Python 实用宝典

Question 1

I C# we do it through reflection. In Javascript it is simple as:

for(var propertyName in objectName)
    var currentPropertyValue = objectName[propertyName];

How to do it in Python?

Question 2

for property, value in vars(theObject).items():
    print(property, ":", value)

Be aware that in some rare cases there’s a __slots__ property, such classes often have no __dict__.

Question 3

See inspect.getmembers(object[, predicate]).

Return all the members of an object in a list of (name, value) pairs sorted by name. If the optional predicate argument is supplied, only members for which the predicate returns a true value are included.

>>> [name for name,thing in inspect.getmembers([])]
['__add__', '__class__', '__contains__', '__delattr__', '__delitem__', 
'__delslice__',    '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', 
'__getitem__', '__getslice__', '__gt__', '__hash__', '__iadd__', '__imul__', '__init__', '__iter__', 
'__le__', '__len__', '__lt__', '__mul__', '__ne__', '__new__', '__reduce__','__reduce_ex__', 
'__repr__', '__reversed__', '__rmul__', '__setattr__', '__setitem__', '__setslice__', 
'__sizeof__', '__str__', '__subclasshook__', 'append', 'count', 'extend', 'index', 
'insert', 'pop', 'remove', 'reverse', 'sort']
>>>

Question 4

dir() is the simple way. See here:

Guide To Python Introspection

Question 5

The __dict__ property of the object is a dictionary of all its other defined properties. Note that Python classes can override getattr and make things that look like properties but are not in__dict__. There’s also the builtin functions vars() and dir() which are different in subtle ways. And __slots__ can replace __dict__ in some unusual classes.

Objects are complicated in Python. __dict__ is the right place to start for reflection-style programming. dir() is the place to start if you’re hacking around in an interactive shell.

Question 6

for one-liners:

print vars(theObject)

Question 7

If you’re looking for reflection of all properties, the answers above are great.

If you’re simply looking to get the keys of a dictionary (which is different from an ‘object’ in Python), use

my_dict.keys()

my_dict = {'abc': {}, 'def': 12, 'ghi': 'string' }
my_dict.keys() 
> ['abc', 'def', 'ghi']

Question 8

This is totally covered by the other answers, but I’ll make it explicit. An object may have class attributes and static and dynamic instance attributes.

class foo:
    classy = 1
    @property
    def dyno(self):
        return 1
    def __init__(self):
        self.stasis = 2

    def fx(self):
        return 3

stasis is static, dyno is dynamic (cf. property decorator) and classy is a class attribute. If we simply do __dict__ or vars we will only get the static one.

o = foo()
print(o.__dict__) #{'stasis': 2}
print(vars(o)) #{'stasis': 2}

So if we want the others __dict__ will get everything (and more). This includes magic methods and attributes and normal bound methods. So lets avoid those:

d = {k: getattr(o, k, '') for k in o.__dir__() if k[:2] != '__' and type(getattr(o, k, '')).__name__ != 'method'}
print(d) #{'stasis': 2, 'classy': 1, 'dyno': 1}

The type called with a property decorated method (a dynamic attribute) will give you the type of the returned value, not method. To prove this let’s json stringify it:

import json
print(json.dumps(d)) #{"stasis": 2, "classy": 1, "dyno": 1}

Had it been a method it would have crashed.

TL;DR. try calling extravar = lambda o: {k: getattr(o, k, '') for k in o.__dir__() if k[:2] != '__' and type(getattr(o, k, '')).__name__ != 'method'} for all three, but not methods nor magic.

Question 9

Given a string of a Python class, e.g. my_package.my_module.MyClass, what is the best possible way to load it?

In other words I am looking for a equivalent Class.forName() in Java, function in Python. It needs to work on Google App Engine.

