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how to get definition of a function in python? [duplicate]

Suppose I have a Python function as defined below:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
I can get the name of the function using foo.func_name. How can I programmatically get its source code, as I typed above?

If the function is from a source file available on the filesystem, then inspect.getsource(foo) might be of help:
If foo is defined as:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
Then:
import inspect
lines = inspect.getsource(foo)
print(lines)
Returns:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
But I believe that if the function is compiled from a string, stream or imported from a compiled file, then you cannot retrieve its source code.

The inspect module has methods for retrieving source code from python objects. Seemingly it only works if the source is located in a file though. If you had that I guess you wouldn't need to get the source from the object.
The following tests inspect.getsource(foo) using Python 3.6:
import inspect
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
source_foo = inspect.getsource(foo) # foo is normal function
print(source_foo)
source_max = inspect.getsource(max) # max is a built-in function
print(source_max)
This first prints:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
Then fails on inspect.getsource(max) with the following error:
TypeError: <built-in function max> is not a module, class, method, function, traceback, frame, or code object

Just use foo?? or ??foo.
If you are using IPython, then you need to type foo?? or ??foo to see the complete source code. To see only the docstring in the function, use foo? or ?foo. This works in Jupyter notebook as well.
In [19]: foo??
Signature: foo(arg1, arg2)
Source:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
File: ~/Desktop/<ipython-input-18-3174e3126506>
Type: function

dis is your friend if the source code is not available:
>>> import dis
>>> def foo(arg1,arg2):
... #do something with args
... a = arg1 + arg2
... return a
...
>>> dis.dis(foo)
3 0 LOAD_FAST 0 (arg1)
3 LOAD_FAST 1 (arg2)
6 BINARY_ADD
7 STORE_FAST 2 (a)
4 10 LOAD_FAST 2 (a)
13 RETURN_VALUE

While I'd generally agree that inspect is a good answer, I'd disagree that you can't get the source code of objects defined in the interpreter. If you use dill.source.getsource from dill, you can get the source of functions and lambdas, even if they are defined interactively.
It also can get the code for from bound or unbound class methods and functions defined in curries... however, you might not be able to compile that code without the enclosing object's code.
>>> from dill.source import getsource
>>>
>>> def add(x,y):
... return x+y
...
>>> squared = lambda x:x**2
>>>
>>> print getsource(add)
def add(x,y):
return x+y
>>> print getsource(squared)
squared = lambda x:x**2
>>>
>>> class Foo(object):
... def bar(self, x):
... return x*x+x
...
>>> f = Foo()
>>>
>>> print getsource(f.bar)
def bar(self, x):
return x*x+x
>>>

To expand on runeh's answer:
>>> def foo(a):
... x = 2
... return x + a
>>> import inspect
>>> inspect.getsource(foo)
u'def foo(a):\n x = 2\n return x + a\n'
print inspect.getsource(foo)
def foo(a):
x = 2
return x + a
EDIT: As pointed out by #0sh this example works using ipython but not plain python. It should be fine in both, however, when importing code from source files.

Since this post is marked as the duplicate of this other post, I answer here for the "lambda" case, although the OP is not about lambdas.
So, for lambda functions that are not defined in their own lines: in addition to marko.ristin's answer, you may wish to use mini-lambda or use SymPy as suggested in this answer.
mini-lambda is lighter and supports any kind of operation, but works only for a single variable
SymPy is heavier but much more equipped with mathematical/calculus operations. In particular it can simplify your expressions. It also supports several variables in the same expression.
Here is how you can do it using mini-lambda:
from mini_lambda import x, is_mini_lambda_expr
import inspect
def get_source_code_str(f):
if is_mini_lambda_expr(f):
return f.to_string()
else:
return inspect.getsource(f)
# test it
def foo(arg1, arg2):
# do something with args
a = arg1 + arg2
return a
print(get_source_code_str(foo))
print(get_source_code_str(x ** 2))
It correctly yields
def foo(arg1, arg2):
# do something with args
a = arg1 + arg2
return a
x ** 2
See mini-lambda documentation for details. I'm the author by the way ;)

You can use inspect module to get full source code for that. You have to use getsource() method for that from the inspect module. For example:
import inspect
def get_my_code():
x = "abcd"
return x
print(inspect.getsource(get_my_code))
You can check it out more options on the below link.
retrieve your python code

to summarize :
import inspect
print( "".join(inspect.getsourcelines(foo)[0]))

