I have this code in Mathematica:
nxBin = Table[{-5 sX + (i - 0.5)*step, nBin[[i]]}, {i, 1, Length[nBin]}]
and I did this in MATLAB:
a=zeros(length(nBin),1);
nxBin=zeros(length(nBin),1);
for i=1:length(nBin)
anew=a*step*(i-0.5) -5*sX;
b=zeros(length(nBin(i)),1);
nxBin(i,:)=[anew , b]
end
but MATLAB says
??? Error using ==> horzcat
CAT arguments dimensions are not consistent.
Error in ==> begin at 52
nxBin(i,:)=[anew , b]
Can anyone tell me why I get this error? Also, can I do this with fewer lines?
You want to catenate the n-by-1 array nBin with steps (probably x-values for a histogram). Thus, you can simply create the "x-vector" and combine them.
nxBin = [ -5*sX + ((1:length(nBin))' - 0.5) * nStep, nBin(:)]
Here's the same step-by-step
%# make a vector with values from 1 to nBin
x = 1:length(nBin);
%# transpose, since it's 1-by-n and we want n-by-1
x = x'; %'#
%# apply the modification to x
x = -5*sX + (x-0.5)*nStep;
%# catenate with nBin (the colon operator guarantees it's n-by-1
nxBin = [x, nBin(:)];
EDIT
In case you want to plot this, you can do
plot(nxBin(:,1),nxBin(:,2),'.')
or, if I guess right and it's a histogram
bar(nxBin(:,1),nxBin(:,2))
Related
Suppose I have a column vector of formulae like this
N =
4*k2 + 5*k3 + k1*x
7*k2 + 8*k3 + k1*y
and a column vector of symbolic variables like this
k =
k1
k2
k3
The formulae are linear with respect to k. I'd like to find a matrix M such that M*k equals N.
I can do this with N/k. However, that gives
[ (4*k2 + 5*k3 + k1*x)/k1, 0, 0]
[ (7*k2 + 8*k3 + k1*y)/k1, 0, 0]
which is correct, but not what I want. What I want is the matrix
x 4 5
y 7 8
which seems to me the simplest answer in that it involves no variables from k.
How do I convince Matlab to factor out the specified variables from a formula or a vector of formulae?
You can use coeffs, specifically the form
C = coeffs(p,vars) returns coefficients of the multivariate polynomial p with respect to the variables vars.
Since the first input needs to be a polynomial, you need to pass each component of N:
coeffs(N(1), k)
coeffs(N(2), k)
Or use a loop and store all results in a symbolic array:
result = sym('result', [numel(N) numel(k)]); % create symbolic array
for m = 1:numel(N)
result(m,:) = coeffs(N(m), k);
end
In your example, this gives
result =
[ 5, 4, x]
[ 8, 7, y]
Based on #LuisMendo's answer, I used coeffs. But there are a couple of problems with coeffs. The first is that its result doesn't include any coefficients that are 0. The second is that it doesn't seem to guarantee that the coefficients are ordered the same way as the variables in its second argument. I came up with the following function to replace coeffs.
Luckily coeffs returns a second result that lists the variables associated with each item in the first result. (It's more complicated if the formula is not linear.)
function m = factorFormula(f, v )
% Pre: f is a 1x1 sym representing a
% linear function of the variables in v.
% Pre: v is a column vector of variables
% Post: m is a row vector such that m*v equals f
% and the formulas in m do not contain the
% variables in v
[cx,tx] = coeffs(f,v)
n = size(v,1)
m = sym(zeros(1,n))
for i = 1:n
j = find(tx==v(i))
if size(j,2) == 1
m(i) = cx(j)
end
end
end
This only works for one formula, but it can be extended to a vector using the loop in #LuisMendo's answer or this equivalent expression in #Sanchises comment there.
cell2sym(arrayfun( #(f)factorFormula(f,k),N,'UniformOutput',false ) )
I hope there is a better answer than this.
