How would I call a function by using a dictionary? - python

I recently found some code that calls a function from a different directory like this:
modules["module_name"].function(parameters)
Then I had a thought about calling a function (but not a module; I already know how to do that) through lists in that format like so:
[function_name](parameters)

You can't call a function like that in a list. I don't know what that syntax you have there. t's not possible, but you can call method references from dicts or variables below
If you're function is stored in a dictionary like so
def add(x, y):
return x + y
some_dict = {
'add' : add
}
Then you can call it like so
some_dict['add'](10, 10)

Calling a function using list... you can have a list of functions and call it out from index
def foo(arg):
print(arg)
list_of_functions = [foo]
list_of_functions[0](3)
This also works with dictionaries, since functions are just pointers, as anything else.
But i believe you wanted to find the function by the name from variable, in that case, probably (idk, never went this route) use globals for that one
def foo(arg):
print(arg)
function_name = 'foo'
globals()[function_name](3)
P.S. I still wonder about 2.7 though

If the list is created in other module :
module.py it should looks like this:
def add(a,b):
return a+b
lista = [add]
And you could call the fuction like this in your own module:
from module import lista
res = lista[0](2,3)
print(res)
The important here is note that the function should be define before to list it and in the same module where the list is created. In other way your code will be broken.
Don't ask why but I have python 2.7 too

Related

Function prototype in python? [duplicate]

