Related
In MATLAB, there is the bwlabel function, that given a binary vector, for instance x=[1 1 0 0 0 1 1 0 0 1 1 1 0] gives (bwlabel(x)):
[1 1 0 0 0 2 2 0 0 3 3 3 0]
but what I want to obtain is
[1 1 2 2 2 3 3 4 4 5 5 5 6]
I know I can negate x to obtain (bwlabel(~x))
[0 0 1 1 1 0 0 2 2 0 0 0 3]
But how can I combine them?
All in one line:
y = cumsum([1,abs(diff(x))])
Namely, abs(diff(x)) spots changes in the binary vector, and you gain the output with the cumulative sum.
You can still do it using bwlabel by vertically concatenating x and ~x, using 4-connected components for the labeling, then taking the maximum down each column:
>> max(bwlabel([x; ~x], 4))
ans =
1 1 2 2 2 3 3 4 4 5 5 5 6
However, the solution from Bentoy13 is probably a bit faster.
x=[1 1 0 0 0 1 1 0 0 1 1 1 0];
A = bwlabel(x);
B = bwlabel(~x);
if x(1)==1
tmp = A>0;
A(tmp) = 2*A(tmp)-1;
tmp = B>0;
B(tmp) = 2*B(tmp);
C = A+B
elseif x(1)==0
tmp = A>0;
A(tmp) = 2*A(tmp);
tmp = B>1;
B(tmp) = 2*B(tmp)-1;
C = A+B
end
C =
1 1 2 2 2 3 3 4 4 5 5 5 6
You know the first index should remain 1, but the second index should go from 1 to 2, the third from 2 to 3 etc; thus even indices should be doubled and odd indices should double minus one. This is given by A+A-1 for odd entries, and B+B for even entries. So a simple check for whether A or B contains the even points is sufficient, and then simply add the two arrays.
I found this function that does exactly what i wanted:
https://github.com/davidstutz/matlab-multi-label-connected-components
So, clone the repository and compile in matlab using mex :
mex sp_fast_connected_relabel.cpp
Then,
labels = sp_fast_connected_relabel(x);
I have an array containing numbers.
A = [1 0 5 6 2 4 5 7 8 8 3 2 1 0 0 1 0 0];
I have calculated peaks and locations of these numbers in an array.
pks = [6 8 1 ]
locs = [4 9 16]
Now I want to update the array with the new peaks value that I have calculated and plot it.
Example.
I have received peaks [6, 8, 1] at locations [4, 9, 16].
I have altered the peaks values e.g. (pks-1).
I want to replace the peak values in the original array with the new values [5, 7, 0].
Like this.
% replace: ↓ ↓ ↓
A = [1 0 5 5 2 4 5 7 7 8 3 2 1 0 0 0 0 0];
Is there any trick to do this in MATLAB?
Thanks a lot.
Example Code
A = [1 0 5 6 2 4 5 7 8 8 3 2 1 0 0 1 0 0];
[pks,locs] = findpeaks(A);
for i=1:length(pks)
if (pks(i)==locs(i))
pks_1(i)=(pks(i)-1);
A_copy(A_copy==pks(i))=pks_1(i);
else
goto if
end
end
You can directly index them, replace your example code with the following:
A = [1 0 5 6 2 4 5 7 8 8 3 2 1 0 0 1 0 0];
% We aren't interested in the actual pks values, so use ~ instead
[~,locs] = findpeaks(A)
% Reduce all values at 'locs' by 1
A(locs) = A(locs) - 1;
Note, there were several errors in your code. For instance,
you are comparing pks(i) == locs(i), have a think about what that's actually comparing because it doesn't find when your loop is at a peak. For that you would need a double loop
for jj = 1:numel(pks)
for ii = 1:numel(A)
if (ii == locs(ii))
% Peak is at index ii
end
end
end
Better would be
for ii = locs
% Peak is at index ii
end
Even better would be the direct indexing I've shown at the top of this answer!
You are also indexing A_copy and pks_1 before they're defined, so that could cause issues.
Also I'm not sure what you think the goto statement is doing?!
