• [ZZ] python 语言技巧


    http://sahandsaba.com/thirty-python-language-features-and-tricks-you-may-not-know.html  感谢原作者

    30 Python Language Features and Tricks You May Not Know About

    Posted on Mar 05, 2014 , last modified on Mar 16, 2014

    By Sahand Saba
     

    1.1   Unpacking

    >>> 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   Unpacking for swapping variables

    >>> a, b = 1, 2
    >>> a, b = b, a
    >>> a, b
    (2, 1)
    

    1.3   Extended unpacking (Python 3 only)

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

    1.4   Negative indexing

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    >>> a[-1]
    10
    >>> a[-3]
    8
    

    1.5   List slices (a[start:end])

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    >>> a[2:8]
    [2, 3, 4, 5, 6, 7]
    

    1.6   List slices with negative indexing

    >>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    >>> a[-4:-2]
    [7, 8]
    

    1.7   List slices with step (a[start:end:step])

    >>> 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   List slices with negative step

    >>> 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   List slice assignment

    >>> 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   Naming slices (slice(start, end, step))

    >>> a = [0, 1, 2, 3, 4, 5]
    >>> LASTTHREE = slice(-3, None)
    >>> LASTTHREE
    slice(-3, None, None)
    >>> a[LASTTHREE]
    [3, 4, 5]
    

    1.11   Zipping and unzipping lists and iterables

    >>> 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   Grouping adjacent list items using zip

    >>> 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   Inverting a dictionary using zip

    >>> 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'}
    

    1.14   Flattening lists:

    >>> 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]
    

    Note: according to Python's documentation on sumitertools.chain.from_iterable is the preferred method for this.

    1.15   Generator expressions

    >>> 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
    

    1.16   Dictionary comprehensions

    >>> 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'}
    

    1.17   Inverting a dictionary using a dictionary comprehension

    >>> 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'}
    

    1.18   Named tuples (collections.namedtuple)

    >>> 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
    

    1.19   Inheriting from named tuples:

    >>> 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)
    

    1.20   Sets and set operations

    >>> 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
    

    1.21   Multisets and multiset operations (collections.Counter)

    >>> 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})
    

    1.22   Most common elements in an iterable (collections.Counter)

    >>> 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)]
    

    1.23   Double-ended queue (collections.deque)

    >>> 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])
    

    1.24   Double-ended queue with maximum length (collections.deque)

    >>> 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
    

    1.25   Ordered dictionaries (collections.OrderedDict)

    >>> 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
    

    1.26   Default dictionaries (collections.defaultdict)

    >>> 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]'
    

    1.27   Using default dictionaries to represent simple trees

    >>> 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"
        }
    }
    

    (See https://gist.github.com/hrldcpr/2012250 for more on this.)

    1.28   Mapping objects to unique counting numbers (collections.defaultdict)

    >>> 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
    

    1.29   Largest and smallest elements (heapq.nlargest and heapq.nsmallest)

    >>> 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]
    

    1.30   Cartesian products (itertools.product)

    >>> 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
    

    1.31   Combinations and combinations with replacement (itertools.combinations anditertools.combinations_with_replacement)

    >>> 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
    

    1.32   Permutations (itertools.permutations)

    >>> 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
    

    1.33   Chaining iterables (itertools.chain)

    >>> 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)
    

    1.34   Grouping rows by a given key (itertools.groupby)

    >>> 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 '
    '.join('	'.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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  • 原文地址:https://www.cnblogs.com/foreveryl/p/3616292.html
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