Copy elements to end of row vector - help needed!

Dear Matlab forum,

For my research I have a full EMG row matrix that includes stance and swing phases.

To make this easy to answer, I'll make an example:

Let's call the EMG matrix, matrix 'a', in which stance and swing phases are included. In matrix 'b', in the rows, the indices are mentioned of where the stance phase in matrix a will start and where the stance phase will stop. Here, that would be from row 1:2, row 4:5, row 7:8 etc.

Simply put I desire to write a row matrix 'c', where all the stance phase part will be pasted. In this example c should look like [10 10 20 30 40 40 50 60]'.

How should I define (???) to let the loop fill [c] as a row vector mentioned above? As I'm a novice in this topic, I hope you can help me out. At this point

[c] = [10   20  40  50      
       10   30  40  60]

and should look like [c] = [10 10 20 30 40 40 50 60];

Thanks,

Jerome

--------------------------------------------------------------------------------------------

a = [10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90]';
b = [1 4 7 10; 
     2 5 8 11]; 
c = []'; 
 
% test   
c1 = a( b(1,1) : b(2,1),1 )
c2 = a( b(1,2) : b(2,2),1 ) 
c3 = a( b(1,3) : b(2,3),1 )
 
for i = 0 : 3 
    ?????????????????????????
    c (:,end+1)  = a( b(1,1+i) : b(2,1+i), 1)
    ???????????????????????
end

Answer 1

[Rik]

I don't fully understand your code, but looking to your results, this code should do.

a = [10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90]';
b = [1 4 7 10; 
     2 5 8 11]; 
 c=a(b);
 c=reshape(c,1,[])

[Jeroen Vermeulen]

Dear Rik, this should also be reachable more efficiently, without reshaping the whole matrix, which will be a huge matrix in my research.

For i=0

a( b(1,1) : (2,1), 1) should be pasted in c as a row. c = [10 10]'

For i=1

a( b(1,2): b(2,2), 1) should be pasted at the end of c as a row c= [10 10 20 30]'

For i=2

a( b(1,3): b(2,3), 1) should be pasted at the end of c as a row

Then c would look like [c] = [10 10 20 30 40 40]

How can I reach this in the loop below?

for i = 0 : 3

?????????????????????????

c (:,end+1) = a( b(1,1+i) : b(2,1+i), 1)

???????????????????????

end

[Rik]

Your code is quite fragile. You're trying to expand you c vector with the end keyword, but you are only creating a single index. The code below generates a large enough c vector first, then loops through b, determining the indices into a and c separately.

a = [10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90]';
b = [1 4 7 10;
    2 5 8 11];
clc
c=NaN(1,sum(1+diff(b)));
counter=0;
for n=1:size(b,2)
    a_idx=b(1,n):b(2,n);
    c_idx=counter+(1:numel(a_idx));
    c(c_idx)=a(a_idx);
    counter=c_idx(end);
end

As a further note: reshape is not very expensive in general, because normally it will only have to change the header of the variable.

[Jeroen Vermeulen]

Thanks for your clear answer Rik. Much appreciated, thanks for your help!

Answer 2

[dpb]

A)

c=c(:).';

B)

" matrix 'a', in which stance and swing phases are included. In matrix 'b', in the rows, the indices are mentioned of where the stance phase in matrix a will start and where the stance phase will stop. Here, that would be from row 1:2, row 4:5, row 7:8 etc."

To extract the rows of 'b' from a' where 'b' is the pair of rows as above, consider that 'b' is everything except rows 3, 6, 9, ... use that fact:

b=a;                % copy the original first
b(3:6:end,:)=[];    % remove unwanted rows

alternatively, can write

i1=[1:3:size(a,1)].;    % vector of first rows of b (1, 4, 7, ...)
i2=i1+1;                % and the next row after (2, 5, 8, ...)
b=a([i1;i2],:);         % select rows from a for b

Note one can dispense with the explicit i2 variable above as well...

b=a([i1;i1+1],:);         % select rows from a for b

creates i2 as temporary w/o explicit name.

Look up addressing operations in the "Getting Started" documentation

[Jeroen Vermeulen]

Thanks for your help, I only don't see how A) solves it!

[dpb]

You said

[c] = [10   20  40  50      
       10   30  40  60]

and should look like ...

[c] = [10 10 20 30 40 40 50 60]; 

That's what it does... (:) creates column vector, (.') transposes to row.