• 技术文章 >常见问题 >Python常见问题

    Python语言的特点技巧有哪些?

    silencementsilencement2019-07-22 10:57:03原创2966

    每个窍门或语言特性只能通过实例来验证,无需过多解释。虽然我已尽力使例子清晰,但它们中的一些仍会看起来有些复杂,这取决于你的熟悉程度。所以如果看过例子后还不清楚的话,标题能够提供足够的信息让你通过Google获取详细的内容。

    列表按难度排序,常用的语言特征和技巧放在前面。

    1.1 分拆

    >>> a, b, c = 1, 2, 3 
    >>> a, b, c  
    (1, 2, 3)  
    >>> a, b, c = [1, 2, 3]  
    >>> a, b, c  
    (1, 2, 3)  
    >>> a, b, c = (2 * i + 1 for i in range(3))  
    >>> a, b, c  
    (1, 3, 5)  
    >>> a, (b, c), d = [1, (2, 3), 4]  
    >>> a  
    1 
    >>> b  
    2 
    >>> c  
    3 
    >>> d  
    4

    1.2 交换变量分拆

    >>> a, b = 1, 2 
    >>> a, b = b, a  
    >>> a, b  
    (2, 1)

    1.3 拓展分拆 (Python 3下适用)

    >>> a, *b, c = [1, 2, 3, 4, 5]  
    >>> a  
    1 
    >>> b  
    [2, 3, 4]  
    >>> c  
    5

    1.4 负索引

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  
    >>> a[-1]  
    10 
    >>> a[-3]  
    8

    1.5 列表切片 (a[start:end])

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  
    >>> a[2:8]  
    [2, 3, 4, 5, 6, 7]

    1.6 使用负索引的列表切片

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  
    >>> a[-4:-2]  
    [7, 8]

    1.7 带步进值的列表切片 (a[start:end:step])

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  
    >>> a[::2]  
    [0, 2, 4, 6, 8, 10]  
    >>> a[::3]  
    [0, 3, 6, 9]  
    >>> a[2:8:2]  
    [2, 4, 6]

    1.8 负步进值得列表切片

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  
    >>> a[::-1]  
    [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]  
    >>> a[::-2]  
    [10, 8, 6, 4, 2, 0]

    1.9 列表切片赋值

    >>> a = [1, 2, 3, 4, 5]  
    >>> a[2:3] = [0, 0]  
    >>> a  
    [1, 2, 0, 0, 4, 5]  
    >>> a[1:1] = [8, 9]  
    >>> a  
    [1, 8, 9, 2, 0, 0, 4, 5]  
    >>> a[1:-1] = []  
    >>> a  
    [1, 5]

    1.10 命名切片 (slice(start, end, step))

    >>> a = [0, 1, 2, 3, 4, 5]  
    >>> LASTTHREE = slice(-3, None)  
    >>> LASTTHREE  
    slice(-3, None, None)  
    >>> a[LASTTHREE]  
    [3, 4, 5]

    1.11 zip打包解包列表和倍数

    >>> a = [1, 2, 3]  
    >>> b = ['a', 'b', 'c']  
    >>> z = zip(a, b)  
    >>> z  
    [(1, 'a'), (2, 'b'), (3, 'c')]  
    >>> zip(*z)  
    [(1, 2, 3), ('a', 'b', 'c')]

    1.12 使用zip合并相邻的列表项

    >>> a = [1, 2, 3, 4, 5, 6]  
    >>> zip(*([iter(a)] * 2))  
    [(1, 2), (3, 4), (5, 6)]  
    >>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))  
    >>> group_adjacent(a, 3)  
    [(1, 2, 3), (4, 5, 6)]  
    >>> group_adjacent(a, 2)  
    [(1, 2), (3, 4), (5, 6)]  
    >>> group_adjacent(a, 1)  
    [(1,), (2,), (3,), (4,), (5,), (6,)]  
    >>> zip(a[::2], a[1::2])  
    [(1, 2), (3, 4), (5, 6)]  
    >>> zip(a[::3], a[1::3], a[2::3])  
    [(1, 2, 3), (4, 5, 6)]  
    >>> group_adjacent = lambda a, k: zip(*(a[i::k] for i in range(k)))  
    >>> group_adjacent(a, 3)  
    [(1, 2, 3), (4, 5, 6)]  
    >>> group_adjacent(a, 2)  
    [(1, 2), (3, 4), (5, 6)]  
    >>> group_adjacent(a, 1)  
    [(1,), (2,), (3,), (4,), (5,), (6,)]

    1.13 使用zip和iterators生成滑动窗口 (n -grams)

    >>> from itertools import islice  
    >>> def n_grams(a, n):  
    ...     z = (islice(a, i, None) for i in range(n))  
    ...     return zip(*z)  
    ...  
    >>> a = [1, 2, 3, 4, 5, 6]  
    >>> n_grams(a, 3)  
    [(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]  
    >>> n_grams(a, 2)  
    [(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]  
    >>> n_grams(a, 4)  
    [(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6)]

    1.14 使用zip反转字典

    >>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}  
    >>> m.items()  
    [('a', 1), ('c', 3), ('b', 2), ('d', 4)]  
    >>> zip(m.values(), m.keys())  
    [(1, 'a'), (3, 'c'), (2, 'b'), (4, 'd')]  
    >>> mi = dict(zip(m.values(), m.keys()))  
    >>> mi  
    {1: 'a', 2: 'b', 3: 'c', 4: 'd'}
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