zip和*在python中的具体例子

The python docs gives this code as the reverse operation of zip:

>>> x2, y2 = zip(*zipped)

In particular "zip() in conjunction with the * operator can be used to unzip a list". Can someone explain to me how the * operator works in this case? As far as I understand, * is a binary operator and can be used for multiplication or shallow copy...neither of which seems to be the case here.

解释一：

zip(*zipped) means "feed each element of zipped as an argument to zip". zip is similar to transposing a matrix in that doing it again will leave you back where you started.

>>> a = [(1, 2, 3), (4, 5, 6)]>>> b = zip(*a)>>> b[(1, 4), (2, 5), (3, 6)]>>> zip(*b)[(1, 2, 3), (4, 5, 6)]

关于*和**的详细解释二：

Although hammar's answer explains how the reversing works in the case of the zip() function, it may be useful to look at argument unpacking in a more general sense. Let's say we have a simple function which takes some arguments:

>>> def do_something(arg1, arg2, arg3):...     print 'arg1: %s' % arg1...     print 'arg2: %s' % arg2...     print 'arg3: %s' % arg3... >>> do_something(1, 2, 3)arg1: 1arg2: 2arg3: 3

Instead of directly specifying the arguments, we can create a list (or tuple for that matter) to hold them, and then tell Python to unpack that list and use its contents as the arguments to the function:

>>> arguments = [42, 'insert value here', 3.14]>>> do_something(*arguments)arg1: 42arg2: insert value herearg3: 3.14

This behaves as normal if you don't have enough arguments (or too many):

>>> arguments = [42, 'insert value here']>>> do_something(*arguments)---------------------------------------------------------------------------TypeError                                 Traceback (most recent call last)/home/blair/<ipython console> in <module>()TypeError: do_something() takes exactly 3 arguments (2 given)

You can use the same construct when defining a function to accept any number of positional arguments. They are given to your function as a tuple:

>>> def show_args(*args):...     for index, value in enumerate(args):...         print 'Argument %d: %s' % (index, value)...>>> show_args(1, 2, 3)Argument 0: 1Argument 1: 2Argument 2: 3

And of course you can combine the two techniques:

>>> show_args(*arguments)Argument 0: 42Argument 1: insert value here

You can do a similar thing with keyword arguments, using a double asterix (**) and a dictionary:

>>> def show_kwargs(**kwargs):...     for arg, value in kwargs.items():...         print '%s = %s' % (arg, value)...>>> show_kwargs(age=24, name='Blair')age = 24name = Blair

And, of course, you can pass keyword arguments through a dictionary:

>>> values = {'name': 'John', 'age': 17}>>> show_kwargs(**values)age = 17name = John

It is perfectly acceptable to mix the two, and you can always have required arguments and optional extra arguments to a function:

>>> def mixed(required_arg, *args, **kwargs):...     print 'Required: %s' % required_arg...     if args:...         print 'Extra positional arguments: %s' % str(args)...     if kwargs:...         print 'Extra keyword arguments: %s' % kwargs...>>> mixed(1)Required: 1>>> mixed(1, 2, 3)Required: 1Extra positional arguments: (2, 3)>>> mixed(1, 2, 3, test=True)Required: 1Extra positional arguments: (2, 3)Extra keyword arguments: {'test': True}>>> args = (2, 3, 4)>>> kwargs = {'test': True, 'func': min}>>> mixed(*args, **kwargs)Required: 2Extra positional arguments: (3, 4)Extra keyword arguments: {'test': True, 'func': <built-in function min>}

If you are taking optional keyword arguments and you want to have default values, remember you are dealing with a dictionary and hence you can use its get() method with a default value to use if the key does not exist:

>>> def take_keywords(**kwargs):...     print 'Test mode: %s' % kwargs.get('test', False)...     print 'Combining function: %s' % kwargs.get('func', all)... >>> take_keywords()Test mode: FalseCombining function: <built-in function all>>>> take_keywords(func=any)Test mode: FalseCombining function: <built-in function any>