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python(29)强大的zip函数
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)
代码一:
>>> name = ("lilei","lihua","zhang") >>> age = ("21","14","34") >>> for a,n in zip(name,age): ... print a,n ... lilei 21 lihua 14 zhang 34 >>>
代码二:
>>> all={"hell":20,"helo":23,"sony":15,"lei":19} >>> for i in all.keys(): ... print i,all[i] ... hell 20 helo 23 lei 19 sony 15 >>>
代码三: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。
>>> aa = ("1","3","5") >>> bb = ("ee","tt") >>> for a,n in zip(aa,bb): ... print a,n ... 1 ee 3 tt >>> zip(aa,bb) [(‘1‘, ‘ee‘), (‘3‘, ‘tt‘)] >>>
代码四:zip()配合*号操作符,可以将已经zip过的列表对象解压
>>> aa (‘1‘, ‘3‘, ‘5‘) >>> bb (‘ee‘, ‘tt‘) >>> ad = zip(aa,bb) >>> zip(*ad) [(‘1‘, ‘3‘), (‘ee‘, ‘tt‘)] >>> zip(ad) [((‘1‘, ‘ee‘),), ((‘3‘, ‘tt‘),)] >>>
参考地址:http://www.cnblogs.com/BeginMan/archive/2013/03/14/2959447.html
扩宽:
* 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‘)]
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‘}
python(29)强大的zip函数
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