Let's say I have a short vector x = [a,b,c,d,e]; What would be the best way to compute all the difference between members of the vector as:
y = [e-d e-c e-b e-a
d-e d-c d-b d-a
c-e c-d c-b c-a
b-e b-d b-c b-a
a-e a-d a-c a-b];
Thanks in advance
To give that exact matrix, try:
x = [1;2;3;4;5]; %# note this is a column vector (matrix of rows in general)
D = squareform( pdist(x,#(p,q)q-p) );
U = triu(D);
L = tril(D);
y = flipud(fliplr( L(:,1:end-1) - U(:,2:end) ))
result in this case:
y =
1 2 3 4
-1 1 2 3
-2 -1 1 2
-3 -2 -1 1
-4 -3 -2 -1
First creat a circulant matrix, then compute the different between the first column and the rest columns. Here is a reference for creating a circulant matrix
Related
Background
I am currently working on a lecture for my Engineering in MATLAB course and stumbled upon a problem I would like to present to the class. I have made many different attempts to solve this problem, but my graphs keep coming out incorrect. I will describe the problem below and all the steps I took to try to solve this problem.
Problem
Find the coefficients of the fourth-degree polynomial
P(x) = ax^4 + bx^3 + cx^2 + dx + e
whose graph goes through the points (0, 1), (1, 1), (-1,3), and
whose slopes at x = -1 is 20 and at x = 1 is 9.
Check your answer visually.
Attempt
I began by creating a matrix of the above x-values that I have derived as follows:
A = [0^4 0^3 0^2 0 1; 1^4 1^3 1^2 1 1; (-1)^4 (-1)^3 (-1)^2 -1 1];
A = [0 0 0 0 1; 1 1 1 1 1; 1 -1 1 -1 1];
This creates a 5 column by 3 row matrix that I may use to plot the polynomial.
My issue is that I am unable to get the last row of x-values, since each row is an equation in the system of equations and there must be as many equations as there are unknowns (4: a, b, c, and d are unknown, but e always equals 1 as you can see).
Ignoring this issue for a moment, I can continue to create a vertical matrix of y-values so that I may solve the system of equations. These y values are already given, so all I have to do is type this code in:
y = [1 1 3]';
Once again, there should be a fourth y-value to go along with the system of equations, but I have been unable to derive it using just the slopes of the points at x = -1 and x = 1.
Once both the x-values and the y-values are derived, we can proceed to using the backslash operator (/) to solve the system of linear equations A*x = y.
p = A\y;
mldivide is more info on the mldivide function for anyone who needs reference.
From here on out, the following code which creates a polynomial from this system of equations and graphs it, should stay the same.
u = -1:.01:1;
v = polyval(p, u);
plot(u,v);
In this code, u is the domain of x-values from -1 to 1 with a 0.01 interval. This is needed by us to use the polyval function, which creates a polynomial from a system of equations we derived at p on the interval u.
Lastly, plot simply graphs our derived polynomial using MATLAB's GUI on the interval u.
As you can see, the only missing pieces I have are one more row of x-values in my matrix A and one y-value in matrix y that I need to find the four unknowns a, b, c, and d. I believe you must use the two slopes given in the problem to find each point. I have tried using the polyder function to get the derivative of the matrix p by doing,
q = polyder(p);
but I am still confused as to how to continue from there. Any help will be greatly appreciated.
I would calculate the derivative of the polynomial:
dP(x) = 4ax^3 + 3bx^2 + 2cx + d
Now, you know that dP(-1)=20 and dP(1)=9 so you have 5 equations with 5 unknowns:
e = 1
a + b + c + d + e = 1
a - b + c - d + e = 3
-4*a + 3*b - 2*c + d = 20
4*a + 3*b + 2*c + d = 9
So you can construct a 5x5 matrix and solve the system, as you did with A\y.
The code to construct this 5x5 matrix is:
A = [0 0 0 0 1 ; 1 1 1 1 1 ; 1 -1 1 -1 1 ; -4 3 -2 1 0 ; 4 3 2 1 0];
y = [1 1 3 20 9]';
You can then check the results on a plot:
p=A\y;
u = -1:.01:1;
v = polyval(p, u);
data_pts = [0, 1; 1, 1;-1, 3]
plot(u,v,data_pts(:,1),data_pts(:,2),'rx')
which gives the following plot:
You can do the same with the derivative and checks it goes through the points (-1,20) and (1,9).
