Python 实用宝典

Question 1

Given a class Foo (whether it is a new-style class or not), how do you generate all the base classes – anywhere in the inheritance hierarchy – it issubclass of?

Question 2

inspect.getmro(cls) works for both new and old style classes and returns the same as NewClass.mro(): a list of the class and all its ancestor classes, in the order used for method resolution.

>>> class A(object):
>>>     pass
>>>
>>> class B(A):
>>>     pass
>>>
>>> import inspect
>>> inspect.getmro(B)
(<class '__main__.B'>, <class '__main__.A'>, <type 'object'>)

Question 3

See the __bases__ property available on a python class, which contains a tuple of the bases classes:

>>> def classlookup(cls):
...     c = list(cls.__bases__)
...     for base in c:
...         c.extend(classlookup(base))
...     return c
...
>>> class A: pass
...
>>> class B(A): pass
...
>>> class C(object, B): pass
...
>>> classlookup(C)
[<type 'object'>, <class __main__.B at 0x00AB7300>, <class __main__.A at 0x00A6D630>]

Question 4

inspect.getclasstree() will create a nested list of classes and their bases. Usage:

inspect.getclasstree(inspect.getmro(IOError)) # Insert your Class instead of IOError.

Question 5

you can use the __bases__ tuple of the class object:

class A(object, B, C):
    def __init__(self):
       pass
print A.__bases__

The tuple returned by __bases__ has all its base classes.

Hope it helps!

Question 6

In python 3.7 you don’t need to import inspect, type.mro will give you the result.

>>> class A:
...   pass
... 
>>> class B(A):
...   pass
... 
>>> type.mro(B)
[<class '__main__.B'>, <class '__main__.A'>, <class 'object'>]
>>>

attention that in python 3.x every class inherits from base object class.

Question 7

According to the Python doc, we can also simply use class.__mro__ attribute or class.mro() method:

>>> class A:
...     pass
... 
>>> class B(A):
...     pass
... 
>>> B.__mro__
(<class '__main__.B'>, <class '__main__.A'>, <class 'object'>)
>>> A.__mro__
(<class '__main__.A'>, <class 'object'>)
>>> object.__mro__
(<class 'object'>,)
>>>
>>> B.mro()
[<class '__main__.B'>, <class '__main__.A'>, <class 'object'>]
>>> A.mro()
[<class '__main__.A'>, <class 'object'>]
>>> object.mro()
[<class 'object'>]
>>> A in B.mro()
True

Question 8

Although Jochen’s answer is very helpful and correct, as you can obtain the class hierarchy using the .getmro() method of the inspect module, it’s also important to highlight that Python’s inheritance hierarchy is as follows:

ex:

class MyClass(YourClass):

An inheriting class

Child class
Derived class
Subclass

ex:

class YourClass(Object):

An inherited class

Parent class
Base class
Superclass

One class can inherit from another – The class’ attributed are inherited – in particular, its methods are inherited – this means that instances of an inheriting (child) class can access attributed of the inherited (parent) class

instance -> class -> then inherited classes

using

import inspect
inspect.getmro(MyClass)

will show you the hierarchy, within Python.

Question 9

Suppose myapp/foo.py contains:

def info(msg):
    caller_name = ????
    print '[%s] %s' % (caller_name, msg)

And myapp/bar.py contains:

import foo
foo.info('Hello') # => [myapp.bar] Hello

I want caller_name to be set to the __name__ attribute of the calling functions’ module (which is ‘myapp.foo’) in this case. How can this be done?

Question 10

Check out the inspect module:

inspect.stack() will return the stack information.

Inside a function, inspect.stack()[1] will return your caller’s stack. From there, you can get more information about the caller’s function name, module, etc.

