Purpose of `numpy.log1p( )`?

The NumPy docs give a hint:

For real-valued input, log1p is accurate also for x so small that 1 + x == 1 in floating-point accuracy.

So for example let’s add a tiny non-zero number and 1.0. Rounding errors make it a 1.0.

>>> 1e-100 == 0.0
False
>>> 1e-100 + 1.0 == 1.0
True

If we try to take the log of that incorrect sum, we get an incorrect result (compare to WolframAlpha):

>>> np.log(1e-100 + 1)
0.0

But if we use log1p(), we get the correct result

>>> np.log1p(1e-100)
1e-100

The same principle holds for exp1m() and logaddexp(): The’re more accurate for small x.

Leave a Comment