整個(gè)集合大概是按照難易程度排序，簡單常見的在前面，比較少見的在最后。

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

''' 遇到問題沒人解答？小編創(chuàng)建了一個(gè)Python學(xué)習(xí)交流QQ群：531509025 尋找有志同道合的小伙伴，互幫互助,群里還有不錯(cuò)的視頻學(xué)習(xí)教程和PDF電子書！ ''' >>> 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

''' 遇到問題沒人解答？小編創(chuàng)建了一個(gè)Python學(xué)習(xí)交流QQ群：531509025 尋找有志同道合的小伙伴，互幫互助,群里還有不錯(cuò)的視頻學(xué)習(xí)教程和PDF電子書！ ''' >>> 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

>>> 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

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 有最大長度的雙端隊(duì)列

''' 遇到問題沒人解答？小編創(chuàng)建了一個(gè)Python學(xué)習(xí)交流QQ群：531509025 尋找有志同道合的小伙伴，互幫互助,群里還有不錯(cuò)的視頻學(xué)習(xí)教程和PDF電子書！ ''' >>> 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'] 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 默認(rèn)字典的簡單樹狀表達(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'] 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

''' 遇到問題沒人解答？小編創(chuàng)建了一個(gè)Python學(xué)習(xí)交流QQ群：531509025 尋找有志同道合的小伙伴，互幫互助,群里還有不錯(cuò)的視頻學(xué)習(xí)教程和PDF電子書！ ''' >>> 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

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