I have this program which calculates the realized covariance for each day in my sample but I have some troubles with storing the output in a matrix.
the program is as follows:
for i=1:66:(2071*66)
vec = realized_covariance(datapa(i:i+65),data(i:i+65),datapo(i:i+65),data(i:i+65),'wall','Fixed',fixedInterval,5)
mat(2,4142) = vec
end
Output:
vec =
1.0e-03 *
0.1353 -0.0283
-0.0283 0.0185
Subscripted assignment dimension mismatch.
I have tried various way to store the output in a matrix like defining a matrix on zeroes to store the output in or let the row dimension of the storing matrix be undefined, but nothing seems to do the job.
I would really appreciate an advice on how to tackle this challenge.
I have used a solution which does the job.
I defined a matrix and then filled in all my output one at the time using the following:
A = zeros(0,0) %before loop, only serve to define the storing matrix
A = [A; vec]%after the calculating function, inside the loop.
Actually mat(2,4142) is a single location in a matrix, you can't assign there four values.
You need to define the exact location inside mat every time you want to assign values into it. Try doing it like that:
mat=zeros(2,2142);
for k=1:66:(2071*66)
vec=realized_covariance(datapa(i:i+65),data(i:i+65),datapo(i:i+65),data(i:i+65),'wall','Fixed',fixedInterval,5)
mat(:,[(((k-1)/66)*2)+1 (((k-1)/66)*2)+2])=vec;
end
You're trying to store a 2 x 2 matrix into a single element. I.e. 4 elements on the right hand side, one on the left. That won't fit. See it like this: you have a garage besides your house where 1 car fits. You've got three friends coming over and they also want to park their car inside. That's a problem though, as you've got only space for one. So you have to buy a bigger garage: assign 4 elements on the left (e.g. mat(ii:ii+1,jj:jj+1) = [1 2;3 4]), or use a cell/structure array.
As Steve suggests in a comment below, you can use a 3D matrix quite easily:
counters = 1:66:(2071*66);
mat = zeros(2,2,numel(counters)); %// initialise output matrix
for ii=1:numel(counters)
vec = realized_covariance(datapa(counters(ii):counters(ii+65)),...
data(counters(ii):counters(ii+65)),datapo(counters(ii):counters(ii+65)),...
data(counters(ii):counters(ii+65)),'wall','Fixed',fixedInterval,5)
mat(:,:,ii) = vec; %// store in a 3D matrix
end
Now mat is 3D, with the first two coordinates being your regular output, i.e.e vec, and the last index is the iteration number. So to access the output of iteration 1032 you'd do mat(:,:,1032), possibly with a squeeze around that to make it 2D instead of 3D.
Related
I have a structure (data) which consist of 322 cells a 296(features)*2000(timepoints). I want a matrix per timepoint which consists of trials^features^timepoints (322 *296*2000). What I am currently doing and what also works fine is using a for-loop:
for k=1:size(data.trial{1,1},2)
for i= 1:length(data.trialinfo)
between=data.trial{1,i}';
data(i,:,k)=between(k,:);
end
end
Can anyone think of a faster way to do that? Because it takes ages as the matrix increases.
Thanks!
Carlos
Try:
data2 = permute(reshape([data.trial{:}],296,2000,322),[3,1,2]);
The issue with reshape is that it thinks in columns, so you need to permute afterwords (ie. 3D transpose), since that's how you set up the output variable in your loop. I used concatenation instead of cell2mat for speed.
I assumed this worked as sample data:
for ii= 1:322
data.trial{1,ii} = rand(296,2000);
end
I have matchcounts (5x5)cell, every cell has a vector of double [106x1]. The vectors of double have zeros and non zero values. I want to find non zero values for every cell, count them and put the result in a matrix.
I tried with this code:
a{i,j}(k,1)=[];
for k=1:106
for i=1:5
for j=1:5
if (matchcounts{i,j}(k,1))~=0
a{i,j}=a{i,j}(k,1)+1;
end
end
end
end
and others but it's not correct! Can you help me? Thanks
While it is possible to fix your answer above, I recommend to change the data structure to have a much simpler solution possible. Instead of having a 2D cell array which holds 1D data, choose a single 3D data structure.
