你可能不知道的30个Python语言的特点技巧

甜思思

　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]
>>> 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 "", line 1, in
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 '\n'.join('\t'.join('{: