标签：
今天看到一个外国人写的一篇关于*args，与**kwargs如何使用的总结，非常有学习价值，尤其是他给出了一个继承之后重写父类方法的例子，可以很好的解决多重继承中参数传递的问题，看完之后，自己也实验了一下。原文链接http://agiliq.com/blog/2012/06/understanding-args-and-kwargs/。
原文如下：
When i started learning Python, i was very confused regarding what args, kwargs, * and ** does. And i feel there are few like me who had this confusion and problem. With this post, i intend to reduce (hopefully i can eliminate) that confusion.
Throughout this post, i will be using ipython and i suggest you to try everything on ipython as well. We will intentionally make some mistakes along the way, so that we can understand this topic better.
Let‘s define a function "fun" which takes three positional arguments.
In [5]: def fun(a, b, c):
...: print a, b, c
...:
...:
Call this function passing three positional arguments
In [7]: fun(1,2,3)
1 2 3 #Output
So, calling this function by passing three positional arguments, prints the three arguments passed to the function.
Let‘s make a list with three integer values in it.
In [8]: l = [1,2,3]
Let‘s use ‘*‘ now.
In [9]: fun(*l)
1 2 3 #Output
It unpacked the values in list ‘l‘ as positional arguments. And then the unpacked values were passed to function ‘fun‘ as positional arguments.
So, unpacking the values in list and changing it to positional arguments meant writing fun(*l) was equivalent to writing fun(1,2,3). Keep in mind that l=[1,2,3]. Let‘s try with some other value of ‘l‘.
In [10]: l=[5,7,9]
In [11]: fun(*l)
5 7 9 #Output
Let‘s make some errors now. Let‘s put four values in "l".
In [12]: l=[3,5,6,9]
Now, try to call function "fun".
In [13]: fun(*l)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
TypeError: fun() takes exactly 3 arguments (4 given)
So, in our last statement which is ‘fun(*l)‘ we did not get a proper output and a TypeError was raised. See the error, it says "TypeError: fun() takes exactly 3 arguments (4 given)".
list ‘l‘ contains four values. So, when we tried ‘fun(*l)‘, ‘l‘ was unpacked so that its value could be sent as positional arguments. But, "l" has four values in it. So, writing ‘fun(*l)‘ was equivalent to writing ‘fun(3,5,6,9)‘. But, ‘fun‘ is defined to take only three positional arguments and hence we got this error. Similary, you can follow same steps with two values in list ‘l‘ and notice the error.
In [14]: l=[5,6]
In [15]: fun(*l)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
TypeError: fun() takes exactly 3 arguments (2 given)
Let‘s mix ‘*l‘ with a positional argument.
In [16]: fun(1, *l)
1 5 6 #Output.
Here we gave one positional argument which is 1 and two values i.e 5 and 6 were unpacked from "l" and hence 1,5 and 6 were passed to ‘fun‘.
Hope you were able to follow what ‘*‘ does when used inside a function call.
Let‘s change the function definition now.
In [18]: def fun(*args):
....: print args
....:
....:
Call this function with one positional argument.
In [19]: fun(1)
(1,) #Output
Now call this function with two positional arguments or any number of positional arguments you wish.
In [20]: fun(1,2,3)
(1, 2, 3)
It recieves a tuple containing the positional arguments beyond the formal parameter list. So, "args" is a tuple. Don‘t worry about the part "formal parameter list" in our explanation, it will be clear with next few examples. In our last example when we printed "args", it printed a tuple which contained all the values we passed while calling the function.
Let‘s mix "*args" with some "formal parameter list". Note that in our last example we didn‘t have any formal parameter list. Let‘s redefine our function.
In [21]: def fun(a, *args):
....: print "a is ", a
....: print "args is ", args
....:
....:
In this function definition, parameter "a" constitue the "formal parameter list". Let‘s call "fun" with four positional argument.
In [22]: fun(1, 2, 3, 4)
a is 1 #Output
args is (2, 3, 4)
So, we can see that ‘a‘ is assigned 1 which was the first positional argument. There is only one parameter "*args" defined after "a". So, "args" received a tuple containing the positional arguments beyond the formal parameter list. So, args received 2, 3 and 4 as a tuple.
We can also call "fun" with just one positional argument. Let‘s do that.
In [23]: fun(1)
a is 1 #Output
args is () #args received an empty tuple
Here, we passed only one argument to the function which was assigned to the formal parameter ‘a‘. So, ‘args‘ received an empty tuple as can be seen from the output.