How can I find all the cells that have the same values in a multi-dimensional array?
I can get it partially to work with result=A(:,:,1)==A(:,:,2) but I'm not sure how to also include A(:,:,3)
I tried result=A(:,:,1)==A(:,:,2)==A(:,:,3) but the results come back as all 0 when there should be 1 correct answer
which is where the number 8 is located in the same cell on all the pages of the array. Note: this is just a test the repeating number could be found multiple times and as different numbers.
PS: I'm using octave 3.8.1 which is like matlab
See code below:
clear all, tic
%graphics_toolkit gnuplot %use this for now it's older but allows zoom
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8]; A(:,:,2)=[9 1 7; 6 5 4; 7 2 8]; A(:,:,3)=[2 4 6; 8 9 1; 3 5 8]
[i j k]=size(A)
for ii=1:k
maxamp(ii)=max(max(A(:,:,ii)))
Ainv(:,:,ii)=abs(A(:,:,ii)-maxamp(ii));%the extra max will get the max value of all values in array
end
%result=A(:,:,1)==A(:,:,2)==A(:,:,3)
result=A(:,:,1)==A(:,:,2)
result=double(result); %turns logical index into double to do find
[row col page] = find(result) %gives me the col, row, page
This is the output it gives me:
>>>A =
ans(:,:,1) =
1 2 3
4 5 6
7 9 8
ans(:,:,2) =
9 1 7
6 5 4
7 2 8
ans(:,:,3) =
2 4 6
8 9 1
3 5 8
i = 3
j = 3
k = 3
maxamp = 9
maxamp =
9 9
maxamp =
9 9 9
result =
0 0 0
0 1 0
1 0 1
row =
3
2
3
col =
1
2
3
page =
1
1
1
Use bsxfun(MATLAB doc, Octave doc) and check to see if broadcasting the first slice is equal across all slices with a call to all(MATLAB doc, Octave doc):
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
If we're playing code golf, a one liner could be:
result = all(bsxfun(#eq, A, A(:,:,1)), 3);
The beauty of the above approach is that you can have as many slices as you want in the third dimension, other than just three.
Example
%// Your data
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8];
A(:,:,2)=[9 1 7; 6 5 4; 7 2 8];
A(:,:,3)=[2 4 6; 8 9 1; 3 5 8];
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
... gives us:
>> result
result =
0 0 0
0 0 0
0 0 1
The above makes sense since the third row and third column for all slices is the only value where every slice shares this same value (i.e. 8).
Here's another approach: compute differences along third dimension and detect when all those differences are zero:
result = ~any(diff(A,[],3),3);
You can do
result = A(:,:,1) == A(:,:,2) & A(:,:,1) == A(:,:,3);
sum the elements along the third dimension and divide it with the number of dimensions. We get back the original value if the values are the same in all dimension. Otherwise a different (e.g. a decimal) value. Then find the location where A and the summation are equal over the third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
or
You could also do
all(A==repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]),3)
where repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]) would highlight the implicit broadcasting of this when compared with A.
or
you skip the broadcasting altogether and just compare it with the first slice of A
A(:,:,1) == sum(A,3)./size(A,3)
Explanation
3 represents the third dimension .
sum(A,3) means that we are taking the sum over the third dimension.
Then we divide that sum by the number of dimensions.
It's basically the average value for that position in the third dimension.
If you add three values and then divide it by three then you get the original value back.
For example, A(3,3,:) is [8 8 8]. (8+8+8)/3 = 8.
If you take another example, i.e. the value above, A(2,3,:) = [6 4 1].
Then (6+4+1)/3=3.667. This is not equal to A(2,3,:).
sum(A,3)./size(A,3)
ans =
4.0000 2.3333 5.3333
6.0000 6.3333 3.6667
5.6667 5.3333 8.0000
Therefore, we know that the elements are not the same
throughout the third dimension. This is just a trick I use
to determine that. You also have to remember that
sum(A,3)./size(A,3) is originally a 3x3x1 matrix
that will be automatically expanded (i.e. broadcasted) to a
3x3x3 matrix when we do the comparison with A (A == sum(A,3)./size(A,3)).
