A = [0,0,1,0,1,0,1,0,0,0;
0,0,0,0,1,0,1,0,0,0;
0,0,1,0,1,0,1,0,0,0];
B = [2,5;
1,6;
3,10];
Expected Output Cell Array:
C = [1,1,1,1; %// 2-3-4, 3-4-5, 4-5, 5
0,0,1,1,1,0; %// 1-2-3, 2-3-4, 3-4-5, 4-5-6, 5-6, 6
1,1,1,1,1,0,0,0]; %// 3-4-5, 4-5-6, 5-6-7, 7-8-9, 8-9-10, 9-10, 10
Matrix B includes which columns should be used to execute the condition on Matrix A. For example, first row of B is 2 and 5; so elements between 2nd 5th column of matrix A should be used to execute the condition. Second row of B is 1 and 6; so elements between 1st 6th column should be used to execute the condition. And so on...
The condition: if sum of successive 3 elements is bigger than or equal to 1 then write 1 to matrix C; otherwise write 0. For example, A includes 0,1,0 as three successive elements (sum is 0+1+0=1), so write 1 to matrix C. Another example, first three elements of A in second row are 0,0,0 (sum is 0), so write 0 to matrix C. And so on...
"Sometimes it can be considered only 1 or 2 successive elements."
For example, condition execution of first row of A ends with 5th column, so only value 5th column should be considered; which is 1. So 1 is written to matrix C.
Explaining the first row of C:
1, since (sum of 2,3,4 elements of A(1,:)) >= 1
1, since (sum of 3,4,5 elements of A(1,:)) >= 1
since max limit is 5, only 2 successive elements are taken here
1, since (sum of 4,5 elements alone of A(1,:)) >= 1
since max limit is 5, only 1 successive element is taken here
1, since (sum of 5th element alone of A(1,:)) >= 1
Without for loop, only with matrix operations, how can I do this complex task? or any trick?
Using mat2cell, cellfun, im2col and any
subMatLen = 3;
%// Converting both A & B matrix to Cell Arrays to perform operations Row-wise
AC = mat2cell(A,ones(1,size(A,1)),size(A,2));
BC = mat2cell(B,ones(1,size(B,1)),size(B,2));
%// Getting only the columns of each rows within the limits specified by Matrix B
%// Also appended with zeros for my own convenience as it wont affect the 'summing' process
out = cellfun(#(x,y) [x(y(1):y(2)),zeros(1,subMatLen-1)],AC, BC, 'uni', 0);
%// Finally taking each 1x3 sliding sub-matrix and returning 1 if `any` of it is non-zero
%// which is equivalent to summing and checking whether they are >= 1
out = cellfun(#(x) any(im2col(x, [1,subMatLen], 'sliding')), out, 'uni', 0);
Your Sample Input:
A = [0,0,1,0,1,0,1,0,0,0;
0,0,0,0,1,0,1,0,0,0;
0,0,1,0,1,0,1,0,0,0];
B = [2,5;
1,6;
3,10];
Output:
>> celldisp(out)
out{1} =
1 1 1 1
out{2} =
0 0 1 1 1 0
out{3} =
1 1 1 1 1 0 0 0
If you want them as a single row or column matrix, you could add this to the bottom of the code:
out = cat(2,out{:})
or
out = (cat(2,out{:})).'
Related
I have a matrix 'Z' sized 100000x2 and imported as an Excel file using readmatrix. I have a created array 'Time' (Time = [-200:0.1:300]'). I would like to compare all values in column 1 of 'Z' to 'Time' and eliminate all values of column 1 of 'Z' that do not equal a value of 'Time', thus shortening my 'Z' matrix to match my desired time values. Column 2 are pressure traces, so this would give me my desired time values and the corresponding pressure trace.
This sort of thing can be done without loops:
x = [1,2,3,4,1,1,2,3,4];
x = [x', (x+1)'] % this is your 'Z' data from the excel file (toy example here)
x =
1 2
2 3
3 4
4 5
1 2
1 2
2 3
3 4
4 5
y = [1,2]; % this is your row of times you want eliminated
z = x(:,1)==y % create a matrix logical arrays indicating the matches in the first column
z =
9×2 logical array
1 0
0 1
0 0
0 0
1 0
1 0
0 1
0 0
0 0
z = z(:,1)+z(:,2); % there is probably another summing technique that is better for your case
b = [x(z~=1,1), x(z~=1,2)] % use matrix operations to extract the desired rows
b =
3 4
4 5
3 4
4 5
All the entries of x where the first column did not equal 1 or 2 are now gone.
x = ismember(Z(:,1),Time); % logical indexes of the rows you want to keep
Z(~x,:) = []; % get rid of the other rows
Or instead of shortening Z you could create a new array to use downstream in your code:
x = ismember(Z(:,1),Time); % logical indexes of the rows you want to keep
Znew = Z(x,:); % the subset you want
You have to loop over all rows, use a nested if statement to check the item, and delete the row if it doesn't match.
