Python 从 SQL 数据库导入表并按日期过滤行时，将 Pandas 列解析为日期时间

Parse a Pandas column to Datetime when importing table from SQL database and filtering rows by date

提问by Nyxynyx

I have a DataFramewith column named date. How can we convert/parse the 'date' column to a DateTimeobject?

我有一个DataFrame名为date. 我们如何将“日期”列转换/解析为DateTime对象？

I loaded the date column from a Postgresql database using sql.read_frame(). An example of the datecolumn is 2013-04-04.

我使用sql.read_frame(). 该date列的一个示例是2013-04-04。

What I am trying to do is to select all rows in a dataframe that has their date columns within a certain period, like after 2013-04-01and before 2013-04-04.

我想要做的是选择数据框中在特定时间段内具有日期列的所有行，例如 after2013-04-01和 before 2013-04-04。

My attempt below gives the error 'Series' object has no attribute 'read'

我在下面的尝试给出了错误 'Series' object has no attribute 'read'

Attempt

试图

import dateutil

df['date'] = dateutil.parser.parse(df['date'])

Error

错误

AttributeError                            Traceback (most recent call last)
<ipython-input-636-9b19aa5f989c> in <module>()
     15 
     16 # Parse 'Date' Column to Datetime
---> 17 df['date'] = dateutil.parser.parse(df['date'])
     18 
     19 # SELECT RECENT SALES

C:\Python27\lib\site-packages\dateutil\parser.pyc in parse(timestr, parserinfo, **kwargs)
    695         return parser(parserinfo).parse(timestr, **kwargs)
    696     else:
--> 697         return DEFAULTPARSER.parse(timestr, **kwargs)
    698 
    699 

C:\Python27\lib\site-packages\dateutil\parser.pyc in parse(self, timestr, default, ignoretz, tzinfos, **kwargs)
    299             default = datetime.datetime.now().replace(hour=0, minute=0,
    300                                                       second=0, microsecond=0)
--> 301         res = self._parse(timestr, **kwargs)
    302         if res is None:
    303             raise ValueError, "unknown string format"

C:\Python27\lib\site-packages\dateutil\parser.pyc in _parse(self, timestr, dayfirst, yearfirst, fuzzy)
    347             yearfirst = info.yearfirst
    348         res = self._result()
--> 349         l = _timelex.split(timestr)
    350         try:
    351 

C:\Python27\lib\site-packages\dateutil\parser.pyc in split(cls, s)
    141 
    142     def split(cls, s):
--> 143         return list(cls(s))
    144     split = classmethod(split)
    145 

C:\Python27\lib\site-packages\dateutil\parser.pyc in next(self)
    135 
    136     def next(self):
--> 137         token = self.get_token()
    138         if token is None:
    139             raise StopIteration

C:\Python27\lib\site-packages\dateutil\parser.pyc in get_token(self)
     66                 nextchar = self.charstack.pop(0)
     67             else:
---> 68                 nextchar = self.instream.read(1)
     69                 while nextchar == '\x00':
     70                     nextchar = self.instream.read(1)

AttributeError: 'Series' object has no attribute 'read'

---

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df['date'].apply(dateutil.parser.parse)gives me the error AttributeError: 'datetime.date' object has no attribute 'read'

df['date'].apply(dateutil.parser.parse)给我错误 AttributeError: 'datetime.date' object has no attribute 'read'

df['date'].truncate(after='2013/04/01')gives the error TypeError: can't compare datetime.datetime to long

df['date'].truncate(after='2013/04/01')给出错误 TypeError: can't compare datetime.datetime to long

df['date'].dtypereturns dtype('O'). Is it already a datetimeobject?

df['date'].dtype返回dtype('O')。它已经是一个datetime对象了吗？

采纳答案by Ryan Saxe

pandas already reads that as a datetimeobject! So what you want is to select rows between two dates and you can do that by masking:

pandas 已经将其读取为datetime对象！所以你想要的是在两个日期之间选择行，你可以通过屏蔽来做到这一点：

df_masked = df[(df.date > '2012-04-01') & (df.date < '2012-04-04')]

