I'm trying to use some kind of if-then-else statement in an anonymous function, which itself is part of cellfun. I have a cell array that contains a number of double matrices. I would like to substitute all positive numbers in all double matrices with +1, and all negative numbers with -1. I am wondering if I can use an anonymous function rather than coding a separate function that I then call from within the cellfun?
Here's the toy example:
mat = [2, 2, 0, -2; -2, 0, 0, 2; -2, 2, -2, 2]
cellarray = repmat({mat}, 3, 1)
I'm looking for something like this:
new_cellarray = cellfun(#(x) if x > 0 then x = 1 elseif x < 0 then x = -1, cellarray, 'UniformOutput', false)
I also tried this, however, apparently I am not allowed to put an equal sign into an anonymous function.
new_cellarray = cellfun(#(x) x(x > 0) = 1, cellarray, 'UniformOutput', false)
new_cellarray = cellfun(#(x) x(x < 0) = -1, cellarray, 'UniformOutput', false)
You can use the built-in function sign, which returns 1, 0, or -1 depending on its input:
mat = [2, 2, 0, -2; -2, 0, 0, 2; -2, 2, -2, 2];
cellarray = repmat({mat}, 3, 1);
new_cellarray = cellfun(#sign, cellarray, 'UniformOutput', false);
Related
I am trying to find an efficient way of extracting groups of n consecutive columns in a matrix. Example:
A = [0, 1, 2, 3, 4; 0, 1, 2, 3, 4; 0, 1, 2, 3, 4];
n = 3;
should produce an output similar to this:
answer = cat(3, ...
[0, 1, 2; 0, 1, 2; 0, 1, 2], ...
[1, 2, 3; 1, 2, 3; 1, 2, 3], ...
[2, 3, 4; 2, 3, 4; 2, 3, 4]);
I know this is possible using a for loop, such as the following code snippet:
answer = zeros([3, 3, 3]);
for i=1:3
answer(:, :, i) = A(:, i:i+2);
endfor
However, I am trying to avoid using a for loop in this case - is there any possibility to vectorize this operation as well (using indexed expressions)?
Using just indexing
ind = reshape(1:size(A,1)*n, [], n) + reshape((0:size(A,2)-n)*size(A,1), 1, 1, []);
result = A(ind);
The index ind is built using linear indexing and implicit expansion.
Using the Image Package / Image Processing Toolbox
result = reshape(im2col(A, [size(A,1) n], 'sliding'), size(A,1), n, []);
Most of the work here is done by the im2col function with the 'sliding' option.
A=2;
for x=0:2:4
A=[A, A*x];
end
A
I'd appreciate any help! The for loop condition as well as the 3rd line and how they work together I can't quite piece together
So, here comes the walktrough.
A = 2;
A is an array of length 1, with 2 as the only element.
for x = 0:2:4
Have a look at the Examples section of the for help. You create an "iteration variable" x, which iterates through an array with the values [0, 2, 4]. See also the Examples section of the : operator help.
A = [A, A*x];
Concatenate array A with the value of A*x (multiplying an array with a scalar results in an array of the same length, in which each element is multiplied by the given scalar), and re-assign the result to A. See also the help on Concatenating Matrices.
Initially, A = [2].
For x = 0: A = [[2], [2] * 0], i.e. A = [2, 0].
For x = 2: A = [[2, 0], [2, 0] * 2], i.e. A = [2, 0, 4, 0].
For x = 4: A = [[2, 0, 4, 0], [2, 0, 4, 0] * 4], i.e. A = [2, 0, 4, 0, 8, 0, 16, 0].
end
End of for loop.
A
Output content of A by implicitly calling the display function by omitting the semicolon at the end of the line, see here for explanation.
Consider, the two matrices:
>> columns = [1,3,2,4]
and
>> WhichSet =
[2, 2, 1, 2;
1, 1, 2, 1;
1, 2, 1, 2;
2, 1, 2, 2]
My intent is to do the following:
>> result = [WhichSet(1,columns(1)), WhichSet(2,columns(2)), WhichSet(3, columns(3)) and WhichSet(4, columns(4))]
result = [2,2,2,2]
without any loops.
Because how indexing works, you can not just plug them as they are now, unless you use linear indexing
Your desired linear indices are:
ind=sub2ind(size(WhichSet),1:size(whichSet,1),columns);
Then
out=WhichSet(ind);
Just came across this problem might be interesting in many applications, for example,
I have a vector A = [2; 5; 10], the values in vector A is sorted and unique.
I have got a matrix (2D or 3D), for example, B = [2, 8, 10; 2, 5, 5; 9, 1, 10];
Want to get a matrix C = [1, 0, 1; 1, 1, 1; 0, 0, 1].
It means if the element in B is also an element of A, we set it to one; otherwise, set the value to zero.
I did this in a for-loop, but for a large 3D matrix, it takes a long time to finish the loop.
Just wondering if there is a smarter method to do this without 'for' loop.
C = zeros(size(B));
for i = 1:size(A,1)
a = A(i);
C(B==a) = 1;
end
This is exactly what ismember does:
A = [2; 5; 10];
B = [2, 8, 10; 2, 5, 5; 9, 1, 10];
C = ismember(B,A)
C =
1 0 1
1 1 1
0 0 1
From the documentation:
ismember(A,B) returns an array containing 1 (true) where the data in A
is found in B. Elsewhere, it returns 0 (false).
Here I got
A = [1, 2, 3]
B = [1, 0, 0, 1, 0, 1]
I want to create a matrix
C = [1, 0, 0, 2, 0, 3]
You can see B is like a mask, The number of ones in B is equal to the number of elements in A. What I want is arrange elements in A to the place where B is 1.
Any method without loop?
Untested, but should be close:
C = zeros(size(B));
C(logical(B)) = A;
This relies on logical indexing.