I am trying to sum in the second dimension a matrix QI in Matlab. The trick is, the columns contain a series of increasing numbers, but not all columns have the same number of elements (i.e. numel(QI(:,1)) ~= numel(QI(:,2)) and so on). For the sake of clarity, I attach a picture of it. Note that I padded the missing areas with 0, so the previous condition becomes nnz(QI(:,1)) ~= nnz(QI(:,2)).
One initial strategy that I thought of was to treat this as an image and construct a mask for each different gradient level, but that seems like a tedious job.
Anyone has a better idea on how to do this? I should also mention that I am able to freely modify how QI is generated, but I'd rather not if there is a solution for this problem.
EDIT:
Hopefully the new colored image should give a better understanding.
FYI, each column was previously stored in a cell array without the trailing zeros. Then I extracted the columns one by one and stored them in a matrix in order to perform the summation, padding the extra zeros whenever the length isn't the same.
Generally these column data should have the same number of rows, but sometimes that's not the case, and even worse, they do not allign properly.
I'm starting to think if it's better to rework the code that generate the cell arrays rather than this matrix. Thoughts?
Thank you,
edit: following you comment, I modified the answer. Be aware that your data cannot be really "aligned" because they have not the same number of value.
A way would be to use a cell as a storage for your measures.
valueMissing = 0; % here you can put the defauld value you want
% transform you matrix in a cell
QICell = arrayfun(#(x) QI(QI(:,x)!=valueMissing,x), 1:size(QI,2),'UniformOutput', false);
Now you can sum the last element of the vectors inside the cell
QIsum = sum(cellfun(#(x) x(end), QICell))
Or reorder the vectors so that your last element are "aligned"
QICellReordered = cellfun(#(x) x(end:-1:1),QICell, 'UniformOutput',false);
Then you can make all possible sums:
m = min(cellfun(#numel, QICellReordered));
QIsum = zeros(m,1);
for i=1:m
QIsum(i) = sum(cellfun(#(x) x(i), QICellReordered));
end
% reorder QISum to your original order
QIsum = QIsum(end:-1:1);
I hope this help !
Related
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 imported a lot of data from an excel spreadsheet so that I have a 1x27 matrix.
I have imported data from excel using this
filename = 'for_matlab.xlsx';
sheet = 27;
xlRange = 'A1:G6';
all_data = {};
for i=1:sheet,
all_data{i} = xlsread(filename, i, xlRange);
end
However each element of this all_data matrix (which is 1x27) contains my data but I'm having trouble accessing individual elements.
i.e.
all_data{1}
Will give me the entire matrix but I need to perform multiplications on individual elements of this data
also
all_data(1)
just gives '5x6 double', i.e. the matrix dimensions.
Does anybody know how I can divide all elements of each row by the third element in each row and do this for all of my 'sub-matrices' (for want of a better word)
Assuming that all_data is a cell array and that each cell contains a matrix (with at least three columns):
result = cellfun(#(x) bsxfun(#rdivide, x, x(:,3)), all_data, 'uniformoutput', 0);
You are mixing terminology in matlab. what you have is 1x27 CELLS each of them containing a matrix.
If you access all_data{1} it will give you the whole matrix stored in the first cell.
If you want to access the elemets of that matrix then you need to do: all_data{1}(2,4). This example access the 2,4 element of the matrix in the first cell.
Definitely Luis Mendo has solved you problem, but be aware of the differences of Cells and matrixes in Matlab!
Okay I have found the answer now.
Basically you have to use both types of brackets because the data types are different
i.e. all_data{1}(1:4) or something like that anyway.
Cheers
I'm currently working on a cellular automata but i keep running into this problem. I have a matrix idxR which contains zero's and/or ones, depending on a probability process:
idxR = ((rRecr>rEmpty)&(rRecr>rAlgae)&(rRecr>rCoral));
Now i want to replace all ones in idxR with unique values and assign it to the variable colonies. I came up with the following:
colonies = idxR;
no = sum(colonies(:)==1)
maxvalue = max(colonies(:));
replace = [1:no]+maxvalue;
ret = reshape(replace,no,1);
colonies(colonies==1) = colonies(colonies==1).*ret;
When i output colonies it gives me a matrix with just ones and zeros and not a matrix where all ones have been replaced with incremental values. I tried this code in a new file and assigned a matrix with random ones and zeros to idxR and then it seems to work. So i guess to problem lies with the matrix idxR in my automata. It might be worth mentioning that idxR is contained in a for loop.
Can somebody tell me how to fix this?
You got the entire logic correct, except one minor flaw. You have idxR as a logical matrix. Hence colonies is a logical matrix too. Therefore, you get the expected output till the second-last line. Problem occurs on the last line, when you try to assign an array of numbers in which each number is greater than 1 (colonies(colonies==1).*ret;) to a logical matrix.
Elements greater than 1 are clipped to one and thus you see only zeros and ones. There is a simple workaround. Change the first line to
colonies = double(idxR);
P.S. The answer was right in front of you, you just didn't spot it. You had written:
I tried this code in a new file and assigned a matrix with random ones and zeros to idxR and then it seems to work.
The idxR matrix must have been of double datatype, if you used randi.
Parag got it right. You have the solution there.
You can use the following code if you are looking for a more "organized" way to get to 'colonies' -
colonies = double(idxR);
maxvalue = max(colonies(:));
ind1 = find(idxR==1);
colonies(ind1)=maxvalue + (1:numel(ind1));
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 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.