How to not use for loop

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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

采纳的回答

Star Strider
Star Strider 2022-9-7
编辑:Star Strider 2022-9-7
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.
.
  4 个评论
Star Strider
Star Strider 2022-9-7
@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 = ["gradient" gradient(f, x(2)-x(1)); "ddx" dfdx]
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
Torsten 2022-9-7
Yes, for linear functions, centered and forward differencing to approximate the derivative give the same result.

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更多回答(1 个)

Torsten
Torsten 2022-9-7
编辑:Torsten 2022-9-7
function dfdx = ddx(f, h)
dfdx = gradient(f,h);
end
  2 个评论
Declan
Declan 2022-9-7
Oh, I didnt realise that there was a gradient function inbuilt. Thanks!

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