I want to get information about the callers of a specific function in python. For example:
class SomeClass():
def __init__(self, x):
self.x = x
def caller(self):
return special_func(self.x)
def special_func(x):
print "My caller is the 'caller' function in an 'SomeClass' class."
Is it possible with python?
Yes, the sys._getframe() function let's you retrieve frames from the current execution stack, which you can then inspect with the methods and documentation found in the inspect module; you'll be looking for specific locals in the f_locals attribute, as well as for the f_code information:
import sys
def special_func(x):
callingframe = sys._getframe(1)
print 'My caller is the %r function in a %r class' % (
callingframe.f_code.co_name,
callingframe.f_locals['self'].__class__.__name__)
Note that you'll need to take some care to detect what kind of information you find in each frame.
sys._getframe() returns a frame object, you can chain through the whole stack by following the f_back reference on each. Or you can use the inspect.stack() function to produce a lists of frames with additional information.
An example:
def f1(a):
import inspect
print 'I am f1 and was called by', inspect.currentframe().f_back.f_code.co_name
return a
def f2(a):
return f1(a)
Will retrieve the "immediate" caller.
>>> f2(1)
I am f1 and was called by f2
And if wasn't called from another you get (in IDLE):
>>> f1(1)
I am f1 and was called by <module>
Thanks to Jon Clements answer I was able to make a function that returns an ordered list of all callers:
def f1():
names = []
frame = inspect.currentframe()
## Keep moving to next outer frame
while True:
try:
frame = frame.f_back
name = frame.f_code.co_name
names.append(name)
except:
break
return names
and when called in a chain:
def f2():
return f1()
def f3():
return f2()
def f4():
return f3()
print f4()
looks like this:
['f2', 'f3', 'f4', '<module>']
In my case I filter out anything at '<module>' and after, and then take the last item to be the name of the originating caller.
Or modify the original loop to bail at the first appearance of any name starting with '<':
frame = frame.f_back
name = frame.f_code.co_name
if name[0] == '<':
break
names.append(name)
Related
This question already has answers here:
Getting the name of a variable as a string
(32 answers)
Closed 4 months ago.
Is it possible to get the original variable name of a variable passed to a function? E.g.
foobar = "foo"
def func(var):
print var.origname
So that:
func(foobar)
Returns:
>>foobar
EDIT:
All I was trying to do was make a function like:
def log(soup):
f = open(varname+'.html', 'w')
print >>f, soup.prettify()
f.close()
.. and have the function generate the filename from the name of the variable passed to it.
I suppose if it's not possible I'll just have to pass the variable and the variable's name as a string each time.
EDIT: To make it clear, I don't recommend using this AT ALL, it will break, it's a mess, it won't help you in any way, but it's doable for entertainment/education purposes.
You can hack around with the inspect module, I don't recommend that, but you can do it...
import inspect
def foo(a, f, b):
frame = inspect.currentframe()
frame = inspect.getouterframes(frame)[1]
string = inspect.getframeinfo(frame[0]).code_context[0].strip()
args = string[string.find('(') + 1:-1].split(',')
names = []
for i in args:
if i.find('=') != -1:
names.append(i.split('=')[1].strip())
else:
names.append(i)
print names
def main():
e = 1
c = 2
foo(e, 1000, b = c)
main()
Output:
['e', '1000', 'c']
To add to Michael Mrozek's answer, you can extract the exact parameters versus the full code by:
import re
import traceback
def func(var):
stack = traceback.extract_stack()
filename, lineno, function_name, code = stack[-2]
vars_name = re.compile(r'\((.*?)\).*$').search(code).groups()[0]
print vars_name
return
foobar = "foo"
func(foobar)
# PRINTS: foobar
Looks like Ivo beat me to inspect, but here's another implementation:
import inspect
def varName(var):
lcls = inspect.stack()[2][0].f_locals
for name in lcls:
if id(var) == id(lcls[name]):
return name
return None
def foo(x=None):
lcl='not me'
return varName(x)
def bar():
lcl = 'hi'
return foo(lcl)
bar()
# 'lcl'
Of course, it can be fooled:
def baz():
lcl = 'hi'
x='hi'
return foo(lcl)
baz()
# 'x'
Moral: don't do it.
Another way you can try if you know what the calling code will look like is to use traceback:
def func(var):
stack = traceback.extract_stack()
filename, lineno, function_name, code = stack[-2]
code will contain the line of code that was used to call func (in your example, it would be the string func(foobar)). You can parse that to pull out the argument
You can't. It's evaluated before being passed to the function. All you can do is pass it as a string.
