标签归档：datetime-format

知识问答

将DataFrame列类型从字符串转换为日期时间，格式为dd / mm / yyyy

问题：将DataFrame列类型从字符串转换为日期时间，格式为dd / mm / yyyy

如何将字符串的DataFrame列（以dd / mm / yyyy格式）转换为日期时间？

点击查看英文原文

How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetimes?

回答 0

最简单的方法是使用to_datetime：

df['col'] = pd.to_datetime(df['col'])

它还dayfirst为欧洲时代提供了依据（但请注意，这并不严格）。

它在起作用：

In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0   2005-05-23 00:00:00
dtype: datetime64[ns]

您可以传递特定格式：

In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0   2005-05-23
dtype: datetime64[ns]
点击查看英文原文

The easiest way is to use to_datetime:

df['col'] = pd.to_datetime(df['col'])

It also offers a dayfirst argument for European times (but beware this isn’t strict).

Here it is in action:

In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0   2005-05-23 00:00:00
dtype: datetime64[ns]

You can pass a specific format:

In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0   2005-05-23
dtype: datetime64[ns]

回答 1

如果您的日期列是格式为’2017-01-01’的字符串，则可以使用pandas astype将其转换为日期时间。

df['date'] = df['date'].astype('datetime64[ns]')

或使用datetime64 [D]（如果您想要“天”精度而不是纳秒）

print(type(df_launath['date'].iloc[0]))

Yield

<class 'pandas._libs.tslib.Timestamp'> 与使用pandas.to_datetime时相同

您可以尝试使用其他格式，然后是’％Y-％m-％d’，但至少可以使用。

点击查看英文原文

If your date column is a string of the format ‘2017-01-01’ you can use pandas astype to convert it to datetime.

df['date'] = df['date'].astype('datetime64[ns]')

or use datetime64[D] if you want Day precision and not nanoseconds

print(type(df_launath['date'].iloc[0]))

yields

<class 'pandas._libs.tslib.Timestamp'> the same as when you use pandas.to_datetime

You can try it with other formats then ‘%Y-%m-%d’ but at least this works.

回答 2

如果要指定复杂的格式，可以使用以下内容：

df['date_col'] =  pd.to_datetime(df['date_col'], format='%d/%m/%Y')

format此处的更多详细信息：

点击查看英文原文

You can use the following if you want to specify tricky formats:

df['date_col'] =  pd.to_datetime(df['date_col'], format='%d/%m/%Y')

More details on format here:

回答 3

如果您的约会中有多种格式，别忘了设定infer_datetime_format=True以简化生活

df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

资料来源：pd.to_datetime

或者，如果您想要定制的方法：

def autoconvert_datetime(value):
    formats = ['%m/%d/%Y', '%m-%d-%y']  # formats to try
    result_format = '%d-%m-%Y'  # output format
    for dt_format in formats:
        try:
            dt_obj = datetime.strptime(value, dt_format)
            return dt_obj.strftime(result_format)
        except Exception as e:  # throws exception when format doesn't match
            pass
    return value  # let it be if it doesn't match

df['date'] = df['date'].apply(autoconvert_datetime)
点击查看英文原文

If you have a mixture of formats in your date, don’t forget to set infer_datetime_format=True to make life easier

df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

Source: pd.to_datetime

or if you want a customized approach:

def autoconvert_datetime(value):
    formats = ['%m/%d/%Y', '%m-%d-%y']  # formats to try
    result_format = '%d-%m-%Y'  # output format
    for dt_format in formats:
        try:
            dt_obj = datetime.strptime(value, dt_format)
            return dt_obj.strftime(result_format)
        except Exception as e:  # throws exception when format doesn't match
            pass
    return value  # let it be if it doesn't match

df['date'] = df['date'].apply(autoconvert_datetime)
知识问答

Python日期时间到没有微秒组件的字符串

问题：Python日期时间到没有微秒组件的字符串

我正在将UTC时间字符串添加到当前仅包含Amsterdam（！）时间字符串的Bitbucket API响应中。为了与其他地方返回的UTC时间字符串保持一致，请使用所需的格式2011-11-03 11:07:04（后跟+00:00，但这不是紧密联系）。

什么是创建这样一个字符串（最好的方式，而不从一微秒组件）datetime的实例与微秒组成部分？

>>> import datetime
>>> print unicode(datetime.datetime.now())
2011-11-03 11:13:39.278026

我会添加出现在我身上的最佳选择作为可能的答案，但是可能会有更优雅的解决方案。

编辑：我应该提一下，我实际上并不是在打印当前时间-我曾经datetime.now提供一个简单的例子。因此，该解决方案不应假定datetime其接收到的任何实例都将包含微秒组件。

点击查看英文原文

I’m adding UTC time strings to Bitbucket API responses that currently only contain Amsterdam (!) time strings. For consistency with the UTC time strings returned elsewhere, the desired format is 2011-11-03 11:07:04 (followed by +00:00, but that’s not germane).

What’s the best way to create such a string (without a microsecond component) from a datetime instance with a microsecond component?

>>> import datetime
>>> print unicode(datetime.datetime.now())
2011-11-03 11:13:39.278026

I’ll add the best option that’s occurred to me as a possible answer, but there may well be a more elegant solution.

Edit: I should mention that I’m not actually printing the current time – I used datetime.now to provide a quick example. So the solution should not assume that any datetime instances it receives will include microsecond components.

