Python 从熊猫的日期和时间变量中删除时间？
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removing time from date&time variable in pandas?
提问by surendra
i have a variable consisting of 300k records with dates and the date look like
2015-02-21 12:08:51
from that date i want to remove time
我有一个变量由 300k 条记录组成，日期和日期看起来像
2015-02-21 12:08:51
从那个日期开始我想删除时间
type of date variable is pandas.core.series.series
日期变量的类型是 pandas.core.series.series
This is the way i tried
这是我尝试的方式
from datetime import datetime,date
date_str = textdata['vfreceiveddate']
format_string = "%Y-%m-%d"
then = datetime.strftime(date_str,format_string)
some Random ERROR
一些随机错误
In the above code textdata is my datasetname and vfreceived date is a variable consisting of dates
How can i write the code to remove the time from the datetime.
在上面的代码中，textdata 是我的数据集名称，而 vfreceived 日期是一个由日期组成的变量
我如何编写代码以从日期时间中删除时间。
采纳答案by EdChum
Assuming all your datetime strings are in a similar format then just convert them to datetime using to_datetimeand then call the dt.dateattribute to get just the date portion:
假设您所有的日期时间字符串都采用类似的格式，那么只需使用将它们转换为日期时间to_datetime，然后调用该dt.date属性即可获取日期部分：
In [37]:
df = pd.DataFrame({'date':['2015-02-21 12:08:51']})
df
Out[37]:
date
0 2015-02-21 12:08:51
In [39]:
df['date'] = pd.to_datetime(df['date']).dt.date
df
Out[39]:
date
0 2015-02-21
EDIT
编辑
If you just want to change the display and not the dtype then you can call dt.normalize:
如果您只想更改显示而不是 dtype，则可以调用dt.normalize：
In[10]:
df['date'] = pd.to_datetime(df['date']).dt.normalize()
df
Out[10]:
date
0 2015-02-21
You can see that the dtype remains as datetime:
您可以看到 dtype 保持为datetime：
In[11]:
df.dtypes
Out[11]:
date datetime64[ns]
dtype: object
回答by Alex Martelli
You're calling datetime.datetime.strftime, which requires as its first argument a datetime.datetimeinstance, because it's an unbound method; but you're passing it a string instead of a datetime instance, whence the obvious error.
您正在调用datetime.datetime.strftime，它需要一个datetime.datetime实例作为它的第一个参数，因为它是一个未绑定的方法；但是您传递给它的是一个字符串而不是日期时间实例，因此出现了明显的错误。
You can work purely at a string level if that's the result you want; with the data you give as an example, date_str.split()[0]for example would be exactly the 2015-02-21string you appear to require.
如果这是您想要的结果，您可以纯粹在字符串级别工作；例如，date_str.split()[0]以您提供的数据为例，这正是2015-02-21您似乎需要的字符串。
Or, you canuse datetime, but then you need to parsethe string first, not formatit -- hence, strptime, notstrftime:
或者，您可以使用datetime，但随后您需要先解析字符串，而不是对其进行格式化——因此，是 str p时间，而不是str f时间：
dt = datetime.strptime(date_str, '%Y-%m-%d %H:%M:%S')
date = dt.date()
if it's a datetime.dateobject you want (but if all you want is the string form of the date, such an approach might be "overkill":-).
如果它是datetime.date你想要的对象（但如果你想要的只是日期的字符串形式，这种方法可能是“矫枉过正”:-)。
