# Help with interpolation

**URL:** <https://discourse.julialang.org/t/help-with-interpolation/45474>\
**Category:** General Usage\
**Created:** [August 24, 2020, 8:48pm UTC](https://discourse.julialang.org/t/help-with-interpolation/45474 "2020-08-24T20:48:39Z")\
**Posts on this page:** 3\
**Page:** 1

<div class="post-metadata">

**Author:** ![acvalerio](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/acvalerio/32/17444_2.png) [@acvalerio](https://discourse.julialang.org/u/acvalerio)\
**Post date:** [August 24, 2020, 8:48pm UTC](https://discourse.julialang.org/t/help-with-interpolation/45474/1 "2020-08-24T20:48:39Z")

</div>

I’m looking for a function similar to interp1 in Matlab, and browsing the web I found the following piece of code that supposedly do the job:

```julia
using Interpolations
function interp1(X, V, Xq)
    knots = (X,)
    itp = interpolate(knots, V, Gridded(Linear()))
    itp[Xq]
end

```

To check if it is working I tried the following example:

```julia
X = [1,2,3,4,5,6,7,8,9,10];
V = [X.^2 X.^3 X.^4];
Xq = [1.5 1.75; 7.5 7.75]
interp1(X,V,Xq)

```

But I get the following error

```julia
ERROR: MethodError: no method matching interpolate(::Tuple{Array{Float64,1}}, ::Array{Float64,2}, ::Gridded{Linear})

```

What I’m doing wrong here? Is there another way to replicate interp1 in Julia?

Source of interp1 function: [http://robblackwell.com/julia/index.html](http://robblackwell.com/julia/index.html)

---

<div class="post-metadata">

**Author:** ![rdeits](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/rdeits/32/286_2.png) [@rdeits](https://discourse.julialang.org/u/rdeits)\
**Post date:** [August 24, 2020, 9:03pm UTC](https://discourse.julialang.org/t/help-with-interpolation/45474/2 "2020-08-24T21:03:16Z")

</div>

`interpolate` expects that your knot locations should be of the same dimension as the samples to be interpolated, but your `X` is a 1-dimensional vector, while `V` is a two-dimensional matrix. You can try this yourself with a 1-dimensional interpolation:

```julia
x = 1:10
v = x.^2
knots = (x,)
itp = interpolate(knots, v, Gridded(Linear()))
xq = [1.5, 1.75, 7.5, 7.75]
itp.(xq)

```

which gives:

```julia
julia> itp.(xq)
4-element Array{Float64,1}:
  2.5
  3.25
 56.5
 60.25

```

Note the use of `itp.(xq)` which uses Julia’s built-in [broadcasting](https://docs.julialang.org/en/v1/manual/functions/#man-vectorized-1) to compute `itp(y)` for each `y` in `xq`.

As for your 2d example, I’m sure there’s a way to get the output you’re looking for, but I’m having trouble figuring out what the intended behavior is from your code. What result are you expected from interpolating a 2D matrix `V` with a 1D vector of knots `X`?

---

<div class="post-metadata">

**Author:** ![acvalerio](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/acvalerio/32/17444_2.png) [@acvalerio](https://discourse.julialang.org/u/acvalerio)\
**Post date:** [August 24, 2020, 9:11pm UTC](https://discourse.julialang.org/t/help-with-interpolation/45474/3 "2020-08-24T21:11:05Z")

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Hi, thank you for your reply.

The example I provided is taken from Matlab. I’m just checking if both functions will produce similar results. The expected output would be a 3D array of size (2,2,3).