What is this error?
Index exceeds matrix dimensions.
Error in evalution (line 5)
binTempX(i,[1,2,3,4,5,6,7,8])=parentXY(i,[1,2,3,4,5,6,7,8]);
function [tempX_Y_FXY] = evalution(parentXY,fXY)
for i=1:6
binTempX(i,[1,2,3,4,5,6,7,8])=parentXY(i,[1,2,3,4,5,6,7,8]);
binTempY(i,[9,10,11,12,13,14,15,16],8)=parentXY(i,[9,10,11,12,13,14,15,16]);
decTempX=bin2dec(binTempX(i,[1,2,3,4,5,6,7,8]));
decTempY=bin2dec(binTempY(i,[9,10,11,12,13,14,15,16]));
tempX_Y_FXY(i,1)=decTempX;
tempX_Y_FXY(i,2)=decTempY;
tempX_Y_FXY(i,3)=fXY(decTempX,decTempY);
end
tempX_Y_FXY=sortrows(tempX_Y_FXY,3);
end
binTempX(i,[1,2,3,4,5,6,7,8])=parentXY(i,[1,2,3,4,5,6,7,8]);
%%% ------- what is this 8 doing here???
binTempY(i,[9,10,11,12,13,14,15,16],**8**)=parentXY(i,[9,10,11,12,13,14,15,16]);
decTempX=bin2dec(binTempX(i,[1,2,3,4,5,6,7,8]));
decTempY=bin2dec(binTempY(i,[9,10,11,12,13,14,15,16]));
tempX_Y_FXY(i,1)=decTempX;
tempX_Y_FXY(i,2)=decTempY;
tempX_Y_FXY(i,3)=fXY(decTempX,decTempY);
Change this to:
binTempX(i,1:8)=parentXY(i,1:8);
% removed the 8, because I think it is typo??
binTempY(i,9:16)=parentXY(i,9:16);
decTempX=bin2dec(binTempX(i,1:8));
decTempY=bin2dec(binTempY(i,9:16));
tempX_Y_FXY(i,1)=decTempX;
tempX_Y_FXY(i,2)=decTempY;
tempX_Y_FXY(i,3)=fXY(decTempX,decTempY);
if you want to select/assign multiple consecutive columns use the 1:8 notation for example. Both should work but in my opinion the second is clearer and easier to maintain.
And like beaker said, check the sizes of your matrices, does the matrices you use have at least 6 rows and 16 columns?
Related
I have a column vector A with dimensions (35064x1) that I want to reshape into a matrix with 720 lines and as many columns as it needs.
In MATLAB, it'd be something like this:
B = reshape(A,720,[])
in which B is my new matrix.
However, if I divide 35604 by 720, there'll be a remainder.
Ideally, MATLAB would go about filling every column with 720 values until the last column, which wouldn't have 720 values; rather, 504 values (48x720+504 = 35064).
Is there any function, as reshape, that would perform this task?
Since I am not good at coding, I'd resort to built-in functions first before going into programming.
reshape preserves the number of elements but you achieve the same in two steps
b=zeros(720*ceil(35604/720),1); b(1:35604)=a;
reshape(b,720,[])
A = rand(35064,1);
NoCols = 720;
tmp = mod(numel(A),NoCols ); % get the remainder
tmp2 = NoCols -tmp;
B = reshape([A; nan(tmp2,1)],720,[]); % reshape the extended column
This first gets the remainder after division, and then subtract that from the number of columns to find the amount of missing values. Then create an array with nan (or zeros, whichever suits your purpose best) to pad the original and then reshape. One liner:
A = rand(35064,1);
NoCols = 720;
B = reshape([A; nan(NoCols-mod(numel(A),NoCols);,1)],720,[]);
karakfa got the right idea, but some error in his code.
Fixing the errors and slightly simplifying it, you end up with:
B=nan(720,ceil(numel(a)/720));
B(1:numel(A))=A;
Create a matrix where A fits in and assingn the elemnent of A to the first numel(A) elements of the matrix.