After we have "args", we can extract the values and do whatever we want. Let‘s redefine "fun".
In [24]: def fun(a, *args):
....: print a
....: print "args can receive a tuple of any number of arguments. Let‘s print all that."
....: for arg in args:
....: print arg
....:
....:
We can call "fun" with any number of arguments.
In [25]: fun(1,5,6,7)
1 #Output
args can receive a tuple of any number of arguments. Let‘s print all that.
5
6
7
Since ‘args‘ is a tuple, we could iterate over it.
Now, let‘s consider a case where we use whatever we saw till here. Here we need to use two functions. First function has to take an arbitrary number of arguments and it has to calculate the sum of all the arguments except the first argument. Also, this function has to make use of another function to calculate the sum. Weird use case, but we just need to recap whatever we did till here. The objective here is to see how we get a variable number of arguments in a function and pass these arguments to another function.
Let‘s first write the function which has to calculate the sum i.e the second function. For our use case, this function will be used in the first function which we are yet to write.
In [26]: def calculate_sum(*args):
....: return sum(args)
....:
Here we make use of ‘sum‘. Function ‘sum‘ is an inbuilt function which takes a tuple or a list and return the sum of all the elements in the tuple. From our function definition we can see that ‘args‘ will receive a tuple containing all the positional arguments passed to this function. So, ‘args‘ will be a tuple and can be directly used as an argument to function ‘sum‘. Let‘s write the other function which takes any number of arguments and uses previous function to calulate the sum of all arguments except the first argument.
In [29]: def ignore_firstargs_calculate_sum(a, *iargs):
....: required_sum = calculate_sum(*iargs)
....: print "sum is", required_sum
....:
....:
We can pass any number of arguments to this function. First argument will be recieved by ‘a‘ which is a formal parameter. All other arguments will be recieved by ‘iargs‘ as a tuple. As per the case we are considering, we want to calculate the sum of all arguments except the first. So, we leave ‘a‘ as it receives the first argument. ‘iargs‘ is the tuple containing all arguments except the first. We will make use of function ‘calculate_sum‘. But ‘calculate_sum‘ expects number of positional arguments to be sent to it which it will receive in ‘args‘ as a tuple. So, in function ‘ignore_firstargs_calculate_sum‘ we need to unpack ‘iargs‘, as it is a tuple, and then send the unpacked positional arguments to ‘calculate_sum‘. Remember, we used ‘*‘ to unpack a list/tuple.
So, we write ‘required_sum=calculate_sum(*iargs)‘.
We can‘t write ‘required_sum=calculate_sum(iargs)‘ because we need to unpack the values in the tuple ‘iargs‘ before sending to ‘calculate_sum‘. Not using ‘*‘ will not unpack the values and hence we won‘t have the desired behaviour. Let‘s use the function we wrote.
In [34]: ignore_firstargs_calculate_sum(3,1,2,6)
sum is 9 #Output
The output is the sum of all arguments except the first argument.
Let‘s consider a simple example first. Let‘s define a function which takes three arguments.
In [35]: def fun(a, b, c):
....: print a, b, c
....:
....:
Let‘s call this function in various ways.
In [36]: fun(1,2,3)
1 2 3 #Output
In [37]: fun(1, b=4, c=6)
1 4 6 #Output
Let‘s use "**" from inside the function call. For this we want a dictionary. Remember, while using "*" in the function call, we required a list/tuple. For using "**" in the function call, we require a dictionary.
In [38]: d={‘b‘:5, ‘c‘:7}
Let‘s call "fun" using "**" in the function call.
In [39]: fun(1, **d)
1 5 7
It unpacked the dictionary used with it, and passed the items in the dictionary as keyword arguments to the function. So writing "fun(1, **d)" was equivalent to writing "fun(1, b=5, c=7)".
Let‘s try some more examples to understand it better.
In [40]: d={‘c‘:3}
In [42]: fun(1,2,**d) #This is equivalent to fun(1,2,c=3)
1 2 3
In [43]: d={‘a‘:7,‘b‘:8,‘c‘:9}
In [44]: fun(**d)
7 8 9 #Output
Let‘s make some errors now.
In [45]: d={‘a‘:1, ‘b‘:2, ‘c‘:3, ‘d‘:4}
In [46]: fun(**d)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
TypeError: fun() got an unexpected keyword argument ‘d‘
Last statement was equivalent to fun(a=1, b=2, c=3, d=4). But, "fun" expected only three arguments and hence we got this error.
