Python语言的特点技巧有哪些
这期内容当中小编将会给大家带来有关Python语言的特点技巧有哪些，文章内容丰富且以专业的角度为大家分析和叙述，阅读完这篇文章希望大家可以有所收获。
创新互联公司专注为客户提供全方位的互联网综合服务，包含不限于成都网站制作、成都做网站、梓潼网络推广、成都小程序开发、梓潼网络营销、梓潼企业策划、梓潼品牌公关、搜索引擎seo、人物专访、企业宣传片、企业代运营等，从售前售中售后，我们都将竭诚为您服务，您的肯定，是我们最大的嘉奖；创新互联公司为所有大学生创业者提供梓潼建站搭建服务，24小时服务热线：18982081108，官方网址：www.cdcxhl.com
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。
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
>>> 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 交换变量分拆
>>> a, b = 1, 2 >>> a, b = b, a >>> a, b (2, 1)
1.3 拓展分拆 (Python 3下适用)
>>> a, *b, c = [1, 2, 3, 4, 5] >>> a 1 >>> b [2, 3, 4] >>> c 5
1.4 负索引
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-1] 10 >>> a[-3] 8
1.5 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]
1.6 使用负索引的列表切片
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-4:-2] [7, 8]
1.7 带步进值的列表切片 (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]
1.8 负步进值得列表切片
>>> 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]
1.9 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]
1.10 命名切片 (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]
1.11 zip打包解包列表和倍数
>>> 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 使用zip合并相邻的列表项
>>> 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 使用zip和iterators生成滑动窗口 (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)]
1.14 使用zip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 摊平列表:
>>> 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]
注意: 根据Python的文档，itertools.chain.from_iterable是***。
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 迭代字典
>>> 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 通过迭代字典反转字典
>>> 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 命名序列 (collections.namedtuple)
>>> 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 2.01.20 命名列表的继承:
>>> 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)1.21 集合及集合操作
>>> 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)) True1.22 多重集及其操作 (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})1.23 迭代中最常见的元素 (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)]1.24 双端队列 (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])
1.25 有***长度的双端队列 (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
1.26 字典排序 (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
1.27 缺省字典 (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]'
1.28 用缺省字典表示简单的树
>>> 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)
1.29 映射对象到唯一的序列数 (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
1.30 ***最小元素 (heapq.nlargest和heapq.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]
1.31 笛卡尔乘积 (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
1.32 组合的组合和置换 (itertools.combinations 和 itertools.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
1.33 排序 (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
1.34 链接的迭代 (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)
1.35 按给定值分组行 (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上述就是小编为大家分享的Python语言的特点技巧有哪些了，如果刚好有类似的疑惑，不妨参照上述分析进行理解。如果想知道更多相关知识，欢迎关注创新互联行业资讯频道。
网站栏目：Python语言的特点技巧有哪些
URL标题：http://shouzuofang.com/article/ieiosp.html