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本篇內(nèi)容主要講解“Python語(yǔ)言特征和編程技巧有哪些”,感興趣的朋友不妨來(lái)看看。本文介紹的方法操作簡(jiǎn)單快捷,實(shí)用性強(qiáng)。下面就讓小編來(lái)帶大家學(xué)習(xí)“Python語(yǔ)言特征和編程技巧有哪些”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] >>> a 1 >>> b 2 >>> c 3 >>> d 4
2 拆箱變量交換
>>> a, b = 1, 2 >>> a, b = b, a >>> a, b (2, 1)
3 擴(kuò)展拆箱(只兼容python3)
>>> a, *b, c = [1, 2, 3, 4, 5] >>> a 1 >>> b [2, 3, 4] >>> c 5
4 負(fù)數(shù)索引
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-1] 10 >>> a[-3] 8
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]
6 負(fù)數(shù)索引切割列表
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> a[-4:-2] [7, 8]
7 指定步長(zhǎng)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 負(fù)數(shù)步長(zhǎng)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 列表切割賦值
>>> 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) >>> LASTTHREE slice(-3, None, None) >>> a[LASTTHREE] [3, 4, 5]
11 列表以及迭代器的壓縮和解壓縮
>>> 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 列表相鄰元素壓縮器
>>> 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 在列表中用壓縮器和迭代器滑動(dòng)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 用壓縮器反轉(zhuǎn)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 列表展開(kāi)
>>> 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 生成器表達(dá)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
17 字典推導(dǎo)
>>> 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 用字典推導(dǎo)反轉(zhuǎn)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 命名元組
>>> 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
20 繼承命名元組
>>> 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)
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, 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
22 操作多重集合
>>> 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})
23 統(tǒng)計(jì)在可迭代器中最常出現(xiàn)的元素
>>> 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)]
24 兩端都可操作的隊(duì)列
>>> 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])
25 有最大長(zhǎng)度的雙端隊(duì)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, 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
26 可排序詞典
>>> 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, 1
27 默認(rèn)詞典
>>> 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 默認(rèn)字典的簡(jiǎn)單樹(shù)狀表達(dá)
>>> 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 對(duì)象到唯一計(jì)數(shù)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'] 1
30 最大和最小的幾個(gè)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 兩個(gè)列表的笛卡爾積
>>> 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
32 列表組合和列表元素替代組合
>>> 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
33 列表元素排列組合
>>> 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
34 可鏈接迭代器
>>> 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 根據(jù)文件指定列類(lèi)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 none young myope no normal soft young myope yes reduced none young myope yes normal hard young hypermetrope no reduced none young hypermetrope no
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