Parameterize function name in Python - python-3.x

I have a generic function for which I would like the name to change based on the value of predefined variable. Not sure this is possible in Python:
n = 'abc'
def f{n}(x):
print(x)
f{n}(3)
In the above, the function name will end up becoming "fabc"?

It is possible using a code-generation + exec pattern. It is also possible by instantiating a types.FunctionType instance directly. But a much simpler solution would be to leave the generic name, and then inject an alias (another reference to the same function object) for the dynamic name into whichever namespace you want.
>>> def generic_function(x):
... print(x ** 2)
...
>>> dynamic_name = "func_abc"
>>> globals()[dynamic_name] = generic_function
>>> func_abc(3)
9
To inject in some other module namespace that would be a setattr call:
setattr(other_mod, dynamic_name, generic_function)
You could also rewrite the function's __name__ attribute if you wanted to, but there's probably not much point.

You could do it with something like the below:
def f(n):
exec(f"def f{n}(x): print(x)")
return eval(f"f{n}")

Related

Extract type hints for object attributes in Python [duplicate]

I want to get the type hints for an object's attributes. I can only get the hints for the class and not an instance of it.
I have tried using foo_instance.__class__ from here but that only shows the class variables.
So in the example how do I get the type hint of bar?
class foo:
var: int = 42
def __init__(self):
self.bar: int = 2
print(get_type_hints(foo)) # returns {'var': <class 'int'>}
I just had the same problem. The python doc isn't that clear since the example is made with what is now officially called dataclass.
Student(NamedTuple):
name: Annotated[str, 'some marker']
get_type_hints(Student) == {'name': str}
get_type_hints(Student, include_extras=False) == {'name': str}
get_type_hints(Student, include_extras=True) == {
'name': Annotated[str, 'some marker']
}
It give the impression that get_type_hints() works on class directly. Turns out get_type_hints() returns hints based on functions, not on class. That way it can be use with both if we know that. A normal class obviously not being instantiated at it's declaration, it does not have any of the variables set within the __init__() method who hasn't yet been called. It couldn't be that way either if we want the possibility to get the type hints from class-wide variables.
So you could either call it on __init__(), that is if variables are passed in arguments though (yes i seen it's not in your example but might help others since i didn't seen this anywhere in hours of search);
class foo:
var: int = 42
def __init__(self, bar: int = 2):
self.bar = int
print(get_type_hints(foo.__init__))
At last for your exact example i believe you have two choices. You could instantiate a temporary object and use del to clean it right after if your logic allows it. Or declare your variables as class ones with or without default values so you can get them with get_type_hints() and assign them later in instantiations.
Maybe this is a hack, and you have to be the creator of your instances, but there are a subset of cases in which using a data class will get you what you want;
Python 3.7+
#dataclass
class Foo:
bar: str = 2
if __name__ == '__main__':
f = Foo()
print(f.bar)
print(get_type_hints(f))
2
{'bar': <class 'str'>}
Hints only exist at the class level — by the time an instance is created the type of its attributes will be that of whatever value has been assigned to them. You can get the type of any instance attribute by using the first form of the built-in type() function — e.g. type(foo_instance.var).
This information isn't evaluated and only exists in the source code.
if you must get this information, you can use the ast module and extract the information from the source code yourself, if you have access to the source code.
You should also ask yourself if you need this information because in most cases reevaluating the source code will be to much effort.

Why private __var can change outside Class? (Python 3)

I have the following block:
class Bank:
def __init__(self, b):
self.__bal = b
def main():
myaccount = Bank(2500)
myaccount.__bal = 8000
print(myaccount.__bal)
main()
and PyCharm prints: 8000.
I want to ask how is it possible to change a private var __bal outside the class?
You're not doing what you think you're doing.
Prefixing an attribute with double underscores performs "name mangling". You're just assigning a value to a new attribute. Observe:
class Bank:
def __init__(self, b):
self.__bal = b
def show_bal(self):
print(self.__bal)
And now, in interactive mode:
>>> b = Bank(23)
>>> b.__bal = 42
>>> b.show_bal()
23
Before you assign something to b.__bal you will also notice that accessing b.__bal doesn't work — because it doesn't exist yet. The actual value is still accessible, but its name is "hidden". Nothing prevents you from overriding "private" attributes (a concept that doesn't really exist in Python):
>>> b._Bank__bal = 99
>>> b.show_bal()
99
If you want to protect an attribute from change, the best way to do that is via properties, but even they will only protect the public interface of your class.
Let me tell you one thing about Python 3.
All members in a Python class are public by default. Any member can be accessed from outside the class environment.
Hence you can make changes to the variable. For more information about Python 3 class access modifiers, go here. Hope it clarifies your doubt.

How can I change the value of variable in a exec() function?

This is the code:
def test():
c = 0
exec('c = 4 + 5')
print('result', c)
The value of c is 0. What can I do to fix it, let the value of c changed in exec() function?
In cpython, you can't do what you want inside a function. That's because a function's local namespace can't be easily edited by other code (editing the dictionary returned by locals() doesn't do it). If instead you call exec from the top level of your module, it will work as you want, but it is still probably a bad idea.
A better approach may be to use a namespace dictionary. You can pass it to exec and the change will appear there, rather than in the function's local namespace or the module's global namespace:
def test():
namespace = {'c': 0}
exec('c = 4 + 5', namespace)
print('result', namespace['c'])

How do I set an instance attribute in Python when the instance is determined by a function?

