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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