# Julia version of Matlab function interp1

**URL:** <https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540>\
**Category:** General Usage\
**Tags:** question\
**Created:** [January 23, 2018, 2:12am UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540 "2018-01-23T02:12:27Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![Yifan\_Liu](https://avatars.discourse-cdn.com/v4/letter/y/4da419/32.png) [@Yifan\_Liu](https://discourse.julialang.org/u/Yifan_Liu)\
**Post date:** [January 23, 2018, 2:12am UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540/1 "2018-01-23T02:12:27Z")

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I am try to find the Julia version of Matlab function interp1.

In Matlab, vq = interp1(x,v,xq) returns interpolated values of a 1-D function at specific query points using linear interpolation. Vector x contains the sample points, and v contains the corresponding values, v(x). Vector xq contains the coordinates of the query points.

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**Author:** ![stillyslalom](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/stillyslalom/32/45687_2.png) [@stillyslalom](https://discourse.julialang.org/u/stillyslalom)\
**Post date:** [January 23, 2018, 4:10am UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540/2 "2018-01-23T04:10:34Z")

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```julia
using Dierckx # could also use Interpolations.jl, but its syntax is less familiar
x = linspace(0,π)
v = sin.(x)
xq = linspace(π/2, 3π/2)

# Create & evaluate spline interpolant
spl = Spline1D(x, v; k=1) # k: order of interpolant; can be between 1-5
vq = spl(xq)

# or equivalently, as a one-liner:
vq = Spline1D(x, v; k=1)(xq)

```

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**Author:** ![andrewning](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/andrewning/32/3661_2.png) [@andrewning](https://discourse.julialang.org/u/andrewning)\
**Post date:** [January 23, 2018, 6:56pm UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540/3 "2018-01-23T18:56:14Z")

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If you want to use Interpolations.jl (a pure julia implementation), but stick with a familiar syntax you can create a reusable wrapper. I’m sure this could be improved, but here is an example that contains the same basic functionality as the matlab version including extrapolation options. Note that Interpolations has way many more options for spline degree, boundary conditions, and grid representation if you want to customize further.

```julia
using Interpolations

function interp1(xpt, ypt, x; method="linear", extrapvalue=nothing)

    if extrapvalue == nothing
        y = zeros(x)
        idx = trues(x)
    else
        y = extrapvalue*ones(x)
        idx = (x .>= xpt[1]) .& (x .<= xpt[end])
    end
    
    if method == "linear"
        intf = interpolate((xpt,), ypt, Gridded(Linear()))
        y[idx] = intf[x[idx]]

    elseif method == "cubic"
        itp = interpolate(ypt, BSpline(Cubic(Natural())), OnGrid())
        intf = scale(itp, xpt)
        y[idx] = [intf[xi] for xi in x[idx]]
    end
    
    return y
end

```

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<div class="post-metadata">

**Author:** ![Yifan\_Liu](https://avatars.discourse-cdn.com/v4/letter/y/4da419/32.png) [@Yifan\_Liu](https://discourse.julialang.org/u/Yifan_Liu)\
**Post date:** [February 4, 2018, 6:37am UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540/4 "2018-02-04T06:37:33Z")

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I tried this function  
`using Interpolations`

```julia
function interp1(xpt, ypt, x; method="linear", extrapvalue=nothing)

    if extrapvalue == nothing
        y = zeros(x)
        idx = trues(x)
    else
        y = extrapvalue*ones(x)
        idx = (x .>= xpt[1]) .& (x .<= xpt[end])
    end
    
    if method == "linear"
        intf = interpolate((xpt,), ypt, Gridded(Linear()))
        y[idx] = intf[x[idx]]

    elseif method == "cubic"
        itp = interpolate(ypt, BSpline(Cubic(Natural())), OnGrid())
        intf = scale(itp, xpt)
        y[idx] = [intf[xi] for xi in x[idx]]
    end
    
    return y
end

interp1(xpt, ypt, x) works fine but
interp1(xpt, ypt, x, method="cubic") got the error message

ERROR: MethodError: no method matching scale(::Interpolations.BSplineInterpolation{Float64,1,Array{Float64,1},Interpolations.BSpline{Interpolations.Cubic{Interpolations.Line}},Interpolations.OnGrid,1}, ::DataArrays.DataArray{Float64,1})

```

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<div class="post-metadata">

**Author:** ![andrewning](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/andrewning/32/3661_2.png) [@andrewning](https://discourse.julialang.org/u/andrewning)\
**Post date:** [February 5, 2018, 5:02pm UTC](https://discourse.julialang.org/t/julia-version-of-matlab-function-interp1/8540/5 "2018-02-05T17:02:27Z")

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This works for me.

```julia
x = 0:pi/4:2*pi
v = sin.(x)
xq = 0:pi/16:1.5*2*pi
vq = interp1(x, v, xq, method="cubic")

```

This is with Interpolations v 0.7.3, Julia v 0.6.2. I just wrote this as a quick example so I’m sure it’s not super robust.