Is it possible to forward-declare a function in Python? I want to sort a list using my own cmp function before it is declared.
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
I've put the definition of cmp_configs method after the invocation. It fails with this error:
NameError: name 'cmp_configs' is not defined
Is there any way to "declare" cmp_configs method before it's used?
Sometimes, it is difficult to reorganize code to avoid this problem. For instance, when implementing some forms of recursion:
def spam():
if end_condition():
return end_result()
else:
return eggs()
def eggs():
if end_condition():
return end_result()
else:
return spam()
Where end_condition and end_result have been previously defined.
Is the only solution to reorganize the code and always put definitions before invocations?
Wrap the invocation into a function of its own so that
foo()
def foo():
print "Hi!"
will break, but
def bar():
foo()
def foo():
print "Hi!"
bar()
will work properly.
The general rule in Python is that a function should be defined before its usage, which does not necessarily mean it needs to be higher in the code.
If you kick-start your script through the following:
if __name__=="__main__":
main()
then you probably do not have to worry about things like "forward declaration". You see, the interpreter would go loading up all your functions and then start your main() function. Of course, make sure you have all the imports correct too ;-)
Come to think of it, I've never heard such a thing as "forward declaration" in python... but then again, I might be wrong ;-)
If you don't want to define a function before it's used, and defining it afterwards is impossible, what about defining it in some other module?
Technically you still define it first, but it's clean.
You could create a recursion like the following:
def foo():
bar()
def bar():
foo()
Python's functions are anonymous just like values are anonymous, yet they can be bound to a name.
In the above code, foo() does not call a function with the name foo, it calls a function that happens to be bound to the name foo at the point the call is made. It is possible to redefine foo somewhere else, and bar would then call the new function.
Your problem cannot be solved because it's like asking to get a variable which has not been declared.
I apologize for reviving this thread, but there was a strategy not discussed here which may be applicable.
Using reflection it is possible to do something akin to forward declaration. For instance lets say you have a section of code that looks like this:
# We want to call a function called 'foo', but it hasn't been defined yet.
function_name = 'foo'
# Calling at this point would produce an error
# Here is the definition
def foo():
bar()
# Note that at this point the function is defined
# Time for some reflection...
globals()[function_name]()
So in this way we have determined what function we want to call before it is actually defined, effectively a forward declaration. In python the statement globals()[function_name]() is the same as foo() if function_name = 'foo' for the reasons discussed above, since python must lookup each function before calling it. If one were to use the timeit module to see how these two statements compare, they have the exact same computational cost.
Of course the example here is very useless, but if one were to have a complex structure which needed to execute a function, but must be declared before (or structurally it makes little sense to have it afterwards), one can just store a string and try to call the function later.
If the call to cmp_configs is inside its own function definition, you should be fine. I'll give an example.
def a():
b() # b() hasn't been defined yet, but that's fine because at this point, we're not
# actually calling it. We're just defining what should happen when a() is called.
a() # This call fails, because b() hasn't been defined yet,
# and thus trying to run a() fails.
def b():
print "hi"
a() # This call succeeds because everything has been defined.
In general, putting your code inside functions (such as main()) will resolve your problem; just call main() at the end of the file.
There is no such thing in python like forward declaration. You just have to make sure that your function is declared before it is needed.
Note that the body of a function isn't interpreted until the function is executed.
Consider the following example:
def a():
b() # won't be resolved until a is invoked.
def b():
print "hello"
a() # here b is already defined so this line won't fail.
You can think that a body of a function is just another script that will be interpreted once you call the function.
Sometimes an algorithm is easiest to understand top-down, starting with the overall structure and drilling down into the details.
You can do so without forward declarations:
def main():
make_omelet()
eat()
def make_omelet():
break_eggs()
whisk()
fry()
def break_eggs():
for egg in carton:
break(egg)
# ...
main()
# declare a fake function (prototype) with no body
def foo(): pass
def bar():
# use the prototype however you see fit
print(foo(), "world!")
# define the actual function (overwriting the prototype)
def foo():
return "Hello,"
bar()
Output:
Hello, world!
No, I don't believe there is any way to forward-declare a function in Python.
Imagine you are the Python interpreter. When you get to the line
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
either you know what cmp_configs is or you don't. In order to proceed, you have to
know cmp_configs. It doesn't matter if there is recursion.
You can't forward-declare a function in Python. If you have logic executing before you've defined functions, you've probably got a problem anyways. Put your action in an if __name__ == '__main__' at the end of your script (by executing a function you name "main" if it's non-trivial) and your code will be more modular and you'll be able to use it as a module if you ever need to.
Also, replace that list comprehension with a generator express (i.e., print "\n".join(str(bla) for bla in sorted(mylist, cmp=cmp_configs)))
Also, don't use cmp, which is deprecated. Use key and provide a less-than function.
Import the file itself. Assuming the file is called test.py:
import test
if __name__=='__main__':
test.func()
else:
def func():
print('Func worked')
TL;DR: Python does not need forward declarations. Simply put your function calls inside function def definitions, and you'll be fine.
def foo(count):
print("foo "+str(count))
if(count>0):
bar(count-1)
def bar(count):
print("bar "+str(count))
if(count>0):
foo(count-1)
foo(3)
print("Finished.")
recursive function definitions, perfectly successfully gives:
foo 3
bar 2
foo 1
bar 0
Finished.
However,
bug(13)
def bug(count):
print("bug never runs "+str(count))
print("Does not print this.")
breaks at the top-level invocation of a function that hasn't been defined yet, and gives:
Traceback (most recent call last):
File "./test1.py", line 1, in <module>
bug(13)
NameError: name 'bug' is not defined
Python is an interpreted language, like Lisp. It has no type checking, only run-time function invocations, which succeed if the function name has been bound and fail if it's unbound.
Critically, a function def definition does not execute any of the funcalls inside its lines, it simply declares what the function body is going to consist of. Again, it doesn't even do type checking. So we can do this:
def uncalled():
wild_eyed_undefined_function()
print("I'm not invoked!")
print("Only run this one line.")
and it runs perfectly fine (!), with output
Only run this one line.
The key is the difference between definitions and invocations.
The interpreter executes everything that comes in at the top level, which means it tries to invoke it. If it's not inside a definition.
Your code is running into trouble because you attempted to invoke a function, at the top level in this case, before it was bound.
The solution is to put your non-top-level function invocations inside a function definition, then call that function sometime much later.
The business about "if __ main __" is an idiom based on this principle, but you have to understand why, instead of simply blindly following it.
There are certainly much more advanced topics concerning lambda functions and rebinding function names dynamically, but these are not what the OP was asking for. In addition, they can be solved using these same principles: (1) defs define a function, they do not invoke their lines; (2) you get in trouble when you invoke a function symbol that's unbound.
Python does not support forward declarations, but common workaround for this is use of the the following condition at the end of your script/code:
if __name__ == '__main__': main()
With this it will read entire file first and then evaluate condition and call main() function which will be able to call any forward declared function as it already read the entire file first. This condition leverages special variable __name__ which returns __main__ value whenever we run Python code from current file (when code was imported as a module, then __name__ returns module name).
"just reorganize my code so that I don't have this problem." Correct. Easy to do. Always works.
You can always provide the function prior to it's reference.
"However, there are cases when this is probably unavoidable, for instance when implementing some forms of recursion"
Can't see how that's even remotely possible. Please provide an example of a place where you cannot define the function prior to it's use.
Now wait a minute. When your module reaches the print statement in your example, before cmp_configs has been defined, what exactly is it that you expect it to do?
If your posting of a question using print is really trying to represent something like this:
fn = lambda mylist:"\n".join([str(bla)
for bla in sorted(mylist, cmp = cmp_configs)])
then there is no requirement to define cmp_configs before executing this statement, just define it later in the code and all will be well.
Now if you are trying to reference cmp_configs as a default value of an argument to the lambda, then this is a different story:
fn = lambda mylist,cmp_configs=cmp_configs : \
"\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
Now you need a cmp_configs variable defined before you reach this line.
[EDIT - this next part turns out not to be correct, since the default argument value will get assigned when the function is compiled, and that value will be used even if you change the value of cmp_configs later.]
Fortunately, Python being so type-accommodating as it is, does not care what you define as cmp_configs, so you could just preface with this statement:
cmp_configs = None
And the compiler will be happy. Just be sure to declare the real cmp_configs before you ever invoke fn.
Python technically has support for forward declaration.
if you define a function/class then set the body to pass, it will have an empty entry in the global table.
you can then "redefine" the function/class later on to implement the function/class.
unlike c/c++ forward declaration though, this does not work from outside the scope (i.e. another file) as they have their own "global" namespace
example:
def foo(): pass
foo()
def foo(): print("FOOOOO")
foo()
foo is declared both times
however the first time foo is called it does not do anything as the body is just pass
but the second time foo is called. it executes the new body of print("FOOOOO")
but again. note that this does not fix circular dependancies. this is because files have their own name and have their own definitions of functions
example 2:
class bar: pass
print(bar)
this prints <class '__main__.bar'> but if it was declared in another file it would be <class 'otherfile.foo'>
i know this post is old, but i though that this answer would be useful to anyone who keeps finding this post after the many years it has been posted for
One way is to create a handler function. Define the handler early on, and put the handler below all the methods you need to call.
Then when you invoke the handler method to call your functions, they will always be available.
The handler could take an argument nameOfMethodToCall. Then uses a bunch of if statements to call the right method.
This would solve your issue.
def foo():
print("foo")
#take input
nextAction=input('What would you like to do next?:')
return nextAction
def bar():
print("bar")
nextAction=input('What would you like to do next?:')
return nextAction
def handler(action):
if(action=="foo"):
nextAction = foo()
elif(action=="bar"):
nextAction = bar()
else:
print("You entered invalid input, defaulting to bar")
nextAction = "bar"
return nextAction
nextAction=input('What would you like to do next?:')
while 1:
nextAction = handler(nextAction)

