On 2013-02-12 13:27, Dave Angel wrote:
On 02/12/2013 06:46 AM, Helmut Jarausch wrote:
Hi,
I've tried but didn't find an answer on the net.
The exec function in Python modifies a copy of locals() only.
How can I transfer changes in that local copy to the locals of my function
** without ** knowing the names of these variables.
E.g. I have a lot of local names.
Doing
_locals= locals()
This doesn't copy everything. But perhaps you know that and you're just
testing us.
expr=compile(input('statements assigning to some local variables '),
'user input','exec')
exec(expr,globals(),_locals)
How can I "copy" the new values within _locals to my current locals.
If I knew that Var1 has changed I could say
Var1 = _locals['Var1'] but what to do in general?
locals()["Var1"] = _locals["Var1"] will set the same Var1 local.
So you might write a loop on _locals.
But beware if someone has deleted one of the "variables" it may not do
what you'd like. You cannot necessarily add back a local with the above
syntax.
The docs for locals() warns """The contents of this dictionary should
not be modified; changes may not affect the values of local and free
variables used by the interpreter."""
That's because local variables in a function are implemented (at least
in CPython) using 'slots' for efficiency reasons.
>>> def example():
a = 1
print("a is {}, locals() is {}".format(a, locals()))
print("Changing a to 2 directly")
a = 2
print("a is {}, locals() is {}".format(a, locals()))
print("Changing a to 3 via locals()")
locals()["a"] = 3
print("a is {}, locals() is {}".format(a, locals()))
locals()["b"] = 0
print("locals() is {}".format(locals()))
print("b is {}".format(b))
>>> example()
a is 1, locals() is {'a': 1}
Changing a to 2 directly
a is 2, locals() is {'a': 2}
Changing a to 3 via locals()
a is 2, locals() is {'a': 2}
locals() is {'b': 0, 'a': 2}
Traceback (most recent call last):
File "<pyshell#39>", line 1, in <module>
example()
File "<pyshell#38>", line 12, in example
print("b is {}".format(b))
NameError: global name 'b' is not defined
Note how attempting to change variable "a" in locals() is ignored, and
how adding variable "b" to locals() doesn't actually make it a local
variable (it raises a NameError).
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