The result of that comparison will be a logical array with 1 for the positions that are the same throughout the third dimension.
A == sum(A,3)./size(A,3)
ans =
ans(:,:,1) =
0 0 0
0 0 0
0 0 1
ans(:,:,2) =
0 0 0
1 0 0
0 0 1
ans(:,:,3) =
0 0 0
0 0 0
0 0 1
Then use all(....,3) to get those. The result is a 3x3x1
matrix where a 1 indicates that the value is the same in the
third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
In MATLAB, say I have a set of square matrices, say A, with trace(A)=0 as follows:
For example,
A = [0 1 2; 3 0 4; 5 6 0]
How can I remove the zeros and then vertically collapse the matrix to become as follow:
A_reduced = [1 2; 3 4; 5 6]
More generally, what if the zeroes can appear anywhere in the column (i.e., not necessarily at the long diagonal)? Assuming, of course, that the total number of zeros for all columns are the same.
The matrix can be quite big (hundreds x hundreds in dimension). So, an efficient way will be appreciated.
To compress the matrix vertically (assuming every column has the same number of zeros):
A_reduced_v = reshape(nonzeros(A), nnz(A(:,1)), []);
To compress the matrix horizontally (assuming every row has the same number of zeros):
A_reduced_h = reshape(nonzeros(A.'), nnz(A(1,:)), []).';
Case #1
Assuming that A has equal number of zeros across all rows, you can compress it horizontally (i.e. per row) with this -
At = A' %//'# transpose input array
out = reshape(At(At~=0),size(A,2)-sum(A(1,:)==0),[]).' %//'# final output
Sample code run -
>> A
A =
0 3 0 2
3 0 0 1
7 0 6 0
1 0 6 0
0 16 0 9
>> out
out =
3 2
3 1
7 6
1 6
16 9
Case #2
If A has equal number of zeros across all columns, you can compress it vertically (i.e. per column) with something like this -
out = reshape(A(A~=0),size(A,1)-sum(A(:,1)==0),[]) %//'# final output
Sample code run -
>> A
A =
0 3 7 1 0
3 0 0 0 16
0 0 6 6 0
2 1 0 0 9
>> out
out =
3 3 7 1 16
2 1 6 6 9
This seems to work, quite fiddly to get the behaviour right with transposing:
>> B = A';
>> C = B(:);
>> reshape(C(~C==0), size(A) - [1, 0])'
ans =
1 2
3 4
5 6
As your zeros are always in the main diagonal you can do the following:
l = tril(A, -1);
u = triu(A, 1);
out = l(:, 1:end-1) + u(:, 2:end)
A correct and very simple way to do what you want is:
A = [0 1 2; 3 0 4; 5 6 0]
A =
0 1 2
3 0 4
5 6 0
A = sort((A(find(A))))
A =
1
2
3
4
5
6
A = reshape(A, 2, 3)
A =
1 3 5
2 4 6
I came up with almost the same solution as Mr E's though with another reshape command. This solution is more universal, as it uses the number of rows in A to create the final matrix, instead of counting the number of zeros or assuming a fixed number of zeros..
B = A.';
B = B(:);
C = reshape(B(B~=0),[],size(A,1)).'
Suppose I have a 3x3x3 3D-array called A and a 3x3 matrix called B, created like the following:
A = zeros(3,3,3);
B = magic(3);
My intention is to turn the elements at the 2nd and 3rd positions of A's 3rd dimension into the matrix B, so it would look something like
A(:,:,1) = A(:,:,2) = A(:,:,3) =
0 0 0 8 1 6 8 1 6
0 0 0 3 5 7 3 5 7
0 0 0 4 9 2 4 9 2
My first try was to make
A(:,:,2:3) = B
But I get the following:
error: A(I,J,...) = X: dimensions mismatch
It feels strange to me, since, for instance,
B(1,1:2) = 10
would produce a correct result.
How could I solve this?
Thanks in advance
Forget it, I solved it with the following:
A(:,:,1:2) = repmat(B, [1 1 2])
:)