这些Python小技巧可以了解下
Python数据分析与人工智能 今天 来源：http://python.jobbole.com/63320/
从我开始学习python的时候，我就开始自己总结一个python小技巧的集合。后来当我什么时候在Stack Overflow或者在某个开源软件里看到一段很酷代码的时候，我就很惊讶：原来还能这么做！，当时我会努力的自己尝试一下这段代码，直到我懂了它的整体思路以后，我就把这段代码加到我的集合里。这篇博客其实就是这个集合整理后一部分的公开亮相。如果你已经是个python大牛，那么基本上你应该知道这里面的大多数用法了，但我想你应该也能发现一些你不知道的新技巧。而如果你之前是一个c,c++,java的程序员，同时在学习python，或者干脆就是一个刚刚学习编程的新手，那么你应该会看到很多特别有用能让你感到惊奇的实用技巧，就像我当初一样。
每一个技巧和语言用法都会在一个个实例中展示给大家，也不需要有其他的说明。我已经尽力把每个例子弄的通俗易懂，但是因为读者对python的熟悉程度不同，仍然可能难免有一些晦涩的地方。所以如果这些例子本身无法让你读懂，至少这个例子的标题在你后面去google搜索的时候会帮到你。 整个集合大概是按照难易程度排序，简单常见的在前面，比较少见的在最后。
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 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)) 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 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 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] 1.16 生成器表达式
g = (x ** 2 for x in xrange(10)) next(g)
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 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 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
