Hello! I type in sage notebook 5.8 on my local computer:
import numpy as np
np.mgrid?
....
example:
>>> np.mgrid[-1:1:5j]
array([-1. , -0.5, 0. , 0.5, 1. ])But when I type in my notebook:
np.mgrid[-1:1:5j]and press shift-enter, I get mistake: TypeError: Unable to convert -0.200000000000000*I to float; use abs() or real_part() as desired
It seems to me it's the bug...
Sage is interpreting this as complex for some reason. But, np.mgrid[-1:1:0.15] seems to be working for me.
In the notebook, the user could also just evaluate all cells in pure Python, if so desired.
You can bypass the Sage preparser by appending r (for raw) to numerical input.
sage: import numpy as np
sage: np.mgrid[-1:1:5jr]
array([-1. , -0.5, 0. , 0.5, 1. ])Note that some preparsing is still going on with -1 and 1 which are converted to Sage integers by the preparser before going to numpy's mgrid. If you want to feed numpy's mgrid with raw Python integers:
sage: np.mgrid[-1r:1r:5jr]
array([-1. , -0.5, 0. , 0.5, 1. ])It's the same result, except less conversion was done on the way.
On a related note, you can get a Python integer by doing
sage: int(1)but when you do this, the preparsing stage means what is really done is
int(Integer(1))so the Python integer 1 is converted to a Sage integer and back to a Python integer.
Bypassing the Sage preparser with
sage: 1rgets you the Python integer straight away.
Asked: 11 years ago
Seen: 1,718 times
Last updated: Jun 27 '19
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That's the preparser. Well, in Python typically `1j**2` becomes `(-1+0j)`; `j` is a complex number ala electrical engineering. What surprises me is that numpy would use j for some other purpose.
Indeed, in the documentation for `mgrid`: "However, if the step length is a **complex number** (e.g. 5j)," so basically this is a numpy hack, in my opinion.