Interpolating the indexes of values in a Vector

hi

i have a Vector lets say a = [2 4 6 8 10 12 14 16]; and i have a value "b" that can change.

I want to write a code that compares the value "b" to every value in "a" starting from index 1 and going to index 8 in this example.

When the value "b" finds a value in "a" so that b >= a(:) and b < a( :+1) . The algorithm will give me the index "c" of that value wich can also be interpolated.

As an example lets say b = 6, then the algorithm will give me c = 3, because the index of 6 in "a" is 3.

but if b = 5 then the algorithm should give me c = 2.5. even tho 5 doesnt exist in "a" but we can know the index with interpolation

and if b = 4.5 then the algorithm should give me c = 2.25

if b = 14.3 then the algorithm should give me c = 7.15 etc...

How can i do this?

Thank you.

Answer 1

[Star Strider]

Use interp1 for this —

a = [2 4 6 8 10 12 14 16];
k = 1:numel(a);

k = 1×8

1 2 3 4 5 6 7 8

interpc = @(b) interp1(a,k,b);
bvector = [4.5 5 6 14.3];
c = interpc(bvector);

Result = [bvector; c]

Result = 2×4

4.5000 5.0000 6.0000 14.3000 2.2500 2.5000 3.0000 7.1500

.

[malik abdelli]

Tahnks this helped,

just for my understanding what does "interpc = @(b) interp1(a,k,b);" do?

[Star Strider]

It is an anonymous function that accepts whatever its argument is and returns the result. Anonymous Functions work like other functions, with a few restrictions based on their simplicity.

[malik abdelli]

ok thank you

[Star Strider]

As always, my pleasure!

Answer 2

[Konrad]

Hi,

use interp1():

a = [2 4 6 8 10 12 14 16];
b = [6, 5, 4.5, 14.3];
interp1(a,1:numel(a),b,'linear') % 'linear' is also the default

ans = 1×4

3.0000 2.5000 2.2500 7.1500

Best, Konrad