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  • 将元素为多个列表的列表元素进行转置、去重处理

    >>> a
    [['0068', '20160823', '571', '698038', 0, 0, 0, 0, 0, 0, 0], ['0068', '20160823', '571', '698039', 0, 0, 0, 0, 0, 0, 0], ['0068', '20160823', '571', '698040', 0, 0, 0, 0, 0, 0, 0], ['0068', '20160823', '571', '698041', 0, 0, 0, 0, 0, 0, 0], ['0068', '20160823', '571', '698042', 0, 0, 0, 0, 0, 0, 0], ['0068', '20160823', '571', '698043', 0, 0, 0, 0, 0, 0, 0]]
    >>> len(a)
    6>>> zip(*a)
    [('0068', '0068', '0068', '0068', '0068', '0068'), ('20160823', '20160823', '20160823', '20160823', '20160823', '20160823'), ('571', '571', '571', '571', '571', '571'), ('698038', '698039', '698040', '698041', '698042', '698043'), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0)]
    >>> set(zip(*a))
    set([(0, 0, 0, 0, 0, 0), ('0068', '0068', '0068', '0068', '0068', '0068'), ('20160823', '20160823', '20160823', '20160823', '20160823', '20160823'), ('571', '571', '571', '571', '571', '571'), ('698038', '698039', '698040', '698041', '698042', '698043')])
    >>> m=map(set,zip(*a))
    >>> m
    [set(['0068']), set(['20160823']), set(['571']), set(['698038', '698039', '698043', '698042', '698041', '698040']), set([0]), set([0]), set([0]), set([0]), set([0]), set([0]), set([0])]
    >>> map(list,m)
    [['0068'], ['20160823'], ['571'], ['698038', '698039', '698043', '698042', '698041', '698040'], [0], [0], [0], [0], [0], [0], [0]]
    >>> len(map(list,m))
    11
    >>> ii= map(list,m)
    >>> ii
    [['0068'], ['20160823'], ['571'], ['698038', '698039', '698043', '698042', '698041', '698040'], [0], [0], [0], [0], [0], [0], [0]]
    >>> ii.count([0])
    7
    >>> 
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  • 原文地址:https://www.cnblogs.com/apple2016/p/5804302.html
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