Dan Bishop wrote:
On Apr 26, 6:17 pm, John Henry <[EMAIL PROTECTED]> wrote:
How do I determine is something a function?
For instance, I don't want to relying on exceptions below:
And why not?
def f1():
print "In f1"
def f3():
print "In f3"
def others():
print "In others"
for i in xrange(1,3):
fct = "f%d()"%(i+1)
try:
exec fct
except:
others()
I wish to say:
if value of fct is a funtion, invoke it, otherwise invoke others().
Thanks,
hasattr(fct, '__call__')
And be careful about using the exec statement.
1. In the OP's code, fct is a string, e.g. "f2()", so hasattr(fct,
'__call__') will never return True.
2. Objects other than functions have a __call__ attribute. Some callable
objects don't have a __call__ attribute.
3. Use __doubleunderscore__ things only when you need them and only when
you know what you are doing. There are high-level built-in functions
such as callable and isinstance that can be used for such tests. The
literal answer to the OP's question is:
from types import FunctionType
if isinstance(fct, FunctionType):
do_something()
Something like the following will do the required trick (find callables
in the global namespace whose name matches a pattern) without using the
dreaded exec. Note I've left out the logic to call others; the OP's code
will call others once for each missing function ... let's leave him to
sort out whether that's a good idea or that should be changed to calling
it only once if all functions are missing.
>>> def f1():
... print "in f1"
...
>>> def f3():
... print "in f3"
...
>>> globdict = globals()
>>> globdict # output prettied manually
{'f1': <function f1 at 0x00BA0470>,
'f3': <function f3 at 0x00BA04F0>,
'__builtins__': <module '__builtin__' (built-in)>,
'__name__': '__main__',
'__doc__': None}
>>> for i in xrange(1, 4): ### upper bound 3 probably a bug
... fname = "f%d" % i
... if fname in globdict:
... func = globdict[fname]
... if callable(func):
... func()
...
in f1
in f3
>>>
Cheers,
John
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