python面试之数据类型(综合)

1.python中tuple和list的相互转换

if __name__ == '__main__':
    # list to tuple
    lis = [1, 2, 3, 4, 5, 6]
    print(type(lis),lis)
    x = tuple(lis)
    print(type(x), x)

    # tuple to list
    tup = (1, 2, 3, 4, 5, 6)
    print(type(tup), tup)
    y = list(tup)
    print(type(y), y)

	结果：
	<class 'list'> [1, 2, 3, 4, 5, 6]
	<class 'tuple'> (1, 2, 3, 4, 5, 6)
	<class 'tuple'> (1, 2, 3, 4, 5, 6)
	<class 'list'> [1, 2, 3, 4, 5, 6]

2.如何对生成器类型的对象实现切片操作？

from itertools import islice
if __name__ == '__main__':
    gen = iter(range(5))  # iter()函数用来生成一个迭代器对象
    print(type(gen))
    # param1迭代器，param2起始索引，param3结束索引
    print(type(islice(gen, 0, 4)))
    for i in islice(gen, 0, 4):
        print(i)

3.将列表转为生成器

if __name__ == '__main__':
    print([i for i in range(3)])
    print((i for i in range(3)))

4.如何把字符串编码成 bytes 类型

if __name__ == '__main__':
    #1
    a = b'hello'
    #2.
    b = bytes("你好", "utf-8")
    #3.
    c = "你好".encode("utf-8")
    print(a, type(a),b, c)

结果:
b'hello' <class 'bytes'> b'\xe4\xbd\xa0\xe5\xa5\xbd' b'\xe4\xbd\xa0\xe5\xa5\xbd'

5.元组a = (1, 2, 3)中的元素能不能修改(不能)

if __name__ == '__main__':
    a = (1, 2, 3)
    a[1] = 2

结果:TypeError: 'tuple' object does not support item assignment

6.元组a = (1, 2, [4 , 4 , 5])中的元素能不能修改(不一定)

if __name__ == '__main__':
    #第一种(能),修改的是元组里面的列表元素
    a = (1, 2, [4 , 4 , 5])
    a[2][0] = 3
    print(a)

结果:(1, 2, [3, 4, 5])

    #第二种(不能),修改的是列表以外的元素
    a[0] = 1
    
结果:TypeError: 'tuple' object does not support item assignment

来源：CSDN

作者：一个喜欢林俊杰的靓仔

链接：https://blog.csdn.net/qq_37680159/article/details/104474632