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For-loop using only last iteration

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Giuliano
Giuliano el 25 de Oct. de 2022
Comentada: Giuliano el 25 de Oct. de 2022
My code is as follows:
clearvars;
zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C = zeros(1,10) + A;
end
disp(C)
100 100 100 100 100 100 100 100 100 100
Now, what I originally wanted to do was add the square of each iteration on the respective i-th column of the zeros(1,10) vector, instead I get i = 10 for every position and I don't understand why it doesn't do it for each iteration but only for the last one.
  3 comentarios
Giuliano
Giuliano el 25 de Oct. de 2022
I‘m trying to add e.g 1^2 as the first column of my zero vector The final result should be 1 4 9 25 …. 100
Torsten
Torsten el 25 de Oct. de 2022
x = (1:10).^2
x = 1×10
1 4 9 16 25 36 49 64 81 100

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Respuesta aceptada

Star Strider
Star Strider el 25 de Oct. de 2022
It is necessary to index into ‘C’, however I do not know what you want as the result.
% clearvars;
C = zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C(i) = A;
end
disp(C)
1 4 9 16 25 36 49 64 81 100
.
  3 comentarios
Matt J
Matt J el 25 de Oct. de 2022
@Giuliano if so, you should Accept-click the answer.
Star Strider
Star Strider el 25 de Oct. de 2022
@Giuliano — As always, my pleasure!
The problem is that you did not index ‘C’, so every iteration overwrote the previous iteration.
I also assigned the zeros call before the loop to ‘C’ since your obvious intent was to pre-allocate it. That should always be done if at all possible, so that is good programming practice there.
@Matt JThank you!

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Jim Riggs
Jim Riggs el 25 de Oct. de 2022
Editada: Jim Riggs el 25 de Oct. de 2022
Inside the loop
C = zeros(1,10) + A;
the function "zeros(1,10)" is assigning zeros to the entire C vector, then adding A.
Also, the command before the for loop "zeros(1,10)" does not save the 10 element vector.
I think what you intended is something like:
C = zeros(1,10); %Create a row vector with 10 zeros
for i=1:1:10
A = [i.^2];
C(i) = A;
end
  3 comentarios
Jim Riggs
Jim Riggs el 25 de Oct. de 2022
Editada: Jim Riggs el 25 de Oct. de 2022
indexing instructs the program to put the value of "A" in the 'ith" position in the C array.
So inside the for loop, when i=1, A is computed as 1^2 = 1, and put in location C(1).
On the next ittereation of the loop, i=2, A=2^2 = 4 and is stored in position 2 in the C vector (i.e. C(2) ).
When i=3, A=3^2 = 9 and is stored in C(3), etc.
So, the way you had the loop coded, C was assigned a value of zeros(1,10), or ten zeros, then it added A to the vector, which adds A to all 10 elements of C.
Each itteration of the loop assigned ten zeros to C, then added A, so only the last itteration retained any values, and they were 10 copies of A.
so in the first itteration of the loop, i=1, A=1 and C = 1 1 1 1 1 1 1 1 1 1
In the second itteration of the loop, i=2, A=4, and C = 4 4 4 4 4 4 4 4 4 4,
etc.
Giuliano
Giuliano el 25 de Oct. de 2022
Thanks for the detailed explanation! Honestly I thought what would happen if I somehow included i in C was that I‘d get ten vectors C that have i^2 as the i-th element and the rest zeros and that I‘d have to somehow add these then vectors then. Might be stupid, but I just started with MatLab (rather any kind of programming) last week so let me be excused! Thanks again.

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