今天在别人的博客上看到一篇非常干货的文章，在此翻译过来分享给大家。。作者的前言介绍就不再赘述，直接上干货代码。。
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 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
2、使用拆箱进行变量交换
>>> a, b = 1, 2 >>> a, b = b, a >>> a, b (2, 1)
3、扩展的拆箱(Python 3支持)
>>> a, *b, c = [1, 2, 3, 4, 5] >>> a 1 >>> b [2, 3, 4] >>> c 5
4、负数索引
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-1] 10 >>> a[-3] 8
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]
6、负数索引的列表切片
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-4:-2] [7, 8]
7、带步数的列表切片(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]
8、负数步数的列表切片
>>> 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]
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]
10、切片命名(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]
11、遍历列表索引和值(enumerate)
>>> a = ['Hello', 'world', '!']
>>> for i, x in enumerate(a):
... print '{}: {}'.format(i, x)
...
0: Hello
1: world
2: !
12、遍历字典的KEY和VALUE(dict.iteritems)
>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> for k, v in m.iteritems():
... print '{}: {}'.format(k, v)
...
a: 1
c: 3
b: 2
d: 4
# 注意：Python 3中要使用dict.items
13、压缩 & 解压列表和可遍历对象
>>> 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')]
14、使用zip分组相邻列表项
>>> a = [1, 2, 3, 4, 5, 6] >>> # Using iterators >>> 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,)] >>> # Using slices >>> from itertools import islice >>> group_adjacent = lambda a, k: zip(*(islice(a, i, None, 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,)]
15、使用zip &iterators实现推拉窗(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)]
16、使用zip反相字典对象
>>> 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'}
17、合并列表
>>> 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] Note: according to Python's documentation on sum, itertools.chain.from_iterable is the preferred method for this.
18、生成器
>>> 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
19、字典解析
>>> 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'}
20、使用字典解析反相字典对象
>>> 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'}
21、命名的tuples(collections.namedtuple)
>>> Point = collections.namedtuple('Point', ['x', 'y'])
>>> p = Point(x=4.0, y=2.0)
>>> p
Point(x=4.0, y=2.0)
>>> p.x
4.0
>>> p.y
2.0
22、继承命名tuples
>>> 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=4.0, y=2.0)
>>> q = Point(x=2.0, y=3.0)
>>> p + q
Point(x=6.0, y=5.0)
23、Set & Set运算
>>> 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
24、Multisets运算(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})
25、列表中出现最多的元素(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)]
26、双向队列(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])
27、限制长度的双向队列(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
28、排序字典(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
29、默认字典(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]'
30、使用defaultdict代表tree
>>> 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"
}
}
# 查看更多：https://gist.github.com/hrldcpr/2012250
31、映射对象到唯一的计数数字(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
32、最大 & 最小元素(heapq.nlargest andheapq.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]
33、笛卡尔积(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
34、组合(itertools.combinations anditertools.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
35、排列(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
36、链接可遍历对象(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)
37、根据给定的KEY分组(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('{: <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 soft
38、在任意目录启动HTTP服务
python -m SimpleHTTPServer 5000 Serving HTTP on 0.0.0.0 port 5000 ...
39、Python之禅
>>> import this The Zen of Python, by Tim Peters Beautiful is better than ugly. Explicit is better than implicit. Simple is better than complex. Complex is better than complicated. Flat is better than nested. Sparse is better than dense. Readability counts. Special cases aren't special enough to break the rules. Although practicality beats purity. Errors should never pass silently. Unless explicitly silenced. In the face of ambiguity, refuse the temptation to guess. There should be one-- and preferably only one --obvious way to do it. Although that way may not be obvious at first unless you're Dutch. Now is better than never. Although never is often better than *right* now. If the implementation is hard to explain, it's a bad idea. If the implementation is easy to explain, it may be a good idea. Namespaces are one honking great idea -- let's do more of those!
40、使用C风格的大括号代替Python缩进来表示作用域
>>> from __future__ import braces http://www.jizhikongbao.com 空包代发