In [47]: d={‘a‘:1, ‘b‘:5, ‘d‘:9}
In [48]: fun(**d)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
TypeError: fun() got an unexpected keyword argument ‘d‘
Last statement was equivalent to fun(a=1, b=5, d=9). Although it passed three arguments which is the number of arguments expected by "fun", but "fun" does not have ‘d‘ in its parameter list. But, ‘d‘ was passed as a keyword argument And hence we got this error.
So, "**" unpacks the dictionary i.e the key values pairs in the dictionary as keyword arguments and these are sent as keyword arguments to the function being called. "*" unpacks a list/tuple i.e the values in the list as positional arguments and these are sent as positional arguments to the function being called.
Let‘s redefine our function "fun".
In [49]: def fun(a, **kwargs):
....: print a, kwargs
....:
....:
So, this function can only take one positional argument since formal parameter list contains only one variable ‘a‘. But with "**kwargs", it can take any number of keyword arguments. Let‘s see some examples.
In [50]: fun(1, b=4, c=5)
1 {‘c‘: 5, ‘b‘: 4} #Output
In [51]: fun(2, b=6, c=7, d=8)
2 {‘c‘: 7, ‘b‘: 6, ‘d‘: 8} #Output
With "**kwargs" in the function definition, kwargs receives a dictionary containing all the keyword arguments beyond the formal parameter list. Remember ‘kwargs‘ will be a dictionary. In our previous two examples, when we printed kwargs, it printed a dictionary containing all the keyword arguments beyond the formal parameter list.
Let‘s again redefine our function.
In [54]: def fun(a, **kwargs):
....: print "a is", a
....: print "We expect kwargs ‘b‘ and ‘c‘ in this function"
....: print "b is", kwargs[‘b‘]
....: print "c is", kwargs[‘c‘]
....:
....:
Let‘s call "fun" now.
In [55]: fun(1, b=3, c=5)
a is 1
We expect kwargs ‘b‘ and ‘c‘ in this function
b is 3
c is 5
Let‘s make some errors now.
In [56]: fun(1, b=3, d=5)
a is 1
We expect kwargs ‘b‘ and ‘c‘ in this function
b is 3
c is---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
/home/akshar/branding_git/netconference/<ipython console> in fun(a, **kwargs)
KeyError: ‘c‘
We were able to call the function. First positional argument was printed. Keyword argument ‘b‘ was printed. But the other keyword argument we passed was ‘d‘. Since function expected a keyword argument ‘c‘ and tried to access it from the dictionary "kwargs". But since we did not pass any keyword argument ‘c‘, we got this error. If we add a keyword argument ‘c‘ in the function call, we won‘t get the error anymore.
In [57]: fun(1, b=3, d=5, c=7)
a is 1
We expect kwargs ‘b‘ and ‘c‘ in this function
b is 3
c is 7
Since having "**kwargs" in the function argument list, we can pass any number of keyword arguments. We passed ‘d‘ but did not make any use of it in the function.
Let‘s make one more error.
In [58]: fun(1, {‘b‘:2, ‘c‘:3})
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/home/akshar/branding_git/netconference/<ipython console> in <module>()
TypeError: fun() takes exactly 1 argument (2 given)
As the error says, Function "fun" expected only one postional argument but was given two. So, although ‘kwargs‘ receives the keyword arguments as a dictionary, you cannot pass a dictionary as a positional argument to ‘kwargs‘. Although you could have done somehing like:
In [59]: fun(1, **{‘b‘:2, ‘c‘:3})
a is 1
We expect kwargs ‘b‘ and ‘c‘ in this function
b is 2
c is 3
Using "**" in front of the dictionary unpacks the dictionary and passes the items in dictionary as keyword arguments.
Whenever we inherit a class and override some of the methods of inherited class, we should use ‘*args‘ and ‘**kwargs‘ and pass the received positional and keyword arguments to the superclass method. Can be better understood with an example.
In [4]: class Model(object):
...: def __init__(self, name):
...: self.name = name
...: def save(self, force_update=False, force_insert=False):
...: if force_update and force_insert:
...: raise ValueError("Cannot perform both operations")
...: if force_update:
...: #Update an existing record
...: print "Updated an existing record"
...: if force_insert:
...: #Create a new record
...: print "Created a new record"
...:
We defined a class. We can create objects of this class and objects of this class have a method "save()". Assume that the objects of this class can be saved in a database which is being done inside the save() method. Depending on the arguments we pass to save() method, it is determined whether new records need to be created in the database or an existing record need to be updated.
