On 12/29/19 10:39 PM, David Mertz wrote:
On Sun, Dec 29, 2019 at 10:18 PM Richard Damon
<rich...@damon-family.org <mailto:rich...@damon-family.org>> wrote:
IEEE total_order puts NaN as bigger than infinity, and -NaN as
less than -inf.
You mean like this?
>>> def total_order(x):
... if math.isnan(x):
... return (math.copysign(1, x), x)
... return (0, x)
...
...
>>> nums = [1, 2, float('-inf'), float('nan'), float('inf'), float('-nan')]
>>> nums
[1, 2, -inf, nan, inf, nan]
>>> sorted(nums, key=total_order)
[nan, -inf, 1, 2, inf, nan]
It's a little weird that -nan has a repr of 'nan', but bracketing
that, my implementation is EXACTLY what you describe.
The difference probably is in places that Python doesn't show, and maybe
isn't that important. (like all nans regardless of sign are just NaN)
In the IEEE total order, +0 and -0 are distinct, which your order
doesn't handle,
For NaNs, the issue is that NaN is NOT a single representation, but each
combination of the sign bit and the 52 mantissa bits (except all zeros
which signify infinity) when the exponent field is all ones is a
different NaN (and the MSB of the mantissa indicates quiet or
signalling). you code makes all these values unordered with respect to
each other as opposed to having a distinct strict order. That probably
isn't that important of a distinction.
Thus your total_order, while not REALLY a total order, is likely good
enough for most purposes. Unless the code has done something to create
specific NaNs, and then does something to recover that information, the
fact that the NaNs aren't properly sorted within themselves probably
isn't an issue. There is a possible issue that depending on the exact
sort algorithm, if the data includes two (or more) NaNs with the same
sign, then their lack of order might give the sort difficulties.
--
Richard Damon
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