Preferably this would be a function that accepts the FQN of the class as a string, and returns a reference to the class:

my_class = load_class('my_package.my_module.MyClass')
my_instance = my_class()

Question 10

From the python documentation, here’s the function you want:

def my_import(name):
    components = name.split('.')
    mod = __import__(components[0])
    for comp in components[1:]:
        mod = getattr(mod, comp)
    return mod

The reason a simple __import__ won’t work is because any import of anything past the first dot in a package string is an attribute of the module you’re importing. Thus, something like this won’t work:

__import__('foo.bar.baz.qux')

You’d have to call the above function like so:

my_import('foo.bar.baz.qux')

Or in the case of your example:

klass = my_import('my_package.my_module.my_class')
some_object = klass()

EDIT: I was a bit off on this. What you’re basically wanting to do is this:

from my_package.my_module import my_class

The above function is only necessary if you have a empty fromlist. Thus, the appropriate call would be like this:

mod = __import__('my_package.my_module', fromlist=['my_class'])
klass = getattr(mod, 'my_class')

Question 11

If you don’t want to roll your own, there is a function available in the pydoc module that does exactly this:

from pydoc import locate
my_class = locate('my_package.my_module.MyClass')

The advantage of this approach over the others listed here is that locate will find any python object at the provided dotted path, not just an object directly within a module. e.g. my_package.my_module.MyClass.attr.

If you’re curious what their recipe is, here’s the function:

def locate(path, forceload=0):
    """Locate an object by name or dotted path, importing as necessary."""
    parts = [part for part in split(path, '.') if part]
    module, n = None, 0
    while n < len(parts):
        nextmodule = safeimport(join(parts[:n+1], '.'), forceload)
        if nextmodule: module, n = nextmodule, n + 1
        else: break
    if module:
        object = module
    else:
        object = __builtin__
    for part in parts[n:]:
        try:
            object = getattr(object, part)
        except AttributeError:
            return None
    return object

It relies on pydoc.safeimport function. Here are the docs for that:

"""Import a module; handle errors; return None if the module isn't found.

If the module *is* found but an exception occurs, it's wrapped in an
ErrorDuringImport exception and reraised.  Unlike __import__, if a
package path is specified, the module at the end of the path is returned,
not the package at the beginning.  If the optional 'forceload' argument
is 1, we reload the module from disk (unless it's a dynamic extension)."""

Question 12

import importlib

module = importlib.import_module('my_package.my_module')
my_class = getattr(module, 'MyClass')
my_instance = my_class()

Question 13

def import_class(cl):
    d = cl.rfind(".")
    classname = cl[d+1:len(cl)]
    m = __import__(cl[0:d], globals(), locals(), [classname])
    return getattr(m, classname)

Question 14

If you’re using Django you can use this. Yes i’m aware OP did not ask for django, but i ran across this question looking for a Django solution, didn’t find one, and put it here for the next boy/gal that looks for it.

# It's available for v1.7+
# https://github.com/django/django/blob/stable/1.7.x/django/utils/module_loading.py
from django.utils.module_loading import import_string

Klass = import_string('path.to.module.Klass')
func = import_string('path.to.module.func')
var = import_string('path.to.module.var')

Keep in mind, if you want to import something that doesn’t have a ., like re or argparse use:

re = __import__('re')

Question 15

Here is to share something I found on __import__ and importlib while trying to solve this problem.

I am using Python 3.7.3.

When I try to get to the class d in module a.b.c,

mod = __import__('a.b.c')

The mod variable refer to the top namespace a.

So to get to the class d, I need to

mod = getattr(mod, 'b') #mod is now module b
mod = getattr(mod, 'c') #mod is now module c
mod = getattr(mod, 'd') #mod is now class d

If we try to do

mod = __import__('a.b.c')
d = getattr(mod, 'd')

we are actually trying to look for a.d.

When using importlib, I suppose the library has done the recursive getattr for us. So, when we use importlib.import_module, we actually get a handle on the deepest module.

mod = importlib.import_module('a.b.c') #mod is module c
d = getattr(mod, 'd') #this is a.b.c.d

Question 16

OK, for me that is the way it worked (I am using Python 2.7):

a = __import__('file_to_import', globals(), locals(), ['*'], -1)
b = a.MyClass()

Then, b is an instance of class ‘MyClass’

Question 17

If you happen to already have an instance of your desired class, you can use the ‘type’ function to extract its class type and use this to construct a new instance:

class Something(object):
    def __init__(self, name):
        self.name = name
    def display(self):
        print(self.name)

one = Something("one")
one.display()
cls = type(one)
two = cls("two")
two.display()

Question 18

module = __import__("my_package/my_module")
the_class = getattr(module, "MyClass")
obj = the_class()

Question 19

In Google App Engine there is a webapp2 function called import_string. For more info see here:https://webapp-improved.appspot.com/api/webapp2.html

So,

import webapp2
my_class = webapp2.import_string('my_package.my_module.MyClass')

For example this is used in the webapp2.Route where you can either use a handler or a string.

