觉得很有用, 摘抄自: http://enml.github.io/site/2014/04/25/python-code/

列表相邻元素压缩器

>>> 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,)]

用压缩器反转字典

>>> 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\'}

列表展开

>>> 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]

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

字典推导

>>> 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\'}

用字典推导反转字典

>>> 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\'}

命名元组

>>> 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

继承命名元组

>>> 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)

有最大长度的双端队列

>>> 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

可排序词典

>>> 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