An alternative implementation which is probably a bit faster but manipulates your variable b
%pads zeros at the end
A(720*ceil(numel(A)/720))=0;
%reshape
B=reshape(A,720,[]);
I'm new to MATLAB and its development. I have a image which is 1134 (rows) X 1134 (columns). I want that image to save 3 (columns) X 3 (rows). In order to do that I need 378 cells. For that I used following code, but it gives me an error.
image=imread('C:\Users\ven\Desktop\test\depth.png');
I=reshape(image,1,1134*1134);
chunk_size = [3 3]; % your desired size of the chunks image is broken into
sc = sz ./ chunk_size; % number of chunks in each dimension; must be integer
% split to chunk_size(1) by chunk_size(2) chunks
X = mat2cell(I, chunk_size(1) * ones(sc(1),1), chunk_size(2) *ones(sc(2),1));
Error:
Error using mat2cell (line 97)
Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [1 1285956].'
Unfortunately your code does not work as you think it would.
The ./ operator performs point wise division of two matrices. Short example:
[12, 8] ./ [4, 2] == [12/4, 8/2] == [3, 4]
In order for it to work both matrices must have exactly the same size. In your case you try to perform such an operation on a 1134x1134 matrix (the image) and a 1x2 matrix (chunk_size).
In other words you can not use it to divide matrices into smaller ones.
However, a solution to your problem is to use the mat2cell function to pick out subsets of the matrix. A explanation of how it is done can be found here (including examples): http://se.mathworks.com/matlabcentral/answers/89757-how-to-divide-256x256-matrix-into-sixteen-16x16-blocks.
Hope it helps :)
Behind the C=A./B command is loop over all elements of A(ii,jj,...) and B(ii,jj,..) and each C(ii,jj,..)=A(ii,jj,...)/B(ii,jj,...).
Therefore martices A and B must be of same dimension.
If you want to split matrix into groups you can use
sc=cell(1134/3,1);
kk=0;ll=0;
for ii=2:3:1133
kk=kk+1;
for jj=2:3:1133
ll=ll+1;
sc{kk,ll}=image(ii-1:ii+1,jj-1:jj+1);
end
end
The code allocates cell array sc for resulting submatrices and arbitrary counters kk and ll. Then it loops over ii and jj with step of 3 representing centers of each submatrices.
Edit
Or you can use mat2cell command (type help mat2cell or doc mat2cell in matlab shell)
sc=mat2cell(image,3,3);
In both cases the result is cell array and its iith and jjth elements (matrices) are accessible by sc{ii,jj}. If you want call iith anr jjth number in kkth and llth matrix, do it via sc{kk,ll}(ii,jj).
In short, you divided a 1134 x 1134 by 2 x 1 matrix. That doesn't work.
The error "Matrix dimensions must agree**" is from the dividing a matrix with another matrix that doesn't have the right dimensions.
You used the scalar divide "./" which divided a matrix by another matrix.
You want something like:
n = 1134 / 3 % you should measure the length of the image
I1=image(1:n,1:n); % first row
I2=image(1:n,n:2n);
I3=image(1:n,2n:3n);
I4=image(n:2n,1:n); % second row
I5=image(n:2n,n:2n);
I6=image(n:2n,2n:3n);
I7=image(2n:3n,1:n); % third row
I8=image(2n:3n,n:2n);
I9=image(2n:3n,2n:3n);
from here:
http://au.mathworks.com/matlabcentral/answers/46699-how-to-segment-divide-an-image-into-4-equal-halves
There would be a nice loop you could do it in, but sometimes thinking is hard.
This question already has answers here:
Find specific value's count in a vector
(4 answers)
Closed 8 years ago.
I have a NxM matrix for example named A. After some processes I want to count the zero elements.
How can I do this in one line code? I tried A==0 which returns a 2D matrix.
There is a function to find the number of nonzero matrix elements nnz. You can use this function on a logical matrix, which will return the number of true.
In this case, we apply nnz on the matrix A==0, hence the elements of the logical matrix are true, if the original element was 0, false for any other element than 0.
A = [1, 3, 1;
0, 0, 2;
0, 2, 1];
nnz(A==0) %// returns 3, i.e. the number of zeros of A (the amount of true in A==0)
The credits for the benchmarking belong to Divarkar.
Benchmarking
Using the following paramters and inputs, one can benchmark the solutions presented here with timeit.