Question 20

I’m wondering how to convert a python ‘type’ object into a string using python’s reflective capabilities.

For example, I’d like to print the type of an object

print "My type is " + type(someObject) # (which obviously doesn't work like this)

Question 21

print type(someObject).__name__

If that doesn’t suit you, use this:

print some_instance.__class__.__name__

Example:

class A:
    pass
print type(A())
# prints <type 'instance'>
print A().__class__.__name__
# prints A

Also, it seems there are differences with type() when using new-style classes vs old-style (that is, inheritance from object). For a new-style class, type(someObject).__name__ returns the name, and for old-style classes it returns instance.

Question 22

>>> class A(object): pass

>>> e = A()
>>> e
<__main__.A object at 0xb6d464ec>
>>> print type(e)
<class '__main__.A'>
>>> print type(e).__name__
A
>>>

what do you mean by convert into a string? you can define your own repr and str_ methods:

>>> class A(object):
    def __repr__(self):
        return 'hei, i am A or B or whatever'

>>> e = A()
>>> e
hei, i am A or B or whatever
>>> str(e)
hei, i am A or B or whatever

or i dont know..please add explainations ;)

Question 23

print("My type is %s" % type(someObject)) # the type in python

or…

print("My type is %s" % type(someObject).__name__) # the object's type (the class you defined)

Question 24

Using str()

 typeOfOneAsString=str(type(1))

Question 25

In case you want to use str() and a custom str method. This also works for repr.

class TypeProxy:
    def __init__(self, _type):
        self._type = _type

    def __call__(self, *args, **kwargs):
        return self._type(*args, **kwargs)

    def __str__(self):
        return self._type.__name__

    def __repr__(self):
        return "TypeProxy(%s)" % (repr(self._type),)

>>> str(TypeProxy(str))
'str'
>>> str(TypeProxy(type("")))
'str'

Question 26

From something like this:

print(get_indentation_level())

    print(get_indentation_level())

        print(get_indentation_level())

I would like to get something like this:

1
2
3

Can the code read itself in this way?

All I want is the output from the more nested parts of the code to be more nested. In the same way that this makes code easier to read, it would make the output easier to read.

Of course I could implement this manually, using e.g. .format(), but what I had in mind was a custom print function which would print(i*' ' + string) where i is the indentation level. This would be a quick way to make readable output on my terminal.

Is there a better way to do this which avoids painstaking manual formatting?

Question 27

If you want indentation in terms of nesting level rather than spaces and tabs, things get tricky. For example, in the following code:

if True:
    print(
get_nesting_level())

the call to get_nesting_level is actually nested one level deep, despite the fact that there is no leading whitespace on the line of the get_nesting_level call. Meanwhile, in the following code:

print(1,
      2,
      get_nesting_level())

the call to get_nesting_level is nested zero levels deep, despite the presence of leading whitespace on its line.

In the following code:

if True:
  if True:
    print(get_nesting_level())

if True:
    print(get_nesting_level())

the two calls to get_nesting_level are at different nesting levels, despite the fact that the leading whitespace is identical.

In the following code:

if True: print(get_nesting_level())

is that nested zero levels, or one? In terms of INDENT and DEDENT tokens in the formal grammar, it’s zero levels deep, but you might not feel the same way.

