本篇内容主要讲解“Python语言特征和编程技巧有哪些”，感兴趣的朋友不妨来看看。本文介绍的方法操作简单快捷，实用性强。下面就让小编来带大家学习“Python语言特征和编程技巧有哪些”吧!
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]>>> a1>>> b2>>> c3>>> d42 拆箱变量交换
>>> a, b = 1, 2>>> a, b = b, a>>> a, b(2, 1)3 扩展拆箱（只兼容python3）
>>> a, *b, c = [1, 2, 3, 4, 5]>>> a1>>> b[2, 3, 4]>>> c54 负数索引
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[-1]10>>> a[-3]85 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]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]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]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]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]10 命名列表切割方式
>>> a = [0, 1, 2, 3, 4, 5]>>> LASTTHREE = slice(-3, None)>>> LASTTHREEslice(-3, None, None)>>> a[LASTTHREE][3, 4, 5]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')]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,)]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)]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'}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]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)40817 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'}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'}19 命名元组
>>> Point = collections.namedtuple('Point', ['x', 'y'])>>> p = Point(x=1.0, y=2.0)>>> pPoint(x=1.0, y=2.0)>>> p.x1.0>>> p.y2.020 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 + qPoint(x=3.0, y=5.0)21 操作集合
>>> A = {1, 2, 3, 3}>>> Aset([1, 2, 3])>>> B = {3, 4, 5, 6, 7}>>> Bset([3, 4, 5, 6, 7])>>> A | Bset([1, 2, 3, 4, 5, 6, 7])>>> A & Bset([3])>>> A - Bset([1, 2])>>> B - Aset([4, 5, 6, 7])>>> A ^ Bset([1, 2, 4, 5, 6, 7])>>> (A ^ B) == ((A - B) | (B - A))True22 操作多重集合
>>> A = collections.Counter([1, 2, 2])>>> B = collections.Counter([2, 2, 3])>>> ACounter({2: 2, 1: 1})>>> BCounter({2: 2, 3: 1})>>> A | BCounter({2: 2, 1: 1, 3: 1})>>> A & BCounter({2: 2})>>> A + BCounter({2: 4, 1: 1, 3: 1})>>> A - BCounter({1: 1})>>> B - ACounter({3: 1})23 统计在可迭代器中最常出现的元素
>>> A = collections.Counter([1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7])>>> ACounter({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)]24 两端都可操作的队列
>>> Q = collections.deque()>>> Q.append(1)>>> Q.appendleft(2)>>> Q.extend([3, 4])>>> Q.extendleft([5, 6])>>> Qdeque([6, 5, 2, 1, 3, 4])>>> Q.pop()4>>> Q.popleft()6>>> Qdeque([5, 2, 1, 3])>>> Q.rotate(3)>>> Qdeque([2, 1, 3, 5])>>> Q.rotate(-3)>>> Qdeque([5, 2, 1, 3])25 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, 10, 1, 21, 2, 32, 3, 43, 4, 54, 5, 65, 6, 76, 7, 87, 8, 926 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, 127 默认词典
>>> m = dict()>>> m['a']Traceback (most recent call last): File "<stdin>", line 1, in <module>KeyError: 'a'>>>>>> m = collections.defaultdict(int)>>> m['a']>>> m['b']>>> 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]'28 默认字典的简单树状表达
>>> 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" }}29 对象到唯一计数的映射
>>> import itertools, collections>>> value_to_numeric_map = collections.defaultdict(itertools.count().next)>>> value_to_numeric_map['a']>>> value_to_numeric_map['b']1>>> value_to_numeric_map['c']2>>> value_to_numeric_map['a']>>> value_to_numeric_map['b']130 最大和最小的几个列表元素
>>> 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]31 两个列表的笛卡尔积
>>> 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)...000000010010001101000101011001111000100110101011110011011110111132 列表组合和列表元素替代组合
>>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):... print ''.join(str(x) for x in c)...123124125134135145234235245345>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):... print ''.join(str(x) for x in c)...11121322233333 列表元素排列组合
>>> for p in itertools.permutations([1, 2, 3, 4]):... print ''.join(str(x) for x in p)...12341243132413421423143221342143231423412413243131243142321432413412342141234132421342314312432134 可链接迭代器
>>> 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)35 根据文件指定列类聚
>>> 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 noneyoung myope no normal softyoung myope yes reduced noneyoung myope yes normal hardyoung hypermetrope no reduced noneyoung hypermetrope no到此，相信大家对“Python语言特征和编程技巧有哪些”有了更深的了解，不妨来实际操作一番吧！这里是编程网网站，更多相关内容可以进入相关频道进行查询，关注我们，继续学习！