Input sizes
Small sized datasize - 1:10:100
Medium sized datasize - 50:50:1000
Large sized datasize - 500:500:4000
Varying % of zeros
~10% of zeros case - A = round(rand(N)*5);
~50% of zeros case - A = rand(N);A(A<=0.5)=0;
~90% of zeros case - A = rand(N);A(A<=0.9)=0;
The results are shown next -
1) Small Datasizes
2. Medium Datasizes
3. Large Datasizes
Observations
If you look closely into the NNZ and SUM performance plots for medium and large datasizes, you would notice that their performances get closer to each other for 10% and 90% zeros cases. For 50% zeros case, the performance gap between SUM and NNZ methods is comparatively wider.
As a general observation across all datasizes and all three fraction cases of zeros,
SUM method seems to be the undisputed winner. Again, an interesting thing was observed here that the general case solution sum(A(:)==0) seems to be better in performance than sum(~A(:)).
some basic matlab to know: the (:) operator will flatten any matrix into a column vector , ~ is the NOT operator flipping zeros to ones and non zero values to zero, then we just use sum:
sum(~A(:))
This should be also about 10 times faster than the length(find... scheme, in case efficiency is important.
Edit: in the case of NaN values you can resort to the solution:
sum(A(:)==0)
I'll add something to the mix as well. You can use histc and compute the histogram of the entire matrix. You specify the second parameter to be which bins the numbers should be collected at. If we just want to count the number of zeroes, we can simply specify 0 as the second parameter. However, if you specify a matrix into histc, it will operate along the columns but we want to operate on the entire matrix. As such, simply transform the matrix into a column vector A(:) and use histc. In other words, do this:
histc(A(:), 0)
This should be equivalent to counting the number of zeroes in the entire matrix A.
Well I don't know if I'm answering well the question but you could code it as follows :
% Random Matrix
M = [1 0 4 8 0 6;
0 0 7 4 8 0;
8 7 4 0 6 0];
n = size(M,1); % Number of lines of M
p = size(M,2); % Number of columns of M
nbrOfZeros = 0; % counter
for i=1:n
for j=1:p
if M(i,j) == 0
nbrOfZeros = nbrOfZeros + 1;
end
end
end
nbrOfZeros
In MATLAB, is there a more concise way to handle discrete conditional indexing by column than using a for loop? Here's my code:
x=[1 2 3;4 5 6;7 8 9];
w=[5 3 2];
q=zeros(3,1);
for i = 1:3
q(i)=mean(x(x(:,i)>w(i),i));
end
q
My goal is to take the mean of the top x% of a set of values for each column. The above code works, but I'm just wondering if there is a more concise way to do it?
You mentioned that you were using the function PRCTILE, which would indicate that you have access to the Statistics Toolbox. This gives you yet another option for how you could solve your problem, using the function NANMEAN. In the following code, all the entries in x less than or equal to the threshold w for a column are set to NaN using BSXFUN, then the mean of each column is computed with NANMEAN:
x(bsxfun(#le,x,w)) = nan;
q = nanmean(x);
I don't know of any way to index the columns the way you want. This may be faster than a for loop, but it also creates a matrix y that is the size of x.
x=[1 2 3;4 5 6;7 8 9];
w=[5 3 2];
y = x > repmat(w,size(x,1),1);
q = sum(x.*y) ./ sum(y)
I don't claim this is more concise.
Here's a way to solve your original problem: You have an array, and you want to know the mean of the top x% of each column.
%# make up some data
data = magic(5);
%# find out how many rows the top 40% are
nRows = floor(size(data,1)*0.4);
%# sort the data in descending order
data = sort(data,1,'descend');
%# take the mean of the top 20% of values in each column
topMean = mean(data(1:nRows,:),1);
I'm just beginning to teach myself MATLAB, and I'm making a 501x6 array. The columns will contain probabilities for flipping 101 sided die, and as such, the columns contain 101,201,301 entries, not 501. Is there a way to 'stretch the column' so that I add 0s above and below the useful data? So far I've only thought of making a column like a=[zeros(200,1);die;zeros(200,1)] so that only the data shows up in rows 201-301, and similarly, b=[zeros(150,1);die2;zeros(150,1)], if I wanted 200 or 150 zeros to precede and follow the data, respectively in order for it to fit in the array.
Thanks for any suggestions.
You can do several thing:
Start with an all-zero matrix, and only modify the elements you need to be non-zero:
A = zeros(501,6);
A(someValue:someOtherValue, 5) = value;
% OR: assign the range to a vector:
A(someValue:someOtherValue, 5) = 1:20; % if someValue:someOtherValue is the same length as 1:20