Updates:
Labels: -NeedsBetterPatch NeedsReview
Comment #38 on issue 1923 by smi...@gmail.com: count_ops doesn't return a
count (by default)
http://code.google.com/p/sympy/issues/detail?id=1923
How would people feel about making the non-visual count_ops just return a
slightly modified tree length? This is much simpler (and about 3X faster)
than trying to get something mathematically correct but still serves the
purpose of providing a metric for comparing expressions.
The visual option tries to give something that is visually correlated to
what you would see when printing an expression:
>>> count_ops(I-1, visual=1) # I PLUS neg. one TIMES 1
Add + Mul
The existing non-visual count directly correlates to that:
>>> count_ops(I-1)
2
But since such a number would likely be used for comparison purposes, the
tree length is quicker to compute
>>> from sympy.utilities.iterables import preorder_traversal as tree
>>> list(tree(I-1))
[-1 + I, -1, I]
>>> count_ops(I-1, quick=1)
4
Note the 4 because of the negative 1. If there is a +1 that has a shorter
op count:
>>> count_ops(I+1, quick=1)
3
And if the 1 is replaced with 1/2 the count goes up:
>>> count_ops(I-S.Half, quick=1)
5
These corrections for Rationals are the only correction to the tree length
made by the quick calculation.
Right now in [ https://github.com/sympy/sympy/pull/23 ] I've added a
`quick` flag to allow the tree length calculation (which is about 3X
faster). My preference is to just make the non-visual return the modified
tree length.
Does anyone else have a preference?
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