函数的返回值
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2024-03-17 16:00:16
...
函数的返回值
In [1]: def add(x, y): # return 除了返回值以外， 还会结束函数， return之后的语句将不会被执行
...: return x + y
...: print('hahaha')
...:
hahaha
In [2]: def add(x, y):
...: return x + y
...: print('hahaha')
...:
In [3]: ret = add(1, 3)
In [4]: ret
Out[4]: 4
In [5]: def guess(x): # 所以一个函数可以有多个return， 执行到哪个return，就由哪个return 返回结果并结束函数
...: if x > 3:
...: return '>3'
...: return '<=3'
...:
In [6]: s = guess(3)
In [7]: s
Out[7]: '<=3'
In [8]: s = guess(5)
In [9]: s
Out[9]: '>3'
In [10]:
In [10]: def fn(x):
...: for i in range(x):
...: if x > 3:
...: return i
...: else:
...: print('not bigger than 3')
...:
In [11]: fn(10)
Out[11]: 0
In [12]: fn(3)
not bigger than 3
In [13]:
In [13]: def fn(): # 如果当函数没有return 语句的时候， 隐式返回了一个None
...: pass
...:
In [14]: ret = fn()
In [15]: type(ret)
Out[15]: NoneType
In [16]: typr(None)
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-16-3565bd08d192> in <module>()
----> 1 typr(None)
NameError: name 'typr' is not defined
In [17]: type(None)
Out[17]: NoneType
In [18]: ret == None
Out[18]: True
In [19]: def fn():
...: return 3, 5
...:
In [20]: ret = fn()
In [21]: ret # 类似于 封装
Out[21]: (3, 5)
In [22]: x, y = ret # 类似于 解构
In [23]: x
Out[23]: 3
In [24]: y
Out[24]: 5
In [27]: def fn():
...: return None
...:
In [28]: ret = fn()
In [29]: ret == None
Out[29]: True
In [30]:
In [30]: def fn():
...: return # 返回None， 可以简写为 return 通常用于结束函数
...:
In [31]:
In [31]: def fn(x):
...: if x > 3:
...: return
...: return '<=3'
...:
In [32]: fn(5)
In [33]: fn(3)
Out[33]: '<=3'
In [34]:
函数的嵌套
In [35]: def outter():
...: def inner():
...: print('inner')
...: print('outter')
...: inner()
...:
In [36]: outter()
outter
inner
In [37]:
一、作用域
In [39]: x = 1
In [40]: def inc():
...: print(x) # 并未参与到运算中
...: y = x + 1 # 只读可见， 不可写
...: return y
...:
In [41]: inc()
1
Out[41]: 2
In [42]:
In [42]: def fn():
...: xx = 1
...: print(xx)
...:
In [43]: fn()
1
In [44]: xx
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-44-ce3af7760f12> in <module>()
----> 1 xx
NameError: name 'xx' is not defined
In [45]:
In [45]: def fn2():
...: peint(xx)
...:
In [46]: fn()
1
In [47]: fn2() # 变量的作用域 为变量同级的作用域， 也就是说 在哪一个级别定义的， 在哪一个级别就可见
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-47-e87bcb52744c> in <module>()
----> 1 fn2()
<ipython-input-45-ff1e6126d39a> in fn2()
1 def fn2():
----> 2 peint(xx)
3
NameError: name 'peint' is not defined
In [48]:
In [49]: def fn():
...: xx = 1
...: print(xx)
...: def inner():
...: print(xx)
...: inner()
...:
In [50]: fn()
1
1
In [51]:
In [51]: def fn():
...: xx = 1
...: print(xx)
...: print(id(xx))
...: def inner():
...: xx = 2
...: print(id(xx))
...: inner()
...: print(xx)
...: print(id(xx))
...:
In [52]: fn()
1
9289856
9289888
1
9289856
