How to generate a specific number of values within a range ?

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Majid
Majid el 13 de Sept. de 2022
Comentada: Majid el 13 de Sept. de 2022
Hi all!
I'm really wondering how to solve this problem: I have an initial matrix Q (100*60), I extracted the first and second non-zero values in each row using this command line
R = size(Q,1);
First_second_ele = zeros(R,2);
for k = 1:R
tmp = nonzeros(qm(k,:)); %extraction first and second element
idx = 1:min(2,numel(tmp));
First_second_ele(k,idx) = tmp(idx);
end
after this , i got two column vectors. Now i want to see 60 elements between the two extracted values.
for example the first two element are :
4.0000 5.3333
i used :
A=linspace(4.0000,5.3333,60);
the result :
A = 4.0000 4.0226 4.0452 4.0678 4.0904 4.1130 4.1356 4.1582 4.1808 4.2034 4.2260 4.2486 4.2712 4.2938 4.3164 4.3390 4.3616 4.3842 4.4068 4.4294 4.4520 4.4746 4.4972 4.5198 4.5424 4.5650 4.5876 4.6102 4.6328 4.6554 4.6779 4.7005 4.7231 4.7457 4.7683 4.7909 4.8135 4.8361 4.8587 4.8813 4.9039 4.9265 4.9491 4.9717 4.9943 5.0169 5.0395 5.0621 5.0847 5.1073 5.1299 5.1525 5.1751 5.1977 5.2203 5.2429 5.2655 5.2881 5.3107 5.3333
How can i do a loop to get the result of all rows?
  2 comentarios
Image Analyst
Image Analyst el 13 de Sept. de 2022
Do you want the 60 values to be linearly spaced, or randomly spaced. Not sure what "the result of all rows" is, but you can access each element in A in a for loop like this
for k = 1 : numel(A)
thisAValue = A(k);
% Now do something with thisAValue.
end
Or do you want to interpolate with interp1 to get values of your other array between the two values and then insert those 60 new values into your existing Q matrix somewhere?
Majid
Majid el 13 de Sept. de 2022
@Image Analyst Hi! the example above is for the first row , i want to do the same for 100 rows, i can't put in each time the two values in "linspace " function.
the result required is new matrix(100,60)
each row contains the 60 elements (not randomly)between the two extracted values as i gave you the first one.

Iniciar sesión para comentar.

Respuesta aceptada

Jan
Jan el 13 de Sept. de 2022
R = size(Q,1);
First_second_ele = zeros(R,2);
Y = zeros(R, 60);
for k = 1:R
tmp = nonzeros(qm(k,:)); %extraction first and second element
idx = 1:min(2,numel(tmp));
First_second_ele(k,idx) = tmp(idx);
if numel(idx) == 2
Y(k, :) = linspace(tmp(1), tmp(2), 60);
end
end

Más respuestas (1)

David Hill
David Hill el 13 de Sept. de 2022
a=sort(rand(10,2),2);
A=[];
for k=1:size(a,1)
A=[A;linspace(a(k,1),a(k,2),60)];
end

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