If you want to do this, you’re going to have to tokenize the whole file up to the point of the call and count INDENT and DEDENT tokens. The tokenize module would be very useful for such a function:

import inspect
import tokenize

def get_nesting_level():
    caller_frame = inspect.currentframe().f_back
    filename, caller_lineno, _, _, _ = inspect.getframeinfo(caller_frame)
    with open(filename) as f:
        indentation_level = 0
        for token_record in tokenize.generate_tokens(f.readline):
            token_type, _, (token_lineno, _), _, _ = token_record
            if token_lineno > caller_lineno:
                break
            elif token_type == tokenize.INDENT:
                indentation_level += 1
            elif token_type == tokenize.DEDENT:
                indentation_level -= 1
        return indentation_level

Question 28

Yeah, that’s definitely possible, here’s a working example:

import inspect

def get_indentation_level():
    callerframerecord = inspect.stack()[1]
    frame = callerframerecord[0]
    info = inspect.getframeinfo(frame)
    cc = info.code_context[0]
    return len(cc) - len(cc.lstrip())

if 1:
    print get_indentation_level()
    if 1:
        print get_indentation_level()
        if 1:
            print get_indentation_level()

Question 29

You can use sys.current_frame.f_lineno in order to get the line number. Then in order to find the number of indentation level you need to find the previous line with zero indentation then be subtracting the current line number from that line’s number you’ll get the number of indentation:

import sys
current_frame = sys._getframe(0)

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()
    current_line_no = current_frame.f_lineno
    to_current = lines[:current_line_no]
    previous_zoro_ind = len(to_current) - next(i for i, line in enumerate(to_current[::-1]) if not line[0].isspace())
    return current_line_no - previous_zoro_ind

Demo:

if True:
    print get_ind_num()
    if True:
        print(get_ind_num())
        if True:
            print(get_ind_num())
            if True: print(get_ind_num())
# Output
1
3
5
6

If you want the number of the indentation level based on the previouse lines with : you can just do it with a little change:

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()

    current_line_no = current_frame.f_lineno
    to_current = lines[:current_line_no]
    previous_zoro_ind = len(to_current) - next(i for i, line in enumerate(to_current[::-1]) if not line[0].isspace())
    return sum(1 for line in lines[previous_zoro_ind-1:current_line_no] if line.strip().endswith(':'))

Demo:

if True:
    print get_ind_num()
    if True:
        print(get_ind_num())
        if True:
            print(get_ind_num())
            if True: print(get_ind_num())
# Output
1
2
3
3

And as an alternative answer here is a function for getting the number of indentation (whitespace):

import sys
from itertools import takewhile
current_frame = sys._getframe(0)

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()
    return sum(1 for _ in takewhile(str.isspace, lines[current_frame.f_lineno - 1]))

Question 30

To solve the ”real” problem that lead to your question you could implement a contextmanager which keeps track of the indention level and make the with block structure in the code correspond to the indentation levels of the output. This way the code indentation still reflects the output indentation without coupling both too much. It is still possible to refactor the code into different functions and have other indentations based on code structure not messing with the output indentation.

#!/usr/bin/env python
# coding: utf8
from __future__ import absolute_import, division, print_function


class IndentedPrinter(object):

    def __init__(self, level=0, indent_with='  '):
        self.level = level
        self.indent_with = indent_with

    def __enter__(self):
        self.level += 1
        return self

    def __exit__(self, *_args):
        self.level -= 1

    def print(self, arg='', *args, **kwargs):
        print(self.indent_with * self.level + str(arg), *args, **kwargs)


def main():
    indented = IndentedPrinter()
    indented.print(indented.level)
    with indented:
        indented.print(indented.level)
        with indented:
            indented.print('Hallo', indented.level)
            with indented:
                indented.print(indented.level)
            indented.print('and back one level', indented.level)


if __name__ == '__main__':
    main()

Output:

0
  1
    Hallo 2
      3
    and back one level 2

Question 31

>>> import inspect
>>> help(inspect.indentsize)
Help on function indentsize in module inspect:

indentsize(line)
    Return the indent size, in spaces, at the start of a line of text.

Question 32

I was wondering how to check whether a variable is a class (not an instance!) or not.

I’ve tried to use the function isinstance(object, class_or_type_or_tuple) to do this, but I don’t know what type would a class will have.

For example, in the following code

class Foo: pass  
isinstance(Foo, **???**) # i want to make this return True.

I tried to substitute “class” with ???, but I realized that class is a keyword in python.

Question 33

Even better: use the inspect.isclass function.

