Can anyone tell what is this line doing

It seems these two first lines two the exact same thing can anyone help

I cannot figure out what they are doing

a = d{1}(f(1,2));
a = f(1,2);

...

f = [1 2 4];
d = {[4 2 1] [4 1 2] [4 3 5]};

[VBBV]

a = f(1,2) % this line is accessing the 1st row , 
% 2nd column element in vector f = [1 2 4]; which is 2
a = d{1}(f(1,2)) % this line is accessing the 1st row of cell array,d, and further 
% substitutes the value of f(1,2) i.e. 2 obtained earlier to access the 2nd element 
% in the row. i.e. again 2  [4 2 1]
                               ^
   

[Kamil Kacer]

still dont get it then why i cannot do

d{1}(f(1,3))

[VBBV]

Then you would get a = 1 as output.

[VBBV]

Of course you can do it if you want a = 1 as output

[Kamil Kacer]

> d{1}(f(1,3))
Index exceeds the number of array elements (3).

[Kamil Kacer]

gives an error

Answer 1

[John D'Errico]

You asked again, why you cannot use f(1,3), instead of f(1,2).

LOOK CAREFULLY AT WHAT YOU HAVE. TAKE IT ONE STEP AT A TIME.

f = [1 2 4];
d = {[4 2 1] [4 1 2] [4 3 5]};

Now, what is f(1,3)? TRY IT!

f(1,3)

ans = 4

It is the number 4. How are you using it? As an index into a cell of the cell array d. So what is d{1}?

d{1}

ans = 1×3

4 2 1

●

Does d{1} have a 4th element? NO.

Why was this not a problem with f(1,2)? Because f(1,2) is 2.

d{1}(f(1,2))

ans = 2

d{1}(f(1,3))

Index exceeds the number of array elements (3).

The latter case fails because you cannot access the 4th element of a vector that has only 3 elements.

[Kamil Kacer]

now i get it thank you so much

[John D'Errico]

Excellent.

For most problems in MATLAB where you need to understand a complicated line of code, you take apart the line from the inside and work out. Figure out what the inside piece is doing. Then look backwards and outwards. In this case, it is perhaps not fully obvious, that (f(1,2)) was an index into what came before.

[Kamil Kacer]

Thats great advice thanks a lot again

Answer 2

[KSSV]

f = [1 2 4];   % this is an array can be access by f (i) where i = 1,2,3
d = {[4 2 1] [4 1 2] [4 3 5]};  % this is a cell array, can be access by d {i} where i = 1,2,3