I am trying to solve this problem:
Write a function called cancel_middle that takes A, an n-by-m
matrix, as an input where both n and m are odd numbers and k, a positive
odd integer that is smaller than both m and n (the function does not have to
check the input). The function returns the input matrix with its center k-by-k
matrix zeroed out.
Check out the following run:
>> cancel_middle(ones(5),3)
ans =
1 1 1 1 1
1 0 0 0 1
1 0 0 0 1
1 0 0 0 1
1 1 1 1 1
My code works only when k=3. How can I generalize it for all odd values of k? Here's what I have so far:
function test(n,m,k)
A = ones(n,m);
B = zeros(k);
A((end+1)/2,(end+1)/2)=B((end+1)/2,(end+1)/2);
A(((end+1)/2)-1,((end+1)/2)-1)= B(1,1);
A(((end+1)/2)-1,((end+1)/2))= B(1,2);
A(((end+1)/2)-1,((end+1)/2)+1)= B(1,3);
A(((end+1)/2),((end+1)/2)-1)= B(2,1);
A(((end+1)/2),((end+1)/2)+1)= B(2,3);
A(((end+1)/2)+1,((end+1)/2)-1)= B(3,1);
A(((end+1)/2)+1,((end+1)/2))= B(3,2);
A((end+1)/2+1,(end+1)/2+1)=B(3,3)
end
You can simplify your code. Please have a look at
Matrix Indexing in MATLAB. "one or both of the row and column subscripts can be vectors", i.e. you can define a submatrix. Then you simply need to do the indexing correct: as you have odd numbers just subtract m-k and n-k and you have the number of elements left from your old matrix A. If you divide it by 2 you get the padding on the left/right, top/bottom. And another +1/-1 because of Matlab indexing.
% Generate test data
n = 13;
m = 11;
A = reshape( 1:m*n, n, m )
k = 3;
% Do the calculations
start_row = (n-k)/2 + 1
start_col = (m-k)/2 + 1
A( start_row:start_row+k-1, start_col:start_col+k-1 ) = zeros( k )
function b = cancel_middle(a,k)
[n,m] = size(a);
start_row = (n-k)/2 + 1;
start_column = (m-k)/2 + 1;
end_row = (n-k)/2 + k;
end_column = (m-k)/2 + k;
a(start_row:end_row,start_column:end_column) = 0;
b = a;
end
I have made a function in an m file called cancel_middle and it basically converts the central k by k matrix as a zero matrix with the same dimensions i.e. k by k.
the rest of the matrix remains the same. It is a general function and you'll need to give 2 inputs i.e the matrix you want to convert and the order of submatrix, which is k.
I have a vector : for example S=(0,3,2,0,1,2,0,1,1,2,3,3,0,1,2,3,0).
I want to count co-occurrence neighbors, for example the first neighbor "o,3" how many times did it happen till the end of the sequence? Then it investigates the next pair"2,0" and similarly do it for other pairs.
Below is a part of my code:
s=size(pic);
S=s(1)*s(2);
V = reshape(pic,1,S);
min= min(V);
Z=zeros(1,S+1);
Z(1)=min;
Z(2:end)=V;
for i=[0:2:length(Z)-1];
contj=0
for j=0;length(Z)-1;
if Z(i,i+1)= Z(j,j+1)
contj=contj+1
end
end
count(i)= contj
end
It gives me this error:
The expression to the left of the equals sign is not a valid target for an assignment
in this line:
if Z(i,i+1)= Z(j,j+1)
I read similar questions and apply the tips on it but they didn't work!