>>> import inspect
>>> class X(object):
...     pass
... 
>>> inspect.isclass(X)
True

>>> x = X()
>>> isinstance(x, X)
True
>>> y = 25
>>> isinstance(y, X)
False

Question 34

The inspect.isclass is probably the best solution, and it’s really easy to see how it’s actually implemented

def isclass(object):
    """Return true if the object is a class.

    Class objects provide these attributes:
        __doc__         documentation string
        __module__      name of module in which this class was defined"""
    return isinstance(object, (type, types.ClassType))

Question 35

>>> class X(object):
...     pass
... 
>>> type(X)
<type 'type'>
>>> isinstance(X,type)
True

Question 36

isinstance(X, type)

Return True if X is class and False if not.

Question 37

This check is compatible with both Python 2.x and Python 3.x.

import six
isinstance(obj, six.class_types)

This is basically a wrapper function that performs the same check as in andrea_crotti answer.

Example:

>>> import datetime
>>> isinstance(datetime.date, six.class_types)
>>> True
>>> isinstance(datetime.date.min, six.class_types)
>>> False

Question 38

class Foo: is called old style class and class X(object): is called new style class.

Check this What is the difference between old style and new style classes in Python? . New style is recommended. Read about “unifying types and classes“

Question 39

simplest way is to use inspect.isclass as posted in the most-voted answer.
the implementation details could be found at python2 inspect and python3 inspect.
for new-style class: isinstance(object, type)
for old-style class: isinstance(object, types.ClassType)
em, for old-style class, it is using types.ClassType, here is the code from types.py:

class _C:
    def _m(self): pass
ClassType = type(_C)

Question 40

Benjamin Peterson is correct about the use of inspect.isclass() for this job. But note that you can test if a Class object is a specific Class, and therefore implicitly a Class, using the built-in function issubclass. Depending on your use-case this can be more pythonic.

from typing import Type, Any
def isclass(cl: Type[Any]):
    try:
        return issubclass(cl, cl)
    except TypeError:
        return False

Can then be used like this:

>>> class X():
...     pass
... 
>>> isclass(X)
True
>>> isclass(X())
False

Question 41

There are some working solutions here already, but here’s another one:

>>> import types
>>> class Dummy: pass
>>> type(Dummy) is types.ClassType
True

Question 42

I’ve seen plenty of examples of people extracting all of the classes from a module, usually something like:

# foo.py
class Foo:
    pass

# test.py
import inspect
import foo

for name, obj in inspect.getmembers(foo):
    if inspect.isclass(obj):
        print obj

Awesome.

But I can’t find out how to get all of the classes from the current module.

# foo.py
import inspect

class Foo:
    pass

def print_classes():
    for name, obj in inspect.getmembers(???): # what do I do here?
        if inspect.isclass(obj):
            print obj

# test.py
import foo

foo.print_classes()

This is probably something really obvious, but I haven’t been able to find anything. Can anyone help me out?

Question 43

Try this:

import sys
current_module = sys.modules[__name__]

In your context:

import sys, inspect
def print_classes():
    for name, obj in inspect.getmembers(sys.modules[__name__]):
        if inspect.isclass(obj):
            print(obj)

And even better:

clsmembers = inspect.getmembers(sys.modules[__name__], inspect.isclass)

Because inspect.getmembers() takes a predicate.

Question 44

What about

g = globals().copy()
for name, obj in g.iteritems():

?

Question 45

I don’t know if there’s a ‘proper’ way to do it, but your snippet is on the right track: just add import foo to foo.py, do inspect.getmembers(foo), and it should work fine.

Question 46

I was able to get all I needed from the dir built in plus getattr.