See the docs for details:

http://docs.python.org/library/inspect.html

Also, Doug Hellmann has a nice writeup of the inspect module in his PyMOTW series:

http://pymotw.com/2/inspect/index.html#module-inspect

EDIT: Here’s some code which does what you want, I think:

import inspect 

def info(msg):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    print '[%s] %s' % (mod.__name__, msg)

Question 11

Confronted with a similar problem, I have found that sys._current_frames() from the sys module contains interesting information that can help you, without the need to import inspect, at least in specific use cases.

>>> sys._current_frames()
{4052: <frame object at 0x03200C98>}

You can then “move up” using f_back :

>>> f = sys._current_frames().values()[0]
>>> # for python3: f = list(sys._current_frames().values())[0]

>>> print f.f_back.f_globals['__file__']
'/base/data/home/apps/apricot/1.6456165165151/caller.py'

>>> print f.f_back.f_globals['__name__']
'__main__'

For the filename you can also use f.f_back.f_code.co_filename, as suggested by Mark Roddy above. I am not sure of the limits and caveats of this method (multiple threads will most likely be a problem) but I intend to use it in my case.

Question 12

I don’t recommend do this, but you can accomplish your goal with the following method:

def caller_name():
    frame=inspect.currentframe()
    frame=frame.f_back.f_back
    code=frame.f_code
    return code.co_filename

Then update your existing method as follows:

def info(msg):
    caller = caller_name()
    print '[%s] %s' % (caller, msg)

Question 13

Given a class C in Python, how can I determine which file the class was defined in? I need something that can work from either the class C, or from an instance off C.

The reason I am doing this, is because I am generally a fan off putting files that belong together in the same folder. I want to create a class that uses a Django template to render itself as HTML. The base implementation should infer the filename for the template based on the filename that the class is defined in.

Say I put a class LocationArtifact in the file “base/artifacts.py”, then I want the default behaviour to be that the template name is “base/LocationArtifact.html”.

Question 14

You can use the inspect module, like this:

import inspect
inspect.getfile(C.__class__)

Question 15

try:

import sys, os
os.path.abspath(sys.modules[LocationArtifact.__module__].__file__)

Question 16

This is the wrong approach for Django and really forcing things.

The typical Django app pattern is:

/project
- /appname
  - models.py
  - views.py
  - /templates
    - index.html
    - etc.

Question 17

Is there a straightforward way to find all the modules that are part of a python package? I’ve found this old discussion, which is not really conclusive, but I’d love to have a definite answer before I roll out my own solution based on os.listdir().

Question 18

Yes, you want something based on pkgutil or similar — this way you can treat all packages alike regardless if they are in eggs or zips or so (where os.listdir won’t help).

import pkgutil

# this is the package we are inspecting -- for example 'email' from stdlib
import email

package = email
for importer, modname, ispkg in pkgutil.iter_modules(package.__path__):
    print "Found submodule %s (is a package: %s)" % (modname, ispkg)

How to import them too? You can just use __import__ as normal:

import pkgutil

# this is the package we are inspecting -- for example 'email' from stdlib
import email

package = email
prefix = package.__name__ + "."
for importer, modname, ispkg in pkgutil.iter_modules(package.__path__, prefix):
    print "Found submodule %s (is a package: %s)" % (modname, ispkg)
    module = __import__(modname, fromlist="dummy")
    print "Imported", module

Question 19

The right tool for this job is pkgutil.walk_packages.

To list all the modules on your system:

import pkgutil
for importer, modname, ispkg in pkgutil.walk_packages(path=None, onerror=lambda x: None):
    print(modname)

Be aware that walk_packages imports all subpackages, but not submodules.

If you wish to list all submodules of a certain package then you can use something like this:

import pkgutil
import scipy
package=scipy
for importer, modname, ispkg in pkgutil.walk_packages(path=package.__path__,
                                                      prefix=package.__name__+'.',
                                                      onerror=lambda x: None):
    print(modname)

iter_modules only lists the modules which are one-level deep. walk_packages gets all the submodules. In the case of scipy, for example, walk_packages returns

scipy.stats.stats

while iter_modules only returns

scipy.stats

The documentation on pkgutil (http://docs.python.org/library/pkgutil.html) does not list all the interesting functions defined in /usr/lib/python2.6/pkgutil.py.

