This is a snippet from a program I am trying to understand. I have changed the variable names for easier understanding. I haven't done much coding in MatLab so I can't really understand what's happening. When I kept a=magic(4) and got the output from this code I thought it was sorting or something but it doesn't seem so with the other input
a = [14 41 4 16;7 12 45 0;12 12 45 17; 3 2 1 15]
b=a(:)
c=zeros(4,4)
a is a 4x4 matrix,
b is a column vector of a,
c is a 4X4 matrix of zeros.`
for kk = 1:length(b)
c(a==b(kk)) = kk;
end
c =
1 5 9 13
2 7 11 14
7 7 11 15
4 8 12 16
if I try a=magic(4), where
a =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
then
c =
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
The answer in the comments is correct - perhaps it serves a function within the program, however out of context the code does seem pretty much pointless.
This is essentially what is happening. Suppose you have a matrix A
A = [3 4;
5 6]
Then B will look like this:
B = [3 4 5 6]
C is then created by comparing the kkth element of A to the kkth element of B. If the two are equal, then the kkth element of C will be kk.
Thus, in the example above, C will look like this:
C = [1 2;
3 4]
If, as you have found out, there are multiple entries of the same number in your original matrix A, then the final matrix C will only have the index of the last unique occurence of that number. So, if
A = [3 3;
5 6]
B = [3 3 5 6]
C = [2 2;
3 4]
The reason your first C has 7 three times is because the last position of 12 is at position 7. The reason your second C is different from your first C is because the A that it was made from is totally unique.
Related
Using MATLAB I want to check 2 vectors, for example:
A = [1 2 3 4 5 6 7 8 9 10]
B = [10 9 8 7 6 11 12 13 14 15]
and write a matrix that checks each element B if it is in A, if it is not in A, then append the element to A. So in the end I should have a new matrix H=[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15]. I want to check vector A from the end. This is the code I have right now:
A=[1 2 3 4 5 6 7 8 9 10];
B=[10 9 8 7 6 11 12 13 14 15];
for i=A(end:-1:1)
for j=B(1:1:end)
if B(j)==A(i)
pass
else
C=B(j);
H=[A,C]; % i want to append the new values at the end of vector A
end
end
end
The error I get is in the if statement: if B(j)==A(i) Index exceeds number of array elements.
Use ismember to find non-similar elements of B and concatenate them with A
H = [A B(~ismember(B,A))];
Your error is that your loop variables i and j contain elements of A and B, not indices into them.
For example, these two loops produce the same output:
A = [5,2,1];
for i=A
disp(i)
end
for i=1:numel(A)
disp(A(i))
end
You should use for i=numel(A):-1:1, not for i=A(end:-1:1).
A slightly simpler alternative to the one-liner by Sardar is this:
H = unique([A,B]);
In this case, H is always sorted.
Consider a row vector A and row vector B. For example:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
A has previously been checked to be a subset of B. By subset, I specifically mean that all elements in A can be found in B. I know that elements in A will not ever repeat. However, the elements in B are free to repeat as many or as few times as they like. I checked this condition using:
is_subset = all(ismember(A,B));
With all that out of the way, I need to know the indices of the elements of A within B including the times when these elements repeat within B. For the example A and B above, the output would be:
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
Use ismember to find the relevant logical indices. Then convert them to linear indices using find.
C = find(ismember(B,A));
You can find the difference of each element of A with B, and get the indices you want. Something like below:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
tol = 10^-3 ;
N = length(A) ;
iwant = cell(N,1) ;
for i = 1:N
idx = abs(A(i)-B)<=tol ;
iwant{i} = find(idx) ;
end
iwant = [iwant{:}] ;
Let's say I have a vector:
A=[1 2 3 6 7 8 11 12 13]
and I'm trying to achieve final output like:
[1 6 11 2 7 12 3 8 13]
Where the vector is rearranged to front every nth column, in this case, 3rd. Using indexing will work, but it requires a loop, which I'm trying to avoid. Any idea how to do it in a vectorized way? Thanks!
nth=3;
for i=1:nth:size(A,2)
A_(:,nth)= A(:,i:nth:end)
end
The suggestion that #jodag posted in the comments works totally fine. Alternatively, this should also do the job... but the constraint is the same, A must be divisible by nth:
nth = 3;
A = [1 2 3 6 7 8 11 12 13];
A_len = numel(A);
A_div = floor(A_len / nth);
seq = repmat(1:nth:A_len,1,A_div);
inc = sort(repmat(0:nth-1,1,A_div));
A = A(seq + inc)
Output:
A =
1 6 11 2 7 12 3 8 13
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.