回答 0

如果要以datetime不同于标准格式的特定格式格式化对象，则最好明确指定该格式：

>>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 18:21:26'

有关指令的说明，请参见的文档datetime.strftime()%。

点击查看英文原文

If you want to format a datetime object in a specific format that is different from the standard format, it’s best to explicitly specify that format:

>>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 18:21:26'

See the documentation of datetime.strftime() for an explanation of the % directives.

回答 1

>>> import datetime
>>> now = datetime.datetime.now()
>>> print unicode(now.replace(microsecond=0))
2011-11-03 11:19:07
点击查看英文原文 
>>> import datetime
>>> now = datetime.datetime.now()
>>> print unicode(now.replace(microsecond=0))
2011-11-03 11:19:07

回答 2

在Python 3.6中：

from datetime import datetime
datetime.datetime.now().isoformat(' ', 'seconds')
'2017-01-11 14:41:33'

https://docs.python.org/3.6/library/datetime.html#datetime.datetime.isoformat

点击查看英文原文

In Python 3.6:

from datetime import datetime
datetime.datetime.now().isoformat(' ', 'seconds')
'2017-01-11 14:41:33'

https://docs.python.org/3.6/library/datetime.html#datetime.datetime.isoformat

回答 3

这就是我做到的方式。ISO格式：

import datetime
datetime.datetime.now().replace(microsecond=0).isoformat()
# Returns: '2017-01-23T14:58:07'

如果您不想使用ISO格式，则可以替换为’T’：

datetime.datetime.now().replace(microsecond=0).isoformat(' ')
# Returns: '2017-01-23 15:05:27'
点击查看英文原文

This is the way I do it. ISO format:

import datetime
datetime.datetime.now().replace(microsecond=0).isoformat()
# Returns: '2017-01-23T14:58:07'

You can replace the ‘T’ if you don’t want ISO format:

datetime.datetime.now().replace(microsecond=0).isoformat(' ')
# Returns: '2017-01-23 15:05:27'

回答 4

另一个选择：

>>> import time
>>> time.strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 11:31:28'

默认情况下，这使用本地时间，如果您需要UTC，则可以使用以下时间：

>>> time.strftime("%Y-%m-%d %H:%M:%S", time.gmtime())
'2011-11-03 18:32:20'
点击查看英文原文

Yet another option:

>>> import time
>>> time.strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 11:31:28'

By default this uses local time, if you need UTC you can use the following:

>>> time.strftime("%Y-%m-%d %H:%M:%S", time.gmtime())
'2011-11-03 18:32:20'

回答 5

通过切片保留所需的前19个字符：

>>> str(datetime.datetime.now())[:19]
'2011-11-03 14:37:50'
点击查看英文原文

Keep the first 19 characters that you wanted via slicing:

>>> str(datetime.datetime.now())[:19]
'2011-11-03 14:37:50'

回答 6

我通常这样做：

import datetime
now = datetime.datetime.now()
now = now.replace(microsecond=0)  # To print now without microsecond.

# To print now:
print(now)

输出：

2019-01-13 14:40:28
点击查看英文原文

I usually do:

import datetime
now = datetime.datetime.now()
now = now.replace(microsecond=0)  # To print now without microsecond.

# To print now:
print(now)

output:

2019-01-13 14:40:28

回答 7

由于并非所有datetime.datetime实例都具有微秒成分（即当它为零时），因此可以将字符串划分为“”。并只取第一项，它将始终有效：

unicode(datetime.datetime.now()).partition('.')[0]
点击查看英文原文

Since not all datetime.datetime instances have a microsecond component (i.e. when it is zero), you can partition the string on a “.” and take only the first item, which will always work:

unicode(datetime.datetime.now()).partition('.')[0]

回答 8

我们可以尝试如下

import datetime

date_generated = datetime.datetime.now()
date_generated.replace(microsecond=0).isoformat(' ').partition('+')[0]
点击查看英文原文

We can try something like below

import datetime

date_generated = datetime.datetime.now()
date_generated.replace(microsecond=0).isoformat(' ').partition('+')[0]

回答 9

我发现这是最简单的方法。

>>> t = datetime.datetime.now()
>>> t
datetime.datetime(2018, 11, 30, 17, 21, 26, 606191)
>>> t = str(t).split('.')
>>> t
['2018-11-30 17:21:26', '606191']
>>> t = t[0]
>>> t
'2018-11-30 17:21:26'
>>>
点击查看英文原文

I found this to be the simplest way.

>>> t = datetime.datetime.now()
>>> t
datetime.datetime(2018, 11, 30, 17, 21, 26, 606191)
>>> t = str(t).split('.')
>>> t
['2018-11-30 17:21:26', '606191']
>>> t = t[0]
>>> t
'2018-11-30 17:21:26'
>>>

回答 10

我之所以使用它，是因为我可以更好地理解并记住它（日期时间格式也可以根据您的选择进行自定义）：-

import datetime
moment = datetime.datetime.now()
print("{}/{}/{} {}:{}:{}".format(moment.day, moment.month, moment.year,
                                 moment.hour, moment.minute, moment.second))
点击查看英文原文

This I use because I can understand and hence remember it better (and date time format also can be customized based on your choice) :-

import datetime
moment = datetime.datetime.now()
print("{}/{}/{} {}:{}:{}".format(moment.day, moment.month, moment.year,
                                 moment.hour, moment.minute, moment.second))