I am trying to use mvregress with the data I have with dimensionality of a couple of hundreds. (3~4). Using 32 gb of ram, I can not compute beta and I get "out of memory" message. I couldn't find any limitation of use for mvregress that prevents me to apply it on vectors with this degree of dimensionality, am I doing something wrong? is there any way to use multivar linear regression via my data?
here is an example of what goes wrong:
dim=400;
nsamp=1000;
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X=randn(dim, nsamp)*sqrt(dataVariance ) + repmat(mixtureCenters,1,nsamp);
N=randn(dim, nsamp)*sqrt(noiseVariance ) + repmat(mixtureCenters,1,nsamp);
A=2*eye(dim);
Y=A*X+N;
%without residual term:
A_hat=mvregress(X',Y');
%wit residual term:
[B, y_hat]=mlrtrain(X,Y)
where
function [B, y_hat]=mlrtrain(X,Y)
[n,d] = size(Y);
Xmat = [ones(n,1) X];
Xmat_sz=size(Xmat);
Xcell = cell(1,n);
for i = 1:n
Xcell{i} = [kron([Xmat(i,:)],eye(d))];
end
[beta,sigma,E,V] = mvregress(Xcell,Y);
B = reshape(beta,d,Xmat_sz(2))';
y_hat=Xmat * B ;
end
the error is:
Error using bsxfun
Out of memory. Type HELP MEMORY for your options.
Error in kron (line 36)
K = reshape(bsxfun(#times,A,B),[ma*mb na*nb]);
Error in mvregress (line 319)
c{j} = kron(eye(NumSeries),Design(j,:));
and this is result of whos command:
whos
Name Size Bytes Class Attributes
A 400x400 1280000 double
N 400x1000 3200000 double
X 400x1000 3200000 double
Y 400x1000 3200000 double
dataVariance 1x1 8 double
dim 1x1 8 double
mixtureCenters 400x1 3200 double
noiseVariance 1x1 8 double
nsamp 1x1 8 double
Okay, I think I have a solution for you, short version first:
dim=400;
nsamp=1000;
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X=randn(dim, nsamp)*sqrt(dataVariance ) + repmat(mixtureCenters,1,nsamp);
N=randn(dim, nsamp)*sqrt(noiseVariance ) + repmat(mixtureCenters,1,nsamp);
A=2*eye(dim);
Y=A*X+N;
[n,d] = size(Y);
Xmat = [ones(n,1) X];
Xmat_sz=size(Xmat);
Xcell = cell(1,n);
for i = 1:n
Xcell{i} = kron(Xmat(i,:),speye(d));
end
[beta,sigma,E,V] = mvregress(Xcell,Y);
B = reshape(beta,d,Xmat_sz(2))';
y_hat=Xmat * B ;
Strangely, I could not access the function's workspace, it did not appear in the call stack. This is why I put the function after the script here.
Here's the explanation that might also help you in the future:
Looking at the kron definition, the result when inserting an m by n and a p by q matrix has size mxp by nxq, in your case 400 by 1001 and 1000 by 1000, that makes a 400000 by 1001000 matrix, which has 4*10^11 elements. Now you have four hundred of them, and each element takes up 8 bytes for double precision, that is a total size of about 1.281 Petabytes of memory (or 1.138 Pebibytes, if you prefer), well out of reach even with your grand 32 Gibibyte.
Seeing that one of your matrices, the eye one, contains mostly zeros, and the resulting matrix contains all possible element product combinations, most of them will be zero, too. For such cases specifically, MATLAB offers the sparse matrix format, which saves a lot of memory depending on the number of zero elements in a matrix by only storing nonzero ones. You can convert a full matrix to a sparse representation with sparse(X), or you get an eye matrix directly by using speye(n), which is what I did above. The sparse property propagates to the result, which you should now have enough memory for (I have with 1/4 of your memory available, and it works).
However, what remains is the problem Matthew Gunn mentioned in a comment. I get an error saying:
Error using mvregress (line 260)
Insufficient data to estimate either full or least-squares models.
