Comments (3)
Ok. I'm looking at that. Note that 1.5d0 is not valid Python syntax: https://docs.python.org/2/reference/lexical_analysis.html. I think you are trying to use Lisp syntax there.
So, you can work around it with something like:
(let (($my-num 1.5d0))
[numpy.sin($my-num)])
Anyway, I'm implementing Python literal float syntax now.
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I've implemented Python float syntax.
You can try things like these now:
BURGLED-BATTERIES> [^numpy.sin(1.5)]
0.9974949866040544d0
BURGLED-BATTERIES> [^numpy.sin(1.5e2)]
-0.7148764296291646d0
BURGLED-BATTERIES> [^numpy.sin(-1.5e2)]
0.7148764296291646d0
BURGLED-BATTERIES> [^numpy.sin(-1.5-e3)]
; Evaluation aborted on Incomplete parse..
BURGLED-BATTERIES> [^numpy.sin(-1.5e-3)]
-0.0014999994505385647d0
Please test.
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That was amazingly fast!
Thanks a lot, works perfectly!
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