I would like to iterate through a selection of class instances and set a member variable equal to a value. I can access the members value with:
for foo in range(1,4): #class members: pv1, pv2, pv3
bar[foo] ='{0}'.format(locals()['pv' + str(foo)+'.data'])
However when I try to set/mutate the values like so:
for foo in range(1,4): #class members:
'{0}'.format(locals()['pv' + str(foo)+'.data']) = bar[foo]
I obviously get the error:
SyntaxError: can't assign to function call
I have tried a few methods to get it done with no success. I am using many more instances than 3 in my actual code(about 250), but my question is hopefully clear. I have looked at several stack overflow questions, such as Automatically setting class member variables in Python -and- dynamically set an instance property / memoized attribute in python? Yet none seem to answer this question. In C++ I would just use a pointer as an intermediary. What's the Pythonic way to do this?
An attr is a valid assignment target, even if it's an attr of the result of an expression.
for foo in range(1,3):
locals()['pv' + str(foo)].data = bar[foo]
Another developer wrote a few lines about setattr(), mostly about how it should be avoided.
setattr is unnecessary unless the attribute name is dynamic.
But they didn't say why. Do you mind elaborating why you switched your answer away from setattr()?
In this case, the attr is data, which never changes, so while
for i in range(1, 3):
setattr(locals()['pv' + str(i)], 'data', bar[i])
does the same thing, setattr isn't required here. The .data = form is both good enough and typically preferred--it's faster and has clearer intent--which is why I changed it. On the other hand, if you needed to change the attr name every loop, you'd need it, e.g.
for i in range(1,3):
setattr(locals()['pv' + str(i)], 'data' + str(i), bar[i])
The above code sets attrs named data1, data2, data3, unrolled, it's equivalent to
pv1.data1 = bar[1]
pv2.data2 = bar[2]
pv3.data3 = bar[3]
I originally thought your question needed to do something like this, which is why I used setattr in the first place. Once I tested it and got it working I just posted it without noticing that the setattr was no longer required.
If the attr name changes at runtime like that (what the other developer meant by "dynamic") then you can't use the dot syntax, since you have a string object rather than a static identifier. Another reason to use setattr might be if you need a side effect in an expression. Unlike in C, assignments are statements in Python. But function calls like setattr are expressions.
Here is an example of creating a class which explicitly allows access through index or attribute calls to change internal variables. This is not generally promoted as 'good programming' though. It does not explicitly define the rules by which people should be expected to interact with the underlying variables.
the definition of __getattr__() function allows for the assignment of (object).a .
the definition of __getitem__() function allows for the assignment of
(object)['b']
class Foo(object):
def __init__(self, a=None,b=None,c=None):
self.a=a
self.b=b
self.c=c
def __getattr__(self, x):
return self.__dict__.get(x, None)
def __getitem__(self, x):
return self.__dict__[x]
print
f1 = Foo(3,2,4)
print 'f1=', f1.a, f1['b'], f1['c']
f2 = Foo(4,6,2)
print 'f2=', f2.a, f2['b'], f2['c']
f3 = Foo(3,5,7)
print 'f3=', f3.a, f3['b'], f3['c']
for x in range(1, 4):
print 'now setting f'+str(x)
locals()['f'+str(x)].a=1
locals()['f'+str(x)].b=1
locals()['f'+str(x)].c=1
print
print 'f1=', f1.a, f1['b'], f1['c']
print 'f2=', f2.a, f2['b'], f2['c']
print 'f3=', f3.a, f3['b'], f3['c']
The result is
f1= 3 2 4
f2= 4 6 2
f3= 3 5 7
now setting f1
now setting f2
now setting f3
f1= 1 1 1
f2= 1 1 1
f3= 1 1 1

Access element of list by variable name

How can I access a list element using the name of the list?
I would like to allow a user to edit the code in determine a single variable to be inputted into a function. For example:
blah = [1,2]
blah2 = 5
toBeChanged = "blah2"
def foo():
print(blah)
def changeVariable():
globals()[toBeChanged] += 1
for time in range(5):
changeVariable()
simulate
This works for blah2 since it is a simple variable, however it will not work for blah[0] since it is part of a list. I've also tried placing my variables into a dictionary as other answers have suggested, but I still am unable to change list elements through a simple string.
Is there a way to do this that I am missing? Thanks!
Rather than using globals() and altering directly it would be much, much better to use a dictionary to store the variables you want the user to alter, and then manipulate that:
my_variables = {
'blah': [1,2]
'blah2': 5
}
toBeChanged = "blah2"
def foo():
print(my_variables['blah'])
def changeVariable():
my_variables[toBeChanged] = my_variables.get(toBeChanged,0) + 1
for time in range(5):
changeVariable()
This has the added advantage that if a user enters a variable that doesn't exist a default is chosen, and doesn't override any variables that might be important for future execution.

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