#Ivo Wetzel's answer works in the case of function call are made in one line, like
e = 1 + 7
c = 3
foo(e, 100, b=c)
In case that function call is not in one line, like:
e = 1 + 7
c = 3
foo(e,
1000,
b = c)
below code works:
import inspect, ast
def foo(a, f, b):
frame = inspect.currentframe()
frame = inspect.getouterframes(frame)[1]
string = inspect.findsource(frame[0])[0]
nodes = ast.parse(''.join(string))
i_expr = -1
for (i, node) in enumerate(nodes.body):
if hasattr(node, 'value') and isinstance(node.value, ast.Call)
and hasattr(node.value.func, 'id') and node.value.func.id == 'foo' # Here goes name of the function:
i_expr = i
break
i_expr_next = min(i_expr + 1, len(nodes.body)-1)
lineno_start = nodes.body[i_expr].lineno
lineno_end = nodes.body[i_expr_next].lineno if i_expr_next != i_expr else len(string)
str_func_call = ''.join([i.strip() for i in string[lineno_start - 1: lineno_end]])
params = str_func_call[str_func_call.find('(') + 1:-1].split(',')
print(params)
You will get:
[u'e', u'1000', u'b = c']
But still, this might break.
You can use python-varname package
from varname import nameof
s = 'Hey!'
print (nameof(s))
Output:
s
Package below:
https://github.com/pwwang/python-varname
For posterity, here's some code I wrote for this task, in general I think there is a missing module in Python to give everyone nice and robust inspection of the caller environment. Similar to what rlang eval framework provides for R.
import re, inspect, ast
#Convoluted frame stack walk and source scrape to get what the calling statement to a function looked like.
#Specifically return the name of the variable passed as parameter found at position pos in the parameter list.
def _caller_param_name(pos):
#The parameter name to return
param = None
#Get the frame object for this function call
thisframe = inspect.currentframe()
try:
#Get the parent calling frames details
frames = inspect.getouterframes(thisframe)
#Function this function was just called from that we wish to find the calling parameter name for
function = frames[1][3]
#Get all the details of where the calling statement was
frame,filename,line_number,function_name,source,source_index = frames[2]
#Read in the source file in the parent calling frame upto where the call was made
with open(filename) as source_file:
head=[source_file.next() for x in xrange(line_number)]
source_file.close()
#Build all lines of the calling statement, this deals with when a function is called with parameters listed on each line
lines = []
#Compile a regex for matching the start of the function being called
regex = re.compile(r'\.?\s*%s\s*\(' % (function))
#Work backwards from the parent calling frame line number until we see the start of the calling statement (usually the same line!!!)
for line in reversed(head):
lines.append(line.strip())
if re.search(regex, line):
break
#Put the lines we have groked back into sourcefile order rather than reverse order
lines.reverse()
#Join all the lines that were part of the calling statement
call = "".join(lines)
#Grab the parameter list from the calling statement for the function we were called from
match = re.search('\.?\s*%s\s*\((.*)\)' % (function), call)
paramlist = match.group(1)
#If the function was called with no parameters raise an exception
if paramlist == "":
raise LookupError("Function called with no parameters.")
#Use the Python abstract syntax tree parser to create a parsed form of the function parameter list 'Name' nodes are variable names
parameter = ast.parse(paramlist).body[0].value
#If there were multiple parameters get the positional requested
if type(parameter).__name__ == 'Tuple':
#If we asked for a parameter outside of what was passed complain
if pos >= len(parameter.elts):
raise LookupError("The function call did not have a parameter at postion %s" % pos)
parameter = parameter.elts[pos]
#If there was only a single parameter and another was requested raise an exception
elif pos != 0:
raise LookupError("There was only a single calling parameter found. Parameter indices start at 0.")
#If the parameter was the name of a variable we can use it otherwise pass back None
if type(parameter).__name__ == 'Name':
param = parameter.id
finally:
#Remove the frame reference to prevent cyclic references screwing the garbage collector
del thisframe
#Return the parameter name we found
return param
If you want a Key Value Pair relationship, maybe using a Dictionary would be better?
...or if you're trying to create some auto-documentation from your code, perhaps something like Doxygen (http://www.doxygen.nl/) could do the job for you?
I wondered how IceCream solves this problem. So I looked into the source code and came up with the following (slightly simplified) solution. It might not be 100% bullet-proof (e.g. I dropped get_text_with_indentation and I assume exactly one function argument), but it works well for different test cases. It does not need to parse source code itself, so it should be more robust and simpler than previous solutions.