Syntax for loops:
for n = 1:100000:
//(operation)//
end
Syntax for if statements:
if x == y
//(operation)//
Syntax for deleting a row: Z(rownum,:) = [];
I have a matrix a in Matlab that looks like the following:
a = zeros(10,3);
a(3:6,1)=2; a(5:9,3)=1; a(5:7,2)=3; a(8:10,1)=2;
a =
0 0 0
0 0 0
2 0 0
2 0 0
2 3 1
2 3 1
0 3 1
2 0 1
2 0 1
2 0 0
I would like to obtain a cell array with the number of times that each number appears in a column. Also, it should be ordered depending on the element value, regardless of the column number. In the example above I would like to obtain the cell:
b = {[5],[4,3],[3]}
Because the number 1 appears once for 5 times, the number 2 twice in blocks of 4 and 3, and the number 3 once for 3 times. As you can see the recurrences are ordered according to the element value and not to the number of the column where the elements appear.
Since you're not concerned with the column, you can string all the columns into a single column vector, padding with zeroes on either end to prevent spans at the start and end of columns from running together:
v = reshape(padarray(a, [1 0]), [], 1);
% Or if you don't have the Image Processing Toolbox function padarray...
v = reshape([zeros(1, size(a, 2)); a; zeros(1, size(a, 2))], [], 1);
Now, assuming spans are always separated by 1 or more zeroes, you can find the length of each span as follows:
endPoints = find(diff(v) ~= 0); % Find where transitions to or from 0 occur
spans = endPoints(2:2:end)-endPoints(1:2:end); % Index of transitions to 0 minus
% index of transitions from 0
And finally you can accumulate the spans based on the value present in those spans:
b = accumarray(v(endPoints(1:2:end)+1), spans, [], #(v) {v(:).'}).';
And for your example:
b =
1×3 cell array
[5] [1×2 double] [3]
Note:
The ordering of values in the resulting cell array is not guaranteed to match the order in spans (i.e. b{2} above is [3 4] instead of [4 3]). If order matters, you'll need to sort your subscripts as per this section of the documentation. Here's how you would change the computation of b:
[vals, index] = sort(v(endPoints(1:2:end)+1));
b = accumarray(vals, spans(index), [], #(v) {v(:).'}).';
The hard part is finding and separating the blocks. diff will find the starting point of any run of numbers, which is the starting point for this solution:
b = [zeros(1,size(a,2)); a; zeros(1,size(a,2))];
idx = diff(b)~=0;
block_values = b(idx);
block_lengths = diff([0; find(idx)]);
Now we have two vectors of the values of each block, and how long they are, and they just need to be captured in the cell array, ignoring the zero blocks
c = accumarray(block_values(block_values~=0), block_lengths(block_values~=0), [], #(x) {x}).';
b = {}
for i = 1:ncolumns
for n = 1:nnumbers
b{i}(n) = sum(a(:,i) == n)
end
end
(Note that this places zeros for numbers at which the count is 0, but otherwise I don't see how else you would be able to recognize which value is being counted)
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
I have a 21x19 matrix B
Each index of the matrix is either 1,0, or -1. I want to count the number of occurrences for each row and column. Performing the column count is easy:
Colcount = sum( B == -1 );
Colcount = sum( B == 0 );
Colcount = sum( B == 1 );
However accessing the other dimension to attain the row counts is proving difficult. It would be great of it could be accessed in one statement.
Then i need to use a fprintf statement to print the results to the screen.
By default sum operates on the columns of a matrix. You can change this by specifying a second argument to sum. For example:
A = [ 1 1 1; 0 1 0];
C = sum(A,2);
C -> [3; 1];
Additionally you can transpose the matrix and get the same result:
A = [ 1 1 1; 0 1 0];
C = sum(A'); % Transpose A, ie convert rows to columns and columns to rows
C -> [3 1]; % note that the result is transposed as well
Then calling fprintf is easy, provide it with a vector and it will produce a string for each index of that vector.
fprintf('The count is %d\n', C)
The count is 3
The count is 1
The second input argument of SUM indicates in which direction you want to sum.
For example, if you want to count the number of occurrences of 1 along rows and columns, respectively, and print the result using fprintf, you can write:
rowCount = sum(B==1,2);
colCount = sum(B==1,1); %# 1 is the default
fprintf('The rowCount of ones is\n%s\nThe colCount of ones is\n%s\n',...
num2str(rowCount'),num2str(colCount))
Note that I use num2str so that you can easily print a vector.
A matrix has m rows and n columns (n being a number not exceeding 10), and the nth column contains either 1 or 0 (binary). I want to use this binary as a decision to take out the associated row (if 1, or otherwise if 0). I understand that this can be done through iteration with the use of the IF conditional.
However, this may become impractical with matrices whose number of rows m gets into the hundreds (up to 1000). What other procedures are available?
You can use logical datatypes for indexing. For example,
M =
1 2 0
4 5 1
7 8 0
M = [1 2 0;4 5 1;7 8 0];
v = (M(:,n) == 1);
M(v,2) = 1;
M =
1 2 0
4 1 1
7 8 0
Now you have set all the elements in column 2 to 1 if the corresponding element in column n is true.
Note that the v = (M(:,n) == 1) converts the nth column to a logical vector. You can accomplish the same with v = logical(M(:,n));
I would recommend this blog entry for a detailed look at logical indexing.
Update:
If you want to erase rows, then use:
M(v,:) = [];