35个不会也要知道的Python小技巧

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。

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] >>> a 1 >>> b [2, 3, 4] >>> c 5

4 负数索引

''' 遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ ''' >>> 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 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) >>> LASTTHREE slice(-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

''' 遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ ''' >>> 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

13 >>> 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

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'}

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) >>> p Point(x=1.0, y=2.0) >>> p.x 1.0 >>> p.y 2.0

20 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 + 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 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]) >>> 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 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]) >>> 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 有最大长度的双端队列

''' 遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ ''' >>> 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

26 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, 1

27 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'] 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]'

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'] 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

30 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

''' 遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ ''' >>> 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)) ... 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 列表元素排列组合

>>> 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 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=lambda r: r[-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

总结

以上是生活随笔为你收集整理的35个不会也要知道的Python小技巧的全部内容，希望文章能够帮你解决所遇到的问题。

如果觉得生活随笔网站内容还不错，欢迎将生活随笔推荐给好友。