qujin_liang 发表于2年前

• 发表于 2年前
• 阅读 1
• 收藏 0
• 评论 0

1.1 拆箱

>>> 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 inrange(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 拆箱变量交换

>>> a, b = 1, 2

>>> a, b = b, a

>>> a, b

(2, 1)

1.3 扩展拆箱（只兼容python3）

>>> a, *b, c = [1, 2, 3, 4, 5]

>>> a

1

>>> b

[2, 3, 4]

>>> c

5

1.4 负数索引

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10]

>>> a[-1]

10

>>> a[-3]

8

1.5 切割列表

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10]

>>> a[2:8]

[2, 3, 4, 5, 6, 7]

1.6 负数索引切割列表

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10]

>>> a[-4:-2]

[7, 8]

1.7指定步长切割列表

>>> 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 负数步长切割列表

>>> 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 列表切割赋值

>>> 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 命名列表切割方式

>>> a = [0, 1, 2, 3, 4, 5]

>>> LASTTHREE = slice(-3, None)

>>> LASTTHREE

slice(-3, None, None)

>>> a[LASTTHREE]

[3, 4, 5]

1.11 列表以及迭代器的压缩和解压缩

>>> 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 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))

[(1, 2, 3), (4, 5, 6)]

[(1, 2), (3, 4), (5, 6)]

[(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)))

[(1, 2, 3), (4, 5, 6)]

[(1, 2), (3, 4), (5, 6)]

[(1,), (2,), (3,), (4,), (5,), (6,)]

1.13 在列表中用压缩器和迭代器滑动取值窗口

>>> def n_grams(a, n):

...     z = [iter(a[i:])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 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'}

1.15 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 forx 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 inx for y in flatten(l)] if type(x) is list else [x]

>>> flatten(a)

[1, 2, 3, 4, 5, 6, 7, 8]

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 字典推导

>>> m = {x: x ** 2 for x inrange(5)}

>>> m

{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

>>> m = {x: 'A' + str(x) for x inrange(10)}

>>> m

{0: 'A0', 1: 'A1', 2: 'A2', 3: 'A3', 4: 'A4',5: 'A5', 6: 'A6', 7: 'A7', 8: 'A8', 9: 'A9'}

1.18 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'}

1.19 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

1.20 继承命名元组

>>> classPoint(collections.namedtuple('PointBase', ['x', 'y'])):

...     __slots__ = ()

...            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 操作集合

>>> 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

×