numpy datetime转为date，pandas的日期类型转为python的daetime

dataframe的数据格式是这样子的：


 
info看一下里面的数据类型：

<class 'pandas.core.frame.DataFrame'>

RangeIndex: 307 entries, 0 to 306

Data columns (total 7 columns):

日期         307 non-null datetime64[ns]

指数         307 non-null float64

成交额(亿元)    307 non-null float64

涨跌         307 non-null float64

涨跌额        307 non-null float64

转债数目       307 non-null float64

剩余规模       307 non-null float64

dtypes: datetime64[ns](1), float64(6)

memory usage: 16.9 KB

日期 307 non-null datetime64[ns]
 
然后转为list看看：
a=list(df['日期'].values)
如果使用上面的方法，返回的是这样的数据：

[numpy.datetime64('2017-12-29T00:00:00.000000000'),

 numpy.datetime64('2018-01-02T00:00:00.000000000'),

 numpy.datetime64('2018-01-03T00:00:00.000000000'),

 numpy.datetime64('2018-01-04T00:00:00.000000000'),

 numpy.datetime64('2018-01-05T00:00:00.000000000'),

 numpy.datetime64('2018-01-08T00:00:00.000000000'),

 numpy.datetime64('2018-01-09T00:00:00.000000000'),

 numpy.datetime64('2018-01-10T00:00:00.000000000'),

 numpy.datetime64('2018-01-11T00:00:00.000000000'),

 numpy.datetime64('2018-01-12T00:00:00.000000000'),

 numpy.datetime64('2018-01-15T00:00:00.000000000'),

 numpy.datetime64('2018-01-16T00:00:00.000000000'),

 numpy.datetime64('2018-01-17T00:00:00.000000000'),

 
如何转化为python的daetime格式呢?
 
可以使用内置的：s.dt.to_pydatetime()
s为df的一列，也就是series数据格式

b=list(df['日期'].dt.to_pydatetime())

得到的是

[datetime.datetime(2017, 12, 29, 0, 0),

 datetime.datetime(2018, 1, 2, 0, 0),

 datetime.datetime(2018, 1, 3, 0, 0),

 datetime.datetime(2018, 1, 4, 0, 0),

 datetime.datetime(2018, 1, 5, 0, 0),

 datetime.datetime(2018, 1, 8, 0, 0),

 datetime.datetime(2018, 1, 9, 0, 0),

 datetime.datetime(2018, 1, 10, 0, 0),

 datetime.datetime(2018, 1, 11, 0, 0),

 datetime.datetime(2018, 1, 12, 0, 0),

 datetime.datetime(2018, 1, 15, 0, 0)

为了不想要小时，分钟，秒的数据，可以清洗一下：
b=[i.strftime('%Y-%m-%d') for i in b]
 
得到：

['2017-12-29',

 '2018-01-02',

 '2018-01-03',

 '2018-01-04',

 '2018-01-05',

 '2018-01-08',

 '2018-01-09',

 '2018-01-10',

 '2018-01-11',

 '2018-01-12',

 '2018-01-15',

 '2018-01-16',

 '2018-01-17',]