# How to not use for loop

5 views (last 30 days)
Declan on 7 Sep 2022
Commented: Torsten on 7 Sep 2022
Hi, I have a function that I am trying to get rid of the for loop and rewrite the function so that it doesnt use any loops. I have looked on various links like the Vector Creation (https://au.mathworks.com/help/matlab/ref/colon.html) and Vectorisation (https://au.mathworks.com/help/matlab/matlab_prog/vectorization.html) but I still cant get it to work. Below I have the function with the for loop.
function dfdx = ddx(f, h)
% Add description, name, date, inputs, outputs
dfdx = nan(size(f));
dfdx(1) = (f(2) - f(1))/h;
for j = 2:length(f)-1;
dfdx(j) = 0.5*(f(j+1) - f(j-1))/h;
end
dfdx(end) = (f(end) - f(end-1))/h;
And here is the code to call the function
format compact
a = randn(2, 1)
x = linspace(-1, 1, 20) % equispaced x
f = a(1) + a(2)*x % function values
dfdx = ddx(f, x(2)-x(1)) % derivatives should be exact for linear
computeError = a(2) - dfdx % should be zeros to 1e-15

Star Strider on 7 Sep 2022
Edited: Star Strider on 7 Sep 2022
Try something like this —
format compact
a = randn(2, 1)
a = 2×1
0.4175 1.4768
x = linspace(-1, 1, 20) % equispaced x
x = 1×20
-1.0000 -0.8947 -0.7895 -0.6842 -0.5789 -0.4737 -0.3684 -0.2632 -0.1579 -0.0526 0.0526 0.1579 0.2632 0.3684 0.4737 0.5789 0.6842 0.7895 0.8947 1.0000
f = a(1) + a(2)*x % function values
f = 1×20
-1.0593 -0.9038 -0.7484 -0.5929 -0.4374 -0.2820 -0.1265 0.0289 0.1844 0.3398 0.4953 0.6507 0.8062 0.9616 1.1171 1.2725 1.4280 1.5835 1.7389 1.8944
dfdx = ddx(f, x(2)-x(1)) % derivatives should be exact for linear
dfdx = 1×20
1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768
computeError = a(2) - dfdx % should be zeros to 1e-15
computeError = 1×20
1.0e-14 * -0.0444 -0.0444 0.3775 -0.0444 0.3331 0.1776 0.0666 0.2220 0.1332 0.1332 0.1776 0.1110 0.1776 0.1776 0.1776 0.1776 0.1776 -0.0444 0.3775 -0.0444
function dfdx = ddx(f,h)
dfdx(1) = (f(2) - f(1))/h;
dfdx(2:numel(f)) = (f(2:end) - f(1:end-1))/h;
end
EDIT — The gradient function already exists to do this, however I’m assuming here that you want to write your own function to do the numerical derivative.
.
Star Strider on 7 Sep 2022
@Declan — As always, my pleasure!
I checked it against the gradient function and both gave the same result.
That was my criterion —
format compact
a = randn(2, 1)
a = 2×1
0.6877 1.4736
x = linspace(-1, 1, 20) % equispaced x
x = 1×20
-1.0000 -0.8947 -0.7895 -0.6842 -0.5789 -0.4737 -0.3684 -0.2632 -0.1579 -0.0526 0.0526 0.1579 0.2632 0.3684 0.4737 0.5789 0.6842 0.7895 0.8947 1.0000
f = a(1) + a(2)*x % function values
f = 1×20
-0.7859 -0.6307 -0.4756 -0.3205 -0.1654 -0.0103 0.1448 0.2999 0.4550 0.6102 0.7653 0.9204 1.0755 1.2306 1.3857 1.5408 1.6959 1.8510 2.0062 2.1613
dfdx = ddx(f, x(2)-x(1)) % derivatives should be exact for linear
dfdx = 1×20
1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736
computeError = a(2) - dfdx % should be zeros to 1e-15
computeError = 1×20
1.0e-14 * 0.0666 0.0666 0.2887 -0.1332 0.3775 0.0666 0.1776 0.1776 0.1776 0.1776 0.0666 0.1776 0.2887 0.0666 0.0666 0.2887 0.2887 -0.1332 0.2887 -0.1332
CompareResults = 2×21 string array
"gradient" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "ddx" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736" "1.4736"
function dfdx = ddx(f,h)
dfdx(1) = (f(2) - f(1))/h;
dfdx(2:numel(f)) = (f(2:end) - f(1:end-1))/h;
end
.
Torsten on 7 Sep 2022
Yes, for linear functions, centered and forward differencing to approximate the derivative give the same result.

Torsten on 7 Sep 2022
Edited: Torsten on 7 Sep 2022
function dfdx = ddx(f, h)
end
Declan on 7 Sep 2022
Oh, I didnt realise that there was a gradient function inbuilt. Thanks!
Stephen23 on 7 Sep 2022
+1 very neat.

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