How to find interpolated values when there are multiple?

I'm looking to find points where a vector crosses a certain threshold. I have been using interp1 to try to accomplish this but it does not recognise multiple solutions.

for example:

a = [1 2 3 5 8 4];

b = interp1(a, 1:length(a), 6);

returns 4.3333

This is one of the correct answers, but I am looking to get this to return both points where the interpolated value would be 6 (4.33 and 5.5). Any help much appreciated!

Answer 1

[John D'Errico]

interp1 assumes a single valued functional relationship. You cannot use interp1 to do what you wish to do.

Instead, use a tool like Doug Schwarz's intersections , downloaded from the file exchange.

a = [1 2 3 5 8 4];
intersections(1:length(a),a,[1 6],[6 6])
ans =
     4.3333
        5.5

[Adam Hockley]

Thanks! got this working perfectly for what I needed!

[Sahil Wani]

if i have to find the interpolation of : a_new = [(0:0.5:8),(8:-0.5:4)];

How should i modify the code?

[Walter Roberson]

@Sahil WANI

Is that your independent variable? If so then it duplicates the range 4:.5:8; what would you like to have happen in that range?

If you want to get both outputs, then I recommend you break the problem up into two interp1 searches,

a1 = 0:0.5:8; a2 = 8:-0.5:4;
b1 = values for 0 to 8; b2 = values for 8 to 4
q = list of points to query

out1 = interp1(a1, b1, q(:));
out2 = interp1(a2, b2, q(:));
out = [out1, out2];

This would have two columns, one for results for 0 to 8, and the other for results for the 8 to 4. q values not present in a range would give nan in the output, which you could detect with isnan();

[Sahil Wani]

Thanks @Walter Roberson! Very useful.