Passing Python functions from YAML file

I have a YAML script where the users can define some parameters. They can also state in the YAML file a custom function they want to apply. For example:
processes:
args1: 10
args2: 5
funct: np.sum
This YAML will be passed to a function like this:
def custom_func(args1, args2, funct):
return funct(args1, args2)
With the YAML example above, I expect custom_func() to execute np.sum(10,5). How can I make the data in YAML file to be parsed as Callable? eval() probably does the job but it might have security issue. Is there any proper way? Thanks!
You can also load the function given its Python dotted path.
However, this only gets the function and calls it, it does not try to figure out calling signature, which is np.sum([5,10]), not np.sum(10,5)
from yaml import safe_load as yload
from importlib import import_module
def load_func(dotpath : str):
""" load function in module. function is right-most segment """
module_, func = dotpath.rsplit(".", maxsplit=1)
m = import_module(module_)
return getattr(m, func)
yaml = """
processes:
args1: 10
args2: 5
#gotta give real package name, not `np`
funct: numpy.sum
"""
di = yload(yaml)
pr = di["processes"]
func = load_func(pr["funct"])
array = [pr["args1"], pr["args2"]]
print (f"{func=} {func(array)=}")
output:
func=<function sum at 0x106645e50> func(array)=15
Yes, eval and exec are both bad practices. Good that you know that :D
You can use dictionaries to do this.
What you can do is define a dictionary and make the key the funct value in YAML. The value will be an uncalled function. This is similar to a variable.
funct_parser = {
"np.sum": np.sum # Notice how this doesn't have ()
}
You can then use funct against the dictionary to call the proper method. Something like the following:
def custom_func(args1, args2, funct):
my_func = funct_parser[str(funct)] # This will cause my_func to be a copy of `np.sum`.
return my_func(args1, args2) # Call my_func
BTW, you don't really have to do custom_func on multiple lines, I just did so to explain it better.
Alternative choice:
If you don't feel like hardcoding every single part in a dictionary, you can use getattr().
# This code assumes you have numpy imported as np (You can always change it)
def custom_func(args1, args2, funct):
funct = funct.split("np.")[0] # Will work for np. You can always change to other modules when needed.
return getattr(np, funct)