Please mind that the accepted answers work only if the lambda is given on a separate line. If you pass it in as an argument to a function and would like to retrieve the code of the lambda as object, the problem gets a bit tricky since inspect will give you the whole line.
For example, consider a file test.py:
import inspect
def main():
x, f = 3, lambda a: a + 1
print(inspect.getsource(f))
if __name__ == "__main__":
main()
Executing it gives you (mind the indention!):
x, f = 3, lambda a: a + 1
To retrieve the source code of the lambda, your best bet, in my opinion, is to re-parse the whole source file (by using f.__code__.co_filename) and match the lambda AST node by the line number and its context.
We had to do precisely that in our design-by-contract library icontract since we had to parse the lambda functions we pass in as arguments to decorators. It is too much code to paste here, so have a look at the implementation of this function.

If you're strictly defining the function yourself and it's a relatively short definition, a solution without dependencies would be to define the function in a string and assign the eval() of the expression to your function.
E.g.
funcstring = 'lambda x: x> 5'
func = eval(funcstring)
then optionally to attach the original code to the function:
func.source = funcstring

Rafał Dowgird's answer states:
I believe that if the function is compiled from a string, stream or imported from a compiled file, then you cannot retrieve its source code.
However, it is possible to retrieve the source code of a function compiled from a string, provided that the compiling code also added an entry to the linecache.cache dict:
import linecache
import inspect
script = '''
def add_nums(a, b):
return a + b
'''
bytecode = compile(script, 'unique_filename', 'exec')
tmp = {}
eval(bytecode, {}, tmp)
add_nums = tmp["add_nums"]
linecache.cache['unique_filename'] = (
len(script),
None,
script.splitlines(True),
'unique_filename',
)
print(inspect.getsource(add_nums))
# prints:
# """
# def add_nums(a, b):
# return a + b
# """
This is how the attrs library creates various methods for classes automatically, given a set of attributes that the class expects to be initialized with. See their source code here. As the source explains, this is a feature primarily intended to enable debuggers such as PDB to step through the code.

I believe that variable names aren't stored in pyc/pyd/pyo files, so you can not retrieve the exact code lines if you don't have source files.

Is there a programmatic way to print the definition of a dynamically created function in python? [duplicate]

Suppose I have a Python function as defined below:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
I can get the name of the function using foo.func_name. How can I programmatically get its source code, as I typed above?

If the function is from a source file available on the filesystem, then inspect.getsource(foo) might be of help:
If foo is defined as:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
Then:
import inspect
lines = inspect.getsource(foo)
print(lines)
Returns:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
But I believe that if the function is compiled from a string, stream or imported from a compiled file, then you cannot retrieve its source code.

The inspect module has methods for retrieving source code from python objects. Seemingly it only works if the source is located in a file though. If you had that I guess you wouldn't need to get the source from the object.
The following tests inspect.getsource(foo) using Python 3.6:
import inspect
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
source_foo = inspect.getsource(foo) # foo is normal function
print(source_foo)
source_max = inspect.getsource(max) # max is a built-in function
print(source_max)
This first prints:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
Then fails on inspect.getsource(max) with the following error:
TypeError: <built-in function max> is not a module, class, method, function, traceback, frame, or code object

Just use foo?? or ??foo.
If you are using IPython, then you need to type foo?? or ??foo to see the complete source code. To see only the docstring in the function, use foo? or ?foo. This works in Jupyter notebook as well.
In [19]: foo??
Signature: foo(arg1, arg2)
Source:
def foo(arg1,arg2):
#do something with args
a = arg1 + arg2
return a
File: ~/Desktop/<ipython-input-18-3174e3126506>
Type: function

dis is your friend if the source code is not available:
>>> import dis
>>> def foo(arg1,arg2):
... #do something with args
... a = arg1 + arg2
... return a
...
>>> dis.dis(foo)
3 0 LOAD_FAST 0 (arg1)
3 LOAD_FAST 1 (arg2)
6 BINARY_ADD
7 STORE_FAST 2 (a)
4 10 LOAD_FAST 2 (a)
13 RETURN_VALUE

