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How to store value of for loop in a array

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Jay Talreja
Jay Talreja on 14 Oct 2020
Commented: Adam Danz on 15 Oct 2020
Hello everyone,
I am trying to store value of A in an array. But some how it stops after 3 itrations at z = 0.03.
Why So? If anybody can have a look where I going wrong.
A = zeros(1,31);
aes = 2;
counter= 1;
for z= 0:0.01:0.3
A(1, counter) = 1/(1+((z/aes)^2));
counter = counter+ 1;
end
A;
Thanks in Advance

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Accepted Answer

Adam Danz
Adam Danz on 14 Oct 2020
Edited: Adam Danz on 14 Oct 2020
"some how it stops after 3 itrations at z = 0.03"
That's actually not true. In your script, z contains 4 values and the loop has 4 iterations
% First 7 values of A from your version
>> A(1:7)
ans =
1 0.99998 0.9999 0.99978 0 0 0
You're initializeing A as a 1x31 vector of zeros which is why there are extra value in the output.
There's nothing wrong with your loop but I find it more intuitive to loop over intergers 1:n rather than looping over elements of a vector directly. Consider this version,
z= 0:0.01:0.03;
A = zeros(size(z));
aes = 2;
counter= 1;
for i= 1:numel(z)
A(i) = 1/(1+((z(i)/aes)^2));
end
Result:
>> A
A =
1 0.99998 0.9999 0.99978
If you intended to have 31 iterations, define z in the first line of my version as
z = linspace(0,.03,31)
% or
z = 0 : 0.001: 0.03;
and then run the rest of my version.
Result:
A =
Columns 1 through 11
1 1 1 1 1 0.99999 0.99999 0.99999 0.99998 0.99998 0.99998
Columns 12 through 22
0.99997 0.99996 0.99996 0.99995 0.99994 0.99994 0.99993 0.99992 0.99991 0.9999 0.99989
Columns 23 through 31
0.99988 0.99987 0.99986 0.99984 0.99983 0.99982 0.9998 0.99979 0.99978

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Jay Talreja
Jay Talreja on 15 Oct 2020
Thanks it worked.
Here in the code the range was from 0.0 to 0.3 with interval of 0.01.
But it worked for me. Thankyou!!!!
Adam Danz
Adam Danz on 15 Oct 2020
Opps, maybe it's time for me to get new glasses. 🤓

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More Answers (2)

David Hill
David Hill on 14 Oct 2020
A = zeros(1,31);
aes = 2;
counter= 1;
for z= 0:0.001:0.03%I believe you want the interval to be .001 (otherwise loop only runs for 4 times)
A(1, counter) = 1/(1+((z/aes)^2));
counter = counter+ 1;
end

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madhan ravi
madhan ravi on 14 Oct 2020
for z = linspace(0, 0.03, 31)
By the way you don’t need a Loop it’s simply:
A = 1 ./ (1 + ((z / aes) .^ 2))

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