Perhaps this means the functions are not part of the “public” interface and are subject to change.

However, at least as of Python 2.6 (and perhaps earlier versions?) pkgutil comes with a walk_packages method which recursively walks through all the modules available.

Question 20

This works for me:

import types

for key, obj in nltk.__dict__.iteritems():
    if type(obj) is types.ModuleType: 
        print key

Question 21

I was looking for a way to reload all submodules that I’m editing live in my package. It is a combination of the answers/comments above, so I’ve decided to post it here as an answer rather than a comment.

package=yourPackageName
import importlib
import pkgutil
for importer, modname, ispkg in pkgutil.walk_packages(path=package.__path__, prefix=package.__name__+'.', onerror=lambda x: None):
    try:
        modulesource = importlib.import_module(modname)
        reload(modulesource)
        print("reloaded: {}".format(modname))
    except Exception as e:
        print('Could not load {} {}'.format(modname, e))

Question 22

Here’s one way, off the top of my head:

>>> import os
>>> filter(lambda i: type(i) == type(os), [getattr(os, j) for j in dir(os)])
[<module 'UserDict' from '/usr/lib/python2.5/UserDict.pyc'>, <module 'copy_reg' from '/usr/lib/python2.5/copy_reg.pyc'>, <module 'errno' (built-in)>, <module 'posixpath' from '/usr/lib/python2.5/posixpath.pyc'>, <module 'sys' (built-in)>]

It could certainly be cleaned up and improved.

EDIT: Here’s a slightly nicer version:

>>> [m[1] for m in filter(lambda a: type(a[1]) == type(os), os.__dict__.items())]
[<module 'copy_reg' from '/usr/lib/python2.5/copy_reg.pyc'>, <module 'UserDict' from '/usr/lib/python2.5/UserDict.pyc'>, <module 'posixpath' from '/usr/lib/python2.5/posixpath.pyc'>, <module 'errno' (built-in)>, <module 'sys' (built-in)>]
>>> [m[0] for m in filter(lambda a: type(a[1]) == type(os), os.__dict__.items())]
['_copy_reg', 'UserDict', 'path', 'errno', 'sys']

NOTE: This will also find modules that might not necessarily be located in a subdirectory of the package, if they’re pulled in in its __init__.py file, so it depends on what you mean by “part of” a package.

Question 23

I have a dict, which I need to pass key/values as keyword arguments.. For example..

d_args = {'kw1': 'value1', 'kw2': 'value2'}
example(**d_args)

This works fine, but if there are values in the d_args dict that are not accepted by the example function, it obviously dies.. Say, if the example function is defined as def example(kw2):

This is a problem since I don’t control either the generation of the d_args, or the example function.. They both come from external modules, and example only accepts some of the keyword-arguments from the dict..

Ideally I would just do

parsed_kwargs = feedparser.parse(the_url)
valid_kwargs = get_valid_kwargs(parsed_kwargs, valid_for = PyRSS2Gen.RSS2)
PyRSS2Gen.RSS2(**valid_kwargs)

I will probably just filter the dict, from a list of valid keyword-arguments, but I was wondering: Is there a way to programatically list the keyword arguments the a specific function takes?

Question 24

A little nicer than inspecting the code object directly and working out the variables is to use the inspect module.

>>> import inspect
>>> def func(a,b,c=42, *args, **kwargs): pass
>>> inspect.getargspec(func)
(['a', 'b', 'c'], 'args', 'kwargs', (42,))

If you want to know if its callable with a particular set of args, you need the args without a default already specified. These can be got by:

def getRequiredArgs(func):
    args, varargs, varkw, defaults = inspect.getargspec(func)
    if defaults:
        args = args[:-len(defaults)]
    return args   # *args and **kwargs are not required, so ignore them.