While I'd generally agree that inspect is a good answer, I'd disagree that you can't get the source code of objects defined in the interpreter. If you use dill.source.getsource from dill, you can get the source of functions and lambdas, even if they are defined interactively.
It also can get the code for from bound or unbound class methods and functions defined in curries... however, you might not be able to compile that code without the enclosing object's code.
>>> from dill.source import getsource
>>>
>>> def add(x,y):
... return x+y
...
>>> squared = lambda x:x**2
>>>
>>> print getsource(add)
def add(x,y):
return x+y
>>> print getsource(squared)
squared = lambda x:x**2
>>>
>>> class Foo(object):
... def bar(self, x):
... return x*x+x
...
>>> f = Foo()
>>>
>>> print getsource(f.bar)
def bar(self, x):
return x*x+x
>>>

To expand on runeh's answer:
>>> def foo(a):
... x = 2
... return x + a
>>> import inspect
>>> inspect.getsource(foo)
u'def foo(a):\n x = 2\n return x + a\n'
print inspect.getsource(foo)
def foo(a):
x = 2
return x + a
EDIT: As pointed out by #0sh this example works using ipython but not plain python. It should be fine in both, however, when importing code from source files.

Since this post is marked as the duplicate of this other post, I answer here for the "lambda" case, although the OP is not about lambdas.
So, for lambda functions that are not defined in their own lines: in addition to marko.ristin's answer, you may wish to use mini-lambda or use SymPy as suggested in this answer.
mini-lambda is lighter and supports any kind of operation, but works only for a single variable
SymPy is heavier but much more equipped with mathematical/calculus operations. In particular it can simplify your expressions. It also supports several variables in the same expression.
Here is how you can do it using mini-lambda:
from mini_lambda import x, is_mini_lambda_expr
import inspect
def get_source_code_str(f):
if is_mini_lambda_expr(f):
return f.to_string()
else:
return inspect.getsource(f)
# test it
def foo(arg1, arg2):
# do something with args
a = arg1 + arg2
return a
print(get_source_code_str(foo))
print(get_source_code_str(x ** 2))
It correctly yields
def foo(arg1, arg2):
# do something with args
a = arg1 + arg2
return a
x ** 2
See mini-lambda documentation for details. I'm the author by the way ;)

You can use inspect module to get full source code for that. You have to use getsource() method for that from the inspect module. For example:
import inspect
def get_my_code():
x = "abcd"
return x
print(inspect.getsource(get_my_code))
You can check it out more options on the below link.
retrieve your python code

to summarize :
import inspect
print( "".join(inspect.getsourcelines(foo)[0]))

Please mind that the accepted answers work only if the lambda is given on a separate line. If you pass it in as an argument to a function and would like to retrieve the code of the lambda as object, the problem gets a bit tricky since inspect will give you the whole line.
For example, consider a file test.py:
import inspect
def main():
x, f = 3, lambda a: a + 1
print(inspect.getsource(f))
if __name__ == "__main__":
main()
Executing it gives you (mind the indention!):
x, f = 3, lambda a: a + 1
To retrieve the source code of the lambda, your best bet, in my opinion, is to re-parse the whole source file (by using f.__code__.co_filename) and match the lambda AST node by the line number and its context.
We had to do precisely that in our design-by-contract library icontract since we had to parse the lambda functions we pass in as arguments to decorators. It is too much code to paste here, so have a look at the implementation of this function.

If you're strictly defining the function yourself and it's a relatively short definition, a solution without dependencies would be to define the function in a string and assign the eval() of the expression to your function.
E.g.
funcstring = 'lambda x: x> 5'
func = eval(funcstring)
then optionally to attach the original code to the function:
func.source = funcstring

Rafał Dowgird's answer states:
I believe that if the function is compiled from a string, stream or imported from a compiled file, then you cannot retrieve its source code.
However, it is possible to retrieve the source code of a function compiled from a string, provided that the compiling code also added an entry to the linecache.cache dict:
import linecache
import inspect
script = '''
def add_nums(a, b):
return a + b
'''
bytecode = compile(script, 'unique_filename', 'exec')
tmp = {}
eval(bytecode, {}, tmp)
add_nums = tmp["add_nums"]
linecache.cache['unique_filename'] = (
len(script),
None,
script.splitlines(True),
'unique_filename',
)
print(inspect.getsource(add_nums))
# prints:
# """
# def add_nums(a, b):
# return a + b
# """
This is how the attrs library creates various methods for classes automatically, given a set of attributes that the class expects to be initialized with. See their source code here. As the source explains, this is a feature primarily intended to enable debuggers such as PDB to step through the code.