Preface
If your regressors are all the same across each regression equation and you're interested in the OLS estimate, you can replace a call to mvregress with a simple call to \.
It appears in the call to mlrtrain you had a matrix transposition error (since corrected). In the language of mvregress, n is the number of observations, d is the number of outcome variables. You generate a matrix Y that is d by n. But THEN when you should call mlrtrain(X', Y') not mlrtrain(X, Y).
If below isn't specifically, what you're looking for, I suggest you precisely define what you're trying to estimate.
What I would have written if I were you
So much that's been said here is completely off base that I'm posting code of what I would have written if I were you. I've reduced the dimensionality to show the equivalence in your special case to simply calling \. I've also written stuff in a more standard way (i.e. having observations run down the rows and not making matrix transposition errors).
dim=5; % These can go way higher but only if you use my code
nsamp=20; % rather than call mvregress
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X = randn(nsamp, dim)*sqrt(dataVariance ) + repmat(mixtureCenters', nsamp, 1); %'
E = randn(nsamp, dim)*sqrt(noiseVariance); %noise should be mean zero
B = 2*eye(dim);
Y = X*B+E;
% without constant:
B_hat = mvregress(X,Y); %<-------- slow, blows up with high dimension
B_hat2 = X \ Y; %<-------- fast, fine with higher dimensions
norm(B_hat - B_hat2) % show numerical equivalent if basically 0
% with constant:
B_constant_hat = mlrtrain(X,Y) %<-------- slow, blows up with high dimension
B_constant_hat2 = [ones(nsamp, 1), X] \ Y; % <-- fast, and fine with higher dimensions
norm(B_constant_hat - B_constant_hat2) % show numerical equivalent if basically 0
Explanation
I'll assume you have:
An nsamp by dim sized data matrix X.
An nsamp by ny sized matrix of outcome variables Y
You want the results from regressing each column of Y on data matrix X. That is, we're doing multivariate regression but there's a common data matrix X.
That is, we're estimating:
y_{ij} = \sum_k b_k * x_{ik} + e_{ijk} for i=1...nsamp, j = 1...ny, k=1...dim
If you're trying to do something different than this, you need to clearly state what you're trying to do!
To regress Y on X you could do:
[beta_mvr, sigma_mvr, resid_mvr] = mvregress(X, Y);
This appears to be horribly slow. The following should match mvregress for the case where you're using the same data matrix for each regression.
beta_hat = X \ Y; % estimate beta using least squares
resid = Y - X * beta_hat; % calculate residual
If you want to construct a new data matrix with a vector of ones, you would do:
X_withones = [ones(nsamp, 1), X];
Further clarification for some that are confused
Let's say we want to run the regression
y_i = \sum_j x_{ij} + e_i i=1...n, j=1...k
We can construct the data matrix n by k datamatrix X and an n by 1 outcome vector y. The OLS estimate is bhat = pinv(X' * X) * X' * y which can also be computed in MATLAB with bhat = X \ y.
If you want to do this multiple times (i.e. run multivariate regression on the same data matrix X), you can construct an outcome matrix Y where EACH column represents a separate outcome variable. Y = [ya, yb, yc, ...]. Trivially, the OLS solution is B = pinv(X'*X)*X'*Y which can be computed as B = X \ Y. The first column of B is the result of regressing Y(:,1) on X. The second column of B is the result of regressing Y(:,2) on X, etc... Under these conditions, this is equivalent to a call to B = mvregress(X, Y)
Even more test code
If regressors are the same and estimation is by simple OLS, there is an equivalence between multivariate regression and equation by equation ordinary least squares.
d = 10;
k = 15;
n = 100;
C = RandomCorr(d + k, 1); %Use any method you like to generate a random correlation matrix
s = randn(d+k , 1) * 10;
S = (s * s') .* C; % generate covariance matrix
mu = randn(d+k,1);
data = mvnrnd(ones(n, 1) * mu', S);
Y = data(:,1:d);
X = data(:,d+1:end);
[b1, sigma] = mvregress(X, Y);
b2 = X \ Y;
norm(b1 - b2)
You will notice b1 and b2 are numerically equivalent. They are equivalent even though sigma is EXTREMELY different from zero.
