How to set and use function parameters in Python

1. Parameters and shared references:

In [56]: def changer(a,b):
  ....:   a=2
  ....:   b[0]='spam'
  ....:   
In [57]: X=1
In [59]: L=[1,2]
In [60]: changer(X,L)
In [61]: X,L
Out[61]: (1, ['spam', 2])

Function parameters are assigned values ​​and passed through variables when calling To implement shared objects, remote modification of variable object parameters in functions can affect the caller.

Avoid variable parameter modification:

In [67]: X=1
In [68]: a=X
In [69]: a=2
In [70]: print(X)
1
In [71]: L=[1,2]
In [72]: b=L
In [73]: b[0]='spam'
In [74]: print(L)
['spam', 2]
In [75]: changer(X,L[:]) 
#不想要函数内部在原处的修改影响传递给它的对象，可以创建一个对象的拷贝

In [77]: changer(a,b)
In [78]: def changer(a,b): 
....:   b=b[:] 
#如果不想改变传入对象，无论函数怎么调用，同样可以在函数内部进行拷贝。
....:   
In [79]: a=2
In [80]: b[0]='spam'

2. Specific parameter matching model:

Function matching syntax:

Example:

Keyword parameters:

In [2]: def f(a,b,c):print (a,b,c)
In [3]: f(1,2,3) #位置参数调用
(1, 2, 3)
In [4]: f(c=3,b=2,a=1) #关键字参数调用
(1, 2, 3)

Default parameters:

In [5]: def f(a,b=2,c=3):print (a,b,c)
In [6]: f(1)  #给a赋值，b,c使用默认赋值 
(1, 2, 3)
In [7]: f(a=1) 
(1, 2, 3)
In [8]: f(1,4) 
(1, 4, 3)
In [9]: f(1,4,5) #不适用默认值
(1, 4, 5)
In [10]: f(1,c=6) #a通过位置得到1，b使用默认值，c通过关键字得到6
(1, 2, 6)

3. Any parameters:

1. Collection parameters:

#*和**出现在函数定义或函数调用中。

In [11]: def f(*args):print (args)
In [12]: f()  #将所有位置相关的参数收集到一个新的元祖中
()
In [13]: f(1)
(1,)
In [14]: f(1,2,3,4)
(1, 2, 3, 4)
In [15]: def f(**args):print (args)
In [16]: f() 
{}
In [17]: f(a=1,b=2) #**只对关键字参数有效
{'a': 1, 'b': 2}

In [19]: def f(a, *pargs,**kargs):print(a,pargs,kargs)
In [20]: f(1,2,3,4,5,6,x=1,y=2,z=3)
(1, (2, 3, 4, 5, 6), {'y': 2, 'x': 1, 'z': 3})

2. Unpacking parameters:

Note: Do not confuse The syntax of */** in the function header or function call: in the header means collecting any number of parameters, and in the call, it connects any number of parameters.

In [21]: def func(a,b,c,d):print(a,b,c,d)
In [22]: args=(1,2)
In [23]: args += (3,4)
In [24]: func(*args)
(1, 2, 3, 4)
In [25]: args={'a':1,'b':2,'c':3}
In [26]: args['d']=4
In [27]: func(**args)
(1, 2, 3, 4)
In [28]: func(*(1,2),**{'d':4,'c':4})
(1, 2, 4, 4)
In [30]: func(1,*(2,3),**{'d':4})
(1, 2, 3, 4)
In [31]: func(1,c=3,*(2,),**{'d':4})
(1, 2, 3, 4)
In [32]: func(1,*(2,3,),d=4)
(1, 2, 3, 4)
In [33]: func(1,*(2,),c=3,**{'d':4})
(1, 2, 3, 4)

3. Application function versatility:

In [34]: def tracer(func,*pargs,**kargs):
  ....: print ('calling:',func.__name__)
  ....: return func(*pargs,**kargs)
  ....: 
In [35]: def func(a,b,c,d):
  ....: return a+b+c+d
  ....: print (tracer(func,1,2,c=3,d=4))
  ....: 
('calling:', 'func')
10

4. The apply built-in function is abandoned in python3.

In [36]: pargs=(1,2)
In [37]: kargs={'a':3,'b':4}
In [41]: def echo(*args,**kargs):print (args,kargs)
In [42]: apply(echo,pargs,kargs)
((1, 2), {'a': 3, 'b': 4})

4. Keyword-only parameters in python3.x

python3.x generalizes the sorting rules of function headers, allowing We specify keyword-only parameters, that is, parameters that are passed according to keywords and will not be filled by a positional parameter; after the parameter *args, the keyword syntax must be called to pass.