I believe that variable names aren't stored in pyc/pyd/pyo files, so you can not retrieve the exact code lines if you don't have source files.

Python: How to pass parameters when it isn't an option?

I have a function - func_main - which has several inputs. One of the inputs is another function - func_mini. The smaller function, func_mini, loads some data, and it needs to be able to load data from different sources depending on the arguments passed to it. The issue is that I am reading the documentation on func_main and it says that it only accepts parameterless functions. How can I get around this?
For example:
def func_main(evaluator = None):
num_list = [1,2,3]
return evaluator(num_list)
def func_mini(data_source = None):
if not data_source:
data_source = config.DATA_SOURCE
return pandas.read_csv(data_source).min
I want do to this:
func_main(func_mini(data_souce='path/to/my/file'))
But func_main documentation says:
"evaluator: A parameterless function that defines and return all data needed in the above model definition."

You can use functools.partial to build a function without parameters from your original func_mini function. Demo:
>>> from functools import partial
>>> def foo(x):
... print(x)
...
>>> foo_argless = partial(foo, 'hi')
>>> foo_argless()
hi

How do I get the return value when using Python exec on the code object of a function?

For testing purposes I want to directly execute a function defined inside of another function.
I can get to the code object of the child function, through the code (func_code) of the parent function, but when I exec it, i get no return value.
Is there a way to get the return value from the exec'ed code?

Yes, you need to have the assignment within the exec statement:
>>> def foo():
... return 5
...
>>> exec("a = foo()")
>>> a
5
This probably isn't relevant for your case since its being used in controlled testing, but be careful with using exec with user defined input.

A few years later, but the following snippet helped me:
the_code = '''
a = 1
b = 2
return_me = a + b
'''
loc = {}
exec(the_code, globals(), loc)
return_workaround = loc['return_me']
print(return_workaround) # 3
exec() doesn't return anything itself, but you can pass a dict which has all the local variables stored in it after execution. By accessing it you have a something like a return.
I hope it helps someone.

While this is the ugliest beast ever seen by mankind, this is how you can do it by using a global variable inside your exec call:
def my_exec(code):
exec('global i; i = %s' % code)
global i
return i
This is misusing global variables to get your data across the border.
>>> my_exec('1 + 2')
3
Needless to say that you should never allow any user inputs for the input of this function in there, as it poses an extreme security risk.

use eval() instead of exec(), it returns result

Something like this can work:
def outer():
def inner(i):
return i + 10
for f in outer.func_code.co_consts:
if getattr(f, 'co_name', None) == 'inner':
inner = type(outer)(f, globals())
# can also use `types` module for readability:
# inner = types.FunctionType(f, globals())
print inner(42) # 52
The idea is to extract the code object from the inner function and create a new function based on it.
Additional work is required when an inner function can contain free variables. You'll have to extract them as well and pass to the function constructor in the last argument (closure).

Here's a way to return a value from exec'd code:
def exec_and_return(expression):
exec(f"""locals()['temp'] = {expression}""")
return locals()['temp']
I'd advise you to give an example of the problem you're trying to solve. Because I would only ever use this as a last resort.

This doesn't get the return value per say, but you can provide an empty dictionary when calling exec to retrieve any variables defined in the code.
# Python 3
ex_locals = {}
exec("a = 'Hello world!'", None, ex_locals)
print(ex_locals['a'])
# Output: Hello world!
From the Python 3 documentation on exec:
The default locals act as described for function locals() below: modifications to the default locals dictionary should not be attempted. Pass an explicit locals dictionary if you need to see effects of the code on locals after function exec() returns.
For more information, see How does exec work with locals?