Because you said that you were getting an error from the string for some reason, try this:

因为您说由于某种原因从字符串中收到错误，请尝试以下操作：

df_masked = df[(df.date > datetime.date(2012,4,1)) & (df.date < datetime.date(2012,4,4))]

回答by herrfz

You probably need apply, so something like:

你可能需要apply，所以像：

df['date'] = df['date'].apply(dateutil.parser.parse)

Without an example of the column I can't guarantee this will work, but something in that direction should help you to carry on.

如果没有专栏的例子，我不能保证这会奏效，但是在这个方向上的一些东西应该可以帮助你继续。

回答by ryzhiy

You should iterate over the items and parse them independently, then construct a new list.

您应该遍历项目并独立解析它们，然后构建一个新列表。

df['date'] = [dateutil.parser.parse(x) for x in df['date']]

回答by Keith

Pandas is aware of the object datetime but when you use some of the import functions it is taken as a string. So what you need to do is make sure the column is set as the datetime type not as a string. Then you can make your query.

Pandas 知道对象日期时间，但是当您使用某些导入函数时，它被视为字符串。因此，您需要做的是确保将列设置为日期时间类型而不是字符串。然后你就可以进行查询了。

df['date']  = pd.to_datetime(df['date'])
df_masked = df[(df['date'] > datetime.date(2012,4,1)) & (df['date'] < datetime.date(2012,4,4))]

回答by jpp

Don't confuse datetime.datewith Pandas pd.Timestamp

不要datetime.date与熊猫混淆pd.Timestamp

A "Pandas datetimeseries" contains pd.Timestampelements, notdatetime.dateelements. The recommended solution for Pandas:

“熊猫datetime系列”包含pd.Timestamp元素，而不是datetime.date元素。推荐的 Pandas 解决方案：

s = pd.to_datetime(s)    # convert series to Pandas
mask = s > '2018-03-10'  # calculate Boolean mask against Pandas-compatible object

The top answers have issues:

顶级答案有问题：

• @RyanSaxe's accepted answer's first attempt doesn't work; the second answer is inefficient.
• As of Pandas v0.23.0, @Keith's highly upvoted answer doesn't work; it gives TypeError.

• @RyanSaxe 接受的答案的第一次尝试不起作用；第二个答案是低效的。
• 从 Pandas v0.23.0 开始，@Keith 高度赞成的答案不起作用；它给TypeError.

Any good Pandas solution mustensure:

任何好的 Pandas 解决方案都必须确保：

1. The series is a Pandas datetimeseries, not objectdtype.
2. The datetimeseries is compared to a compatible object, e.g. pd.Timestamp, or string in the correct format.

1. 该系列是 Pandasdatetime系列，而不是objectdtype。
2. 将该datetime系列与兼容对象（例如pd.Timestamp，或正确格式的字符串）进行比较。

Here's a demo with benchmarking, demonstrating that the one-off cost of conversion can be immediately offset by a single operation:

这是一个带有基准测试的演示，演示了一次性的转换成本可以通过单个操作立即抵消：

from datetime import date

L = [date(2018, 1, 10), date(2018, 5, 20), date(2018, 10, 30), date(2018, 11, 11)]
s = pd.Series(L*10**5)

a = s > date(2018, 3, 10)             # accepted solution #2, inefficient
b = pd.to_datetime(s) > '2018-03-10'  # more efficient, including datetime conversion

assert a.equals(b)                    # check solutions give same result

%timeit s > date(2018, 3, 10)                  # 40.5 ms
%timeit pd.to_datetime(s) > '2018-03-10'       # 33.7 ms

s = pd.to_datetime(s)

%timeit s > '2018-03-10'                       # 2.85 ms