• [转载]你可能不知道的 30 个 Python 语言的特点技巧


    [转载地址:http://www.oschina.net/translate/thirty-python-language-features-and-tricks-you-may-not-know]

    从我开始学习Python时我就决定维护一个经常使用的“窍门”列表。不论何时当我看到一段让我觉得“酷,这样也行!”的代码时(在一个例子中、在StackOverflow、在开源码软件中,等等),我会尝试它直到理解它,然后把它添加到列表中。这篇文章是清理过列表的一部分。如果你是一个有经验的Python程序员,尽管你可能已经知道一些,但你仍能发现一些你不知道的。如果你是一个正在学习Python的C、C++或Java程序员,或者刚开始学习编程,那么你会像我一样发现它们中的很多非常有用。
    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。
    Garfielt
    Garfielt
    翻译于 1年前
    2人顶
    顶 翻译的不错哦!
    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'}

    1.15 摊平列表:

    >>> a = [[1, 2], [3, 4], [5, 6]]
    >>> list(itertools.chain.from_iterable(a))
    [1, 2, 3, 4, 5, 6]

    >>> sum(a, [])
    [1, 2, 3, 4, 5, 6]

    >>> [x for l in a for x in l]
    [1, 2, 3, 4, 5, 6]

    >>> a = [[[1, 2], [3, 4]], [[5, 6], [7, 8]]]
    >>> [x for l1 in a for l2 in l1 for x in l2]
    [1, 2, 3, 4, 5, 6, 7, 8]

    >>> a = [1, 2, [3, 4], [[5, 6], [7, 8]]]
    >>> flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
    >>> flatten(a)
    [1, 2, 3, 4, 5, 6, 7, 8]

    注意: 根据Python的文档,itertools.chain.from_iterable是首选。
    1.16 生成器表达式

    >>> g = (x ** 2 for x in xrange(10))
    >>> next(g)
    0
    >>> next(g)
    1
    >>> next(g)
    4
    >>> next(g)
    9
    >>> sum(x ** 3 for x in xrange(10))
    2025
    >>> sum(x ** 3 for x in xrange(10) if x % 3 == 1)
    408

    1.17 迭代字典

    >>> m = {x: x ** 2 for x in range(5)}
    >>> m
    {0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

    >>> m = {x: 'A' + str(x) for x in range(10)}
    >>> m
    {0: 'A0', 1: 'A1', 2: 'A2', 3: 'A3', 4: 'A4', 5: 'A5', 6: 'A6', 7: 'A7', 8: 'A8', 9: 'A9'}

    1.18 通过迭代字典反转字典

    >>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
    >>> m
    {'d': 4, 'a': 1, 'b': 2, 'c': 3}
    >>> {v: k for k, v in m.items()}
    {1: 'a', 2: 'b', 3: 'c', 4: 'd'}

    1.19 命名序列 (collections.namedtuple)

    >>> Point = collections.namedtuple('Point', ['x', 'y'])
    >>> p = Point(x=1.0, y=2.0)
    >>> p
    Point(x=1.0, y=2.0)
    >>> p.x
    1.0
    >>> p.y
    2.0

    1.20 命名列表的继承:

    >>> class Point(collections.namedtuple('PointBase', ['x', 'y'])):
    ... __slots__ = ()
    ... def __add__(self, other):
    ... return Point(x=self.x + other.x, y=self.y + other.y)
    ...
    >>> p = Point(x=1.0, y=2.0)
    >>> q = Point(x=2.0, y=3.0)
    >>> p + q
    Point(x=3.0, y=5.0)

    1.21 集合及集合操作

    >>> A = {1, 2, 3, 3}
    >>> A
    set([1, 2, 3])
    >>> B = {3, 4, 5, 6, 7}
    >>> B
    set([3, 4, 5, 6, 7])
    >>> A | B
    set([1, 2, 3, 4, 5, 6, 7])
    >>> A & B
    set([3])
    >>> A - B
    set([1, 2])
    >>> B - A
    set([4, 5, 6, 7])
    >>> A ^ B
    set([1, 2, 4, 5, 6, 7])
    >>> (A ^ B) == ((A - B) | (B - A))
    True

    1.22 多重集及其操作 (collections.Counter)

    >>> A = collections.Counter([1, 2, 2])
    >>> B = collections.Counter([2, 2, 3])
    >>> A
    Counter({2: 2, 1: 1})
    >>> B
    Counter({2: 2, 3: 1})
    >>> A | B
    Counter({2: 2, 1: 1, 3: 1})
    >>> A & B
    Counter({2: 2})
    >>> A + B
    Counter({2: 4, 1: 1, 3: 1})
    >>> A - B
    Counter({1: 1})
    >>> B - A
    Counter({3: 1})

    1.23 迭代中最常见的元素 (collections.Counter)

