Python 使用pandas to_datetime时如何定义格式？
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How to define format when use pandas to_datetime?
提问by ju.
I want to plot RESULT vs TIME based on a testresult.csvfile that has following format, and I have trouble to get the TIME column's datatype defined properly.
我想根据testresult.csv具有以下格式的文件绘制 RESULT 与 TIME 的关系图，但无法正确定义 TIME 列的数据类型。
TIME,RESULT
03/24/2016 12:27:11 AM,2
03/24/2016 12:28:41 AM,76
03/24/2016 12:37:23 AM,19
03/24/2016 12:38:44 AM,68
03/24/2016 12:42:02 AM,44
...
To read the csv file, this is the code I wrote:
raw_df = pd.read_csv('testresult.csv', index_col=None, parse_dates=['TIME'], infer_datetime_format=True)
This code works, but it is extremely slow, and I assume that the infer_datetime_formattakes time. So I tried to read in the csv by default first, and then convert the object dtype 'TIME' to datetime dtype by using to_datetime(), and I hope by defining the format, it might expedite the speed.
要读取 csv 文件，这是我编写的代码：
raw_df = pd.read_csv('testresult.csv', index_col=None, parse_dates=['TIME'], infer_datetime_format=True)
此代码有效，但速度非常慢，我认为这infer_datetime_format需要时间。所以我尝试先默认读取csv，然后使用 将对象dtype 'TIME' 转换为datetime dtype to_datetime()，我希望通过定义格式，可以加快速度。
raw_df = pd.read_csv('testresult.csv')
raw_df.loc['NEWTIME'] = pd.to_datetiem(raw_df['TIME'], format='%m/%d%Y %-I%M%S %p')
This code complained error:
此代码抱怨错误：
"ValueError: '-' is a bad directive in format '%m/%d%Y %-I%M%S %p'"
"ValueError: '-' is a bad directive in format '%m/%d%Y %-I%M%S %p'"
Please any suggestion or hint will be helpful.
请任何建议或提示都会有所帮助。
Thanks
谢谢
回答by Andy
The format you are passing is invalid. The dash between the %and the Iis not supposed to be there.
您传递的格式无效。%和之间的破折号I不应该在那里。
df['TIME'] = pd.to_datetime(df['TIME'], format="%m/%d/%Y %I:%M:%S %p")
This will convert your TIMEcolumn to a datetime.
这会将您的TIME列转换为日期时间。
Alternatively, you can adjust your read_csvcall to do this:
或者，您可以调整您的read_csv通话来执行此操作：
pd.read_csv('testresult.csv', parse_dates=['TIME'],
date_parser=lambda x: pd.to_datetime(x, format='%m/%d/%Y %I:%M:%S %p'))
Again, this uses the appropriate format with out the extra -, but it also passes in the format to the date_parserparameter instead of having pandas attempt to guess it with the infer_datetime_formatparameter.
同样，这使用了不带 extra 的适当格式-，但它也会将格式传递给date_parser参数，而不是让 Pandas 尝试使用infer_datetime_format参数来猜测它。
回答by MaxU
you can try this:
你可以试试这个：
In [69]: df = pd.read_csv(fn, parse_dates=[0],
date_parser=lambda x: pd.to_datetime(x, format='%m/%d/%Y %I:%M:%S %p'))
In [70]: df
Out[70]:
TIME RESULT
0 2016-03-24 00:27:11 2
1 2016-03-24 00:28:41 76
2 2016-03-24 00:37:23 19
3 2016-03-24 00:38:44 68
4 2016-03-24 00:42:02 44