#!/usr/bin/env python3
import inspect
from executing import Source
def func(var):
callFrame = inspect.currentframe().f_back
callNode = Source.executing(callFrame).node
source = Source.for_frame(callFrame)
expression = source.asttokens().get_text(callNode.args[0])
print(expression, '=', var)
i = 1
f = 2.0
dct = {'key': 'value'}
obj = type('', (), {'value': 42})
func(i)
func(f)
func(s)
func(dct['key'])
func(obj.value)
Output:
i = 1
f = 2.0
s = string
dct['key'] = value
obj.value = 42
Update: If you want to move the "magic" into a separate function, you simply have to go one frame further back with an additional f_back.
def get_name_of_argument():
callFrame = inspect.currentframe().f_back.f_back
callNode = Source.executing(callFrame).node
source = Source.for_frame(callFrame)
return source.asttokens().get_text(callNode.args[0])
def func(var):
print(get_name_of_argument(), '=', var)
If you want to get the caller params as in #Matt Oates answer answer without using the source file (ie from Jupyter Notebook), this code (combined from #Aeon answer) will do the trick (at least in some simple cases):
def get_caller_params():
# get the frame object for this function call
thisframe = inspect.currentframe()
# get the parent calling frames details
frames = inspect.getouterframes(thisframe)
# frame 0 is the frame of this function
# frame 1 is the frame of the caller function (the one we want to inspect)
# frame 2 is the frame of the code that calls the caller
caller_function_name = frames[1][3]
code_that_calls_caller = inspect.findsource(frames[2][0])[0]
# parse code to get nodes of abstract syntact tree of the call
nodes = ast.parse(''.join(code_that_calls_caller))
# find the node that calls the function
i_expr = -1
for (i, node) in enumerate(nodes.body):
if _node_is_our_function_call(node, caller_function_name):
i_expr = i
break
# line with the call start
idx_start = nodes.body[i_expr].lineno - 1
# line with the end of the call
if i_expr < len(nodes.body) - 1:
# next expression marks the end of the call
idx_end = nodes.body[i_expr + 1].lineno - 1
else:
# end of the source marks the end of the call
idx_end = len(code_that_calls_caller)
call_lines = code_that_calls_caller[idx_start:idx_end]
str_func_call = ''.join([line.strip() for line in call_lines])
str_call_params = str_func_call[str_func_call.find('(') + 1:-1]
params = [p.strip() for p in str_call_params.split(',')]
return params
def _node_is_our_function_call(node, our_function_name):
node_is_call = hasattr(node, 'value') and isinstance(node.value, ast.Call)
if not node_is_call:
return False
function_name_correct = hasattr(node.value.func, 'id') and node.value.func.id == our_function_name
return function_name_correct
You can then run it as this:
def test(*par_values):
par_names = get_caller_params()
for name, val in zip(par_names, par_values):
print(name, val)
a = 1
b = 2
string = 'text'
test(a, b,
string
)
to get the desired output:
a 1
b 2
string text
Since you can have multiple variables with the same content, instead of passing the variable (content), it might be safer (and will be simpler) to pass it's name in a string and get the variable content from the locals dictionary in the callers stack frame. :
def displayvar(name):
import sys
return name+" = "+repr(sys._getframe(1).f_locals[name])
If it just so happens that the variable is a callable (function), it will have a __name__ property.
E.g. a wrapper to log the execution time of a function:
def time_it(func, *args, **kwargs):
start = perf_counter()
result = func(*args, **kwargs)
duration = perf_counter() - start
print(f'{func.__name__} ran in {duration * 1000}ms')
return result
I want to get information about the callers of a specific function in python. For example:
class SomeClass():
def __init__(self, x):
self.x = x
def caller(self):
return special_func(self.x)
def special_func(x):
print "My caller is the 'caller' function in an 'SomeClass' class."
Is it possible with python?
Yes, the sys._getframe() function let's you retrieve frames from the current execution stack, which you can then inspect with the methods and documentation found in the inspect module; you'll be looking for specific locals in the f_locals attribute, as well as for the f_code information:
import sys
def special_func(x):
callingframe = sys._getframe(1)
print 'My caller is the %r function in a %r class' % (
callingframe.f_code.co_name,
callingframe.f_locals['self'].__class__.__name__)
Note that you'll need to take some care to detect what kind of information you find in each frame.
sys._getframe() returns a frame object, you can chain through the whole stack by following the f_back reference on each. Or you can use the inspect.stack() function to produce a lists of frames with additional information.
