Anonymous function or for loop?

Assume I have a simple electrical function of V=C*sin(t) where C is a constant which varies based on t intervals.

I have used 2 methods to generate the C and by using tic/toc to check, Method 2 seems to be faster than Method 1.

For efficiency and speed wise, should I always use anonymous function instead of for loop? Or is my loop not being coded efficient enough?

t = 0 : 0.1 : 20;

c1 = 53;  % (0 <= t <= 10)
c2 = 68;  % (10 < t <= 20)

% Method 1 (For Loop)
tic
n = length(t);
a = ones(1,n);
for i = 1:n
    if t(i)>=0 && t(i)<=10
        A = c1;
    else
        A = c2;
    end
    a(i) = A;
end
toc

% Method 2 (Anonymous Function)
tic
C = @(t) (t>=0 & t<=10).*c1 + (t>10).*c2;
c = C(t);
toc

v = sin(t);
v2 = c.*v;
v3 = a.*v;

% Checking
a==c;
v2==v3;

Answer 1

[KSSV]

Off course, you can achieve the same without using anonymous function.

t = 0 : 0.1 : 20;

c1 = 53;  % (0 <= t <= 10)
c2 = 68;  % (10 < t <= 20)

tic
a = c1*ones(size(t)) ; 
a(t>10) = c2 ; 
toc

Elapsed time is 0.009947 seconds.

% Method 2 (Anonymous Function)
tic
C = @(t) (t>=0 & t<=10).*c1 + (t>10).*c2;
c = C(t);
toc

Elapsed time is 0.005566 seconds.

v = sin(t);
v3 = a.*v;
v4 = a.*v ; 


% Checking
isequal(a,c)

ans = logical
   1

isequal(v3,v4) 

ans = logical
   1

[Amanda Liu]

Ohh..thank you again! ^^

Answer 2

[KSSV]

The second version is a vectorised version and this will be fast compared to loop.

Your loop is implemented.

You check the finals values uisng:

% Checking
isequal(a,c)
isequal(v2,v3)

[Amanda Liu]

I see, thank you!

Answer 3

[Sulaymon Eshkabilov]

Your second method is much faster. Try to avoid for or while loop if feasible.

At the same time, you may improve your 2nd method by computing the values of C directly for the predefined values of t, e.g:

% Method 2 (Anonymous Function)
t1 =0:0.1:10; t2 =10.01:.1:20;  % t is split up into two ranges
tic
C =t1*c1 + t2*c2;
toc
v = sin(t);
v2 = C.*v;
v3 = a.*v;
% Checking
a==c;
v2==v3;

[Amanda Liu]

Ohh..so I wouldn't need an anonymous function anymore.

Btw, I think you meant

t2 = 10.1 : .1 : 20   % instead of 10.01 : .1 : 20

I'm getting dimension error in this line:

C = t1*c1 + t2*c2 % (t1 is of size 1x101 while t2 is 1x100)

What I understand from the line above is that C is the sum of 2 scalar values meaning that even if I change t1 and t2 to be of the same size, there will still only be a total of 101 Cs' even though it should be 201 Cs'.

Therefore, I changed it into:

C = [t1*c1 t2*c2];

However if i check C & a using isequal(a,C), it returns false because t itself should not be multiplied into the constants.

Anyways, thank you very much for helping me! I will avoid for/while loop if feasible.