Unfortunately my programming skills are not that advanced and I really need to vectorize some loops to finish my thesis.
I tried to make things really clear and simple and I have the following two questions in matlab:
1.
If we have a 5x5 matrix A and we want to set the diagonal elements of this matrix to the diagonal of a matrix B, apart from diag(A)=diag(B) we could use :
for i=1:5
B(i,i)=A(i,i)
end
Now if I want to vectorize this I can not use:
i=1:5
B(i,i)=A(i,i)
In that way we assign each combination from 1:5. So, in the end we asign each element of A equal to B and not the diagonal.
Is there some way that we could assign each identical pair of (i,i)?
I tried :
i=1:5
j=1:5
B(i,find(j==i))=A(i,find(j==i))
But still does not work. I repeat I know the diag property but Im only interested on the particular problem.
2.
A similar problem is the fillowing.
b=[ones(2,2) ones(2,2)*2 ones(2,2)*3 ones(2,2)*4] ;
a = zeros(8,12);
for i=1:4
a((i-1)*2+1:(i)*2,(i-1)*3+1:(i)*3) = [8*ones(2,1) b(:,[2*(i-1)+1 2*i])];
end
Thank you for your time and for your help.
Let's bring some mask magic, shall we!
Problem #1
mask = eye(size(A))==1
A(mask) = B(mask)
For creating the mask, you can also use bsxfun -
N = size(A,1)
bsxfun(#eq,[1:N]',1:N)
Or finally, you can use linear indexing -
N = size(A,1)
A(1:N+1:N^2) = B(1:N+1:N^2)
Sample run -
>> A
A =
5 2 9 6 5
9 1 6 2 2
9 7 5 3 9
4 5 8 8 7
7 5 8 1 8
>> B
B =
5 5 2 8 2
1 1 6 5 2
7 8 5 4 4
1 8 9 8 8
1 7 6 1 8
>> mask = eye(size(A))==1;
>> A(mask) = B(mask)
A =
5 2 9 6 5
9 1 6 2 2
9 7 5 3 9
4 5 8 8 7
7 5 8 1 8
Problem #2
%// Append 8's at the start of every (2,2) block in b
b1 = reshape([8*ones(2,4) ; reshape(b,4,[])],2,[])
%// Mask where b1 values are to be put in an otherwise zeros filled array
mask = kron(eye(4,4),ones(2,3))==1
%// Initialize output arraya and set values from b1 into masked places
out = zeros(size(mask))
out(mask) = b1
For your first problem. Use logical indexing:
index = diag(ones(1,size(B,1))
B(index) = A(index)
Related
How do I rotate a matrix to create a spiral order of values?
For example,
12 4 2
8 3 11
6 7 2
I am supposed to to display 12 4 2 11 2 7 6 8 3 but I don't know how to terminate at the 1st row and rotate the function 90 degrees. Thanks in advance for the help.
Hint:
Check the spiral function:
spiral(n) is an n-by-n matrix with elements ranging
from 1 to n^2 in a rectangular spiral pattern.
Use its output to build an index into the original values. You may also need sort, as well as fliplr to reverse the order of values.
See the code after you've given it a try.
x = [12 4 2; 8 3 11; 6 7 2];
t = fliplr(spiral(sqrt(numel(x))));
[~, ind] = sort(t(:));
result = fliplr(x(ind).');
A =[12 4 2;...
8 3 11;...
6 7 2];
B=[];
for ii=1:5
B = [B A(1,:)];
A(1,:)=[];
A=rot90(A);
end
B
B =
12 4 2 11 2 7 6 8 3
I would like to align and count vectors with different time stamps to count the corresponding bins.
Let's assume I have 3 matrix from [N,edges] = histcounts in the following structure. The first row represents the edges, so the bins. The second row represents the values. I would like to sum all values with the same bin.
A = [0 1 2 3 4 5;
5 5 6 7 8 5]
B = [1 2 3 4 5 6;
2 5 7 8 5 4]
C = [2 3 4 5 6 7 8;
1 2 6 7 4 3 2]
Now I want to sum all the same bins. My final result should be:
result = [0 1 2 3 4 5 6 7 8;
5 7 12 16 ...]