An example:
def f1(a):
import inspect
print 'I am f1 and was called by', inspect.currentframe().f_back.f_code.co_name
return a
def f2(a):
return f1(a)
Will retrieve the "immediate" caller.
>>> f2(1)
I am f1 and was called by f2
And if wasn't called from another you get (in IDLE):
>>> f1(1)
I am f1 and was called by <module>
Thanks to Jon Clements answer I was able to make a function that returns an ordered list of all callers:
def f1():
names = []
frame = inspect.currentframe()
## Keep moving to next outer frame
while True:
try:
frame = frame.f_back
name = frame.f_code.co_name
names.append(name)
except:
break
return names
and when called in a chain:
def f2():
return f1()
def f3():
return f2()
def f4():
return f3()
print f4()
looks like this:
['f2', 'f3', 'f4', '<module>']
In my case I filter out anything at '<module>' and after, and then take the last item to be the name of the originating caller.
Or modify the original loop to bail at the first appearance of any name starting with '<':
frame = frame.f_back
name = frame.f_code.co_name
if name[0] == '<':
break
names.append(name)
I am trying to get a list of functions from a module using inspect.getmembers in the order of source code.
Below is the code
def get_functions_from_module(app_module):
list_of_functions = dict(inspect.getmembers(app_module,
inspect.isfunction))
return list_of_functions.values
The current code will not return the list of function objects in order of the source code and I'm wondering if it would be possible to order it.
Thank you!
I think I came up with a solution.
def get_line_number_of_function(func):
return func.__code__.co_firstlineno
def get_functions_from_module(app_module):
list_of_functions = dict(inspect.getmembers(app_module,
inspect.isfunction))
return sorted(list_of_functions.values(), key=lambda x:
get_line_number_of_function(x))
You can sort by line numbers using inspect.findsource. Docstring from the source code of that function:
def findsource(object):
"""Return the entire source file and starting line number for an object.
The argument may be a module, class, method, function, traceback, frame,
or code object. The source code is returned as a list of all the lines
in the file and the line number indexes a line in that list. An OSError
is raised if the source code cannot be retrieved."""
Here's an example in Python 2.7:
import ab.bc.de.t1 as t1
import inspect
def get_functions_from_module(app_module):
list_of_functions = inspect.getmembers(app_module, inspect.isfunction)
return list_of_functions
fns = get_functions_from_module(t1)
def sort_by_line_no(fn):
fn_name, fn_obj = fn
source, line_no = inspect.findsource(fn_obj)
return line_no
for name, fn in sorted(fns, key=sort_by_line_no):
print name, fn
My ab.bc.de.t1 is defined as follows:
class B(object):
def a():
print 'test'
def c():
print 'c'
def a():
print 'a'
def b():
print 'b'
And the output I get when I try retrieving sorted functions is below:
c <function c at 0x00000000362517B8>
a <function a at 0x0000000036251438>
b <function b at 0x0000000036251668>
>>>
Python: How to get the caller's method name in the called method?
Assume I have 2 methods:
def method1(self):
...
a = A.method2()
def method2(self):
...
If I don't want to do any change for method1, how to get the name of the caller (in this example, the name is method1) in method2?
inspect.getframeinfo and other related functions in inspect can help:
>>> import inspect
>>> def f1(): f2()
...
>>> def f2():
... curframe = inspect.currentframe()
... calframe = inspect.getouterframes(curframe, 2)
... print('caller name:', calframe[1][3])
...
>>> f1()
caller name: f1
this introspection is intended to help debugging and development; it's not advisable to rely on it for production-functionality purposes.
Shorter version:
import inspect
def f1(): f2()
def f2():
print 'caller name:', inspect.stack()[1][3]
f1()
(with thanks to #Alex, and Stefaan Lippen)
This seems to work just fine:
import sys
print sys._getframe().f_back.f_code.co_name
I would use inspect.currentframe().f_back.f_code.co_name. Its use hasn't been covered in any of the prior answers which are mainly of one of three types:
Some prior answers use inspect.stack but it's known to be too slow.
Some prior answers use sys._getframe which is an internal private function given its leading underscore, and so its use is implicitly discouraged.
One prior answer uses inspect.getouterframes(inspect.currentframe(), 2)[1][3] but it's entirely unclear what [1][3] is accessing.
import inspect
from types import FrameType
from typing import cast
def demo_the_caller_name() -> str:
"""Return the calling function's name."""
