My answer is by no means better than the other two. Pd.to_datetime(c, format='%H:%M') - pd.to_datetime(c, format='%H:%M').dt.normalize()ĭt.normalize() basically sets all time component to 00:00:00, and it will only display the date while keeping the datetime64 data format, thereby making it possible to do calculations with it. My method is just subtract the date from the to_datetime result: c = pd.Series() So if you need any further comparison or calculation operations on the result columns, it's better to avoid such data formats. I can barely do any comparison or calculation on these objects.
Also, be mindful of the fact that midnight is at 12:00: 00 AM in 12-hour format and 00:00:00 in 24-hour format, and noon is at 12:00: 00 PM in 12-hour format and 12:00:00 in 24. For example, ' 12:35:20' is a string representing date and time. Just like EdChum illustrated, using dt.hour or dt.time will give you a datetime.time object, which is probably only good for display. Given time in 12-hour AM/PM format, we aim to convert it into the 24-hour(military) format. For example, the default string representation of the date is 'YYYY-MM-DD', for time 'HH:MM:SS' and 'YYYY-MM-DD HH:MM:SS' is for the datetime. If you want to extract the time from the datetime object: t parser.parse ('18:36').time () which will give you a time object (if that's of more help to you). My method is close to EdChum's method and the result is the same as YOBEN_S's answer. Dateutil should be available for your python installation no need for something large like pandas. I recently also struggled with this problem.
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