
You can create the moving window from the answer of raryeng for the question Matrix with sliding window elements and then appyl cumprod on the columns and take the max with its index.
myvec = [1 2 2 1 3 1];
n = 3;
ind = bsxfun(#plus, 1:n, (0:1:length(myvec)-n).')';
M = cumprod(myvec(ind));
[val,its_ind] = max(M(end,:));
You can encapsulate this with an if condition checking whether the length of myvec is larger than
You can compute the element-wise logarithm of your vector, so multiplication becomes addition; and sliding addition is just convolution with a window of ones:
function [m, p] = f(v, n)
if numel(v) < n
m = -1;
p = 0;
else
c = conv(log(v(:)), ones(n,1), 'valid'); % convolution with vector of n ones
[~, m] = max(c); % starting index of maximizing window
p = prod(v(m+(0:n-1))); % corresponding product
end
Examples:
>> v = [1 2 3 4 5 6]; n = 3;
>> [m, p] = f(v,n)
m =
4
p =
120
>> v = [1 2 3 4 5 6]; n = 7;
>> [m, p] = f(v,n)
m =
-1
p =
0
>> v = [1 4 6 2 5 3]; n = 3;
>> [m, p] = f(v,n)
m =
3
p =
60
MATLAB:
I am trying to do basis expansion of a huge matrix(1000x15).
For example,
X =
x1 x2
1 4
2 5
3 6
I want to build a new matrix.
Y =
x1 x2 x1*x1 x1*x2 x2*x2
1 4 1 4 16
2 5 4 10 25
3 6 9 18 36
Could any one please suggest a easier way to do this
% your input
A = [1 4; 2 5; 3 6];
% generate pairs
[p,q] = meshgrid(1:size(A,2), 1:size(A,2));
% only retain unique pairs
ii = tril(p) > 0;
% perform element wise multiplication
res = [A A(:,p(ii)) .* A(:,q(ii))];
Using the one-liner from this answer to get the 2-combinations of the indices, you can generate the matrix without the interleaved ordering with
function Y = columnCombo(Y)
comb = nchoosek(1:size(Y,2),2);
Y = [Y , Y.^2 , Y(:,comb(:,1)).*Y(:,comb(:,2))];
end
For the interleaved ordering, I came up with this, possibly sub-optimal, solution:
function Y = columnCombo(Y)
[m,n] = size(Y);
comb = nchoosek(1:n,2);
Y = [Y,zeros(m,n + size(comb,1))];
col = n+1;
for k = 1:n-1
Y(:,col) = Y(:,k).*Y(:,k) ;
ms = comb(comb(:,1)==k,:) ;
ncol = size(ms,1) ;
Y(:,col+(1:ncol)) = Y(:,ms(:,1)).*Y(:,ms(:,2)) ;
col = col + ncol + 1 ;
end
Y(:,end) = Y(:,n).^2;
end
I'm trying find all combinations of an n-by-n matrix without repetitions.
For example, I have a matrix like this:
A = [321 319 322; ...
320 180 130; ...
299 100 310];
I want the following result:
(321 180 310)
(321 130 100)
(319 320 310)
(319 139 299)
(322 320 100)
(322 180 299)
I have tried using ndgrid, but it takes the row or the column twice.
Here's a simpler (native) solution with perms and meshgrid:
N = size(A, 1);
X = perms(1:N); % # Permuations of column indices
Y = meshgrid(1:N, 1:factorial(N)); % # Row indices
idx = (X - 1) * N + Y; % # Convert to linear indexing
C = A(idx) % # Extract combinations
The result is a matrix, each row containing a different combination of elements:
C =
321 180 310
319 320 310
321 130 100
319 130 299
322 320 100
322 180 299
This solution can also be shortened to:
C = A((perms(1:N) - 1) * N + meshgrid(1:N, 1:factorial(N)))
ALLCOMB is the key to your question
E.g. I am not front of a MATLAB machine so, I took a sample from web.
x = allcomb([1 3 5],[-3 8],[],[0 1]) ;
ans
1 -3 0
1 -3 1
1 8 0
...
5 -3 1
5 8 0
5 8 1
You can use perms to permute the columns as follows:
% A is given m x n matrix
row = 1:size( A, 1 );
col = perms( 1:size( A, 2 ) );
B = zeros( size( col, 1 ), length( row )); % Allocate memory for storage
% Simple for-loop (this should be vectorized)
% for c = 1:size( B, 2 )
% for r = 1:size( B, 1 )
% B( r, c ) = A( row( c ), col( r, c ));
% end
% end
% Simple for-loop (further vectorization possible)
r = 1:size( B, 1 );
for c = 1:size( B, 2 )
B( r, c ) = A( row( c ), col( r, c ));
end