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Python中zip()函数用法

时间:2015-07-19 14:51:19      阅读:150      评论:0      收藏:0      [点我收藏+]

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定义:zip([iterable, …])
zip()是Python的一个内建函数,它接受一系列可迭代的对象作为参数,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。利用*号操作符,可以将list unzip(解压),看下面的例子就明白了:

>>> a = [1,2,3,4]
>>> b = [5,6,7,8]
>>> c = [5,6,7,8,9,10]
>>> test_zip = zip(a,b)
>>> test_zip
[(1, 5), (2, 6), (3, 7), (4, 8)]
>>> test_zip1 = zip(a,c)
>>> test_zip1
[(1, 5), (2, 6), (3, 7), (4, 8)]
>>> test_zip2 = zip(b,c)
>>> test_zip2
[(5, 5), (6, 6), (7, 7), (8, 8)]
>>> zip(*test_zip)
[(1, 2, 3, 4), (5, 6, 7, 8)]
>>> 
>>> zip(a,b,c)
[(1, 5, 5), (2, 6, 6), (3, 7, 7), (4, 8, 8)]
>>> 

 

 

例子2:

>>> name
(jack, beginman, sony, pcky)
>>> age
(2001, 2003, 2005, 2000)
>>> for n,a in zip(name, age):
...     print n ,a
... 
jack 2001
beginman 2003
sony 2005
pcky 2000
>>> 

再看一例:

>>> all={"jack":2001,"beginman":2003,"sony":2005,"pcky":2000}
>>> for i in all.keys():
...     print i, all[i]
... 
sony 2005
pcky 2000
jack 2001
beginman 2003
>>> 

 

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)

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

 

zip()配合*号操作符,可以将已经zip过的列表对象解压:

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

 

更近一层的了解:
内容来源:http://www.cnblogs.com/diyunpeng/archive/2011/09/15/2177028.html

* 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)]

 

其它高级运用:

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}

 

Python中zip()函数用法

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原文地址:http://www.cnblogs.com/blogofwyl/p/4658571.html

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