In [53]:
In [54]: def fn():
...: xx = 1
...: print(xx)
...: def inner():
...: xx = 2
...: inner()
...: print(xx)
...:
In [55]: xx = 1
In [56]: def fn():
...: global xx
...: xx += 1
...:
In [57]: fn()
In [58]: xx
Out[58]: 2
In [59]: fn()
In [60]: xx
Out[60]: 3
In [61]:
In [61]: def fn():
...: global yy
...: yy = 3
...:
In [62]: fn()
In [63]: yy
Out[63]: 3
In [64]: def fn():
...: global zz
...:
In [65]: zz
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-65-15546de8c3b0> in <module>()
----> 1 zz
NameError: name 'zz' is not defined
In [66]:
金玉良言： 除非你清楚的知道，global 会带来什么， 并明确的知道非global不行，否则不要使用global
二、闭包
In [66]: def counter():
...: x = 0
...: def inc():
...: global x
...: x += 1
...: return x
...: return inc
...:
In [67]: f = counter()
In [68]: f()
Out[68]: 2
In [69]: f()
Out[69]: 3
In [70]: f = counter()
In [71]: f()
Out[71]: 4
In [72]: x
Out[72]: 4
In [73]: f()
Out[73]: 5
In [74]: x
Out[74]: 5
In [75]: x = 0
In [76]: f()
Out[76]: 1
In [77]: x
Out[77]: 1In [79]: def counter():
...: c = [0]
...: def inc():
...: c[0] += 1
...: return c[0]
...: return inc
...:
# 函数执行完毕之后，仍有部分函数变量的引用存在
In [80]: f = counter()
In [81]: f()
Out[81]: 1
In [82]: c = 100
In [83]: f()
Out[83]: 2
In [84]: lst = [1]
In [85]: id(lst)
Out[85]: 140023278144840
In [86]: lst[0] = 2
In [87]: id(lst)
Out[87]: 140023278144840
In [88]: f()
Out[88]: 3
In [89]: f = counter()
In [90]: f()
Out[90]: 1
In [91]: f()
Out[91]: 2
In [92]:
1、nonlocal
In [92]: def counter():
...: x = 0
...: def inc():
...: x += 1
...: return x
...: return inc
...:
In [93]: f = counter()
In [94]: f()
---------------------------------------------------------------------------
UnboundLocalError Traceback (most recent call last)
<ipython-input-94-c43e34e6d405> in <module>()
----> 1 f()
<ipython-input-92-91cd34f02038> in inc()
2 x = 0
3 def inc():
----> 4 x += 1
5 return x
6 return inc
UnboundLocalError: local variable 'x' referenced before assignment
In [95]:
In [95]: def counter(): # 通过 nonlocal 关键字，可以对上级作用域的变量，进行 可读可写
...: x = 0
...: def inc():
...: nonlocal x
...: x += 1
...: return x
...: return inc
...:
In [96]: f = counter()
In [97]: f()
Out[97]: 1
In [98]: f()
Out[98]: 2
In [99]:
赋值既定义
2、默认参数作用域
In [100]: def fn(x=[]):
...: x.append(1)
...: print(x)
...:
In [101]: fn()
[1]
In [102]: fn()
[1, 1]
In [103]: fn
Out[103]: <function __main__.fn>
In [104]: fn.__defaults__
Out[104]: ([1, 1],)
In [105]:
In [105]: fn()
[1, 1, 1]
In [106]: fn.__defaults__
Out[106]: ([1, 1, 1],)
In [107]: fn.__defaults__ = ([], )
In [108]: fn()
[1]
In [109]: fn.__defaults__
Out[109]: ([1],)
In [110]:
In [110]: def gn(a=2, b=1):
...: print(a + b)
...:
In [111]: gn.__defaults__
Out[111]: (2, 1)
In [112]:
全局作用域。什么时候会销毁？
- 重新定义
- del 关键字删除
- 程序结束推出
局部作用域。什么时候销毁？
- 重新定义
- del 关键字删除
- 上级作用域被销毁
In [112]: def fn(aa=[], bb=[]):
