zip函数的作用:
zip函数接受任意多个可迭代对象作为参数,将对象中对应的元素打包成一个tuple
,然后返回一个可迭代的zip对象.
这个可迭代对象可以使用循环的方式列出其元素
若多个可迭代对象的长度不一致,则所返回的列表与长度最短的可迭代对象相同.
用法1:对多个列表生成一个zip对象,并打印
例1
>>> a1=[1,2,3] >>> a2=[4,5,6] >>> a3=[7,8,9] >>> a4=["a","b","c","d"] >>> zip1=zip(a1,a2,a3) >>> print(zip1) <zip object at 0x7f5a22651c08> >>> for i in zip1: ... print(i) ... (1, 4, 7) (2, 5, 8) (3, 6, 9)
例2
>>> zip2=zip(a1,a2,a4) >>> print(zip2) <zip object at 0x7f5a22651d48> >>> for j in zip2: ... print(j) ... (1, 4, 'a') (2, 5, 'b') (3, 6, 'c')
例3
>>> zip3=zip(a4) >>> print(zip3) <zip object at 0x7f5a22651d08> >>> for i in zip3: ... print(i) ... ('a',) ('b',) ('c',) ('d',)
例4
>>> zip4=zip(*a4 *3) >>> >>> print(zip4) <zip object at 0x7f5a22651f08> >>> for j in zip4: ... print(j) ... ('a', 'b', 'c', 'd', 'a', 'b', 'c', 'd', 'a', 'b', 'c', 'd')
用法2:二维矩阵变换(行、列互换)
matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]
使用*号拆分list,zip对每个list打包成tuple
>>> list(zip(*matrix)) [ (1, 4, 7), (2, 5, 8), (3, 6, 9) ]
用列表推导式行列互换
>>> l1=[ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] >>> print([[j[i] for j in l1] for i in range(len(l1[0])) ]) [ [1, 4, 7], [2, 5, 8], [3, 6, 9] ]