Python create custom NaN

Question

is it possible to create a custom NaN value in Python?

I know a bit about floating point representation and that NaN is a FP value with some special exponent. Therefore, each FP value with that special exponent is a NaN.

To be precise, what I want to do is:

x = make_my_special_NaN(int_seed)
assert math.isnan(x)
assert test_my_special_NaN(x, int_seed)
assert not test_my_special_NaN(float("nan"), int_seed)

The function which I called make_my_special_NaN should take an integer seed which can be used to identify the special NaN value I have (if this is possible).

Having this, it should be possible to test if a given NaN value is the special NaN given that integer seed.

Thanks in advance

Need to know exactly what you need it to work for. numpy?? math(implied by your post) if you need math.isnan looks like it would take monkeypatching and test_my_specail_NaN would be fairly easy. — Phil Cooper
– Phil Cooper, Commented Jul 2, 2012 at 23:26

Ignacio Vazquez-Abrams · Accepted Answer · 2012-07-02 22:54:59Z

1

The f and d format specifiers in struct will allow you to use any bit pattern you like.

answered Jul 2, 2012 at 22:54

Ignacio Vazquez-Abrams

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D.Shawley Over a year ago

more specifically, struct.unpack('!f', '\x7F\x80\x00\x01') will unpack to a float NaN with a significand of 1. Replace the trailing 23 bits with anything but zero and you get a NaN. Same for double except the leading portion is `'\x7F\xF' and the significand is 52 bits.

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