[R] approxfun asymmetry

[Duncan Murdoch]

I think you're just seeing undefined behaviour.  In most cases it 
doesn't make sense to interpolate linearly between the points (-Inf, y0) 
and (-1, y1).  In your example with y0 == y1 it's fine, but if you had

   approxfun(x=c(-Inf, 0), y = c(1, 2))

what would you expect the result to be when evaluated at x = -1?  It 
will depend on how you evaluate the linear interpolation formula:

  f(x) = y0 + (x - x0)*(y1 - y0)/(x1 - x0) = 1 + Inf*1/Inf
       = y1 - (x1 - x)*(y1 - y0)/(x1 - x0) = 2 - 1*1/Inf
       = [y0*(x1 - x) + y1*(x - x0)]/(x1 - x0) = [1*1 + 2*Inf]/Inf

The second formula gives 2 (which is a reasonable answer), but the first 
and third are undefined because of the Inf/Inf.

Duncan Murdoch


On 2025-07-04 6:50 a.m., Timothy Earl (Cefas) via R-help wrote:
> Hi,
> 
> I'm seeing an asymmetry in how approxfun treats -Inf and Inf, which I don't understand from reading the help file.
> 
> e.g. a simple step function
> tmp <- approxfun(x=c(-Inf, -0.2, 0.2,  Inf), y=c(-1, -1, 1, 1))
> tmp(c(-1, 0, 1))
> # [1] NaN   0   1
> I expected: # [1] -1   0   1
> 
> Clearly I can work round this by using type=2, and not specifying the end points (or by replacing -Inf by a large negative number), but I was wondering what I was missing about the different behaviour between positive and negative infinity.
> 
> To simplify further:
> approxfun(x=c(-1000,  Inf), y=c(1, 1))(0)
> # [1] 1
> 
> approxfun(x=c(-Inf,  1000), y=c(1, 1))(0)
> # [1] NaN
> 
> Sessioninfo:
> R version 4.3.0 (2023-04-21 ucrt) -- "Already Tomorrow"
> Copyright (C) 2023 The R Foundation for Statistical Computing
> Platform: x86_64-w64-mingw32/x64 (64-bit)
> 
> Thanks,
> 
> Tim
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