Hi guys I want to solve sodoku puzzles in matlab. My problem is that I should find same value in every row and every column and every 3*3 sub array.
Our 2d array is 9*9 and populated with value 1-9 randomly.
I wrote this for finding same value in rows, but I don't know how I should do it for columns and 3*3 sub arrays.
conflict_row = 0;
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
Sorry I'm a newbie.
Find values that are present in all columns:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,1), [3 1 2])),1),2));
Find values that are present in all rows:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,2), [3 1 2])),2),1));
Find values that are present in all 3x3 blocks: reshape the matrix as in this answer by A. Donda to transform each block into a 3D-slice; then reshape each block into a column; and apply 1:
m = 3; %// columns per block
n = 3; %// rows per block
B = permute(reshape(permute(reshape(A, size(A, 1), n, []), [2 1 3]), n, m, []), [2 1 3]);
B = reshape(B,m*n,[]);
v = find(all(any(bsxfun(#eq, B, permute(1:size(B,1), [3 1 2])),1),2));
Probably not the fastest solution but why don't you make a function of it and use it once for rows and once for columns
[conflict_row ] = get_conflict(A)
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
And then you call it twice
conflict_row = get_conflict(A); % Rows
Transpose A to get the columns
Convert the columns to rows and use the same code as before
conflict_col = get_conflict(A.');
If you want to work within the same column then you should do something like this (also sorry this is in C# I don't know what language you are working in):
int currentCol = 0;
foreach (var item in myMultiArray)
{
int currentColValue = item[currentCol];
}
This works because myArray is a array of arrays thus to select a specific column can easily be picked out by just allowing the foreach to perform your row iteration, and you just have to select the column you need with the currentCol value.
Related
I want to calculate the mean value of 6 rows each of a matrix 600*9 dimension. The new matrix should be of size 100*9. Can someone help me ?
I saw this code for calculating mean of 2 rows each and it is working fine foe 2 rows :-
x = rand(1028, 18);
result1 = zeros(1028/2, 18);
for ii = 1:1028/2;
result1(ii,:) = mean(x((2*ii-1):(2*ii),:));
end;
The output matrix will have the mean of first 6 rows ( of the input matrix) as the first row and next 6 rows as the 2nd row and so on
Let the data be defined as
x = rand(600, 9); % example data
N = 6; % group size
The desired result can be obtained very easily without loops:
Reshape the data matrix as a 3-D array, where the size along the first dimension is the desired group size;
Compute the mean along the first dimension;
Remove the first dimension, which is now a singleton.
result = reshape(mean(reshape(x, N, [], size(x,2)), 1), [], size(x,2));
Note how this single line contains the three steps described above:
reshape(x, N, [], size(x,2)) % step 1
mean( , 1) % step 2
result = reshape( , [], size(x,2)); % step 3
Using mean and a for loop:
data = rand(600,9);
num_groups = 6;
group_size = size(data,1)/num_groups;
mean_by_group = NaN(num_groups, size(data,2));
for k = 1:num_groups
mean_by_group(k,:) = mean(data((k-1)*group_size+(1:group_size),:),1)
end
I have matrix A = 50x2
How to convert the data into cell array.
Should be I have 10 cell which each cell contain data [5x2].
Thank you for help.
That is what mat2cell does:
A = rand(50,2); % example matrix
N = 10; % number of cells in which to split the first dimension
result = mat2cell(A, repmat(size(A,1)/N, 1, N), size(A,2));
One can use num2cell:
N_ROWS = 5; N_COLS = 2;
A = rand(50,2);
B = num2cell(reshape(A,N_ROWS,N_COLS,[]),[1,2]); % B is 1x1x10 cell array
What this does is turn your input array into 5x2 "slices" stacked along the 3rd dimension.
You can add a squeeze(B), B = B(:) or a reshape(B,[],1) at the end if you need the output as a column vector.
I have matrices:
a= 0.8147 0.1270 0.6324
0.9058 0.9134 0.0975
b= 0.2785 0.9649 0.9572
0.5469 0.1576 0.4854
0.9575 0.9706 0.8003
c = 0.1419 0.7922
0.4218 0.9595
0.9157 0.6557
and also I have another matrix
I= 1 3 1 1
2 1 3 2
I want to get d matrix such that
d= a(1,3) b(3,1) c(1,1)
a(2,1) b(1,3) c(3,2)
where indices come as two consecutive entries of I matrix.
This is one example I get. However, I get different size matrices for a,b,c,.. and I.
Added: I is m x (n+3) which includes indices, and other (n+2) matrices which have corresponding entries are X,A1,A2,...,An,Y. When n is given, A1,A2,...,An matrices are generated.
Can someone please help me to write Matlab code for this task?
You can do it with varargin. Assuming that your matrices are constructed such that you can form your desired output in the way you want (Updated according to Carmine's answer):
function out = IDcombiner(I, varargin)
out = zeros(size(I, 1), nargin-1);
idx = #(m, I, ii) (sub2ind(size(m), I(:, ii), I(:, ii+1)));
for ii = 1:1:nargin-1
out(:, ii) = varargin{ii}(idx(varargin{ii}, I, ii));
end
Now using this function you can make your selection on a flexible number of inputs:
out = IDcombiner(I, a, b, c)
out =
0.6324 0.9575 0.1419
0.9058 0.9572 0.6557
There is also a one-liner solution, which I do not recommend, since it dramatically decreases the readability of the code and doesn't help you gain much:
IDcombiner = #(I,varargin) ...
cell2mat(arrayfun(#(x) varargin{x}(sub2ind(size(varargin{x}), ...