General Question about function and returning variable - python [duplicate]

Is it possible to forward-declare a function in Python? I want to sort a list using my own cmp function before it is declared.
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
I've put the definition of cmp_configs method after the invocation. It fails with this error:
NameError: name 'cmp_configs' is not defined
Is there any way to "declare" cmp_configs method before it's used?
Sometimes, it is difficult to reorganize code to avoid this problem. For instance, when implementing some forms of recursion:
def spam():
if end_condition():
return end_result()
else:
return eggs()
def eggs():
if end_condition():
return end_result()
else:
return spam()
Where end_condition and end_result have been previously defined.
Is the only solution to reorganize the code and always put definitions before invocations?
Wrap the invocation into a function of its own so that
foo()
def foo():
print "Hi!"
will break, but
def bar():
foo()
def foo():
print "Hi!"
bar()
will work properly.
The general rule in Python is that a function should be defined before its usage, which does not necessarily mean it needs to be higher in the code.
If you kick-start your script through the following:
if __name__=="__main__":
main()
then you probably do not have to worry about things like "forward declaration". You see, the interpreter would go loading up all your functions and then start your main() function. Of course, make sure you have all the imports correct too ;-)
Come to think of it, I've never heard such a thing as "forward declaration" in python... but then again, I might be wrong ;-)
If you don't want to define a function before it's used, and defining it afterwards is impossible, what about defining it in some other module?
Technically you still define it first, but it's clean.
You could create a recursion like the following:
def foo():
bar()
def bar():
foo()
Python's functions are anonymous just like values are anonymous, yet they can be bound to a name.
In the above code, foo() does not call a function with the name foo, it calls a function that happens to be bound to the name foo at the point the call is made. It is possible to redefine foo somewhere else, and bar would then call the new function.
Your problem cannot be solved because it's like asking to get a variable which has not been declared.
I apologize for reviving this thread, but there was a strategy not discussed here which may be applicable.
Using reflection it is possible to do something akin to forward declaration. For instance lets say you have a section of code that looks like this:
# We want to call a function called 'foo', but it hasn't been defined yet.
function_name = 'foo'
# Calling at this point would produce an error
# Here is the definition
def foo():
bar()
# Note that at this point the function is defined
# Time for some reflection...
globals()[function_name]()
So in this way we have determined what function we want to call before it is actually defined, effectively a forward declaration. In python the statement globals()[function_name]() is the same as foo() if function_name = 'foo' for the reasons discussed above, since python must lookup each function before calling it. If one were to use the timeit module to see how these two statements compare, they have the exact same computational cost.
Of course the example here is very useless, but if one were to have a complex structure which needed to execute a function, but must be declared before (or structurally it makes little sense to have it afterwards), one can just store a string and try to call the function later.
If the call to cmp_configs is inside its own function definition, you should be fine. I'll give an example.
def a():
b() # b() hasn't been defined yet, but that's fine because at this point, we're not
# actually calling it. We're just defining what should happen when a() is called.
a() # This call fails, because b() hasn't been defined yet,
# and thus trying to run a() fails.
def b():
print "hi"
a() # This call succeeds because everything has been defined.
In general, putting your code inside functions (such as main()) will resolve your problem; just call main() at the end of the file.
There is no such thing in python like forward declaration. You just have to make sure that your function is declared before it is needed.
Note that the body of a function isn't interpreted until the function is executed.
Consider the following example:
def a():
b() # won't be resolved until a is invoked.
def b():
print "hello"
a() # here b is already defined so this line won't fail.
You can think that a body of a function is just another script that will be interpreted once you call the function.
Sometimes an algorithm is easiest to understand top-down, starting with the overall structure and drilling down into the details.
You can do so without forward declarations:
def main():
make_omelet()
eat()
def make_omelet():
break_eggs()
whisk()
fry()
def break_eggs():
for egg in carton:
break(egg)
# ...
main()
# declare a fake function (prototype) with no body
def foo(): pass
def bar():
# use the prototype however you see fit
print(foo(), "world!")
# define the actual function (overwriting the prototype)
def foo():
return "Hello,"
bar()
Output:
Hello, world!
No, I don't believe there is any way to forward-declare a function in Python.
Imagine you are the Python interpreter. When you get to the line
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
either you know what cmp_configs is or you don't. In order to proceed, you have to
know cmp_configs. It doesn't matter if there is recursion.
You can't forward-declare a function in Python. If you have logic executing before you've defined functions, you've probably got a problem anyways. Put your action in an if __name__ == '__main__' at the end of your script (by executing a function you name "main" if it's non-trivial) and your code will be more modular and you'll be able to use it as a module if you ever need to.
Also, replace that list comprehension with a generator express (i.e., print "\n".join(str(bla) for bla in sorted(mylist, cmp=cmp_configs)))
Also, don't use cmp, which is deprecated. Use key and provide a less-than function.
Import the file itself. Assuming the file is called test.py:
import test
if __name__=='__main__':
test.func()
else:
def func():
print('Func worked')