# Ref: https://stackoverflow.com/a/57712700/
return cast(FrameType, cast(FrameType, inspect.currentframe()).f_back).f_code.co_name
if __name__ == '__main__':
def _test_caller_name() -> None:
assert demo_the_caller_name() == '_test_caller_name'
_test_caller_name()
Note that cast(FrameType, frame) is used to satisfy mypy.
Acknowlegement: comment by 1313e for an answer.
I've come up with a slightly longer version that tries to build a full method name including module and class.
https://gist.github.com/2151727 (rev 9cccbf)
# Public Domain, i.e. feel free to copy/paste
# Considered a hack in Python 2
import inspect
def caller_name(skip=2):
"""Get a name of a caller in the format module.class.method
`skip` specifies how many levels of stack to skip while getting caller
name. skip=1 means "who calls me", skip=2 "who calls my caller" etc.
An empty string is returned if skipped levels exceed stack height
"""
stack = inspect.stack()
start = 0 + skip
if len(stack) < start + 1:
return ''
parentframe = stack[start][0]
name = []
module = inspect.getmodule(parentframe)
# `modname` can be None when frame is executed directly in console
# TODO(techtonik): consider using __main__
if module:
name.append(module.__name__)
# detect classname
if 'self' in parentframe.f_locals:
# I don't know any way to detect call from the object method
# XXX: there seems to be no way to detect static method call - it will
# be just a function call
name.append(parentframe.f_locals['self'].__class__.__name__)
codename = parentframe.f_code.co_name
if codename != '<module>': # top level usually
name.append( codename ) # function or a method
## Avoid circular refs and frame leaks
# https://docs.python.org/2.7/library/inspect.html#the-interpreter-stack
del parentframe, stack
return ".".join(name)
Bit of an amalgamation of the stuff above. But here's my crack at it.
def print_caller_name(stack_size=3):
def wrapper(fn):
def inner(*args, **kwargs):
import inspect
stack = inspect.stack()
modules = [(index, inspect.getmodule(stack[index][0]))
for index in reversed(range(1, stack_size))]
module_name_lengths = [len(module.__name__)
for _, module in modules]
s = '{index:>5} : {module:^%i} : {name}' % (max(module_name_lengths) + 4)
callers = ['',
s.format(index='level', module='module', name='name'),
'-' * 50]
for index, module in modules:
callers.append(s.format(index=index,
module=module.__name__,
name=stack[index][3]))
callers.append(s.format(index=0,
module=fn.__module__,
name=fn.__name__))
callers.append('')
print('\n'.join(callers))
fn(*args, **kwargs)
return inner
return wrapper
Use:
#print_caller_name(4)
def foo():
return 'foobar'
def bar():
return foo()
def baz():
return bar()
def fizz():
return baz()
fizz()
output is
level : module : name
--------------------------------------------------
3 : None : fizz
2 : None : baz
1 : None : bar
0 : __main__ : foo
You can use decorators, and do not have to use stacktrace
If you want to decorate a method inside a class
import functools
# outside ur class
def printOuterFunctionName(func):
#functools.wraps(func)
def wrapper(self):
print(f'Function Name is: {func.__name__}')
func(self)
return wrapper
class A:
#printOuterFunctionName
def foo():
pass
you may remove functools, self if it is procedural
An alternative to sys._getframe() is used by Python's Logging library to find caller information. Here's the idea:
raise an Exception
immediately catch it in an Except clause
use sys.exc_info to get Traceback frame (tb_frame).
from tb_frame get last caller's frame using f_back.
from last caller's frame get the code object that was being executed in that frame.
In our sample code it would be method1 (not method2) being executed.
From code object obtained, get the object's name -- this is caller method's name in our sample.
Here's the sample code to solve example in the question:
def method1():
method2()
def method2():
try:
raise Exception
except Exception:
frame = sys.exc_info()[2].tb_frame.f_back
print("method2 invoked by: ", frame.f_code.co_name)
# Invoking method1
method1()
Output:
method2 invoked by: method1
Frame has all sorts of details, including line number, file name, argument counts, argument type and so on. The solution works across classes and modules too.