Then a function to tell what you are missing from your particular dict is:

def missingArgs(func, argdict):
    return set(getRequiredArgs(func)).difference(argdict)

Similarly, to check for invalid args, use:

def invalidArgs(func, argdict):
    args, varargs, varkw, defaults = inspect.getargspec(func)
    if varkw: return set()  # All accepted
    return set(argdict) - set(args)

And so a full test if it is callable is :

def isCallableWithArgs(func, argdict):
    return not missingArgs(func, argdict) and not invalidArgs(func, argdict)

(This is good only as far as python’s arg parsing. Any runtime checks for invalid values in kwargs obviously can’t be detected.)

Question 25

This will print names of all passable arguments, keyword and non-keyword ones:

def func(one, two="value"):
    y = one, two
    return y
print func.func_code.co_varnames[:func.func_code.co_argcount]

This is because first co_varnames are always parameters (next are local variables, like y in the example above).

So now you could have a function:

def getValidArgs(func, argsDict):
    '''Return dictionary without invalid function arguments.'''
    validArgs = func.func_code.co_varnames[:func.func_code.co_argcount]
    return dict((key, value) for key, value in argsDict.iteritems() 
                if key in validArgs)

Which you then could use like this:

>>> func(**getValidArgs(func, args))

EDIT: A small addition: if you really need only keyword arguments of a function, you can use the func_defaults attribute to extract them:

def getValidKwargs(func, argsDict):
    validArgs = func.func_code.co_varnames[:func.func_code.co_argcount]
    kwargsLen = len(func.func_defaults) # number of keyword arguments
    validKwargs = validArgs[-kwargsLen:] # because kwargs are last
    return dict((key, value) for key, value in argsDict.iteritems() 
                if key in validKwargs)

You could now call your function with known args, but extracted kwargs, e.g.:

func(param1, param2, **getValidKwargs(func, kwargsDict))

This assumes that func uses no *args or **kwargs magic in its signature.

Question 26

In Python 3.0:

>>> import inspect
>>> import fileinput
>>> print(inspect.getfullargspec(fileinput.input))
FullArgSpec(args=['files', 'inplace', 'backup', 'bufsize', 'mode', 'openhook'],
varargs=None, varkw=None, defaults=(None, 0, '', 0, 'r', None), kwonlyargs=[], 
kwdefaults=None, annotations={})

Question 27

For a Python 3 solution, you can use inspect.signature and filter according to the kind of parameters you’d like to know about.

Taking a sample function with positional or keyword, keyword-only, var positional and var keyword parameters:

def spam(a, b=1, *args, c=2, **kwargs):
    print(a, b, args, c, kwargs)

You can create a signature object for it:

from inspect import signature
sig =  signature(spam)

and then filter with a list comprehension to find out the details you need:

>>> # positional or keyword
>>> [p.name for p in sig.parameters.values() if p.kind == p.POSITIONAL_OR_KEYWORD]
['a', 'b']
>>> # keyword only
>>> [p.name for p in sig.parameters.values() if p.kind == p.KEYWORD_ONLY]
['c']

and, similarly, for var positionals using p.VAR_POSITIONAL and var keyword with VAR_KEYWORD.

In addition, you can add a clause to the if to check if a default value exists by checking if p.default equals p.empty.

Question 28

Extending DzinX’s answer:

argnames = example.func_code.co_varnames[:func.func_code.co_argcount]
args = dict((key, val) for key,val in d_args.iteritems() if key in argnames)
example(**args)

Question 29

What does the / mean in Python 3.4’s help output for range before the closing parenthesis?

>>> help(range)
Help on class range in module builtins:

class range(object)
 |  range(stop) -> range object
 |  range(start, stop[, step]) -> range object
 |  
 |  Return a virtual sequence of numbers from start to stop by step.
 |  
 |  Methods defined here:
 |  
 |  __contains__(self, key, /)
 |      Return key in self.
 |  
 |  __eq__(self, value, /)
 |      Return self==value.

                                        ...

