CODEWITHSUNDEEP
pythonnumpy
Sahim
Sahim

Reputation: 95

How to deal with exponent overflow of 64float precision in python?

I am a newbie in python sorry for the simple question.

In the following code, I want to calculate the exponent and then take the log.

Y=numpy.log(1+numpy.exp(1000))

The problem is that when I take the exponent of 710 or larger numbers the numpy.exp() function returns 'inf' even if I print it with 64float it prints 'inf'.

any help regarding the problem will be appreciated.

Upvotes: 2

Views: 3487

Answers (3)

Robert Kern
Robert Kern

Reputation: 13430

You can use the function np.logaddexp() to do such operations. It computes logaddexp(x1, x2) == log(exp(x1) + exp(x2)) without explicitly computing the intermediate exp() values. This avoids the overflow. Since exp(0.0) == 1, you would compute np.logaddexp(0.0, 1000.0) and get the result of 1000.0, as expected.

Upvotes: 3

Jean-François Fabre
Jean-François Fabre

Reputation: 140168

Check this out:

>>> x = numpy.exp(100)
>>> y = x+1
>>> y==x
True

so even with 100 (which computes all right), adding 1 (or even a very big number), the lowest value is absorbed and has absolutely no effect in the addition. Both values are strictly equal.

Playing with sys.float_info.epsilon I tested that:

>>> numpy.log(1e20+numpy.exp(100))==numpy.log(numpy.exp(100))
True
>>> numpy.log(1e30+numpy.exp(100))==numpy.log(numpy.exp(100))
False

so even a value like 1e20 is absorbed by exp(100) ...

So you would get exactly 1000.0 as your result even if it worked.

Upvotes: 1

Dwaxe
Dwaxe

Reputation: 119

Use the decimal library:

>>> import numpy as np
>>> np.exp(1000)
inf
>>> from decimal import Decimal
>>> x = Decimal(1000)
>>> np.exp(x)
Decimal('1.970071114017046993888879352E+434')

Upvotes: 2

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