如何获取Python函数的源代码？

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.

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

If you are using IPython, then you need to type “foo??”

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

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.

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

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 ;)

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

to summarize :

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

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.