As fuglede says, the issue here is that np.float64
can't handle a number as large as exp(1234.1)
. Try using np.float128
instead:
>>> cc = np.array([[0.120,0.34,-1234.1]], dtype=np.float128)
>>> cc
array([[ 0.12, 0.34, -1234.1]], dtype=float128)
>>> 1 / (1 + np.exp(-cc))
array([[ 0.52996405, 0.58419052, 1.0893812e-536]], dtype=float128)
Note however, that there are certain quirks with using extended precision. It may not work on Windows; you don't actually get the full 128 bits of precision; and you might lose the precision whenever the number passes through pure python. You can read more about the details here.
For most practical purposes, you can probably approximate 1 / (1 + <a large number>)
to zero. That is to say, just ignore the warning and move on. Numpy takes care of the approximation for you (when using np.float64
):
>>> 1 / (1 + np.exp(-cc))
/usr/local/bin/ipython3:1: RuntimeWarning: overflow encountered in exp
#!/usr/local/bin/python3.4
array([[ 0.52996405, 0.58419052, 0. ]])
If you want to suppress the warning, you could use scipy.special.expit
, as suggested by WarrenWeckesser in a comment to the question:
>>> from scipy.special import expit
>>> expit(cc)
array([[ 0.52996405, 0.58419052, 0. ]])
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