2013-09-29 02:34:44
到09-29-2013
我有一个包含Created_date列的数据框:
Name: Created_Date, Length: 1162549, dtype: datetime64[ns]`
我尝试在此系列中应用.date()
方法,例如:df.Created_Date.date()
,但我收到错误AttributeError: 'Series' object has no attribute 'date'
有人能帮助我吗?
答案 0 :(得分:32)
map
元素:
In [239]: from operator import methodcaller
In [240]: s = Series(date_range(Timestamp('now'), periods=2))
In [241]: s
Out[241]:
0 2013-10-01 00:24:16
1 2013-10-02 00:24:16
dtype: datetime64[ns]
In [238]: s.map(lambda x: x.strftime('%d-%m-%Y'))
Out[238]:
0 01-10-2013
1 02-10-2013
dtype: object
In [242]: s.map(methodcaller('strftime', '%d-%m-%Y'))
Out[242]:
0 01-10-2013
1 02-10-2013
dtype: object
您可以通过调用构成datetime.date
的{{1}}元素的date()
方法来获取原始Timestamp
对象:
Series
然而另一种方式可以通过调用未绑定的In [249]: s.map(methodcaller('date'))
Out[249]:
0 2013-10-01
1 2013-10-02
dtype: object
In [250]: s.map(methodcaller('date')).values
Out[250]:
array([datetime.date(2013, 10, 1), datetime.date(2013, 10, 2)], dtype=object)
方法来实现:
Timestamp.date
这种方法是最快的,而恕我直言最具可读性。可以在顶级In [273]: s.map(Timestamp.date)
Out[273]:
0 2013-10-01
1 2013-10-02
dtype: object
模块中访问Timestamp
,如下所示:pandas
。我已将其直接导入以用于说明目的。
pandas.Timestamp
个对象的date
属性执行类似操作,但返回DatetimeIndex
对象数组:
numpy
对于较大的In [243]: index = DatetimeIndex(s)
In [244]: index
Out[244]:
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-10-01 00:24:16, 2013-10-02 00:24:16]
Length: 2, Freq: None, Timezone: None
In [246]: index.date
Out[246]:
array([datetime.date(2013, 10, 1), datetime.date(2013, 10, 2)], dtype=object)
datetime64[ns]
个对象,调用Series
的速度比Timestamp.date
更快,这比operator.methodcaller
略快:
lambda
请注意,In [263]: f = methodcaller('date')
In [264]: flam = lambda x: x.date()
In [265]: fmeth = Timestamp.date
In [266]: s2 = Series(date_range('20010101', periods=1000000, freq='T'))
In [267]: s2
Out[267]:
0 2001-01-01 00:00:00
1 2001-01-01 00:01:00
2 2001-01-01 00:02:00
3 2001-01-01 00:03:00
4 2001-01-01 00:04:00
5 2001-01-01 00:05:00
6 2001-01-01 00:06:00
7 2001-01-01 00:07:00
8 2001-01-01 00:08:00
9 2001-01-01 00:09:00
10 2001-01-01 00:10:00
11 2001-01-01 00:11:00
12 2001-01-01 00:12:00
13 2001-01-01 00:13:00
14 2001-01-01 00:14:00
...
999985 2002-11-26 10:25:00
999986 2002-11-26 10:26:00
999987 2002-11-26 10:27:00
999988 2002-11-26 10:28:00
999989 2002-11-26 10:29:00
999990 2002-11-26 10:30:00
999991 2002-11-26 10:31:00
999992 2002-11-26 10:32:00
999993 2002-11-26 10:33:00
999994 2002-11-26 10:34:00
999995 2002-11-26 10:35:00
999996 2002-11-26 10:36:00
999997 2002-11-26 10:37:00
999998 2002-11-26 10:38:00
999999 2002-11-26 10:39:00
Length: 1000000, dtype: datetime64[ns]
In [269]: timeit s2.map(f)
1 loops, best of 3: 1.04 s per loop
In [270]: timeit s2.map(flam)
1 loops, best of 3: 1.1 s per loop
In [271]: timeit s2.map(fmeth)
1 loops, best of 3: 968 ms per loop
的目标之一是在pandas
之上提供一个图层,以便(大多数情况下)您不必处理低级细节numpy
的。{因此,在数组中获取原始ndarray
对象的用途有限,因为它们与datetime.date
支持的任何numpy.dtype
不对应(pandas
仅支持{{1}那是[纳秒] dtypes)。也就是说,有时你需要这样做。
答案 1 :(得分:3)
也许这只是最近出现的,但有内置的方法。尝试:
In [27]: s = pd.Series(pd.date_range(pd.Timestamp('now'), periods=2))
In [28]: s
Out[28]:
0 2016-02-11 19:11:43.386016
1 2016-02-12 19:11:43.386016
dtype: datetime64[ns]
In [29]: s.dt.to_pydatetime()
Out[29]:
array([datetime.datetime(2016, 2, 11, 19, 11, 43, 386016),
datetime.datetime(2016, 2, 12, 19, 11, 43, 386016)], dtype=object)
答案 2 :(得分:2)
您可以尝试使用.dt.date
datetime64[ns]
上的dataframe
。
例如df['Created_date'] = df['Created_date'].dt.date
输入dataframe
,名为test_df
:
print(test_df)
结果:
Created_date
0 2015-03-04 15:39:16
1 2015-03-22 17:36:49
2 2015-03-25 22:08:45
3 2015-03-16 13:45:20
4 2015-03-19 18:53:50
检查dtypes
:
print(test_df.dtypes)
结果:
Created_date datetime64[ns]
dtype: object
提取date
并更新Created_date
列:
test_df['Created_date'] = test_df['Created_date'].dt.date
print(test_df)
结果:
Created_date
0 2015-03-04
1 2015-03-22
2 2015-03-25
3 2015-03-16
4 2015-03-19
答案 3 :(得分:0)
pdTime =pd.date_range(timeStamp, periods=len(years), freq="D")
pdTime[i].strftime('%m-%d-%Y')