從我開(kāi)始學(xué)習(xí)python的時(shí)候，我就開(kāi)始自己總結(jié)一個(gè)python小技巧的集合。后來(lái)當(dāng)我什么時(shí)候在Stack Overflow或者在某個(gè)開(kāi)源軟件里看到一段很酷代碼的時(shí)候，我就很驚訝：原來(lái)還能這么做！當(dāng)時(shí)我會(huì)努力的自己嘗試一下這段代碼，直到我懂了它的整體思路以后，我就把這段代碼加到我的集合里。這篇博客其實(shí)就是這個(gè)集合整理后一部分的公開(kāi)亮相。如果你已經(jīng)是個(gè)python大牛，那么基本上你應(yīng)該知道這里面的大多數(shù)用法了，但我想你應(yīng)該也能發(fā)現(xiàn)一些你不知道的新技巧。而如果你之前是一個(gè)c,c++,java的程序員，同時(shí)在學(xué)習(xí)python，或者干脆就是一個(gè)剛剛學(xué)習(xí)編程的新手，那么你應(yīng)該會(huì)看到很多特別有用能讓你感到驚奇的實(shí)用技巧，就像我當(dāng)初一樣。
每一個(gè)技巧和語(yǔ)言用法都會(huì)在一個(gè)個(gè)實(shí)例中展示給大家，也不需要有其他的說(shuō)明。我已經(jīng)盡力把每個(gè)例子弄的通俗易懂，但是因?yàn)樽x者對(duì)python的熟悉程度不同，仍然可能難免有一些晦澀的地方。所以如果這些例子本身無(wú)法讓你讀懂，至少這個(gè)例子的標(biāo)題在你后面去Google搜索的時(shí)候會(huì)幫到你。
整個(gè)集合大概是按照難易程度排序，簡(jiǎn)單常見(jiàn)的在前面，比較少見(jiàn)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
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>>> 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)
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(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
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
在學(xué)習(xí)過(guò)程中有什么不懂得可以加我的
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>>> 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 "
", 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]'
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']
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
>>> 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)
如果你依然在編程的世界里迷茫，可以加入我們的Python學(xué)習(xí)扣qun：784758214，看看前輩們是如何學(xué)習(xí)的。交流經(jīng)驗(yàn)。從基礎(chǔ)的python腳本到web開(kāi)發(fā)、爬蟲(chóng)、django、數(shù)據(jù)挖掘等，零基礎(chǔ)到項(xiàng)目實(shí)戰(zhàn)的資料都有整理。送給每一位python的小伙伴！分享一些學(xué)習(xí)的方法和需要注意的小細(xì)節(jié)，點(diǎn)擊加入我們的 python學(xué)習(xí)者聚集地
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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