For an optimal solution you would change your previous code code to directly write the 3D-matrix, instead of converting it. To get started, this code converts it so you can already see how the data structure should look like:
%convert to matrix
for idx=1:numel(matchcounts)
matchcounts{idx}=permute(matchcounts{idx},[3,2,1]);
end
matchcounts=cell2mat(matchcounts);
And finding the nonzero elements:
a=(matchcounts~=0)
To index the result, instead of a{k,l}(m,1) you use a(k,l,m)
To give you some rule to avoid complicated data structures in the future. Use cell arrays only for string data and data of different size. Whenever you have a cell array which contains only vectors or matrices of the same size, it should be a multidimensional matrix.
I have a problem with the following code. I want to store all the values I am creating in the for loop below so that I can make a plot of it. I have tried several things, but nothing works. Does anyone know a simple method to create a vector of the results and then plot them?
dx=0.1;
t=1;
e=1;
for x=-1:dx:1
lower_bound=-100;
upper_bound=x/(sqrt(4*t*e));
e=1;
u=(1/sqrt(pi))*quad(#integ,lower_bound,upper_bound);
plot(x,u)
hold on
end
hold off
I would like to use as much of this matlab code as possible.
dx=0.1;
t=1;
e=1;
xval=[-1:dx:1].';
upper_bound = zeros(numel(xval),1);
u = zeros(numel(xval),1);
for ii=1:numel(xval)
x = xval(ii)
lower_bound=-100;
upper_bound(ii,1)=x/(sqrt(4*t*e));
u(ii,1)=(1/sqrt(pi))*quad(#integ,lower_bound,upper_bound(ii));
end
figure;
plot(xval,u)
by adding the (ii) behind your statements it saves your variables in an array. I did not use that on your lower_bound since it is a constant.
Note that I first created an array xval and called that with integers in ii, since subscriptindices must be positive integers in MATLAB. I also initialised both upper_bound and u by creating a zero matrix before the loop executes. This is handy since extending an existing vector is very memory and time consuming in MATLAB and since you know how big they will get (same number of elements as xval) you might as well use that.
I also got the plot call outside the loop, to prevent you from plotting 21 blue lines in 1 plot.
I'm attempting to run this simple diffusion case (I understand that it isn't ideal generally), and I'm doing fine with getting the inside of the solid, but need some help with the outer edges.
global M
size=100
M=zeros(size,size);
M(25,25)=50;
for diffusive_steps=1:500
oldM=M;
newM=zeros(size,size);
for i=2:size-1;
for j=2:size-1;
%we're considering the ij-th pixel
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
It's a pretty simple piece of code, and I know that. I'm not very good at using Octave yet (chemist by trade), so I'd appreciate any help!
If you have concerns about the border of your simulation you could pad your matrix with NaN values, and then remove the border after the simulation has completed. NaN stands for not a number and is often used to denote blank data. There are many MATLAB functions work in a useful way with these values.
e.g. finding the mean of an array which has blanks:
nanmean([0 nan 5 nan 10])
ans =
5
In your case, I would start by adding a border of NaNs to your M matrix. I'm using 'n' instead of 'size', since size is an important function in MATLAB, and using it as a variable can lead to confusing errors.
n=100;
blankM=zeros(n+2,n+2);
blankM([1,end],:) = nan;
blankM(:, [1,end]) = nan;
Now we can define 'M'. N.B that the first column and row will be NaNs so we need to add an offset (25+1):
M = blankM;
M(26,26)=50;
Run the simulation through,
m = size(blankM, 1);
n = size(blankM, 2);
for diffusive_steps=1:500
oldM = M;
newM = blankM;
for i=2:m-1;
for j=2:n-1;
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
and then extract the area of interest
finalResult = M(2:end-1, 2:end-1);
One simple change you might make is to add a boundary of ghost cells, or halo, around the domain of interest. Rather than mis-use the name size I've used a variable called sz. Replace:
M=zeros(sz,sz)
with
M=zeros(sz+2,sz+2)
and then compute your diffusion over the interior of this augmented matrix, ie over cells (2:sz+1,2:sz+1). When it comes to considering the results, discard or just ignore the halo.