Here's a solution with a simple code:
# -*- coding: utf-8 -*-
import math
x = [0]
exec("x[0] = 3*2")
print(x[0]) # 6

Since Python 3.7, dictionary are ordered. So you no longer need to agree on a name, you can just say "last item that got created":
>>> d = {}
>>> exec("def addone(i): return i + 1", d, d)
>>> list(d)
['__builtins__', 'addone']
>>> thefunction = d[list(d)[-1]]
>>> thefunction
<function addone at 0x7fd03123fe50>

if we need a function that is in a file in another directory, eg
we need the function1 in file my_py_file.py
located in /home/.../another_directory
we can use the following code:
def cl_import_function(a_func,py_file,in_Dir):
... import sys
... sys.path.insert(0, in_Dir)
... ax='from %s import %s'%(py_file,a_func)
... loc={}
... exec(ax, globals(), loc)
... getFx = loc[afunc]
... return getFx
test = cl_import_function('function1',r'my_py_file',r'/home/.../another_directory/')
test()
(a simple way for newbies...)

program = 'a = 5\nb=10\nprint("Sum =", a + b)'
program = exec(program)
print(program)

Executing a function in reverse in Python

I have a function that looks something like this:
def f():
call_some_function_A()
call_some_function_B()
[...]
call_some_function_Z()
I'd like the function to be executed in reverse; that is, the execution must look like:
def f'():
call_some_function_Z()
[...]
call_some_function_B()
call_some_function_A()
(f will always be such that it is logically possible to reverse it; i.e. there are no variable declarations or anything like that).
How can I accomplish this?
I can't just write a function f' that calls the statements from f in reverse, because I don't want to have to update f' every time f is changed.
I also can't modify f.
(Please don't tell me that I shouldn't try to do that, or redesign my code, or anything like that- it's not a possibility.)

If your f() consists entirely of these function calls, you can remake it into a list:
functions = [
call_some_function_A,
call_some_function_B,
# [...]
call_some_function_Z,
]
And then use it to call the functions in (reversed) order.
def f():
for func in functions:
func()
def f_():
for func in reversed(functions):
func()

Please don't do this.
If your f() consists entirely of these function calls:
def f():
call_some_function_A()
call_some_function_B()
# [...]
call_some_function_Z()
...you can hack into it and get all the names it references:
names = f.__code__.co_names
# ('call_some_function_A', 'call_some_function_B', 'call_some_function_Z')
But you still need to get the corresponding functions.
If the functions are in some other module or anything similar, just do this:
functions = [getattr(some_module, name) for name in names]
If the functions are defined in the same file as globals, do this:
functions = [globals()[name] for name in names]
# [<function __main__.call_some_function_A>, <function __main__.call_some_function_B>, <function __main__.call_some_function_Z>]
Then all you need to do is call them in reverse order:
def f_():
for func in reversed(functions):
func()
Alternatively, you can obtain the function's source code, parse it, reverse the the abstract syntax tree, compile it back, execute it... and you will have yourself the reversed function.
Let's consider this example:
def f():
call_some_function_A()
if whatever:
call_some_function_B()
call_some_function_C()
call_some_function_D()
import inspect
import ast
original_f = f
source = inspect.getsource(f)
tree = ast.parse(source)
# tree is a Module, with body consisting of 1 FunctionDef
# tree.body[0] is a FunctionDef, with body consisting of Exprs
tree.body[0].body.reverse()
# top level expressions will be reversed
# compile the modified syntax tree to a code object as a module and execute it
exec(compile(tree, '<unknown>', 'exec'))
# f will be overwritten because the function name stays the same
# now f will be equivalent to:
# def f():
# call_some_function_D()
# if test:
# call_some_function_B()
# call_some_function_C()
# call_some_function_A()
f_ = f
f = original_f
So yes, this method is a bit better. It is even possible to recursively reverse all the bodys and achieve the reversal of ...B and ...C as well, but if even the simplest logic code is introduced, you will run into bad problems.

I hacked together this small function which assumes that the function is a simple list of one line statements. It uses exec which is another form of eval and so it makes it hard to compile the code but if you can live with evaluated code here it is:
import inspect
# sample function that will be reversed
def f():
print "first statement"
print "2nd statement"
print "last statement"
def makeReversedFunctionSrc(newName, f):
src = inspect.getsource(f)
srcLines = src.split("\n")
srcLines = srcLines[1:] # get rid of the old function definition
srcLines.reverse() # reverse function body
# hack together new function definition with reversed lines
newSrc = "def " + newName + "():\n"
for line in srcLines:
if line.strip() != "":
newSrc += line + "\n"
return newSrc
# get the code as a string
reverseCode = makeReversedFunctionSrc("reversedF", f)
# execute the string as if it was python (I heard thats evil as in eval)
exec(reverseCode)
# now lets call our new function
reversedF()