    >>> A = collections.Counter([1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7])
    >>> A
    Counter({3: 4, 1: 2, 2: 2, 4: 1, 5: 1, 6: 1, 7: 1})
    >>> A.most_common(1)
    [(3, 4)]
    >>> A.most_common(3)
    [(3, 4), (1, 2), (2, 2)]

    1.24 双端队列 (collections.deque)

    >>> Q = collections.deque()
    >>> Q.append(1)
    >>> Q.appendleft(2)
    >>> Q.extend([3, 4])
    >>> Q.extendleft([5, 6])
    >>> Q
    deque([6, 5, 2, 1, 3, 4])
    >>> Q.pop()
    4
    >>> Q.popleft()
    6
    >>> Q
    deque([5, 2, 1, 3])
    >>> Q.rotate(3)
    >>> Q
    deque([2, 1, 3, 5])
    >>> Q.rotate(-3)
    >>> Q
    deque([5, 2, 1, 3])

    1.25 有最大长度的双端队列 (collections.deque)

    >>> last_three = collections.deque(maxlen=3)
    >>> for i in xrange(10):
    ... last_three.append(i)
    ... print ', '.join(str(x) for x in last_three)
    ...
    0
    0, 1
    0, 1, 2
    1, 2, 3
    2, 3, 4
    3, 4, 5
    4, 5, 6
    5, 6, 7
    6, 7, 8
    7, 8, 9

    1.26 字典排序 (collections.OrderedDict)

    >>> m = dict((str(x), x) for x in range(10))
    >>> print ', '.join(m.keys())
    1, 0, 3, 2, 5, 4, 7, 6, 9, 8
    >>> m = collections.OrderedDict((str(x), x) for x in range(10))
    >>> print ', '.join(m.keys())
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9
    >>> m = collections.OrderedDict((str(x), x) for x in range(10, 0, -1))
    >>> print ', '.join(m.keys())
    10, 9, 8, 7, 6, 5, 4, 3, 2, 1

    1.27 缺省字典 (collections.defaultdict)

    >>> m = dict()
    >>> m['a']
    Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    KeyError: 'a'
    >>>
    >>> m = collections.defaultdict(int)
    >>> m['a']
    0
    >>> m['b']
    0
    >>> m = collections.defaultdict(str)
    >>> m['a']
    ''
    >>> m['b'] += 'a'
    >>> m['b']
    'a'
    >>> m = collections.defaultdict(lambda: '[default value]')
    >>> m['a']
    '[default value]'
    >>> m['b']
    '[default value]'

    1.28 用缺省字典表示简单的树

    >>> import json
    >>> tree = lambda: collections.defaultdict(tree)
    >>> root = tree()
    >>> root['menu']['id'] = 'file'
    >>> root['menu']['value'] = 'File'
    >>> root['menu']['menuitems']['new']['value'] = 'New'
    >>> root['menu']['menuitems']['new']['onclick'] = 'new();'
    >>> root['menu']['menuitems']['open']['value'] = 'Open'
    >>> root['menu']['menuitems']['open']['onclick'] = 'open();'
    >>> root['menu']['menuitems']['close']['value'] = 'Close'
    >>> root['menu']['menuitems']['close']['onclick'] = 'close();'
    >>> print json.dumps(root, sort_keys=True, indent=4, separators=(',', ': '))
    {
    "menu": {
    "id": "file",
    "menuitems": {
    "close": {
    "onclick": "close();",
    "value": "Close"
    },
    "new": {
    "onclick": "new();",
    "value": "New"
    },
    "open": {
    "onclick": "open();",
    "value": "Open"
    }
    },
    "value": "File"
    }
    }

    (到https://gist.github.com/hrldcpr/2012250查看详情)
    1.29 映射对象到唯一的序列数 (collections.defaultdict)

    >>> import itertools, collections
    >>> value_to_numeric_map = collections.defaultdict(itertools.count().next)
    >>> value_to_numeric_map['a']
    0
    >>> value_to_numeric_map['b']
    1
    >>> value_to_numeric_map['c']
    2
    >>> value_to_numeric_map['a']
    0
    >>> value_to_numeric_map['b']
    1

    1.30 最大最小元素 (heapq.nlargest和heapq.nsmallest)

    >>> a = [random.randint(0, 100) for __ in xrange(100)]
    >>> heapq.nsmallest(5, a)
    [3, 3, 5, 6, 8]
    >>> heapq.nlargest(5, a)
    [100, 100, 99, 98, 98]

    1.31 笛卡尔乘积 (itertools.product)

    >>> for p in itertools.product([1, 2, 3], [4, 5]):
    (1, 4)
    (1, 5)
    (2, 4)
    (2, 5)
    (3, 4)
    (3, 5)
    >>> for p in itertools.product([0, 1], repeat=4):
    ... print ''.join(str(x) for x in p)
    ...
    0000
    0001
    0010
    0011
    0100
    0101
    0110
    0111
    1000
    1001
    1010
    1011
    1100
    1101
    1110
    1111