TL;DR: Python does not need forward declarations. Simply put your function calls inside function def definitions, and you'll be fine.
def foo(count):
print("foo "+str(count))
if(count>0):
bar(count-1)
def bar(count):
print("bar "+str(count))
if(count>0):
foo(count-1)
foo(3)
print("Finished.")
recursive function definitions, perfectly successfully gives:
foo 3
bar 2
foo 1
bar 0
Finished.
However,
bug(13)
def bug(count):
print("bug never runs "+str(count))
print("Does not print this.")
breaks at the top-level invocation of a function that hasn't been defined yet, and gives:
Traceback (most recent call last):
File "./test1.py", line 1, in <module>
bug(13)
NameError: name 'bug' is not defined
Python is an interpreted language, like Lisp. It has no type checking, only run-time function invocations, which succeed if the function name has been bound and fail if it's unbound.
Critically, a function def definition does not execute any of the funcalls inside its lines, it simply declares what the function body is going to consist of. Again, it doesn't even do type checking. So we can do this:
def uncalled():
wild_eyed_undefined_function()
print("I'm not invoked!")
print("Only run this one line.")
and it runs perfectly fine (!), with output
Only run this one line.
The key is the difference between definitions and invocations.
The interpreter executes everything that comes in at the top level, which means it tries to invoke it. If it's not inside a definition.
Your code is running into trouble because you attempted to invoke a function, at the top level in this case, before it was bound.
The solution is to put your non-top-level function invocations inside a function definition, then call that function sometime much later.
The business about "if __ main __" is an idiom based on this principle, but you have to understand why, instead of simply blindly following it.
There are certainly much more advanced topics concerning lambda functions and rebinding function names dynamically, but these are not what the OP was asking for. In addition, they can be solved using these same principles: (1) defs define a function, they do not invoke their lines; (2) you get in trouble when you invoke a function symbol that's unbound.
Python does not support forward declarations, but common workaround for this is use of the the following condition at the end of your script/code:
if __name__ == '__main__': main()
With this it will read entire file first and then evaluate condition and call main() function which will be able to call any forward declared function as it already read the entire file first. This condition leverages special variable __name__ which returns __main__ value whenever we run Python code from current file (when code was imported as a module, then __name__ returns module name).
"just reorganize my code so that I don't have this problem." Correct. Easy to do. Always works.
You can always provide the function prior to it's reference.
"However, there are cases when this is probably unavoidable, for instance when implementing some forms of recursion"
Can't see how that's even remotely possible. Please provide an example of a place where you cannot define the function prior to it's use.
Now wait a minute. When your module reaches the print statement in your example, before cmp_configs has been defined, what exactly is it that you expect it to do?
If your posting of a question using print is really trying to represent something like this:
fn = lambda mylist:"\n".join([str(bla)
for bla in sorted(mylist, cmp = cmp_configs)])
then there is no requirement to define cmp_configs before executing this statement, just define it later in the code and all will be well.
Now if you are trying to reference cmp_configs as a default value of an argument to the lambda, then this is a different story:
fn = lambda mylist,cmp_configs=cmp_configs : \
"\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
Now you need a cmp_configs variable defined before you reach this line.
[EDIT - this next part turns out not to be correct, since the default argument value will get assigned when the function is compiled, and that value will be used even if you change the value of cmp_configs later.]
Fortunately, Python being so type-accommodating as it is, does not care what you define as cmp_configs, so you could just preface with this statement:
cmp_configs = None
And the compiler will be happy. Just be sure to declare the real cmp_configs before you ever invoke fn.
Python technically has support for forward declaration.
if you define a function/class then set the body to pass, it will have an empty entry in the global table.
you can then "redefine" the function/class later on to implement the function/class.
unlike c/c++ forward declaration though, this does not work from outside the scope (i.e. another file) as they have their own "global" namespace
example:
def foo(): pass
foo()
def foo(): print("FOOOOO")
foo()
foo is declared both times
however the first time foo is called it does not do anything as the body is just pass
but the second time foo is called. it executes the new body of print("FOOOOO")
but again. note that this does not fix circular dependancies. this is because files have their own name and have their own definitions of functions
example 2:
class bar: pass
print(bar)
this prints <class '__main__.bar'> but if it was declared in another file it would be <class 'otherfile.foo'>
i know this post is old, but i though that this answer would be useful to anyone who keeps finding this post after the many years it has been posted for
One way is to create a handler function. Define the handler early on, and put the handler below all the methods you need to call.
Then when you invoke the handler method to call your functions, they will always be available.
The handler could take an argument nameOfMethodToCall. Then uses a bunch of if statements to call the right method.
This would solve your issue.
def foo():
print("foo")
#take input
nextAction=input('What would you like to do next?:')
return nextAction
def bar():
print("bar")
nextAction=input('What would you like to do next?:')
return nextAction
def handler(action):
if(action=="foo"):
nextAction = foo()
elif(action=="bar"):
nextAction = bar()
else:
print("You entered invalid input, defaulting to bar")
nextAction = "bar"
return nextAction
nextAction=input('What would you like to do next?:')
while 1:
nextAction = handler(nextAction)

