今日内容
参数
作用域
函数嵌套
知识点回顾
函数基本结果
def func(name,age,email):
# 函数体(保持缩进一致)
a = 123
print(a)
return 1111#函数中,出现return后,后面的语句不再执行。
b = 456
print(b)
result = func(1,2,3)
# 函数默认返回值:None
参数
def func(n1,n2):
print(n1,n2)
func(1,2)
func(1,[11,22,3])
func({'k1':'k'},[11,22,3])
# 严格按照顺序传参数:位置方式传参。
# 实际参数可以是任意类型。
返回值
函数内部执行过程中遇到return,就终止。
def func1():
return "完成" # 函数每次执行到此,就返回;所以下面代码永远不执行。
for i in range(10):
print(i)
func1()
def func2():
for i in range(10):
print(i)
return "完成"
print(666)
func2()
return 可以返回任意类型
def func():
return (1,2,3)
v = func()
print(v)
# 特殊:返回元组
def func():
return 5,8,"alex"
v = func()
print(v)
昨日作业题
def func(data_list):
val = data_list[1::2]
return list(val)
v1 = func([1,2,3,4,5,5])
print(v1)
v2 = func((1,2,3,4,5,5))
print(v2)
# 其他类型可以通过强制转换为列表
# v1 = (1,2,3,4)
# v1 = {1,2,3,4}
# v1 = {'k1':'v1','k2':'v2'}
v1 = "asdfasdfasdf"
v2 = list(v1)
print(v2)
def func(arg):
if len(arg) > 5:
return True
else:
return False
data = func([1111,22,3,42,12])
data = func((1111,22,3,42,12)
def func(a1,a2):
if a1 > a2:
return a1
else:
return a2
v1 = func(1,2)
v2 = func(11,2)
def func(a1,a2):
return a1 if a1 > a2 else a2
v1 = func(1,2)
v2 = func(11,2)
def func(name,gender,age,edu):
# template = "%s*%s*%s*%s" %(name,gender,age,edu,)
# return template
data_list = [name,gender,age,edu]
return "*".join(data_list)
n1 = input('>')
n2 = input('>')
n3 = input('>')
n4 = input('>')
result = func(n1,n2,n3,n4)
print(result)
def func(max_range):
result = [1,1]
while True:
val = result[-1] + result[-2]
if val > max_range:
break
result.append(val)
return result
v = func(100)
print(v)
def func(name):
with open('data.txt',mode='r',encoding='utf-8') as obj:
# 方式一
content = obj.read()
flag = False
row_list = content.split('\n') # ['1|alex|123123','2|eric|rwerwe','3|wupeiqi|pppp']
for row in row_list:
v= row.split('|')
if v[1] == name:
flag = True
break
return flag
func('alex')
func('eric')
# ############################################
def func(name):
with open('data.txt',mode='r',encoding='utf-8') as obj:
# 方式一
content = obj.read()
row_list = content.split('\n') # ['1|alex|123123','2|eric|rwerwe','3|wupeiqi|pppp']
for row in row_list:
v= row.split('|')
if v[1] == name:
return True
v1 = func('alex')
if v1:
print('存在')
else:
print('不存在')
今日内容详细
参数
任意类型
def func(a1,a2,a3):
print(a1,a2,a3)
位置传参(调用函数并传入参数)【执行】
def func(a1,a2):
print(a1,a2)
func(1,3)
关键字传参【执行】
def func(a1, a2):
print(a1, a2)
func(a2=99,a1=2)
# 关键字传参数和位置传参可以混合使用(位置传入的参数 > 关键字参数在后 = 总参数个数)
def func1(a1, a2, a3):
print(a1, a2, a3)
# func(1, 2, a3=9)
# func(1, a2=2, a3=9)
# func(a1=1, a2=2, a3=9)
# func(a1=1, 2,3) # 错误
默认参数【定义】
def func(a1,a2,a3=9,a4=10):
print(a1,a2,a3,a4)
func(11,22)
func(11,22,10)
func(11,22,10,100)
func(11,22,10,a4=100)
func(11,22,a3=10,a4=100)
func(11,a2=22,a3=10,a4=100)
func(a1=11,a2=22,a3=10,a4=100)
调用函数无
def func(args):
print(args)
func(1,2,3,4)
(1, 2, 3, 4)
调用函数有
def func(args):
print(args)
func(*(1,2,3,4))
func(*[1,2,3,4])
(1, 2, 3, 4)