I'm new to MATLAB, and programming in general, and I am having difficulty accomplishing what I am sure is a very, very simple task:
I have a list of vectors v_i for i from 1 to n (n in some number), all of the same size k. I would like to create a vector v that is a "concatenation" (don't know if this is the correct terminology) of these vectors in increasing order: what I mean by this is that the first k entries of v are the k entries of v_1, the k+1 to 2k entries of v are the k entries of v_2 etc. etc. Thus v is a vector of length nk.
How should I create v?
To put this into context, here is function I've began writing (rpeakindex will just a vector, roughq would be the vector v I mentioned before):
function roughq = roughq(rpeakindex)
for i from 1 to size(rpeakindex) do
v_i = [rpeakindex(i)-30:1:rpeakindex(i)+90]
end
Any help is appreciated
Let's try two things.
First, for concatenating vectors there are a couple of methods here, but the simplest would be
h = horzcat(v_1, v_2);
The bigger problem is to enumerate all vectors with a "for" loop. If your v_n vectors are in a cell array, and they are in fact v{i}, then
h= [];
for j=1:n
h = horzcat(h, v{i});
end
Finally, if they only differ by name, then call them with
h=[];
for j=1:n
h= horzcat(h, eval(sprintf('v_%d',j));
end
Let the arrays (vectors) be:
v_1=1:10;
v_2=11:20;
v_3=21:30;
v_4=31:40;
and so on.
If they are few (e. g. 4), you can directly set then as input in the cat function:
v=cat(2,v_1,v_2,v_3,v_4)
or the horzcat function
v=horzcat(v_1,v_2,v_3,v_4)
otherwise you can use the eval function within a loop
v1=[];
for i=1:4
eval(['v1=[v1 v_' num2str(i) ']'])
end
Hope this helps.
Concatenating with horzcat is definitely an option, but since these vectors are being created in a function, it would be better to concatenate these vectors automatically in the function itself rather than write out horzcat(v1,v2,....vn) manually.
Given the function mentioned in the question, I would suggest something like this:
function v = roughq(rpeakindex)
v = zeros(121,length(rpeakindex)); %// create a 2D array of all zeros
for i = 1:size(rpeakindex)
v(:,i) = [rpeakindex(i)-30:1:rpeakindex(i)+90]; %// set result to ith column of v
end
v = v(:)'; %'//reshape v to be a single vector with the columns concatenated
end
Here's a simplified example of what's going on:
N = 3;
v = zeros(5,N);
for i = 1:N
v(:,i) = (1:5)*i;
end
v = v(:)';
Output:
v =
1 2 3 4 5 2 4 6 8 10 3 6 9 12 15
You may want to read up on MATLAB's colon operator to understand the v(:) syntax.
If you mean 2d matrix, you are using for holding vectors and each row hold vector v then you can simply use the reshape command in matlab like below:
V = [] ;
for i = 1:10
V(i,:) = randi (10,1 ,10) ;
end
V_reshpae = reshape (V, 1, numel(V)) ;
So I'm trying to implement the Simpson method in Matlab, this is my code:
function q = simpson(x,f)
n = size(x);
%subtracting the last value of the x vector with the first one
ba = x(n) - x(1);
%adding all the values of the f vector which are in even places starting from f(2)
a = 2*f(2:2:end-1);
%adding all the values of the f vector which are in odd places starting from 1
b = 4*f(1:2:end-1);
%the result is the Simpson approximation of the values given
q = ((ba)/3*n)*(f(1) + f(n) + a + b);
This is the error I'm getting:
Error using ==> mtimes
Inner matrix dimensions must agree.
