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。

1.15 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

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