Extraction of elements of an array inside the other array

Suppose i have

period(t1,t2,t3) =time(p)

where t1, t2, t3 are n dimensional arrays say

t1 = (1,3,4,6,8,9,0,5,4)
t2 = (3)
t3 = (4,56,7,8,5,1)

i want to show the 5th element of t1 as the output.

Output should be 8

how do i do that ? Kindly help me :)

Answer 1

[Walter Roberson]

period(t1,t2,t3) =time(p)

So t1, t2, t3 are subscripts, and time(p) is either a function or array that results in either a scalar or else a length(t1) by length(t2) by length(t3) array. Afterwards, period will be max(t1) by max(t2) by max(t3) in size.

t1 = (1,3,4,6,8,9,0,5,4,)

0 is not a valid index, so t1 cannot be used as a subscript for an array.

For the sake of continuing the discussion, let us instead say

t1 = [1,3,4,6,8,9,10,5,4]
t2 = [3]
t3 = [4,56,7,8,5,1]

and now we know that whatever time(p) is, it must be either a scalar or else a 9 x 1 x 6 array

i want to show the 5th element of t1 as the output.

t1(5)

but what does that have to do with "inside another array" ??

[Ganesh Kini]

I will try to explain this way

Period (5,1,3) = 12

so the value of time is 12

Now i have trace back by taking the value of time i.e. 12, i need to get the value of the 5th element of t1

is it possible ?

[Ganesh Kini]

Can someone help me out ?

[Ganesh Kini]

a = (1,3,4,6,8,9,0,5,4,)
b = (3)
c = (4,56,7,8,5,1)

a, b,c are saved in .vec extension

for t1 =1:1:length(a)
    for t2=1:1:length(b)
	for  t3=1:1:length(c)
		period_fun(a,b,c)=time(p);
		p=p+1;
	end
    end
end

So this code will give me the time

period fun(5,1,3) which means it will fetch

5th element of a = 8, 1st element of b = 3, 3rd element of c = 7, and the output is time = 1.1

Now just by looking the output time 1.1 i want to extract the 5th element of a or the 1st element of b or 3rd element of c and get that element as output

is it possible ?

[Image Analyst]

Try ismember()

[Ganesh Kini]

Thank you, is it possible to give me some more details ?

[Walter Roberson]

(Sorry for the image of a reply instead of the text of a reply; my session froze but I was able to rescue a picture of it.)

[Walter Roberson]

What I suspect you are looking for:

for t1 =1:1:length(a)
    for t2=1:1:length(b)
	for  t3=1:1:length(c)
		period_fun(t1,t2,t3)=time(p);
		p=p+1;
	end
    end
end

if time is a vector or array of values then you can instead use:

na = length(a);
nb = length(b);
nc = length(c);
period_fun = permute( reshape(time(1:na*nb*nc), nc, nb, na), [3 2 1] );

with no loop.

After that I suspect you want:

target = 1.1;
[found, idx] = ismembertol(target, period_fun);
[aidx, bidx, cidx] = ind2sub([na,nb,nc], idx);

corresponding_a = a(aidx).';
corresponding_b = b(bidx).';
corresponding_c = c(cidx).';

Each of those could be a vector if there is more than one entry in target that is close enough to the target value.

[Ganesh Kini]

Thank you.

If you have spare time, could you please explain me the above code ?

[Ganesh Kini]

Hi,

after looking at the existing code and analysis.

period_fun(a,b,c)=time(p);

period_fun is not a function, its just an array.

Could you please let me know how to access the elements passed to it ?