Interpolate with t=Infinity returns NaN instead of Infinity
severo opened this issue · 6 comments
d3.interpolateNumber(0,1)(Infinity)
returns NaN
, while I would expect it to return Infinity
.
Note that replacing
Line 3 in e8cddfd
with
return (b - a) * t + a;
would generate a new NaN
issue that is not a problem with the current version:
d3.interpolateNumber(Infinity, Infinity)(0.5)
Two possible implementations as discussed on https://d3js.zulipchat.com/#narrow/stream/273725-D3/topic/d3-scale
isFinite(t) ? a * (1 - t) + b * t : (b - a) * t + a
and
(t * t !== Infinity) ? a * (1 - t) + b * t : (b - a) * t + a
It’s doing exactly what it’s documented to do, so I’m not sure I consider this a bug. 0 * Infinity is NaN. Why is it useful to support Infinity?
I found the case while wanting to transform a domain extent (that accepts -Infinity and Infinity as valid) to a range extent, something like:
[-Infinity, Infinity].map(d3.scaleLinear())
and I expected having [-Infinity, Infinity] as the output, but got [NaN, NaN]
instead.
It's not difficult to test for this case in my application, but I was wondering if it was the "normal" behavior. If it is, OK for me to close the issue.
For the record, another possible implementation by Jacob Rus:
lerp = (a, b, t) => {
var y = a * (1 - t) + b * t;
return (y === y) ? y : (b - a) * t
}
It’s doing exactly what it’s documented to do, so I’m not sure I consider this a bug.
0 * Infinity
isNaN
. Why is it useful to supportInfinity
?
This scale is a way of defining a certain linear function f. In the event that the slope is not 0, f(∞) should theoretically be ±∞, not undefined.
(Whether it is important or useful to support this edge case is debatable.)