In [43]: echo(*pargs,**kargs)
((1, 2), {'a': 3, 'b': 4})
In [44]: echo(0,c=5,*pargs,**kargs)
((0, 1, 2), {'a': 3, 'c': 5, 'b': 4})

1. Sorting rules:

** cannot appear alone in the parameters. The following are incorrect usages :

In [1]: def kwonly(a,*b,c):
  ...: print(a,b,c) 
In [2]: kwonly(1,2,c=3)
1 (2,) 3
In [3]: kwonly(a=1,c=3)
1 () 3
In [4]: kwonly(1,2,3) #c必须按照关键字传递
TypeError: kwonly() missing 1 required keyword-only argument: 'c'

In [6]: def kwonly(a,*,b,c):print(a,b,c)
In [7]: kwonly(1,c=3,b=2)
1 2 3
In [8]: kwonly(c=3,b=2,a=1)
1 2 3
In [9]: kwonly(1,2,3)
TypeError: kwonly() takes 1 positional argument but 3 were given

That is to say, in a function header, keyword-only parameters must be written before any keyword form of *args, or appear before or after args. , and may be included in **args.

In [11]: def kwonly(a,**pargs,b,c):
  ....: 
 File "<ipython-input-11-177c37879903>", line 1
def kwonly(a,**pargs,b,c):  ^
SyntaxError: invalid syntax

In [13]: def kwonly(a,**,b,c):
  ....: 
 File "<ipython-input-13-46041ada2700>", line 1
def kwonly(a,**,b,c):
  ^
SyntaxError: invalid syntax

2. Why use keyword-only parameters?

It is easy to allow a function to accept any number of Handling positional arguments, which also accept configuration options passed as keywords, can reduce code that would otherwise have to use *args and **args and manually check keywords.

3. Min calls

to write a function that can calculate the minimum value in any parameter set and any object data type set.

Method 1: Use slicing

In [14]: def f(a,*b,**d,c=6):print(a,b,c,d)
 File "<ipython-input-14-43c901fce151>", line 1
def f(a,*b,**d,c=6):print(a,b,c,d)
 ^
SyntaxError: invalid syntax
In [15]: def f(a,*b,c=6,**d):print(a,b,c,d) #keyword-only在*args之后，**args之前
In [16]: f(1,2,3,x=4,y=5)
1 (2, 3) 6 {&#39;x&#39;: 4, &#39;y&#39;: 5}

In [20]: f(1,c=7,*(2,3),**dict(x=4,y=5)) #keyword-only在
1 (2, 3) 7 {&#39;x&#39;: 4, &#39;y&#39;: 5}
In [21]: f(1,*(2,3),**dict(x=4,y=5,c=7))
1 (2, 3) 7 {&#39;x&#39;: 4, &#39;y&#39;: 5}

Method 2: Let python obtain it automatically to avoid slicing.

In [23]: def min(*args):
  ....: res=args[0]
  ....: for arg in args[1:]:
  ....: if arg < res:
  ....: res = arg
  ....: return res
  ....:

Method 3: Call the built-in function list, convert the ancestor into a list, and then call the built-in sort method of list. Note: Because the python sort routine is written in C and uses a highly optimized algorithm, the running speed is much faster than the first two.

In [28]: def min2(first,*rest):
  ....: for arg in rest:
  ....: if arg < first:
  ....: first = arg
  ....: return first
  ....:

5. Example:

1. Simulate the general set function:

Write a function to return the common part of the two sequences, and write the inter2.py file as follows:

In [32]: def min3(*args):
  ....: tmp=list(args)
  ....: tmp.sort()
  ....: return tmp[0]
  ....:

In [29]: min2(3,*(1,2,3,4))
Out[29]: 1
In [31]: min(*(5,6,6,2,2,7))
Out[31]: 2
In [33]: min3(3,4,5,5,2)
Out[33]: 2

Test:

#!/usr/bin/python3
def intersect(*args):
  res=[]
  for x in args[0]:
    for other in args[1:]:
      if x not in other: break
    else:
      res.append(x)
  return res
def union(*args):
  res=[]
  for seq in args:
    for x in seq:
      if not x in res:
        res.append(x)
  return res

2. Simulate python 3.x print function

Write the file python30.py

(1) Use *args Interaction results with **args method

environment python2.7

In [3]: from inter2 import intersect,union
In [4]: s1,s2,s3="SPAM","SCAM","SLAM"
In [5]: intersect(s1,s2),union(s1,s2)
Out[5]: (['S', 'A', 'M'], ['S', 'P', 'A', 'M', 'C'])
In [6]: intersect([1,2,3],(1,4))
Out[6]: [1]
In [7]: intersect(s1,s2,s3)
Out[7]: ['S', 'A', 'M']
In [8]: union(s1,s2,s3)
Out[8]: ['S', 'P', 'A', 'M', 'C', 'L']

:

#!/usr/bin/python
import sys
def print30(*args,**kargs):
  sep = kargs.get('sep',' ')
  end = kargs.get('end','\n')
  file = kargs.get('file',sys.stdout)
  if kargs:raise TypeError('extra keywords: %s' %kargs)
  output = ''
  first = True
  for arg in args:
    output += ('' if first else sep)+str(arg)
    first = False
  file.write(output + end)

(2) Use the keyword-only method to achieve the same effect as method one:

In [5]: print30(1,2,3)
1 2 3
In [6]: print30(1,2,3,sep='')
123
In [7]: print30(1,2,3,sep='...')
1...2...3
In [8]: print30(1,[2],(3,),sep='...')
1...[2]...(3,)
In [9]: print30(4,5,6,sep='',end='')
456
In [11]: print30(1,2,3)
1 2 3

In [12]: print30()

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