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Interpolating the indexes of values in a Vector

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malik abdelli
malik abdelli am 19 Sep. 2023
Kommentiert: Star Strider am 19 Sep. 2023
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.

Akzeptierte Antwort

Star Strider
Star Strider am 19 Sep. 2023
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
.
  4 Kommentare
malik abdelli
malik abdelli am 19 Sep. 2023
ok thank you
Star Strider
Star Strider am 19 Sep. 2023
As always, my pleasure!

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Weitere Antworten (1)

Konrad
Konrad am 19 Sep. 2023
Hi,
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

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