I could loop over all numbers, but I would like to have it fast.
You can use accumarray:
H = [A B C].'; %//' Concatenate the histograms and make them column vectors
V = [unique(H(:,1)) accumarray(H(:,1)+1, H(:,2))].'; %//' Find unique values and accumulate
V =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
Note: The H(:,1)+1 is to force the bin values to be positive, otherwise MATLAB will complain. We still use the actual bins in the output V. To avoid this, as #Daniel says in the comments, use the third output of unique (See: https://stackoverflow.com/a/27783568/2732801):
H = [A B C].'; %//' stupid syntax highlighting :/
[U, ~, IU] = unique(H(:,1));
V = [U accumarray(IU, H(:,2))].';
If you're only doing it with 3 variables as you've shown then there likely aren't going to be any performance hits with looping it.
But if you are really averse to the looping idea, then you can do it using arrayfun.
rng = 0:8;
output = arrayfun(#(x)sum([A(2,A(1,:) == x), B(2,B(1,:) == x), C(2,C(1,:) == x)]), rng);
output = cat(1, rng, output);
output =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
This can be beneficial for particularly large A, B, and C variables as there is no copying of data.
I've a vector that I would like to split into overlapping subvectors of size cs in shifts of sh. Imagine the input vector is:
v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13]
given a chunksize of 4 (cs=4) and shift of 2 (sh=2), the result should look like:
[1 2 3 4]
[3 4 5 6]
[5 6 7 8]
[7 8 9 10]
[9 10 11 12]
note that the input vector is not necessarily divisible by the chunksize and therefore some subvectors are discarded. Is there any fast way to compute that, without the need of using e.g. a for loop?
In a related post I found how to do that but when considering non-overlapping subvectors.
You can use the function bsxfun in the following manner:
v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13]
cs=4;
sh=2;
A = v(bsxfun(#plus,(1:cs),(0:sh:length(v)-cs)'));
Here is how it works. bsxfun applies some basic functions on 2 arrays and performs some repmat-like if the sizes of inputs do not fit. In this case, I generate the indexes of the first chunk, and add the offset of each chunck. As one input is a row-vector and the other is a column-vector, the result is a matrix. Finally, when indexing a vector with a matrix, the result is a matrix, that is precisely what you expect.
And it is a one-liner, (almost) always fun :).
Do you have the signal processing toolbox? Then the command is buffer. First look at the bare output:
buffer(v, 4, 2)
ans =
0 1 3 5 7 9 11
0 2 4 6 8 10 12
1 3 5 7 9 11 13
2 4 6 8 10 12 0
That's clearly the right idea, with only a little tuning necessary to give you exactly the output you want:
[y z] = buffer(v, 4, 2, 'nodelay');
y.'
ans =
1 2 3 4
3 4 5 6
5 6 7 8
7 8 9 10
9 10 11 12
That said, consider leaving the vectors columnwise, as that better matches most use cases. For example, the mean of each window is just mean of the matrix, as columnwise is the default.
I suppose the simplest way is actually with a loop.
A vectorizes solution can be faster, but if the result is properly preallocated the loop should perform decently as well.
v = 1:13
cs = 4;
sh = 2;
myMat = NaN(floor((numel(v) - cs) / sh) + 1,cs);
count = 0;
for t = cs:sh:numel(v)
count = count+1;
myMat(count,:) = v(t-cs+1:t);
end
You can accomplish this with ndgrid:
>> v=1:13; cs=4; sh=2;
>> [Y,X]=ndgrid(1:(cs-sh):(numel(v)-cs+1),0:cs-1)
>> chunks = X+Y
chunks =
1 2 3 4
3 4 5 6
5 6 7 8
7 8 9 10
9 10 11 12
The nice thing about the second syntax of the colon operator (j:i:k) is that you don't have to calculate k exactly (e.g. 1:2:6 gives [1 3 5]) if you plan to discard the extra entries, as in this problem. It automatically goes to j+m*i, where m = fix((k-j)/i);
Different test:
>> v=1:14; cs=5; sh=2; % or v=1:15 or v=1:16
>> [Y,X]=ndgrid(1:(cs-sh):(numel(v)-cs+1),0:cs-1); chunks = X+Y
chunks =
1 2 3 4 5
4 5 6 7 8
7 8 9 10 11
10 11 12 13 14
And a new row will form with v=1:17. Does this handle all cases as needed?
