I have an array in Matlab
A = [1 2 3 4 5 6 7 8 9;
67 67 67 86 86 86 86 67 67]';
where every point in the first row of A corresponds to a "code" either 67 or 86. I am trying to extract these blocks of "67s" and "86s" such that every time a block starts the corresponding elements are put into the 3rd dimension of a different array called X, where the .
So for e.g. in A I have 3 different blocks, so I would like to end up with an array X of size 1x9x3. And for e.g. the first 67 block I would like to have X
X(1,:,1) = [1 2 3];
I understand that I would "fill up" this vector X using a for loop
for i=1:size(A,2)
for j=1:size(A,2) %actually j should be up till the number of blocks present
X(1,i,j) = A(1,i)
end
end
But this isn't correct or complete of course because firstly I'm unsure how to separate out the "blocks" and how to correctly "fill in" the j's in X(1,i,j). Secondly how can I get the code to recognise how many blocks there are?
Can anyone help?
Thanks
One possible approach, based on this answer:
>> B = accumarray([0; cumsum(diff(A(:,2)) ~= 0)] + 1, A(:,1), [], #(x) {x}, [])
Now you have this:
>> B{1}
ans =
1
2
3
>> B{2}
ans =
4
5
6
7
>> B{3}
ans =
8
9
Related
Suppose I have a matrix of dimension [4x4], and a vector of [16x1], I need to multiply every 4 element in the vector in one element in the matrix, (instead of multiplying element in row by element in vector), how can I do that using loop ?
For example here below, the results of the first four elements in the resulted vector as shown in the below example, then the same thing for the second, third and fourth rows in the matrix. :
So the results in that example is supposed to be [16x1]
Thank you
Using kron you can use this one-liner:
%A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
%v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
sum(kron(A,ones(4,4)).'.*v).'/4
I use the kronecker tensor product to "replicate" 4x4 time the A matrice. After that it's pure algebra.
This is just matrix multiplication in disguise... If your tall vector was a matrix of the same size as the matrix shown, where each highlighted block is a row, it's matrix multiplication. We can set this up, then reshape back into a vector.
You can use indexing to turn this into simple matrix multiplication. A question I answered already today details how the below indexing works using bsxfun, then we just reshape at the end:
% Setup
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
% Matrix mutliplication
r = numel(v)/size(A,1);
b = A * v( bsxfun( #plus, (1:r:numel(v)).', 0:r-1 ) );
% Make result a column vector
b = reshape( b.', [], 1 );
See if this is what you want:
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
r = reshape(sum(bsxfun(#times, permute(A, [3 2 1]), permute(reshape(v, 1, [], size(A,2)), [2 3 1])), 2), [], 1);
which gives
r =
17
17
17
17
41
41
41
41
65
65
65
65
89
89
89
89
There are details that I assumed, but this shoudl do the trick:
A=reshape(1:16,4,4).';
b=repelem([2,0,1,3],1,4).';
c=[];
for row=1:size(A,1)
c=[ c; sum(reshape(repelem(A(row,:),4).*b.',4,[]),2)];
end
I am assuming here that your demo for the vector is just a bad example and that you wont have repeated values, otherwise an easier version can be achieved by just not doing 3/4ths of the multiplications.
If you do not have access to repelem, have a look at alterative codes that do the same thing:Element-wise array replication in Matlab
I have an array :
Z = [1 24 3 4 52 66 77 8 21 100 101 120 155];
I have another array:
deletevaluesatindex=[1 3; 6 7;10 12]
I want to delete the values in array Z at indices (1 to 3, 6 to 7, 10 to 12) represented in the array deletevaluesatindex
So the result of Z is:
Z=[4 52 8 21 155];
I tried to use the expression below, but it does not work:
X([deletevaluesatindex])=[]
Another solution using bsxfun and cumsum:
%// create index matrix
idx = bsxfun(#plus , deletevaluesatindex.', [0; 1])
%// create mask
mask = zeros(numel(Z),1);
mask(idx(:)) = (-1).^(0:numel(idx)-1)
%// extract unmasked elements
out = Z(~cumsum(mask))
out = 4 52 8 21 155
This will do it:
rdvi= size(deletevaluesatindex,1); %finding rows of 'deletevaluesatindex'
temp = cell(1,rdvi); %Pre-allocation
for i=1:rdvi
%making a cell array of elements to be removed
temp(i)={deletevaluesatindex(i,1):deletevaluesatindex(i,2)};
end
temp = cell2mat(temp); %Now temp array contains the elements to be removed
Z(temp)=[] % Removing the elements
If you control how deletevaluesatindex is generated, you can instead directly generate the ranges using MATLAB's colon operator and concatenate them together using
deletevaluesatindex=[1:3 6:7 10:12]
then use the expression you suggested
Z([deletevaluesatindex])=[]
If you have to use deletevaluesatindex as it is given, you can generate the concatenated range using a loop or something like this
lo = deletevaluseatindex(:,1)
up = deletevaluseatindex(:,2)
x = cumsum(accumarray(cumsum([1;up(:)-lo(:)+1]),[lo(:);0]-[0;up(:)]-1)+1);
deleteat = x(1:end-1)
Edit: as in comments noted this solution only works in GNU Octave
with bsxfun this is possible:
Z=[1 24 3 4 52 66 77 8 21 100 101 120 155];
deletevaluesatindex = [1 3; 6 7;10 12];
idx = 1:size(deletevaluesatindex ,1);
idx_rm=bsxfun(#(A,B) (A(B):deletevaluesatindex (B,2))',deletevaluesatindex (:,1),idx);
Z(idx_rm(idx_rm ~= 0))=[]
I have the following vector A:
A = [34 35 36 5 6 7 78 79 7 9 10 80 81 82 84 85 86 102 3 4 6 103 104 105 106 8 11 107 201 12 202 203 204];
For n = 2, I counted the elements larger or equal to 15 within A:
D = cellfun(#numel, regexp(char((A>=15)+'0'), [repmat('0',1,n) '+'], 'split'));
The above expression gives the following output as duration values:
D = [3 2 7 4 6] = [A(1:3) **stop** A(7:8) **stop** A(12:18) **stop** A(22:25) **stop** A(28:33)];
The above algorithm computes the duration values by counting the elements larger or equal to 15. The counting also allows less than 2 consecutive elements smaller than 15 (n = 2). The counter stops when there are 2 or more consecutive elements smaller than 15 and starts over at the next substring within A.