Access variables in functions of imported classes

I have an imported module in my code. The module contains a function defined in it. The function does not return any value. In my code, I want to access a variable declared in that function. Is there a way to do it?
Edit:
So, there is a module inside an OSS which is generating a speech sequence. But the speech sequence variable is not global, neither the function returns the speech sequence. I want to access the speech sequence in my addon.
If your function looks like this:
def function():
...
my_variable = "foo"
Then you can add a parameter that would force that variable to be returned along with anything that is being returned. So the function would look like this.
def function(get_variable=False):
...
return_variable = "bar"
my_variable = "foo"
if get_variable:
return (my_variable, return_variable)
else:
return return_variable
And then access it like this:
import module
my_variable, return_variable = module.function(get_variable=True)
You don't need this sort of hack if your function doesn't return anything at all. Just make it return the variable you want and assign it to something inside your main code's scope.

List of Python functions, in their order of definition in the module

For a test-driven pedagogical module, I need to check doctests in a precise order.
Is there a way to grab all callables in the current module, in their order of definition?
What I tried:
Loop on globals and check if the object is a callable. The problem is that globals is a dict and thus not ordered.
Using the doctests directly is not convenient because the "stop at first error" won't work for me as I have several functions to test.
Each function object has a code object which stores the first line number, so you can use:
import inspect
ordered = sorted(inspect.getmembers(moduleobj, inspect.isfunction),
key=lambda kv: kv[1].__code__.co_firstlineno)
to get a sorted list of (name, function) pairs. For Python 2.5 and older, you'll need to use .func_code instead of .__code__.
You may need to further filter on functions that were defined in the module itself and have not been imported; func.__module__ == moduleobj.__name__ should suffice there.
Thanks to Martijn, I eventually found.
This is a complete snippet for Python3.
import sys
import inspect
def f1():
"f1!"
pass
def f3():
"f3!"
pass
def f2():
"f2!"
pass
funcs = [elt[1] for elt in inspect.getmembers(sys.modules[__name__],
inspect.isfunction)]
ordered_funcs = sorted(funcs, key=lambda f: f.__code__.co_firstlineno)
for f in ordered_funcs:
print(f.__doc__)

Categories

Resources