I have a 3D large Matrix, the first index (x) represents frequency and the second and third indexes (y and z) are the indexes of the data. I want to print the data for each index for all the frequencies, then print a | char and a new line char. I do this in the following way:
% S is a 3D matrix of size (x,y,z), where y=z and x>>y
N=size(S,2);
MM=real(S);
for mi=1:N
for mj=1:N
fprintf(fid,"%.16g ",MM(:,mi,mj));
fprintf(fid,"|\n");
end
end
But for large matrices this is very slow. Is there a way to speed up the process?
This is done in octave, which means that a Matlab solution will work as well.
If I understood your input data correct, your loop is equivalent to:
fprintf(fid,[repmat('%.16g ',1,size(S,1)) '|\n'],permute(real(S),[1,3,2]));
fprintf starts over using the format string when it reaches the end. The permute is necessary to preserve the order from your code.
Here is a general approach to speed things up:
Calculate everything you want to calculate, store each result as a number or a string
Print the result to file in 1 go
m=zeros(3,4,5);
[x,y,z]=size(m);
for k=1:x
for l=1:y
dlmwrite('D:\test.txt',m(k,l,:),'-append','delimiter',' ','newline','pc');
end
dlmwrite('D:\test.txt',[' '],'-append','newline','pc');
end
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'm trying to split an audio file into 30 millisecond disjoint intervals using Matlab. I have the following code at the moment:
clear all
close all
% load the audio file and get its sampling rate
[y, fs] = audioread('JFK_ES156.wav');
for m = 1 : 6000
[t(m), fs] = audioread('JFK_ES156.wav', [(m*(0.03)*fs) ((m+1)*(0.03)*fs)]);
end
But the problem is that I get the following error:
In an assignment A(I) = B, the number of elements in B and I
must be the same.
Error in splitting (line 12)
[t(m), fs] = audioread('JFK_ES156.wav', [(m*(0.03)*fs)
((m+1)*(0.03)*fs)]);
I don't see why there's a mismatch in the number of elements in B and I and how to solve this. How can I get past this error? Or is there just an easier way to split the audio file (maybe another function I don't know about or something)?
I think the easiest way to split audio is to just load it and use the vec2mat function. so you would have something like this;
[X,Fs] = audioread('JFK_ES156.wav');
%Calculate how many samples you need to capture 30ms of audio
matSize = Fs*0.3;
%Pay attention to that apostrophe. Makes sure samples are stored in columns
%rather than rows.
output = vec2mat(x,matSize)';
%You can now have your audio split up into the different columns of your matrix.
%You can call them by using the column calling command for matrices.
%Plot first 30ms of audio
plot(output(:,1));
%You can join the audio back together using this command.
output = output(:);
Hope that helps. Another good thing about this method is that it keeps all your data in one place!
Edit : One thing I thought of, you may get a problem with this depending on your vector size. But I think vec2mat actually zeroPads your vector. Not a big thing, but if you're moving back and forth between the two, then it might be a good idea to have another variable that stores the original length of your signal.
You should just use the variable y and reshape it to form your split audio. For example,
chunk_size = fs*0.03;
y_chunks = reshape(y, chunk_size, 6000);
That will give you a matrix with each column a 30 ms chunk. This code will also be faster than reading small segments from file in a loop.
As hiandbaii suggested you could also use cell array. Make sure you clear your existing variables before that. Not clearing the array t is probably the reason you got the error "Cell contents assignment to a non-cell array object."
Your original error is because you cannot assign a vector with scalar indexing. That is, 'm' is a scalar, but your audioread call is returning a vector. This is what the error says about mismatch in size of I and B. You could also fix that by making t a 2-D array and use an assignment like
[t(m,:), fs] =
It appears that each 30 ms segment is not equal to one sample. That would be the only case where your code works. i.e. 0.03*fs != 1.
You could try using cells instead.. i.e. replace t(m) with t{m}
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.
I want to apply a function to all columns in a matrix with MATLAB. For example, I'd like to be able to call smooth on every column of a matrix, instead of having smooth treat the matrix as a vector (which is the default behaviour if you call smooth(matrix)).
I'm sure there must be a more idiomatic way to do this, but I can't find it, so I've defined a map_column function:
function result = map_column(m, func)
result = m;
for col = 1:size(m,2)
result(:,col) = func(m(:,col));
end
end
which I can call with:
smoothed = map_column(input, #(c) (smooth(c, 9)));
Is there anything wrong with this code? How could I improve it?
The MATLAB "for" statement actually loops over the columns of whatever's supplied - normally, this just results in a sequence of scalars since the vector passed into for (as in your example above) is a row vector. This means that you can rewrite the above code like this:
function result = map_column(m, func)
result = [];
for m_col = m
result = horzcat(result, func(m_col));
end
If func does not return a column vector, then you can add something like
f = func(m_col);
result = horzcat(result, f(:));
to force it into a column.
Your solution is fine.
Note that horizcat exacts a substantial performance penalty for large matrices. It makes the code be O(N^2) instead of O(N). For a 100x10,000 matrix, your implementation takes 2.6s on my machine, the horizcat one takes 64.5s. For a 100x5000 matrix, the horizcat implementation takes 15.7s.
If you wanted, you could generalize your function a little and make it be able to iterate over the final dimension or even over arbitrary dimensions (not just columns).
Maybe you could always transform the matrix with the ' operator and then transform the result back.
smoothed = smooth(input', 9)';
That at least works with the fft function.
A way to cause an implicit loop across the columns of a matrix is to use cellfun. That is, you must first convert the matrix to a cell array, each cell will hold one column. Then call cellfun. For example:
A = randn(10,5);
See that here I've computed the standard deviation for each column.
cellfun(#std,mat2cell(A,size(A,1),ones(1,size(A,2))))
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Of course, many functions in MATLAB are already set up to work on rows or columns of an array as the user indicates. This is true of std of course, but this is a convenient way to test that cellfun worked successfully.
std(A,[],1)
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Don't forget to preallocate the result matrix if you are dealing with large matrices. Otherwise your CPU will spend lots of cycles repeatedly re-allocating the matrix every time it adds a new row/column.
If this is a common use-case for your function, it would perhaps be a good idea to make the function iterate through the columns automatically if the input is not a vector.
This doesn't exactly solve your problem but it would simplify the functions' usage. In that case, the output should be a matrix, too.
You can also transform the matrix to one long column by using m(:,:) = m(:). However, it depends on your function if this would make sense.