For some reason even if I set q to be
q = f(n)
As a result I get:
q =
0 1
Instead of
q =
0
When I set q to be
q = f(1)
I get:
q =
0
q =
0
I can't explain this behavior, that's probably why I get the error mentioned above. So why does q have two values instead of one?
edit: x = linspace(0,pi/2,12);
f = sin(x);
size(x) returns the size of the array. This will be a vector with all the dimensions of the matrix. There must be at least two dimensions.
In your case n=size(x) will give n=[N, 1], not just the length of the array as you desire. This will mean than ba will have 2 elements.
You can fix this be using length(x) which returns the longest dimension rather than size (or numel(x) or size(x, 1) or 2 depending on how x is defined which returns only the numbered dimension).
Also you want to sum over in a and b whereas now you just create an vector with these elements in. try changing it to a=2*sum(f(...)) and similar for b.
The error occurs because you are doing matrix multiplication of two vectors with different dimensions which isn't allowed. If you change the code all the values should be scalars so it should work.
To get the correct answer (3*n) should also be in brackets as matlab doesn't prefer between / and * (http://uk.mathworks.com/help/matlab/matlab_prog/operator-precedence.html). Your version does (ba/3)*n which is wrong.
I have a 256x256 matrix from that I want to create block of 8x8 matrices. I have a below code which shows just one block i want all the blocks then subtract a number from each element of the matrix. When i do the subtract 128 from each element it doesn't show negative values rather showing only 0.
[x,y] = size(M);
for i=1:8
for j=1:8
k(i,j) = M(i,j);
end
end
disp(k);
for a=1:size(k)
for b=1:size(k)
A(a,b) = k(a,b) - 128;
end
end
disp(A);
well, if you have to subtract a fixed value it's better to do it as
M = M - 128;
It's faster.
You said you get 0 values instead of negative ones; it is likely due to the type of the matrix which is unsigned (i.e. the guys are just positive). A cast to general integer is necessary. Try:
M = int16(M) - 128;
To obtain the partition I propose the following, there can be more efficient ways, btw:
r = rand(256,256); %// test case, substitute with your matrix M
[i j] = meshgrid(1:8:256); %// lattice of indices
idx = num2cell([ i(:) , j(:) ],2); %// cell version
matr = cellfun( #(i) r(i(1):i(1)+7,i(2):i(2)+7), idx, 'UniformOutput',false); %// blocks
matr will contain all the sub-matrices in lexicographical order. e.g.
matr{2}
ans =
0.4026 0.3141 0.4164 0.5005 0.6952 0.1955 0.9803 0.5097
0.8186 0.9280 0.1737 0.6133 0.8562 0.7405 0.8766 0.0975
0.2704 0.8333 0.1892 0.7661 0.5168 0.3856 0.1432 0.9958
0.9973 0.8488 0.6937 0.2630 0.1004 0.5842 0.1844 0.5206
0.4052 0.0629 0.6982 0.1530 0.9234 0.1271 0.7317 0.3541
0.2984 0.3633 0.1510 0.0297 0.0225 0.7945 0.2925 0.0396
0.5097 0.0802 0.8744 0.1032 0.8523 0.6150 0.4845 0.5703
0.8635 0.0194 0.1879 0.5017 0.5297 0.6319 0.2406 0.5125
Explanation: In matlab you can efficiently operate on matrices and slices of them. For instance, you can easily copy submatrices, e.g. the first 8x8 matrix is
sub = M(1:8,1:8);
You want all the submatrices, thus you need kind of a lattice of indices to get
sub_ij = M(1+8*(i-1) : 7 * 8*(i-1) , 1+8*(j-1) : 7 * 8*(j-1))
i.e. you need the lattice
(1+8*(i-1) : 7 * 8*(i-1) , 1+8*(j-1) : 7 * 8*(j-1)) % // for all i,j
you can use meshgrid for that.
Finally you have to cut off the pieces, that is what the last two instruction do. particularly the first one generates the indices (try idx{1},...), and the second one generates the submatrices (try matr{1},...).
Hope this code helps you. It has 16x16 image to divide into 4x4 blocks. You may change the value and get the required output as you like.