# Works on pretty much everything, but be mindful that 
# you get lists of strings back

print dir(myproject)
print dir(myproject.mymodule)
print dir(myproject.mymodule.myfile)
print dir(myproject.mymodule.myfile.myclass)

# But, the string names can be resolved with getattr, (as seen below)

Though, it does come out looking like a hairball:

def list_supported_platforms():
    """
        List supported platforms (to match sys.platform)

        @Retirms:
            list str: platform names
    """
    return list(itertools.chain(
        *list(
            # Get the class's constant
            getattr(
                # Get the module's first class, which we wrote
                getattr(
                    # Get the module
                    getattr(platforms, item),
                    dir(
                        getattr(platforms, item)
                    )[0]
                ),
                'SYS_PLATFORMS'
            )
            # For each include in platforms/__init__.py 
            for item in dir(platforms)
            # Ignore magic, ourselves (index.py) and a base class.
            if not item.startswith('__') and item not in ['index', 'base']
        )
    ))

Question 47

import pyclbr
print(pyclbr.readmodule(__name__).keys())

Note that the stdlib’s Python class browser module uses static source analysis, so it only works for modules that are backed by a real .py file.

Question 48

If you want to have all the classes, that belong to the current module, you could use this :

import sys, inspect
def print_classes():
    is_class_member = lambda member: inspect.isclass(member) and member.__module__ == __name__
    clsmembers = inspect.getmembers(sys.modules[__name__], is_class_member)

If you use Nadia’s answer and you were importing other classes on your module, that classes will be being imported too.

So that’s why member.__module__ == __name__ is being added to the predicate used on is_class_member. This statement checks that the class really belongs to the module.

A predicate is a function (callable), that returns a boolean value.

Question 49

Another solution which works in Python 2 and 3:

#foo.py
import sys

class Foo(object):
    pass

def print_classes():
    current_module = sys.modules[__name__]
    for key in dir(current_module):
        if isinstance( getattr(current_module, key), type ):
            print(key)

# test.py
import foo
foo.print_classes()

Question 50

This is the line that I use to get all of the classes that have been defined in the current module (ie not imported). It’s a little long according to PEP-8 but you can change it as you see fit.

import sys
import inspect

classes = [name for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass) 
          if obj.__module__ is __name__]

This gives you a list of the class names. If you want the class objects themselves just keep obj instead.

classes = [obj for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass)
          if obj.__module__ is __name__]

This is has been more useful in my experience.

Question 51

import Foo 
dir(Foo)

import collections
dir(collections)

Question 52

I think that you can do something like this.

class custom(object):
    __custom__ = True
class Alpha(custom):
    something = 3
def GetClasses():
    return [x for x in globals() if hasattr(globals()[str(x)], '__custom__')]
print(GetClasses())`

if you need own classes

Question 53

I frequently find myself writing command line utilities wherein the first argument is meant to refer to one of many different classes. For example ./something.py feature command —-arguments, where Feature is a class and command is a method on that class. Here’s a base class that makes this easy.

The assumption is that this base class resides in a directory alongside all of its subclasses. You can then call ArgBaseClass(foo = bar).load_subclasses() which will return a dictionary. For example, if the directory looks like this:

arg_base_class.py
feature.py

Assuming feature.py implements class Feature(ArgBaseClass), then the above invocation of load_subclasses will return { 'feature' : <Feature object> }. The same kwargs (foo = bar) will be passed into the Feature class.

#!/usr/bin/env python3
import os, pkgutil, importlib, inspect

class ArgBaseClass():
    # Assign all keyword arguments as properties on self, and keep the kwargs for later.
    def __init__(self, **kwargs):
        self._kwargs = kwargs
        for (k, v) in kwargs.items():
            setattr(self, k, v)
        ms = inspect.getmembers(self, predicate=inspect.ismethod)
        self.methods = dict([(n, m) for (n, m) in ms if not n.startswith('_')])

    # Add the names of the methods to a parser object.
    def _parse_arguments(self, parser):
        parser.add_argument('method', choices=list(self.methods))
        return parser

    # Instantiate one of each of the subclasses of this class.
    def load_subclasses(self):
        module_dir = os.path.dirname(__file__)
        module_name = os.path.basename(os.path.normpath(module_dir))
        parent_class = self.__class__
        modules = {}
        # Load all the modules it the package:
        for (module_loader, name, ispkg) in pkgutil.iter_modules([module_dir]):
            modules[name] = importlib.import_module('.' + name, module_name)

        # Instantiate one of each class, passing the keyword arguments.
        ret = {}
        for cls in parent_class.__subclasses__():
            path = cls.__module__.split('.')
            ret[path[-1]] = cls(**self._kwargs)
        return ret

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