Issue
Does anyone know why the below doesn't equal 0?
import numpy as np
np.sin(np.radians(180))
or:
np.sin(np.pi)
When I enter it into python it gives me 1.22e-16.
Solution
The number π
cannot be represented exactly as a floating-point number. So, np.radians(180)
doesn't give you π
, it gives you 3.1415926535897931
.
And sin(3.1415926535897931)
is in fact something like 1.22e-16
.
So, how do you deal with this?
You have to work out, or at least guess at, appropriate absolute and/or relative error bounds, and then instead of x == y
, you write:
abs(y - x) < abs_bounds and abs(y-x) < rel_bounds * y
(This also means that you have to organize your computation so that the relative error is larger relative to y
than to x
. In your case, because y
is the constant 0
, that's trivial—just do it backward.)
Numpy provides a function that does this for you across a whole array, allclose
:
np.allclose(x, y, rel_bounds, abs_bounds)
(This actually checks abs(y - x) < abs_ bounds + rel_bounds * y)
, but that's almost always sufficient, and you can easily reorganize your code when it's not.)
In your case:
np.allclose(0, np.sin(np.radians(180)), rel_bounds, abs_bounds)
So, how do you know what the right bounds are? There's no way to teach you enough error analysis in an SO answer. Propagation of uncertainty at Wikipedia gives a high-level overview. If you really have no clue, you can use the defaults, which are 1e-5
relative and 1e-8
absolute.
Answered By - abarnert
0 comments:
Post a Comment
Note: Only a member of this blog may post a comment.