Even simpler would be to simply take what you already have and ignore the cells in your existing matrix which are on the N,S,E,W edges.
This technique is widely used in problems such as, and similar to, yours and avoids the need to write code which deals with the computations on cells which don't have a full complement of neighbours. Setting the appropriate value for the contents of the halo cells is a problem-dependent matter, 0 isn't always the right value.
i have two problems in mathematica and want to do them in matlab:
measure := RandomReal[] - 0.5
m = 10000;
data = Table[measure, {m}];
fig1 = ListPlot[data, PlotStyle -> {PointSize[0.015]}]
Histogram[data]
matlab:
measure =# (m) rand(1,m)-0.5
m=10000;
for i=1:m
data(:,i)=measure(:,i);
end
figure(1)
plot(data,'b.','MarkerSize',0.015)
figure(2)
hist(data)
And it gives me :
??? The following error occurred
converting from function_handle to
double: Error using ==> double
If i do :
measure =rand()-0.5
m=10000;
data=rand(1,m)-0.5
then, i get the right results in plot1 but in plot 2 the y=axis is wrong.
Also, if i have this in mathematica :
steps[m_] := Table[2 RandomInteger[] - 1, {m}]
steps[20]
Walk1D[n_] := FoldList[Plus, 0, steps[n]]
LastPoint1D[n_] := Fold[Plus, 0, steps[n]]
ListPlot[Walk1D[10^4]]
I did this :
steps = # (m) 2*randint(1,m,2)-1;
steps(20)
Walk1D =# (n) cumsum(0:steps(n)) --> this is ok i think
LastPointold1D= # (n) cumsum(0:steps(n))
LastPoint1D= # (n) LastPointold1D(end)-->but here i now i must take the last "folding"
Walk1D(10)
LastPoint1D(10000)
plot(Walk1D(10000),'b')
and i get an empty matrix and no plot..
Since #Itamar essentially answered your first question, here is a comment on the second one. You did it almost right. You need to define
Walk1D = # (n) cumsum(steps(n));
since cumsum is a direct analog of FoldList[Plus,0,your-list]. Then, the plot in your code works fine. Also, notice that, either in your Mathematica or Matlab code, it is not necessary to define LastPoint1D separately - in both cases, it is the last point of your generated list (vector) steps.
EDIT:
Expanding a bit on LastPoint1D: my guess is that you want it to be a last point of the walk computed by Walk1D. Therefore, it would IMO make sense to just make it a function of a generated walk (vector), that returns its last point. For example:
lastPoint1D = #(walk) (walk(end));
Then, you use it as:
walk = Walk1D(10000);
lastPoint1D(walk)
HTH
You have a few errors/mistakes translating your code to Matlab:
If I am not wrong, the line data = Table[measure, {m}]; creates m copies of measure, which in your case will create a random vector of size (1,m). If that is true, in Matlab it would simply be data = measure(m);
The function you define gets a single argument m, therefor it makes no sense using a matrix notation (the :) when calling it.
Just as a side-note, if you insert data into a matrix inside a for loop, it will run much faster if you allocate the matrix in advance, otherwise Matlab will re-allocate memory to resize the matrix in each iteration. You do this by data = zeros(1,m);.
What do you mean by "in plot 2 the y=axis is wrong"? What do you expect it to be?
EDIT
Regarding your 2nd question, it would be easier to help you if you describe in words what you want to achieve, rather than trying to read your (error producing) code. One thing which is clearly wrong is using expression like 0:steps(n), since you use m:n with two scalars m and n to produce a vector, but steps(n) produces a vector, not a scalar. You probably get an empty matrix since the first value in the vector returned by steps(n) might be -1, and 0:-1 produces an empty vector.