Code:
#!/usr/bin/env python
import inspect
called=lambda: inspect.stack()[1][3]
def caller1():
print "inside: ",called()
def caller2():
print "inside: ",called()
if __name__=='__main__':
caller1()
caller2()
Output:
shahid#shahid-VirtualBox:~/Documents$ python test_func.py
inside: caller1
inside: caller2
shahid#shahid-VirtualBox:~/Documents$
I found a way if you're going across classes and want the class the method belongs to AND the method. It takes a bit of extraction work but it makes its point. This works in Python 2.7.13.
import inspect, os
class ClassOne:
def method1(self):
classtwoObj.method2()
class ClassTwo:
def method2(self):
curframe = inspect.currentframe()
calframe = inspect.getouterframes(curframe, 4)
print '\nI was called from', calframe[1][3], \
'in', calframe[1][4][0][6: -2]
# create objects to access class methods
classoneObj = ClassOne()
classtwoObj = ClassTwo()
# start the program
os.system('cls')
classoneObj.method1()
Hey mate I once made 3 methods without plugins for my app and maybe that can help you, It worked for me so maybe gonna work for you too.
def method_1(a=""):
if a == "method_2":
print("method_2")
if a == "method_3":
print("method_3")
def method_2():
method_1("method_2")
def method_3():
method_1("method_3")
method_2()
Why can two functions with the same id value have differing attributes like __doc__ or __name__?
Here's a toy example:
some_dict = {}
for i in range(2):
def fun(self, *args):
print i
fun.__doc__ = "I am function {}".format(i)
fun.__name__ = "function_{}".format(i)
some_dict["function_{}".format(i)] = fun
my_type = type("my_type", (object,), some_dict)
m = my_type()
print id(m.function_0)
print id(m.function_1)
print m.function_0.__doc__
print m.function_1.__doc__
print m.function_0.__name__
print m.function_1.__name__
print m.function_0()
print m.function_1()
Which prints:
57386560
57386560
I am function 0
I am function 1
function_0
function_1
1 # <--- Why is it bound to the most recent value of that variable?
1
I've tried mixing in a call to copy.deepcopy (not sure if the recursive copy is needed for functions or it is overkill) but this doesn't change anything.
You are comparing methods, and method objects are created anew each time you access one on an instance or class (via the descriptor protocol).
Once you tested their id() you discard the method again (there are no references to it), so Python is free to reuse the id when you create another method. You want to test the actual functions here, by using m.function_0.__func__ and m.function_1.__func__:
>>> id(m.function_0.__func__)
4321897240
>>> id(m.function_1.__func__)
4321906032
Method objects inherit the __doc__ and __name__ attributes from the function that they wrap. The actual underlying functions are really still different objects.
As for the two functions returning 1; both functions use i as a closure; the value for i is looked up when you call the method, not when you created the function. See Local variables in Python nested functions.
The easiest work-around is to add another scope with a factory function:
some_dict = {}
for i in range(2):
def create_fun(i):
def fun(self, *args):
print i
fun.__doc__ = "I am function {}".format(i)
fun.__name__ = "function_{}".format(i)
return fun
some_dict["function_{}".format(i)] = create_fun(i)
Per your comment on ndpu's answer, here is one way you can create the functions without needing to have an optional argument:
for i in range(2):
def funGenerator(i):
def fun1(self, *args):
print i
return fun1
fun = funGenerator(i)
fun.__doc__ = "I am function {}".format(i)
fun.__name__ = "function_{}".format(i)
some_dict["function_{}".format(i)] = fun
#Martjin Pieters is perfectly correct. To illustrate, try this modification
some_dict = {}
for i in range(2):
def fun(self, *args):
print i
fun.__doc__ = "I am function {}".format(i)
fun.__name__ = "function_{}".format(i)
some_dict["function_{}".format(i)] = fun
print "id",id(fun)
my_type = type("my_type", (object,), some_dict)
m = my_type()
print id(m.function_0)
print id(m.function_1)
print m.function_0.__doc__
print m.function_1.__doc__
print m.function_0.__name__
print m.function_1.__name__
print m.function_0()
print m.function_1()
c = my_type()
print c
print id(c.function_0)
You see that the fun get's a different id each time, and is different from the final one. It's the method creation logic that send's it pointing to the same location, as that's where the class's code is stored. Also, if you use the my_type as a sort of class, instances created with it have the same memory address for that function
This code gives:
id 4299601152
id 4299601272
4299376112
4299376112
I am function 0
I am function 1
function_0
function_1
1
None
1
None
<main.my_type object at 0x10047c350>
4299376112
You should save current i to make this:
1 # <--- Why is it bound to the most recent value of that variable?
1
work, for example by setting default value to function argument:
for i in range(2):
def fun(self, i=i, *args):
print i
# ...
or create a closure:
for i in range(2):
def f(i):
def fun(self, *args):
print i
return fun
fun = f(i)
# ...