How can I get the source code of a Python function

Question

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

To expand on runeh s answer    gt  gt  gt  def foo a          x   2        return x   a   gt  gt  gt  import inspect   gt  gt  gt  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

User · Answer

to summarize    import inspect print     join inspect getsourcelines foo  0

User · Answer

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 gt  5  func   eval funcstring    then optionally to attach the original code to the function    func source   funcstring

User · Answer

dis is your friend if the source code is not available    gt  gt  gt  import dis  gt  gt  gt  def foo arg1 arg2            do something with args         a   arg1   arg2         return a      gt  gt  gt  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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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  lt ipython-input-18-3174e3126506 gt  Type       function

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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    gt  gt  gt  from dill source import getsource  gt  gt  gt    gt  gt  gt  def add x y         return x y       gt  gt  gt  squared   lambda x x  2  gt  gt  gt    gt  gt  gt  print getsource add  def add x y     return x y   gt  gt  gt  print getsource squared  squared   lambda x x  2   gt  gt  gt    gt  gt  gt  class Foo object         def bar self  x           return x x x       gt  gt  gt  f   Foo    gt  gt  gt    gt  gt  gt  print getsource f bar  def bar self  x       return x x x   gt  gt  gt

User · Answer

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    gt  gt  gt  from dill source import getsource  gt  gt  gt    gt  gt  gt  def add x y         return x y       gt  gt  gt  squared   lambda x x  2  gt  gt  gt    gt  gt  gt  print getsource add  def add x y     return x y   gt  gt  gt  print getsource squared  squared   lambda x x  2   gt  gt  gt    gt  gt  gt  class Foo object         def bar self  x           return x x x       gt  gt  gt  f   Foo    gt  gt  gt    gt  gt  gt  print getsource f bar  def bar self  x       return x x x   gt  gt  gt

User · Answer

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

User · Answer

to summarize    import inspect print     join inspect getsourcelines foo  0

User · Answer

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

User · Answer

dis is your friend if the source code is not available    gt  gt  gt  import dis  gt  gt  gt  def foo arg1 arg2            do something with args         a   arg1   arg2         return a      gt  gt  gt  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

User · Answer

To expand on runeh s answer    gt  gt  gt  def foo a          x   2        return x   a   gt  gt  gt  import inspect   gt  gt  gt  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

User · Answer

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

User · Answer

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  lt ipython-input-18-3174e3126506 gt  Type       function

User · Answer

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

User · Answer

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 gt  5  func   eval funcstring    then optionally to attach the original code to the function    func source   funcstring

[python] How can I get the source code of a Python function?

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