Question 30

It signifies the end of the positional only parameters, parameters you cannot use as keyword parameters. Before Python 3.8, such parameters could only be specified in the C API.

It means the key argument to __contains__ can only be passed in by position (range(5).__contains__(3)), not as a keyword argument (range(5).__contains__(key=3)), something you can do with positional arguments in pure-python functions.

Also see the Argument Clinic documentation:

To mark all parameters as positional-only in Argument Clinic, add a / on a line by itself after the last parameter, indented the same as the parameter lines.

and the (very recent addition to) the Python FAQ:

A slash in the argument list of a function denotes that the parameters prior to it are positional-only. Positional-only parameters are the ones without an externally-usable name. Upon calling a function that accepts positional-only parameters, arguments are mapped to parameters based solely on their position.

The syntax is now part of the Python language specification, as of version 3.8, see PEP 570 – Python Positional-Only Parameters. Before PEP 570, the syntax was already reserved for possible future inclusion in Python, see PEP 457 – Syntax For Positional-Only Parameters.

Positional-only parameters can lead to cleaner and clearer APIs, make pure-Python implementations of otherwise C-only modules more consistent and easier to maintain, and because positional-only parameters require very little processing, they lead to faster Python code.

Question 31

I asked this question myself. :) Found out that / was originally proposed by Guido in here.

Alternative proposal: how about using ‘/’ ? It’s kind of the opposite of ‘*’ which means “keyword argument”, and ‘/’ is not a new character.

Then his proposal won.

Heh. If that’s true, my ‘/’ proposal wins:
 def foo(pos_only, /, pos_or_kw, *, kw_only): ...

I think the very relevant document covering this is PEP 570. Where recap section looks nice.

Recap

The use case will determine which parameters to use in the function definition:
 def f(pos1, pos2, /, pos_or_kwd, *, kwd1, kwd2):
As guidance:

Use positional-only if names do not matter or have no meaning, and there are only a few arguments which will always be passed in the same order. Use keyword-only when names have meaning and the function definition is more understandable by being explicit with names.

If the function ends with /

def foo(p1, p2, /)

This means all functional arguments are positional.

Question 32

Forward Slash (/) indicates all arguments prior to it are positional only argument. Positional only arguments feature was added in python 3.8 after PEP 570 was accepted. Initially this notation was defined in PEP 457 – Notation for Notation For Positional-Only Parameters

Parameters in function definition prior Foraward slash (/) are positional only and parameters followed by slash(/) can be of any kind as per syntax. Where arguments are mapped to positional only parameters solely based on their position upon calling a function. Passing positional-only parameters by keywords(name) is invalid.

Let’s take following example

def foo(a, b, / , x, y):
   print("positional ", a, b)
   print("positional or keyword", x, y)

Here in the above function definition parameters a and b are positional-only, while x or y can be either positional or keyword.

Following function calls are valid

foo(40, 20, 99, 39)
foo(40, 3.14, "hello", y="world")
foo(1.45, 3.14, x="hello", y="world")

But, following function call is not valid which raises an exception TypeError since a, b are not passed as positional arguments instead passed as keyword

foo(a=1.45, b=3.14, x=1, y=4)

TypeError: foo() got some positional-only arguments passed as keyword arguments: ‘a, b’

Many built in function in python accept positional only arguments where passing arguments by keyword doesn’t make sense. For example built-in function len accepts only one positional(only) argument, Where calling len as len(obj=”hello world”) impairs readability, check help(len).

>>> help(len)
Help on built-in function len in module builtins:

len(obj, /)
    Return the number of items in a container.

Positional only parameters make underlying c/library functions easy to maintain. It allows parameters names of positional only parameters to be changes in future without risk of breaking client code that uses API

Last but not least, positional only parameters allow us to use their names to be used in variable length keyword arguments. Check following example

>>> def f(a, b, /, **kwargs):
...     print(a, b, kwargs)
...
>>> f(10, 20, a=1, b=2, c=3)         # a and b are used in two ways
10 20 {'a': 1, 'b': 2, 'c': 3}

Positional only parameters is better Explained here at Types of function arguments in python: Positional Only Parameters

Positional-only parameters syntax was officially added to python3.8. Checkout what’s new python3.8 – positional only arguments

PEP Related: PEP 570 — Python Positional-Only Parameters

Question 33

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?

