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  • Python零碎知识(2):强大的zip

    转载:http://www.cnblogs.com/BeginMan/archive/2013/03/14/2959447.html
     

    一、代码引导

    首先看这一段代码:

     1 >>> name=('jack','beginman','sony','pcky')
     2 >>> age=(2001,2003,2005,2000)
     3 >>> for a,n in zip(name,age):
     4     print a,n
     5 
     6 输出:
     7 jack 2001
     8 beginman 2003
     9 sony 2005
    10 pcky 2000
     

    再看这一段代码:

    1 all={"jack":2001,"beginman":2003,"sony":2005,"pcky":2000}
    2 for i in all.keys():
    3     print i,all[i]
    4 
    5 输出:
    6 sony 2005
    7 pcky 2000
    8 jack 2001
    9 beginman 2003

    发现它们之间的区别么?

    最显而易见的是:第一种简洁、灵活、而且能顺序输入。

    二、zip()函数

    它是Python的内建函数,(与序列有关的内建函数有:sorted()、reversed()、enumerate()、zip()),其中sorted()和zip()返回一个序列(列表)对象,reversed()、enumerate()返回一个迭代器(类似序列)

    1 >>> type(sorted(s))
    2 <type 'list'>
    3 >>> type(zip(s))
    4 <type 'list'>
    5 >>> type(reversed(s))
    6 <type 'listreverseiterator'>
    7 >>> type(enumerate(s))
    8 <type 'enumerate'>

    那么什么是zip()函数 呢?

    我们help(zip)看看:

    1 >>> help(zip)
    2 Help on built-in function zip in module __builtin__:
    3 
    4 zip(...)
    5     zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]
    6     
    7     Return a list of tuples, where each tuple contains the i-th element
    8     from each of the argument sequences.  The returned list is truncated
    9     in length to the length of the shortest argument sequence.

    提示:不懂的一定多help

    定义:zip([seql, ...])接受一系列可迭代对象作为参数,将对象中对应的元素打包成一个个tuple(元组),然后返回由这些tuples组成的list(列表)。若传入参数的长度不等,则返回list的长度和参数中长度最短的对象相同。

     1 >>> z1=[1,2,3]
     2 >>> z2=[4,5,6]
     3 >>> result=zip(z1,z2)
     4 >>> result
     5 [(1, 4), (2, 5), (3, 6)]
     6 >>> z3=[4,5,6,7]
     7 >>> result=zip(z1,z3)
     8 >>> result
     9 [(1, 4), (2, 5), (3, 6)]
    10 >>> 

    zip()配合*号操作符,可以将已经zip过的列表对象解压

    1 >>> zip(*result)
    2 [(1, 2, 3), (4, 5, 6)]

    更近一层的了解:
    内容来源:http://www.cnblogs.com/diyunpeng/archive/2011/09/15/2177028.html   (博客园人才真多!)

    * 二维矩阵变换(矩阵的行列互换)
    比如我们有一个由列表描述的二维矩阵
    a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
    通过python列表推导的方法,我们也能轻易完成这个任务
    print [ [row[col] for row in a] for col in range(len(a[0]))]
    [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
    另外一种让人困惑的方法就是利用zip函数:
    >>> a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
    >>> zip(*a)
    [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
    >>> map(list,zip(*a))
    [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
     
    zip函数接受任意多个序列作为参数,将所有序列按相同的索引组合成一个元素是各个序列合并成的tuple的新序列,新的序列的长度以参数中最短的序列为准。另外(*)操作符与zip函数配合可以实现与zip相反的功能,即将合并的序列拆成多个tuple。
    ①tuple的新序列
    >>>>x=[1,2,3],y=['a','b','c']
    >>>zip(x,y)
    [(1,'a'),(2,'b'),(3,'c')]
    
    ②新的序列的长度以参数中最短的序列为准.
    >>>>x=[1,2],y=['a','b','c']
    >>>zip(x,y)
    [(1,'a'),(2,'b')]
    
    ③(*)操作符与zip函数配合可以实现与zip相反的功能,即将合并的序列拆成多个tuple。
    >>>>x=[1,2,3],y=['a','b','c']
    >>>>zip(*zip(x,y))
    [(1,2,3),('a','b','c')]

     其他高级应用:

    1.zip打包解包列表和倍数
    >>> 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')]
    
    2. 使用zip合并相邻的列表项
    
    >>> 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,)]
    
    3.使用zip和iterators生成滑动窗口 (n -grams) 
    >>> from itertools import islice
    >>> def n_grams(a, n):
    ...     z = (islice(a, i, None) 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)]
    
    4.使用zip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'}
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  • 原文地址:https://www.cnblogs.com/sunshishi/p/4763489.html
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