python基础考试一整理

3、s = [1, "h", 2, "e", [1, 2, 3], "l", (4, 5), "l", {1: "111"}, "o"], 将s中的5个字符提取出来并拼接成字符串。

s = [1, "h", 2, "e", [1, 2, 3], "l", (4, 5), "l", {1: "111"}, "o"]
print("".join([i for i in s if isinstance(i, str)]))  # hello

6.浅拷贝

a = [1, 2, [3, "hello"], {"egon": "aigan"}]
b = a[:]  # 此时发生了一个浅拷贝，嵌套列表的 内存地址是相同的
a[0] = 5
a[2][0] = 6666
print(a)  # [5, 2, [6666, 'hello'], {'egon': 'aigan'}]
print(b)  # [1, 2, [6666, 'hello'], {'egon': 'aigan'}]

11.用str.center打印菱形

#    *
#   ***
#  *****
# *******
#  *****
#   ***
#    *
def draw(n):
    for i in range(1,n,2):
        print(("*"*i).center(n))
    for i in range(n,0,-2):
        print(("*"*i).center(n))
draw(*)

12.nonlocal 往外包一层

def outer():
    count = 10

    def inner():
        nonlocal count
        count = 20
        print(count)

    inner()
    print(count)
outer()

思考题

# 思考题
# 有如下类，请重写方法判断，只要name和sex相同我们就认为是一个对象
class Person:
    def __init__(self,name,age,sex,weight):
        self.name = name
        self.sex = sex
        self.age = age
        self.weight = weight

    def __eq__(self, obj):
        if self.name == obj.name and self.sex == obj.sex:
            return True

    def __hash__(self):
        return (self.name, self.sex).__hash__()


a1 = Person("alex", 18, "male", 60)
a2 = Person("alex", 30, "male", 60)
a2 = Person("alex1", 30, "male", 60)

s = set([a1,a2])
print(s)