Comment #4 on issue 1902 by alberthilbert: 'has' method fails when argument
is 'Symbol' or 'Wild'
http://code.google.com/p/sympy/issues/detail?id=1902
From what I can see from source code, 'has' method accepts 'BasicType'
object as
parameter and return 'True' if 'self' contains some instance of that
type, 'False' if
doesn't.
Some working examples:
x, y, z = symbols('xyz')
f = Function('f')
(f(x)).has(Function)
True
(f(x)).has(Piecewise)
False
(f(x)).has(Pow)
False
(x + y).has(Add)
True
(x*y + 1).has(Mul)
True
I think there is no reason why this should not work when you pass 'Symbol'
or 'Wild'
as parameter, so to have:
(x + y).has(Wild)
False
(x**y + 1).has(Symbol)
True
p = Wild('p')
(x + p).has(Wild)
True
'atoms' is supposed to have a completely different behaviour, it list all
the
instances of a certain type present in an expression.
But in a lot of circumstances, you need only to know if at least one of
them is present.
In those cases you can use something like:
bool(expr.atoms(Symbol))
but this is by far less elegant than:
expr.has(Symbol)
Moreover, it is a waste of resources, walking the entire expression tree to
collect
all the informations needed by 'atoms' when you can stop at the first
occurrence of a
specific object.
To be honest, as now, there is not such resource advantage, because of the
implementation of 'has' that use 'bool(self.atoms(type))' to calculate its
answer,
but that may change in future... ;)
Raffaele
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