必看的的30个Python语言的特点技巧(2)

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 "", 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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