Question 34

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.

Question 35

Shorter version:

import inspect

def f1(): f2()

def f2():
    print 'caller name:', inspect.stack()[1][3]

f1()

(with thanks to @Alex, and Stefaan Lippen)

Question 36

This seems to work just fine:

import sys
print sys._getframe().f_back.f_code.co_name

Question 37

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)

Question 38

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 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 caller_name() == '_test_caller_name'
    _test_caller_name()

Note that cast(FrameType, frame) is used to satisfy mypy.

Acknowlegement: prior comment by 1313e for an answer.

Question 39

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

Question 40

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()

Question 41

#!/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()
shahid@shahid-VirtualBox:~/Documents$ python test_func.py 
inside:  caller1
inside:  caller2
shahid@shahid-VirtualBox:~/Documents$

Question 42

Given the Python function:

def a_method(arg1, arg2):
    pass

How can I extract the number and names of the arguments. I.e., given that I have a reference to func, I want the func.[something] to return ("arg1", "arg2").

The usage scenario for this is that I have a decorator, and I wish to use the method arguments in the same order that they appear for the actual function as a key. I.e., how would the decorator look that printed "a,b" when I call a_method("a", "b")?

Question 43

Take a look at the inspect module – this will do the inspection of the various code object properties for you.

>>> inspect.getfullargspec(a_method)
(['arg1', 'arg2'], None, None, None)

The other results are the name of the *args and **kwargs variables, and the defaults provided. ie.

>>> def foo(a, b, c=4, *arglist, **keywords): pass
>>> inspect.getfullargspec(foo)
(['a', 'b', 'c'], 'arglist', 'keywords', (4,))

Note that some callables may not be introspectable in certain implementations of Python. For Example, in CPython, some built-in functions defined in C provide no metadata about their arguments. As a result, you will get a ValueError if you use inspect.getfullargspec() on a built-in function.

Since Python 3.3, you can use inspect.signature() to see the call signature of a callable object:

>>> inspect.signature(foo)
<Signature (a, b, c=4, *arglist, **keywords)>

Question 44

In CPython, the number of arguments is

a_method.func_code.co_argcount

and their names are in the beginning of

a_method.func_code.co_varnames

These are implementation details of CPython, so this probably does not work in other implementations of Python, such as IronPython and Jython.

One portable way to admit “pass-through” arguments is to define your function with the signature func(*args, **kwargs). This is used a lot in e.g. matplotlib, where the outer API layer passes lots of keyword arguments to the lower-level API.

Question 45

In a decorator method, you can list arguments of the original method in this way:

import inspect, itertools 

def my_decorator():

        def decorator(f):

            def wrapper(*args, **kwargs):

                # if you want arguments names as a list:
                args_name = inspect.getargspec(f)[0]
                print(args_name)

                # if you want names and values as a dictionary:
                args_dict = dict(itertools.izip(args_name, args))
                print(args_dict)

                # if you want values as a list:
                args_values = args_dict.values()
                print(args_values)

If the **kwargs are important for you, then it will be a bit complicated:

        def wrapper(*args, **kwargs):

            args_name = list(OrderedDict.fromkeys(inspect.getargspec(f)[0] + kwargs.keys()))
            args_dict = OrderedDict(list(itertools.izip(args_name, args)) + list(kwargs.iteritems()))
            args_values = args_dict.values()

Example:

@my_decorator()
def my_function(x, y, z=3):
    pass


my_function(1, y=2, z=3, w=0)
# prints:
# ['x', 'y', 'z', 'w']
# {'y': 2, 'x': 1, 'z': 3, 'w': 0}
# [1, 2, 3, 0]

Question 46

I think what you’re looking for is the locals method –


In [6]: def test(a, b):print locals()
   ...: 

In [7]: test(1,2)              
{'a': 1, 'b': 2}

Question 47

The Python 3 version is:

def _get_args_dict(fn, args, kwargs):
    args_names = fn.__code__.co_varnames[:fn.__code__.co_argcount]
    return {**dict(zip(args_names, args)), **kwargs}

The method returns a dictionary containing both args and kwargs.

