python中map()与zip()操作方法

1091次阅读  |  发布于5年以前

对于map()它的原型是:map(function,sequence),就是对序列sequence中每个元素都执行函数function操作。
比如之前的a,b,c = map(int,raw_input().split()),意思就是说把输入的a,b,c转化为整数。再比如:


    a = ['1','2','3','4']
    print map(list,a)
    print map(int,a)

第一个map是把列表a中每个元素转化为列表,第二个map是把a中每个元素转化为整数。
而对于zip(),原型是zip(list),list是一个列表,zip(list)返回的是一个元组,比如:


    list = [[1,2,3],[4,5,6],[7,8,9]]
    t = zip(*list)
    print t

输出:[(1, 4, 7), (2, 5, 8), (3, 6, 9)]


    x = [1,2,3,4,5]
    y = [6,7,8,9,10]
    a = zip(x,y)
    print a

输出:[(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]

下面是一些补充:


    [python] 
    >>> list = [[0,1,2],[3,1,4]] 
    >>> [sum(x) for x in list] 
    [3, 8] 
    >>> map(sum,list) 
    [3, 8] 

如果要得到每列之和,需要用zip(*list)先unzip list,得到一个元组list,其中第i个元组包含了每行的第i个元素:


    [python] 
    >>> list = [[0,1,2],[3,1,4]] 
    >>> zip(*list) 
    [(0, 3), (1, 1), (2, 4)] 
    >>> [sum(x) for x in zip(*list)] 
    [3, 2, 6] 
    >>> map(sum,zip(*list)) 
    [3, 2, 6] 

下面的例子是关于zip和unzip(其实是zip和*一起用)如何work的:


    [python] 
    >>> x=[1,2,3] 
    >>> y=[4,5,6] 
    >>> zipped = zip(x,y) 
    >>> zipped 
    [(1, 4), (2, 5), (3, 6)] 
    >>> x2,y2=zip(*zipped) 
    >>> x2 
    (1, 2, 3) 
    >>> y2 
    (4, 5, 6) 
    >>> x3,y3=map(list,zip(*zipped)) 
    >>> x3 
    [1, 2, 3] 
    >>> y3 
    [4, 5, 6] 

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