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35個(gè)Python編程小技巧

 更新時(shí)間：2014年04月01日 12:01:59   作者：  
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。

1.1 拆箱

復(fù)制代碼 代碼如下:

>>> 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

1.2 拆箱變量交換
復(fù)制代碼 代碼如下:
>>> a, b = 1, 2
>>> a, b = b, a
>>> a, b
(2, 1)

1.3 擴(kuò)展拆箱（只兼容python3）
復(fù)制代碼 代碼如下:
>>> a, *b, c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
[2, 3, 4]
>>> c
5

1.4 負(fù)數(shù)索引
復(fù)制代碼 代碼如下:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-1]
10
>>> a[-3]
8

1.5 切割列表
復(fù)制代碼 代碼如下:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[2:8]
[2, 3, 4, 5, 6, 7]

1.6 負(fù)數(shù)索引切割列表
復(fù)制代碼 代碼如下:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-4:-2]
[7, 8]

1.7指定步長(zhǎng)切割列表
復(fù)制代碼 代碼如下:
>>> 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]

1.8 負(fù)數(shù)步長(zhǎng)切割列表
復(fù)制代碼 代碼如下:
>>> 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]

1.9 列表切割賦值
復(fù)制代碼 代碼如下:
>>> 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]

1.10 命名列表切割方式
復(fù)制代碼 代碼如下:
>>> a = [0, 1, 2, 3, 4, 5]
>>> LASTTHREE = slice(-3, None)
>>> LASTTHREE
slice(-3, None, None)
>>> a[LASTTHREE]
[3, 4, 5]

1.11 列表以及迭代器的壓縮和解壓縮
復(fù)制代碼 代碼如下:
>>> 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')]

1.12 列表相鄰元素壓縮器
復(fù)制代碼 代碼如下:
>>> 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,)]

1.13 在列表中用壓縮器和迭代器滑動(dòng)取值窗口
復(fù)制代碼 代碼如下:
>>> 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)]

1.14 用壓縮器反轉(zhuǎn)字典
復(fù)制代碼 代碼如下:
>>> 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'}

1.15 列表展開(kāi)
復(fù)制代碼 代碼如下:
>>> 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]

1.16 生成器表達(dá)式
復(fù)制代碼 代碼如下:
>>> 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

1.17 字典推導(dǎo)
復(fù)制代碼 代碼如下:
>>> 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'}

1.18 用字典推導(dǎo)反轉(zhuǎn)字典
復(fù)制代碼 代碼如下:
>>> 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'}

1.19 命名元組
復(fù)制代碼 代碼如下:
>>> 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
1.20 繼承命名元組
復(fù)制代碼 代碼如下:
>>> 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)

1.21 操作集合
復(fù)制代碼 代碼如下:
>>> 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

1.22 操作多重集合
復(fù)制代碼 代碼如下:
>>> 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})

1.23 統(tǒng)計(jì)在可迭代器中最常出現(xiàn)的元素
復(fù)制代碼 代碼如下:
>>> 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)]

1.24 兩端都可操作的隊(duì)列
復(fù)制代碼 代碼如下:
>>> 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])

1.25 有最大長(zhǎng)度的雙端隊(duì)列
復(fù)制代碼 代碼如下:
>>> 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

1.26 可排序詞典
復(fù)制代碼 代碼如下:
>>> 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

1.27 默認(rèn)詞典
復(fù)制代碼 代碼如下:
>>> 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]'

1.28 默認(rèn)字典的簡(jiǎn)單樹(shù)狀表達(dá)
復(fù)制代碼 代碼如下:
>>> 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"
    }
}

1.29 對(duì)象到唯一計(jì)數(shù)的映射
復(fù)制代碼 代碼如下:
>>> 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

1.30 最大和最小的幾個(gè)列表元素
復(fù)制代碼 代碼如下:
>>> 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]

1.31 兩個(gè)列表的笛卡爾積
復(fù)制代碼 代碼如下:
>>> 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

1.32 列表組合和列表元素替代組合
復(fù)制代碼 代碼如下:
>>> 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

1.33 列表元素排列組合
復(fù)制代碼 代碼如下:
>>> 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

1.34 可鏈接迭代器
復(fù)制代碼 代碼如下:
>>> 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)

1.35 根據(jù)文件指定列類(lèi)聚
復(fù)制代碼 代碼如下:
>>> 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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