What about this? First I generate the starting-indices based on cs and sh for slicing the single vectors out of the full-length vector, then I delete all indices for which idx+cs would exceed the vector length and then I'm slicing out the single sub-vectors via arrayfun and afterwards converting them into a matrix:
v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13]
cs=4;
sh=2;
idx = 1:(cs-sh):length(v);
idx = idx(idx+cs-1 <= length(v))
A = arrayfun(#(i) v(i:(i+cs-1)), idx, 'UniformOutput', false);
cell2mat(A')
E.g. for cs=5; sh=3; this would give:
idx =
1 3 5 7
ans =
1 2 3 4 5
3 4 5 6 7
5 6 7 8 9
7 8 9 10 11
Depending on where the values cs; sh come from, you'd probably want to introduce a simple error-check so that cs > 0; as well as sh < cs. sh < 0 would be possible theoretically if you'd want to leave some values out in between.
EDIT: Fixed a very small bug, should be running for different combinations of sh and cs now.
I'm trying to start the circshift at the specific index of a number using the find command how can I do this? See example code below
%test find and circshift
a=[3:2:11]
%find index of number and start there
a_ind=find(a==9)
b=circshift(a,[0 a_ind])
I get a =[3 5 7 9 11]
a_ind = 4
b = [ 5 7 9 11 3]
I'm trying to get the circshift (b) to start at 9 and have
b = [9 11 3 5 7]
Please note a_ind will vary so I just can't have circshift starting at 2 each time
Here's another option that's good for vectors:
a=[3:2:11];
shift = find(a==9);
circshift(a(:), -shift + 1)'
a(:) guarantees you a column vector and circshift shifts on the row dimension i.e. it needs a column vector. Then just transpose again at the end to recover your row vector. You want to shift left so you must specify a negative shift.
I think you want
>> circshift(a,[0 (length(a)-a_ind+1)])
ans =
9 11 3 5 7
If I try with a different vector a:
>> a=[3:1:11]
a =
3 4 5 6 7 8 9 10 11
>> a_ind=find(a==9)
a_ind = 7
>> circshift(a,[0 (length(a)-a_ind+1)])
ans =
9 10 11 3 4 5 6 7 8
This is the input matrix
7 9 6
8 7 9
7 6 7
Based on the frequency their appearance in the matrix (Note. these values are for explanation purpose. I didn't pre-calculate them in advance. That why I ask this question)
number frequency
6 2
7 4
8 1
9 2
and the output I expect is
4 2 2
1 4 2
4 2 4
Is there a simple way to do this?
Here's a three-line solution. First prepare the input:
X = [7 9 6;8 7 9;7 6 7];
Now do:
[a m n] = unique(X);
b = hist(X(:),a);
c = reshape(b(n),size(X));
Which gives this value for c:
4 2 2
1 4 2
4 2 4
If you also wanted the frequency matrix, you can get it with this code:
[a b']
Here is a code with for-loop (a is input matrix, freq - frequency matrix with 2 columns):
weight = zeros(size(a));
for k = 1:size(freq,1)
weight(a==freq(k,1)) = freq(k,2);
end
Maybe it can be solved without loops, but my code looks like:
M = [7 9 6 ;
8 7 9 ;
7 6 7 ;];
number = unique(M(:));
frequency = hist(M(:), number)';
map = containers.Map(number, frequency);
[height width] = size(M);
result = zeros(height, width); %allocate place
for i=1:height
for j=1:width
result(i,j) = map(M(i,j));
end
end