I(:,x), I(:,x+1))), 1:nargin-1, 'UniformOutput', false));
Normally a matrix is not interpreted as a list of indices, but you can have this if you use sub2ind. To use it you need the size of the matrix you are addressing. Let's make an example starting with a:
a(sub2ind(size(a), I(:,1), I(:,2)))
The code does not change if you first assign the newly generated matrices to a variable name.
will use the column I(:,1) as rows and I(:,2) as columns.
To make the code more readable you can define an anonymous function that does this, let's call it idx:
idx = #(m,I,i)(sub2ind(size(m), I(:,i), I(:,i+1)))
So finally the code will be
d = [a(idx(a,I,1)), b(idx(b,I,2)), c(idx(c,I,3))]
The code does not change if you first assign the newly generated matrices to a variable name.
Other details
Let's make an example with 2 central matrices:
a = rand(3,1) % 3 rows, 1 column
b = rand(3,3) % 3 rows, 3 columns
c = rand(3,3) % another squared matrix
d = rand(3,1) % 3 rows, 1 column
The definition of the anonymous function is the same, you just change the definition of the output vector:
output = [a(idx(a,I,1)), b(idx(b,I,2)), c(idx(c,I,3)), d(idx(d,I,3))]
Keep in mind that following that pattern you always need a I matrix with (n_matrices + 1) columns.
Generalization
Let's generalize this code for a number n of central matrices of size rxr and for "side matrices" of size rxc. I will use some values of those parameters for this example, but you can use what you want.
Let me generate an example to use:
r = 3;
c = 4;
n = 3;
a = rand(r,c); % 2D array
b = rand(r,r,n); % 3D array, along z = 1:n you have 2D matrices of size rxr
c = rand(r,c);
I = [1 3 1 2 1 3; 2 1 3 1 1 1];
The code I wrote can easily be extended using cat to append matrices (note the 2 in the function tells MATLAB to append on the direction of the columns) and a for cycle:
idx = #(m,I,i)(sub2ind(size(m), I(:,i), I(:,i+1)))
d = a(idx(a,I,1));
for i = 1:n
temp = b(:,:,i);
d = cat(2,d,temp(idx(tmp,I,i+1)));
end
d = cat(2,d,c(idx(c,I,n+1)));
If you really don't want to address anything "by hand", you can use cell arrays to put all the matrices together and then cyclically apply the anonymous function to each matrix in the cell array.
I am sure this question must be answered somewhere else but I can't seem to find the answer.
Given a matrix M, what is the most efficient/succinct way to return two matrices respectively containing the row and column indices of the elements of M.
E.g.
M = [1 5 ; NaN 2]
and I want
MRow = [1 1; 2 2]
MCol = [1 2; 1 2]
One way would be to do
[MRow, MCol] = find(ones(size(M)))
MRow = reshape(MRow, size(M))
MCol = reshape(MCol, size(M))
But this does not seem particular succinct nor efficient.
This essentially amounts to building a regular grid over possible values of row and column indices. It can be achieved using meshgrid, which is more effective than using find as it avoids building the matrix of ones and trying to "find" a result that is essentially already known.
M = [1 5 ; NaN 2];
[nRows, nCols] = size(M);
[MCol, MRow] = meshgrid(1:nCols, 1:nRows);
Use meshgrid:
[mcol, mrow] = meshgrid(1:size(M,2),1:size(M,1))
I have 64 characters in a 4*4 matrix.I need to convert it into a cell array such that cell has 4 characters.For eg
Consider A=[TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT]
i need to get a cell array as below
B={[TCTG][CTCT][CGGT][TATA]
[TACA][CTGC][CCAG][AACA]
[CGTC][AACA][AGGC][CAGT]
[GTAT][CCTT][CTTT][GTGT]}
i tried using the mat2cell function but im not able to understand it.please help.
Using a for-loop:
clc
clear
A = 'TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT';
B = cell(4,4);
currentIdx = 0; % Use index to increment by steps of 4 when going through A
for k = 1:16
B{k} = A(currentIdx+1:currentIdx+4);
currentIdx = currentIdx+4;
end
B = B'
B =
'TCTG' 'CTCT' 'CGGT' 'TATA'
'TACA' 'CTGC' 'CCAG' 'AACA'
'CGTC' 'AACA' 'AGGC' 'CAGT'
'GTAT' 'CCTT' 'CTTT' 'GTGT'
You are starting with a 1xN matrix and want to convert it to a 1xN/4 cell array of 1x4 matrices. Your command should then be:
N = length(A);
M = 4;
B = mat2cell(A,1,ones(1,N/M)*M);
The first dimension is the 1, the second dimension is a string of 4's the size of the output cell array. The result:
B =
Columns 1 through 12
'TCTG' 'CTCT' 'CGGT' 'TATA' 'TACA' 'CTGC' 'CCAG' 'AACA' 'CGTC' 'AACA' 'AGGC' 'CAGT'
Columns 13 through 16
'GTAT' 'CCTT' 'CTTT' 'GTGT'
You can use method vec2mat that breaks your input vector to matrix
M = vec2mat(A, numberOfColumns)
(In your case numberOfColumns would be 16) and then use mat2cell. In your case, it would be:
C = mat2cell(M, [1,1,1,1], [4,4,4,4])
It means that all cels will have one row and 4 columns).
Effect of function c = mat2cell(x, [10, 20, 30], [25, 25]) would be:
The image shows why you have to convert vector to matrix. (example from matlab documentation)
You can also (ab)use the very versatile accumarray for this task:
A = 'TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT';
n = 4;
B = accumarray(ceil(1/n:1/n:numel(A)/n).', A(:), [], #(x) {x.'}).'