Question 48

Here is something I think will work for what you want, using a decorator.

class LogWrappedFunction(object):
    def __init__(self, function):
        self.function = function

    def logAndCall(self, *arguments, **namedArguments):
        print "Calling %s with arguments %s and named arguments %s" %\
                      (self.function.func_name, arguments, namedArguments)
        self.function.__call__(*arguments, **namedArguments)

def logwrap(function):
    return LogWrappedFunction(function).logAndCall

@logwrap
def doSomething(spam, eggs, foo, bar):
    print "Doing something totally awesome with %s and %s." % (spam, eggs)


doSomething("beans","rice", foo="wiggity", bar="wack")

Run it, it will yield the following output:

C:\scripts>python decoratorExample.py
Calling doSomething with arguments ('beans', 'rice') and named arguments {'foo':
 'wiggity', 'bar': 'wack'}
Doing something totally awesome with beans and rice.

Question 49

Python 3.5+:

DeprecationWarning: inspect.getargspec() is deprecated, use inspect.signature() instead

So previously:

func_args = inspect.getargspec(function).args

Now:

func_args = list(inspect.signature(function).parameters.keys())

To test:

'arg' in list(inspect.signature(function).parameters.keys())

Given that we have function ‘function’ which takes argument ‘arg’, this will evaluate as True, otherwise as False.

Example from the Python console:

Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 07:18:10) [MSC v.1900 32 bit (Intel)] on win32
>>> import inspect
>>> 'iterable' in list(inspect.signature(sum).parameters.keys())
True

Question 50

In Python 3.+ with the Signature object at hand, an easy way to get a mapping between argument names to values, is using the Signature’s bind() method!

For example, here is a decorator for printing a map like that:

import inspect

def decorator(f):
    def wrapper(*args, **kwargs):
        bound_args = inspect.signature(f).bind(*args, **kwargs)
        bound_args.apply_defaults()
        print(dict(bound_args.arguments))

        return f(*args, **kwargs)

    return wrapper

@decorator
def foo(x, y, param_with_default="bars", **kwargs):
    pass

foo(1, 2, extra="baz")
# This will print: {'kwargs': {'extra': 'baz'}, 'param_with_default': 'bars', 'y': 2, 'x': 1}

Question 51

Here is another way to get the function parameters without using any module.

def get_parameters(func):
    keys = func.__code__.co_varnames[:func.__code__.co_argcount][::-1]
    sorter = {j: i for i, j in enumerate(keys[::-1])} 
    values = func.__defaults__[::-1]
    kwargs = {i: j for i, j in zip(keys, values)}
    sorted_args = tuple(
        sorted([i for i in keys if i not in kwargs], key=sorter.get)
    )
    sorted_kwargs = {
        i: kwargs[i] for i in sorted(kwargs.keys(), key=sorter.get)
    }   
    return sorted_args, sorted_kwargs


def f(a, b, c="hello", d="world"): var = a
    

print(get_parameters(f))

Output:

(('a', 'b'), {'c': 'hello', 'd': 'world'})

Question 52

Returns a list of argument names, takes care of partials and regular functions:

def get_func_args(f):
    if hasattr(f, 'args'):
        return f.args
    else:
        return list(inspect.signature(f).parameters)

Question 53

Update for Brian’s answer:

If a function in Python 3 has keyword-only arguments, then you need to use inspect.getfullargspec:

def yay(a, b=10, *, c=20, d=30):
    pass
inspect.getfullargspec(yay)

yields this:

