python容器及容器的访问使用

╰+哭是因爲堅強的太久メ 2022-03-09 14:22 207阅读 0赞

*  列表切片

li = list(range(10))
    print(li)
          
    # 切片 [start:end:steps]  >= start & < end
    print(li[2:5])  # [3, 4, 5]
    print(li[:4])   # [0, 1, 2, 3]
    print(li[5:])   # [6, 7, 8, 9]
    print(li[0:20:3])   # [0, 3, 6, 9]
    
    # 负值怎么处理？
    print(li[5: -2])    # [5, 6, 7]
    print(li[9:0:-1])   # [9, 8, 7, 6, 5, 4, 3, 2,1]
    print(li[9::-1])   # [9 ... 0]
    print(li[::-2]) # [9, 7, 5, 3, 1]
    
    # 切片生成一个新的对象
    print(li)   # 还是保持原样
    
    re_li = li[::-1]
    print(re_li)

*  列表的访问，增加，删除

li = [1, 2, 3, '456', [1, 2, 3], {1: 'one', 2: 'two'}]
    print(type(list))
    print(type(li))
    
    # 元素访问
    print(li[0])
    print(li[-1])   # li[len(li) - 1]
    print(li[-2])   # li[len(li) - 2]
    
    # 查找元素位置
    print(li.index('456'))
    print(li.index([1, 2, 3]))
    # print(li.index(-1))
    
    # 添加元素
    l_a = [1, 2, 3]
    l_a.append(4)
    l_a.append(5)
    l_b = [6, 7, 8]
    l_a.extend(l_b) # 试下用append是什么结果
    print(l_a)
    #删除元素
    del(l_a[-1])    #删除最后一个元素
    del(l_a[0])    #删除第一个元素
    print(l_a)
    del(l_a[1])    #删除第二个元素
    print(l_a)
    
    l_a = []
    if not l_a:
        print('Empty')  # not XX和is None不是一回事
    
    if len(l_a) == 0:
        print('Empty')
    
    for i in li:
        print(i)
    for i in range(len(li)):
        print(li[i])
    
    t = (1, 2, 3, '456')
    print(type(t))
    # t[0] = 'a'
    # t.append('x')
    
    del(li[-1]) # del(list[index])
    del(li[1])
    del(li[-2])
    print(li)

*  字典

d={'name':'kang','age':22,'sex':'boy'}
    for key in d.keys():    #遍历
        print(d[key])
    for key, value in d.items():    #遍历
        print(key,value)
    print('name' in d)    #判断元素是否在字典里
    a = set([1,2,3,6,5,2,3,6,5,9])
    b = set([2,2,6,5,9,8,5,2,4])
    print(a)
    print(5 in a)
    print(10 in a)

*  生成器

print(type(range(10)))
    
    # 平方表
    square_table = []
    for i in range(5000):
        square_table.append(i * i)
    for i in range(5):
        print(square_table[i])
        
    square_generator = (x * x for x in range(50000))
    print(type(square_generator))
    for i in range(5):
        print(next(square_generator))
    
    def fib(limit):
        n, a, b = 0, 0, 1
        while n < limit:
            yield b
            a, b = b, a + b
            n += 1
        return 'done'
    
    import traceback
    f = fib(5)
    print(type(f))
    print(next(f))
    print(next(f))
    print(next(f))
    print(next(f))
    print(next(f))
    try:
        print(next(f))
    except StopIteration:
        traceback.print_exc()
    for i in fib(5):
        print(i)