(1, 2, 3, 4)
只能用位置传参
def func(*args):
print(args)
# func(1)
# func(1,2)
func(1,2) # args=(1, 2)
func((11,22,33,44,55))
# args=((11,22,33,44,55),),不带星号,相当于把当当成一个整体,当作元组的一个元素。
func(*(11,22,33,44,55)) # args=(11,22,33,44,55)
(1,)
(1, 2)
(1, 2)
((11, 22, 33, 44, 55),)
(11, 22, 33, 44, 55)
调用函数无
def func(kwargs):
print(kwargs)
func(k1=1,k2="alex")
{'k1': 1, 'k2': 'alex'}
综合应用:无敌 + 无敌 => 真无敌
def func(*args,**kwargs):
print(args,kwargs)
# func(1,2,3,4,5,k1=2,k5=9,k19=999)
func(*[1,2,3],k1=2,k5=9,k19=999)
func(*[1,2,3],**{'k1':1,'k2':3})
func(111,222,*[1,2,3],k11='alex',**{'k1':1,'k2':3})
(1, 2, 3, 4, 5) {'k1': 2, 'k5': 9, 'k19': 999}
(1, 2, 3) {'k1': 2, 'k5': 9, 'k19': 999}
(1, 2, 3) {'k1': 1, 'k2': 3}
(111, 222, 1, 2, 3) {'k11': 'alex', 'k1': 1, 'k2': 3}
参数相关重点:定义函数
def func1(a1,a2):
pass
def func2(a1,a2=None):
pass
def func3(*args,**kwargs):
pass
调用函数
位置参数 > 关键字参数
作用域
python中:
函数:局部作用域
a = 1
def s1():
x1 = 666
print(x1)
print(a)
print(b)
b = 2
print(a)
s1()
a = 88888
def s2():
print(a,b)
s1()
s2()
======
1
666
1
2
88888 2
666
88888
2
作用域中查找数据规则:优先在自己的作用域找数据,自己没有就去 "父级" -> "父级" -> 直到全局,全部么有就报错。注意:父级作用域中的值到底是什么?
x = 10
def func():
x = 9
print(x)
func()
练习题:
#!/usr/bin/env python
# -- coding:utf-8 --
# x = 10
# def func():
# x = 9
# print(x)
# def x1():
# x = 999
# print(x)
#
# func()
# x = 10
# def func():
# x = 9
# print(x)
# def x1():
# x = 999
# print(x)
# x1()
#
# func()
# x = 10
# def func():
# x = 9
# print(x)
# def x1():
# x = 999
# print(x)
# print(x)
# x1()
#
# func()
# x = 10
# def func():
# x = 8
# print(x)
# def x1():
# x = 999
# print(x)
# x1()
# print(x)
#
# func()
# x = 10
# def func():
# x = 8
# print(x)
# def x1():
# print(x)
# x1()
# print(x)
#
# func()
# x = 10
# def func():
# x = 8
# print(x)
# def x1():
# print(x)
# x = 9
# x1()
# x = 10
# print(x)
#
# func()
#
# x = 10
# def func():
# x = 8
# print(x)
# def x1():
# print(x)
#
# x1()
# x = 9
# x1()
# x = 10
# print(x)
#
# func()
子作用域中只能 找到父级中的值 ,默认无法重新为父级的变量进行赋值。(global/nonlocal可以强制做)
# #####################
name = ‘oldboy‘
def func():
name = ‘alex‘ # 在自己作用域再创建一个这样的值。
print(name)
func()
print(name)
# #####################
name = [1,2,43]
def func():
name.append(999)#列表为可变类型,可以修改。这里是修改,不是重新赋值。
print(name)
func()
print(name)
# ###################### 如果非要对全局的变量进行赋值
# 示例一
name = ["老男孩",'alex']
def func():
global name
name = '我'
func()
print(name)
# 示例一
name = "老男孩"
def func():
name = 'alex'
def inner():
global name
name = 999
inner()
print(name)
func()
print(name)
name = "老男孩"
def func():
name = 'alex'
def inner():
global name
name = 999
inner()
print(name)
func()
print(name)
# ############################## nonlocal
name = "老男孩"
def func():
name = 'alex'
def inner():
nonlocal name # 找到上一级的name
name = 999
inner()
print(name)
func()
print(name)
总结
补充
全部变量以后必须全部是大写
USER_LIST = [11,22,3]
def func():
name = 'asdf'
USER_LIST.append(12)
USER_LIST.append(name)
func()
print(USER_LIST)
原文:https://www.cnblogs.com/cuiyongchao007/p/12261391.html