Eventually, I want a way to find the median position points of the duration events A(1:3), A(7:8), A(12:18), A(22:25) and A(28:33), which are correctly computed. The result should look like this:
a1 = round(median(A(1:3))) = 2;
a2 = round(median(A(7:8))) = 8;
a3 = round(median(A(12:18))) = 15;
a4 = round(median(A(22:25))) = 24;
a5 = round(median(A(28:33))) = 31;
I edited the question to make it more clear, because the solution that was provided here assigns the last number within the row of 2 or more consecutive numbers smaller than 15 (3 in this case) after A(1:3) to the next substring A(7:8)and the same with the other substrings, therefore generating wrong duration values and in consequence wrong median position points of the duration events when n = 2 or for any given even n.
Anyone has any idea how to achieve this?
If there is a vector like this,
T = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16]
(the size of vector T can be flexible)
How can I get a array of 'sum of divisions'?
For example,
fn(T, 5) = [ (1+2+3+4+5) , (6+7+8+9+10), (11+12+13+14+15) , 16]
One option, which doesn't require the padding of zeros on the original array, is the use of accumarray and ceil:
div = 5;
out = accumarray(ceil((1:numel(T))/div).',T(:))
Another option using cumsum and diff instead:
div = 5;
T(ceil(numel(T)/div)*div) = 0;
cs = cumsum(T)
out = diff( [0 cs(div:div:end) ] )
Edit: once the padding is done, cumsum and diff are a little overkill and one should proceed as in Bentoy's answer.
Another way, close to the 2nd option of thewaywewalk:
div = 5;
T(ceil(numel(T)/div)*div) = 0;
out = sum(reshape(T,div,[])).'; % transpose if you really want a column vector
Also, one one-liner solution (I prefer this one):
out = blockproc(T,[1 5], #(blk) sum(blk.data), 'PadPartialBlocks',true);
Don't forget to set the parameter 'PadPartialBlocks', this is the key of avoiding explicit padding.
There is an in-built function vec2mat in Communications System Toolbox to convert a vector into a 2D matrix that cuts off after every N elements and puts into separate rows, padding the leftover places at the end with zeros to maintain 2D size . So, after using vec2mat, summing all the rows would be enough to give you the desired output. Here's the implementation -
sum(vec2mat(T,5),2)
Sample run -
>> T = 1:16;
>> vec2mat(T,5)
ans =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 0 0 0 0
>> sum(vec2mat(T,5),2)
ans =
15
40
65
16
suppose that we have following array :
a=[12 21 23 10 34 54 10 9 5 6 7 8]
a =
12 21 23 10 34 54 10 9 5 6 7 8
length(a)=
length(a)
ans =
12
now i want to create following vector b ,which b(1),b(2)...b(6) are following
b(1)=sqrt(a(1)^2+a(2)^2)
b(2)=sqrt(a(3)^2+a(4)^2)
b(3)=sqrt(a(5)^2+a(6)^2))
b(4)=sqrt(a(7)^2+a(8)^2)
b(5)=sqrt(a(9)^2+a(10)^2))
b(6)=sqrt(a(11)^2+a(12)^2)
i have wrote following code
or i=2:2:length(a)
b(i/2)=sqrt(a(i-1)^2+a(i)^2);
end
>> b
b =
24.1868 25.0799 63.8122 13.4536 7.8102 10.6301
but i am not sure if it is correct,pleas help me to clarify if everything is ok in my code
In matlab, loops are quite slow. Using vectors is much faster. I suggest therefore a solution without a loop:
a_1 = a(1:2:end);
a_2 = a(2:2:end);
b = sqrt(a_1.^2 + a_2.^2);
first, you create a vector a_1 containing all elements with odd indices of a and a vector a_2 containing all elements with even indices.
Then you square them element wise (.^) and take the square of the sum.
For you example of a, this is 75 times faster. As you increase the size of the array, you will save even more time.