FullArgSpec(args=['a', 'b'], varargs=None, varkw=None, defaults=(10,), kwonlyargs=['c', 'd'], kwonlydefaults={'c': 20, 'd': 30}, annotations={})

Question 54

In python 3, below is to make *args and **kwargs into a dict (use OrderedDict for python < 3.6 to maintain dict orders):

from functools import wraps

def display_param(func):
    @wraps(func)
    def wrapper(*args, **kwargs):

        param = inspect.signature(func).parameters
        all_param = {
            k: args[n] if n < len(args) else v.default
            for n, (k, v) in enumerate(param.items()) if k != 'kwargs'
        }
        all_param .update(kwargs)
        print(all_param)

        return func(**all_param)
    return wrapper

Question 55

inspect.signature is very slow. Fastest way is

def f(a, b=1, *args, c, d=1, **kwargs):
   pass

f_code = f.__code__
f_code.co_varnames[:f_code.co_argcount + f_code.co_kwonlyargcount]  # ('a', 'b', 'c', 'd')

Question 56

To update a little bit Brian’s answer, there is now a nice backport of inspect.signature that you can use in older python versions: funcsigs. So my personal preference would go for

try:  # python 3.3+
    from inspect import signature
except ImportError:
    from funcsigs import signature

def aMethod(arg1, arg2):
    pass

sig = signature(aMethod)
print(sig)

For fun, if you’re interested in playing with Signature objects and even creating functions with random signatures dynamically you can have a look at my makefun project.

Question 57

What about dir() and vars() now?

Seems doing exactly what is being asked super simply…

Must be called from within the function scope.

But be wary that it will return all local variables so be sure to do it at the very beginning of the function if needed.

Also note that, as pointed out in the comments, this doesn’t allow it to be done from outside the scope. So not exactly OP’s scenario but still matches the question title. Hence my answer.

Question 58

I’m starting to code in various projects using Python (including Django web development and Panda3D game development).

To help me understand what’s going on, I would like to basically ‘look’ inside the Python objects to see how they tick – like their methods and properties.

So say I have a Python object, what would I need to print out its contents? Is that even possible?

Question 59

Python has a strong set of introspection features.

Take a look at the following built-in functions:

type() and dir() are particularly useful for inspecting the type of an object and its set of attributes, respectively.

Question 60

object.__dict__

Question 61

First, read the source.

Second, use the dir() function.

Question 62

I’m surprised no one’s mentioned help yet!

In [1]: def foo():
   ...:     "foo!"
   ...:

In [2]: help(foo)
Help on function foo in module __main__:

foo()
    foo!

Help lets you read the docstring and get an idea of what attributes a class might have, which is pretty helpful.

Question 63

If this is for exploration to see what’s going on, I’d recommend looking at IPython. This adds various shortcuts to obtain an objects documentation, properties and even source code. For instance appending a “?” to a function will give the help for the object (effectively a shortcut for “help(obj)”, wheras using two ?’s (“func??“) will display the sourcecode if it is available.

There are also a lot of additional conveniences, like tab completion, pretty printing of results, result history etc. that make it very handy for this sort of exploratory programming.

For more programmatic use of introspection, the basic builtins like dir(), vars(), getattr etc will be useful, but it is well worth your time to check out the inspect module. To fetch the source of a function, use “inspect.getsource” eg, applying it to itself:

>>> print inspect.getsource(inspect.getsource)
def getsource(object):
    """Return the text of the source code for an object.

    The argument may be a module, class, method, function, traceback, frame,
    or code object.  The source code is returned as a single string.  An
    IOError is raised if the source code cannot be retrieved."""
    lines, lnum = getsourcelines(object)
    return string.join(lines, '')

inspect.getargspec is also frequently useful if you’re dealing with wrapping or manipulating functions, as it will give the names and default values of function parameters.

Question 64

If you’re interested in a GUI for this, take a look at objbrowser. It uses the inspect module from the Python standard library for the object introspection underneath.

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