    1.32 组合的组合和置换 (itertools.combinations 和 itertools.combinations_with_replacement)

    >>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):
    ... print ''.join(str(x) for x in c)
    ...
    123
    124
    125
    134
    135
    145
    234
    235
    245
    345
    >>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):
    ... print ''.join(str(x) for x in c)
    ...
    11
    12
    13
    22
    23
    33

    1.33 排序 (itertools.permutations)

    >>> for p in itertools.permutations([1, 2, 3, 4]):
    ... print ''.join(str(x) for x in p)
    ...
    1234
    1243
    1324
    1342
    1423
    1432
    2134
    2143
    2314
    2341
    2413
    2431
    3124
    3142
    3214
    3241
    3412
    3421
    4123
    4132
    4213
    4231
    4312
    4321

    1.34 链接的迭代 (itertools.chain)

    >>> a = [1, 2, 3, 4]
    >>> for p in itertools.chain(itertools.combinations(a, 2), itertools.combinations(a, 3)):
    ... print p
    ...
    (1, 2)
    (1, 3)
    (1, 4)
    (2, 3)
    (2, 4)
    (3, 4)
    (1, 2, 3)
    (1, 2, 4)
    (1, 3, 4)
    (2, 3, 4)
    >>> for subset in itertools.chain.from_iterable(itertools.combinations(a, n) for n in range(len(a) + 1))
    ... print subset
    ...
    ()
    (1,)
    (2,)
    (3,)
    (4,)
    (1, 2)
    (1, 3)
    (1, 4)
    (2, 3)
    (2, 4)
    (3, 4)
    (1, 2, 3)
    (1, 2, 4)
    (1, 3, 4)
    (2, 3, 4)
    (1, 2, 3, 4)

    1.35 按给定值分组行 (itertools.groupby)

    >>> from operator import itemgetter
    >>> import itertools
    >>> with open('contactlenses.csv', 'r') as infile:
    ... data = [line.strip().split(',') for line in infile]
    ...
    >>> data = data[1:]
    >>> def print_data(rows):
    ... print ' '.join(' '.join('{: <16}'.format(s) for s in row) for row in rows)
    ...

    >>> print_data(data)
    young myope no reduced none
    young myope no normal soft
    young myope yes reduced none
    young myope yes normal hard
    young hypermetrope no reduced none
    young hypermetrope no normal soft
    young hypermetrope yes reduced none
    young hypermetrope yes normal hard
    pre-presbyopic myope no reduced none
    pre-presbyopic myope no normal soft
    pre-presbyopic myope yes reduced none
    pre-presbyopic myope yes normal hard
    pre-presbyopic hypermetrope no reduced none
    pre-presbyopic hypermetrope no normal soft
    pre-presbyopic hypermetrope yes reduced none
    pre-presbyopic hypermetrope yes normal none
    presbyopic myope no reduced none
    presbyopic myope no normal none
    presbyopic myope yes reduced none
    presbyopic myope yes normal hard
    presbyopic hypermetrope no reduced none
    presbyopic hypermetrope no normal soft
    presbyopic hypermetrope yes reduced none
    presbyopic hypermetrope yes normal none

    >>> data.sort(key=itemgetter(-1))
    >>> for value, group in itertools.groupby(data, lambda r: r[-1]):
    ... print '-----------'
    ... print 'Group: ' + value
    ... print_data(group)
    ...
    -----------
    Group: hard
    young myope yes normal hard
    young hypermetrope yes normal hard
    pre-presbyopic myope yes normal hard
    presbyopic myope yes normal hard
    -----------
    Group: none
    young myope no reduced none
    young myope yes reduced none
    young hypermetrope no reduced none
    young hypermetrope yes reduced none
    pre-presbyopic myope no reduced none
    pre-presbyopic myope yes reduced none
    pre-presbyopic hypermetrope no reduced none
    pre-presbyopic hypermetrope yes reduced none
    pre-presbyopic hypermetrope yes normal none
    presbyopic myope no reduced none
    presbyopic myope no normal none
    presbyopic myope yes reduced none
    presbyopic hypermetrope no reduced none
    presbyopic hypermetrope yes reduced none
    presbyopic hypermetrope yes normal none
    -----------
    Group: soft
    young myope no normal soft
    young hypermetrope no normal soft
    pre-presbyopic myope no normal soft
    pre-presbyopic hypermetrope no normal soft
    presbyopic hypermetrope no normal soft

    “If you give someone a program, you will frustrate them for a day; if you teach them how to program, you will frustrate them for a lifetime.”
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  • 原文地址:https://www.cnblogs.com/Scorpio989/p/4774262.html
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