1 介绍

　　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。

　　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]>>> a1>>> b2>>> c3>>> d4

　　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]>>> a1>>> b[2, 3, 4]>>> c5

　　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)>>> LASTTHREEslice(-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)>>> pPoint(x=1.0, y=2.0)>>> p.x1.0>>> p.y2.0

　　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 + qPoint(x=3.0, y=5.0)

　　1.21   集合及集合操作

>>> A = {1, 2, 3, 3}>>> Aset([1, 2, 3])>>> B = {3, 4, 5, 6, 7}>>> Bset([3, 4, 5, 6, 7])>>> A | Bset([1, 2, 3, 4, 5, 6, 7])>>> A & Bset([3])>>> A - Bset([1, 2])>>> B - Aset([4, 5, 6, 7])>>> A ^ Bset([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])>>> ACounter({2: 2, 1: 1})>>> BCounter({2: 2, 3: 1})>>> A | BCounter({2: 2, 1: 1, 3: 1})>>> A & BCounter({2: 2})>>> A + BCounter({2: 4, 1: 1, 3: 1})>>> A - BCounter({1: 1})>>> B - ACounter({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])>>> ACounter({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])>>> Qdeque([6, 5, 2, 1, 3, 4])>>> Q.pop()4>>> Q.popleft()6>>> Qdeque([5, 2, 1, 3])>>> Q.rotate(3)>>> Qdeque([2, 1, 3, 5])>>> Q.rotate(-3)>>> Qdeque([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)...00, 10, 1, 21, 2, 32, 3, 43, 4, 54, 5, 65, 6, 76, 7, 87, 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)...0000000100100011010001010110011110001001101010111100110111101111

　　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)...123124125134135145234235245345>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):...     print ‘‘.join(str(x) for x in c)...111213222333

　　1.33   排序 (itertools.permutations)

>>> for p in itertools.permutations([1, 2, 3, 4]):...     print ‘‘.join(str(x) for x in p)...123412431324134214231432213421432314234124132431312431423214324134123421412341324213423143124321

　　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 = http://www.mamicode.com/[line.strip().split(‘,‘) for line in infile]>
　　原文地址：http://sahandsaba.com/thirty-python-language-features-and-tricks-you-may-not-know.html
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