We want a new class where we want ‘Model‘ behaviour but we only want to have save the objects of this class after we have checked some conditions. So let‘s subclass ‘Model‘ and override ‘save()‘ of ‘Model‘.
In [6]: class ChildModel(Model):
...: def save(self, *args, **kwargs):
...: if self.name==‘abcd‘:
...: super(ChildModel, self).save(*args, **kwargs)
...: else:
...: return None
...:
Actual saving of object(as per our assumption, connecting with database and creating/updating) happens in the "save" method of "Model". So we need to call the "save()" method of superclass from save() method of ChildModel. Also, save() method of subclass i.e ChildModel should be able to accept any arguments that save() of superclass accepts and must pass through these arguments to the superclass save(). So we have "*args" and "**kwargs" in the argument list of subclass save() method to receive any positional arguments or keyword arguments beyond the formal parameter list.
Let‘s create an instance of ChildModel and save it.
In [7]: c=ChildModel(‘abcd‘)
So, we created an instance of ChildModel with name=‘abcd‘.
In [9]: c.save(force_insert=True)
Created a new record #Output
Here, we passed a keyword argument to save() of the object. The save() we called is the subclass save(). It received a dictionary containing the keyword argument in "kwargs". Then it used "**" to unpack this dictionary as keyword arguments and then passed it to the superclass save(). So, superclass save() got a keyword argument ‘force_insert‘ and acted accordingly.
Let‘s try to pass another keyword argument.
In [10]: c.save(force_update=True)
Updated an existing record #Output
That was all about it. Hope you liked the post.
========================================================================================================
当我们继承一个写好的类的时候，尤其是这个类中内容会经常变化的时候（比如某个经常修改的公共类库），最好的方式就是使用*args，**kwargs作为派生类中方法的参数，最大程度的提高灵活性，父类中的方法做出修改之后，派生类的代码不必修改。可以写一个小的例子程序，验证一下。
class A(): def __init__(self, va, vb): self.va = va self.vb = vb def fun(self, vva, vvb): print("Method in A is invoking:", vva," ", vvb) class B(A): def __init__(self, vc, *args, **kwargs): self.vc = vc super().__init__(*args, **kwargs) def fun(self, vvc, *args, **kwargs): print("Method in B is invoking:"," ", vvc) super().fun(*args, **kwargs) b = B("C", "A", "B") b.fun("VVC", "VVA", "VVB")
运行结果：
Method in B is invoking: VVC Method in A is invoking: VVA VVB
class A(): def __init__(self, va): self.va = va def fun(self, vva): print("Method in A is invoking:", vva)
则B中的代码不必跟着变化。
我们只需要改变程序初始化的时候传入的参数就行。
b = B("C", "A") b.fun("VVC", "VVA")
运行结果：
Method in B is invoking: VVC Method in A is invoking: VVA
在，设计多继承的类结构的时候，重写方法，使用super的时候，方法的参数，使用*args，**kwargs，可以避免出现，因为程序跳转到兄弟类的代码，出现传值错误。
class A(): def __init__(self, name): self.name = name def fun(self, va): print("Method in A:", va) class B(A): def __init__(self, age, *args, **kwargs): self.age = age super().__init__(*args, **kwargs) def fun(self, vb, *args, **kwargs): print("Method in B:", vb) super().fun(*args, **kwargs) class C(A): def __init__(self, grade, *args, **kwargs): self.grade = grade super().__init__(*args, **kwargs) def fun(self, vc, *args, **kwargs): print("Method in C:", vc) super().fun(*args, **kwargs) class D(B, C): def __init__(self, sex, *args, **kwargs): self.sex = sex super().__init__(*args, **kwargs) def fun(self, vd, *args, **kwargs): print("Method in D:", vd) super().fun(*args, **kwargs) d = D(‘male‘, 19, 3, "Mark") print("d:",d.name, d.age, d.grade, d.sex) d.fun("VD", "VB", "VC", "VA") print("===========================") dd = D(sex="female", age=20, grade=3, name="Alice") # Unordered parameters dd.fun(vd="DD", va="AA", vc="CC", vb="BB") # Unordered parameters print("===========================") bb = B(18, "Lucy") bb.fun("BB", "AA")
运行结果：
d: Mark 19 3 male Method in D: VD Method in B: VB Method in C: VC Method in A: VA =========================== Method in D: DD Method in B: BB Method in C: CC Method in A: AA =========================== Method in B: BB Method in A: AA
标签：
原文地址：http://www.cnblogs.com/dotxxx/p/4190607.html
