你可能不知道的30个Python语言的特点技巧

jopen 10年前

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。

20140411085000039252.jpg

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

</tr> </tbody> </table> </div> </div>

1.2   交换变量分拆

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.3   拓展分拆 (Python 3下适用)

1
2
3
4
</td>
>>> a, b = 1, 2
>>> a, b = b, a
>>> a, b
(2, 1)
</tr> </tbody> </table> </div> </div>

1.4   负索引

1
2
3
4
5
6
7
</td>
>>> a, *b, c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
[2, 3, 4]
>>> c
5
</tr> </tbody> </table> </div> </div>

1.5   列表切片 (a[start:end])

1
2
3
4
5
</td>
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-1]
10
>>> a[-3]
8
</tr> </tbody> </table> </div> </div>

1.6   使用负索引的列表切片

1
2
3
</td>
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[2:8]
[2, 3, 4, 5, 6, 7]
</tr> </tbody> </table> </div> </div>

1.7   带步进值的列表切片 (a[start:end:step])

1
2
3
</td>
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-4:-2]
[7, 8]
</tr> </tbody> </table> </div> </div>

1.8   负步进值得列表切片

1
2
3
4
5
6
7
</td>
>>> 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]
</tr> </tbody> </table> </div> </div>

1.9   列表切片赋值

1
2
3
4
5
</td>
>>> 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]
</tr> </tbody> </table> </div> </div>

1.10   命名切片 (slice(start, end, step))

1
2
3
4
5
6
7
8
9
10
</td>
>>> 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]
</tr> </tbody> </table> </div> </div>

1.11   zip打包解包列表和倍数

1
2
3
4
5
6
</td>
>>> a = [0, 1, 2, 3, 4, 5]
>>> LASTTHREE = slice(-3, None)
>>> LASTTHREE
slice(-3, None, None)
>>> a[LASTTHREE]
[3, 4, 5]
</tr> </tbody> </table> </div> </div>

1.12   使用zip合并相邻的列表项

1
2
3
4
5
6
7
</td>
>>> 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')]
</tr> </tbody> </table> </div> </div>

1.13  使用zip和iterators生成滑动窗口 (n -grams) 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
</td>
>>> 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,)]
</tr> </tbody> </table> </div> </div>

1.14   使用zip反转字典

1
2
3
4
5
6
7
8
9
10
11
12
</td>
>>> 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)]
</tr> </tbody> </table> </div> </div>

1.15   摊平列表:

1
2
3
4
5
6
7
8
</td>
>>> 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'}
</tr> </tbody> </table> </div> </div>

1.16   生成器表达式

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
</td>
>>> 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。
</tr> </tbody> </table> </div> </div>

1.17   迭代字典

1
2
3
4
5
6
7
8
9
10
11
12
13
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.18   通过迭代字典反转字典

1
2
3
4
5
6
7
</td>
>>> 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'}
</tr> </tbody> </table> </div> </div>

1.19   命名序列 (collections.namedtuple)

1
2
3
4
5
</td>
>>> 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'}
</tr> </tbody> </table> </div> </div>

1.20   命名列表的继承:

1
2
3
4
5
6
7
8
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.21   集合及集合操作

1
2
3
4
5
6
7
8
9
</td>
>>> 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)
</tr> </tbody> </table> </div> </div>

1.22   多重集及其操作 (collections.Counter)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
</td>
>>> 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})
</tr> </tbody> </table> </div> </div>

1.24   双端队列 (collections.deque)

1
2
3
4
5
6
7
</td>
>>> 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)]
</tr> </tbody> </table> </div> </div>

1.25   有最大长度的双端队列 (collections.deque)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
</td>
>>> 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])
</tr> </tbody> </table> </div> </div>

1.26   字典排序 (collections.OrderedDict)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.27   缺省字典 (collections.defaultdict)

1
2
3
4
5
6
7
8
9
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.28   用缺省字典表示简单的树

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
</td>
>>> 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]'
</tr> </tbody> </table> </div> </div>

1.29   映射对象到唯一的序列数 (collections.defaultdict)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
</td>
>>> 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)
</tr> </tbody> </table> </div> </div>

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

1
2
3
4
5
6
7
8
9
10
11
12
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.31   笛卡尔乘积 (itertools.product)

1
2
3
4
5
</td>
>>> 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]
</tr> </tbody> </table> </div> </div>

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.33   排序 (itertools.permutations)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.34   链接的迭代 (itertools.chain)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
</td>
>>> 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
</tr> </tbody> </table> </div> </div>

1.35   按给定值分组行 (itertools.groupby)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
</td>
>>> 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
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
>>> 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('{: <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   

来自：http://simpleblog.cn/qxl#article_id=82

相关经验