...: aa.append(1)
...: bb.append(2)
...: print(aa, bb)
...:
In [113]: fn()
[1] [2]
In [114]: fn()
[1, 1] [2, 2]
In [115]: fn.__defaults__
Out[115]: ([1, 1], [2, 2])
In [116]:
In [116]: def fn(x=0, y=0): # int 是不可变类型
...: print(id(x))
...: x = 3
...: print(id(x))
...: x = 4
...: print(id(x))
...: y = 3
...: print(x, y)
...:
In [117]: fn()
9289824
9289920
9289952
4 3
In [118]: fn.__defaults__ # int 是不可变类型
Out[118]: (0, 0)
In [119]:
如何去接解决list的问题？
- 使用不可变类型作为默认值
- 函数体内不改变默认值
In [120]: def fn(lst=None):
...: if lst is None:
...: lst = []
...: lst.append(3)
...: print(lst)
...:
In [121]: fn.__defaults__
Out[121]: (None,)
In [122]: fn()
[3]
In [123]:
In [123]:
In [123]: def fn(lst=[]):
...: print(id(lst))
...: lst = lst[:]
...: print(id(lst))
...: lst.append(1)
...: print(lst)
...:
In [124]: fn()
140023278495752
140023268626120
[1]
In [125]: fn.__defaults__
Out[125]: ([],)
In [126]: fn()
140023278495752
140023278218120
[1]
In [127]: fn.__defaults__
Out[127]: ([],)
In [128]:
In [128]:
In [128]: def fn(lst=None):
...: if lst is None:
...: lst = []
...: else:
...: lst = lst[:]
...: lst.append(3)
...: print(lst)
...:
In [129]: fn()
[]
In [130]: fn()
[]
In [131]: fn.__defaults__
Out[131]: (None,)
In [132]: fn(1)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-132-454eedcabd9c> in <module>()
----> 1 fn(1)
<ipython-input-128-3bf22a286480> in fn(lst)
3 lst = []
4 else:
----> 5 lst = lst[:]
6 lst.append(3)
7 print(lst)
TypeError: 'int' object is not subscriptable
In [133]: fn([1, 2, 3])
[1, 2, 3, 3]
In [134]:
函数的执行流程：
main
f1()
f2()
f3()
f4()
f5()
main
三、递归函数
程序调用自身的编程技巧，称为 递归。
归纳法：通过一类事物的部分对象具有某种性质，推出这类事物的所有对象都具有这种性质的推理方法。
例如，斐波那契数列
fib(n) = 1 if n = 0
fib(n) = 1 if n = 1
fib(n) = fib(n-1) + fib(n-2)
In [146]: def fib(n):
...: if n == 0:
...: return 1
...: if n == 1:
...: return 1
...: return fib(n-1) + fib(n-2)
...:
In [147]: fib(0)
Out[147]: 1
In [148]: fib(1)
Out[148]: 1
In [149]: fib(2)
Out[149]: 2
In [150]: fib(3)
Out[150]: 3
In [151]: fib(4)
Out[151]: 5
In [152]: fib(5)
Out[152]: 8
In [153]:
In [153]:
In [153]: def fib(n):
...: return fib(n-1) + fib(n-2)
...:
In [154]: fib(4)
---------------------------------------------------------------------------
RecursionError Traceback (most recent call last)
<ipython-input-154-252d5fa3ed3f> in <module>()
----> 1 fib(4)
<ipython-input-153-3e11de167e36> in fib(n)
1 def fib(n):
----> 2 return fib(n-1) + fib(n-2)
... last 1 frames repeated, from the frame below ...
<ipython-input-153-3e11de167e36> in fib(n)
1 def fib(n):
----> 2 return fib(n-1) + fib(n-2)
RecursionError: maximum recursion depth exceeded
In [155]: import sys
In [156]: sys.getrecursionlimit()
Out[156]: 3000
转载于:https://my.oschina.net/u/3552459/blog/1563172
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