If pairs are defined without overlapping (according to comments):
S = [0,3,2,0,1,2,0,1,1,2,3,3,0,1,2,3]; %// define data
S2 = reshape(S,2,[]).'; %'// arrange in pairs: one pair in each row
[~, jj, kk] = unique(S2,'rows'); %// get unique labels for pairs
pairs = S2(jj,:); %// unique pairs
counts = accumarray(kk, 1); %// count of each pair. Or use histc(kk, 1:max(kk))
Example: with S as above (I introduce blanks to make pairs stand out),
S = [0,3, 2,0, 1,2, 0,1, 1,2, 3,3, 0,1, 2,3];
the result is
pairs =
0 1
0 3
1 2
2 0
2 3
3 3
counts =
2
1
2
1
1
1
If pairs are defined without overlapping but counted with overlapping:
S = [0,3,2,0,1,2,0,1,1,2,3,3,0,1,2,3]; %// define data
S2 = reshape(S,2,[]).'; %'// arrange in pairs: one pair in each row
[~, jj] = unique(S2,'rows'); %// get unique labels for pairs
pairs = S2(jj,:); %// unique pairs
P = [S(1:end-1).' S(2:end).']; %// all pairs, with overlapping
counts = sum(pdist2(P,pairs,'hamming')==0);
If you don't have pdist2 (Statistics Toolbox), replace last line by
counts = sum(all(bsxfun(#eq, pairs.', permute(P, [2 3 1]))), 3);
Result:
>> pairs
pairs =
0 1
0 3
1 2
2 0
2 3
3 3
>> counts
counts =
3 1 3 2 2 1
do it using sparse command
os = - min(S) + 1; % convert S into indices
% if you want all pairs, i.e., for S = (2,0,1) count (2,0) AND (0,1):
S = sparse( S(1:end-1) + os, S(2:end) + os, 1 );
% if you don't want all pairs, i.e., for S = (2,0,1,3) count (2,0) and (1,3) ONLY:
S = sparse( S(1:2:end)+os, S(2:2:end) + os, 1 );
[f s c] = find(S);
f = f - os; % convert back
s = s - os;
co-occurences and their count are in the pairs (f,s) - c
>> [f s c]
ans =
2 0 2 % i.e. the pair (2,0) appears twice in S...
3 0 2
0 1 3
1 1 1
1 2 3
3 2 1
0 3 1
2 3 2
3 3 1
Let's take two vectors:
a = [1 ; 2; 3]
b = [0 ; 9 ; -5]
If I want minimum value of the vector and it's position I can simply:
[x, ix] = min(a)
I can also compare two vectors and get minimum values:
> min(a, b)
ans =
0
2
-5
But it is impossible to get positions of min values of two vectors:
> [x, ix] = min(a, b)
x =
0
2
-5
error: element number 2 undefined in return list
Why? How to get them? Is there a simple method?
here's how to do that:
[v id]=min([a,b]')
It's a matter of having the right insight:
[x,ix] = min([a b],[],2)
You must think about what the intended output of ix is.
This shows you in which vector the minimum is:
ix=a<b;
x=a.*ix+b.*not(ix);
I have a vector a=[1 2 3 1 4 2 5]'
I am trying to create a new vector that would give for each row, the occurence number of the element in a. For instance, with this matrix, the result would be [1 1 1 2 1 2 1]': The fourth element is 2 because this is the first time that 1 is repeated.
The only way I can see to achieve that is by creating a zero vector whose number of rows would be the number of unique elements (here: c = [0 0 0 0 0] because I have 5 elements).
I also create a zero vector d of the same length as a. Then, going through the vector a, adding one to the row of c whose element we read and the corresponding number of c to the current row of d.
Can anyone think about something better?
This is a nice way of doing it
C=sum(triu(bsxfun(#eq,a,a.')))
My first suggestion was this, a not very nice for loop
for i=1:length(a)
F(i)=sum(a(1:i)==a(i));
end
This does what you want, without loops:
m = max(a);
aux = cumsum([ ones(1,m); bsxfun(#eq, a(:), 1:m) ]);
aux = (aux-1).*diff([ ones(1,m); aux ]);
result = sum(aux(2:end,:).');
My first thought:
M = cumsum(bsxfun(#eq,a,1:numel(a)));
v = M(sub2ind(size(M),1:numel(a),a'))
on a completely different level, you can look into tabulate to get info about the frequency of the values. For example:
tabulate([1 2 4 4 3 4])
Value Count Percent
1 1 16.67%
2 1 16.67%
3 1 16.67%
4 3 50.00%
Please note that the solutions proposed by David, chappjc and Luis Mendo are beautiful but cannot be used if the vector is big. In this case a couple of naïve approaches are:
% Big vector
a = randi(1e4, [1e5, 1]);
a1 = a;
a2 = a;
% Super-naive solution
tic
x = sort(a);
x = x([find(diff(x)); end]);
for hh = 1:size(x, 1)
inds = (a == x(hh));
a1(inds) = 1:sum(inds);
end
toc
% Other naive solution
tic
x = sort(a);
y(:, 1) = x([find(diff(x)); end]);
y(:, 2) = histc(x, y(:, 1));
for hh = 1:size(y, 1)
a2(a == y(hh, 1)) = 1:y(hh, 2);
end
toc
% The two solutions are of course equivalent:
all(a1(:) == a2(:))
Actually, now the question is: can we avoid the last loop? Maybe using arrayfun?