dateutil.parser._parser.ParserError: hour must be in 0..23:

November 28, 2023

The dataframe is on my drive:

url = 'https://drive.google.com/file/d/12R6nMvN81GJHSBEElP8NiXXkwtuv04wf/view?usp=sharing'
url = 'https://drive.google.com/uc?id=' + url.split('/')[-2]
df = pd.read_csv(url)

                sym     time  online_price
0     1000FLOKIUSDT  00:04.9       0.02567
1     1000FLOKIUSDT  00:31.7       0.02527
2     1000FLOKIUSDT  59:23.4       0.03638
3     1000FLOKIUSDT  59:23.4       0.03554
4     1000FLOKIUSDT  58:59.2       0.03552
...             ...      ...           ...
3640        ZRXUSDT  00:58.8       0.30240
3641        ZRXUSDT  00:19.7       0.42730
3642        ZRXUSDT  00:19.7       0.42920
3643        ZRXUSDT  00:19.7       0.43130
3644        ZRXUSDT  00:19.7       0.42810

I want to convert column time to datetime. I have tried several ways but all of them leads to the error that is the title of the post.

My desired format for date is: format='%Y-%m-%d %H:%M:%S'

These are the ways that I have tried:

df['time'] = pd.to_datetime(df.time)
df['time'] = pd.to_datetime(df.time, format='%Y-%m-%d %H:%M:%S')
df['time'] = pd.to_datetime(df.time, infer_datetime_format=True)

>Solution :

The time column looks like this:

00:04.9
00:31.7
59:23.4

Those aren’t timestamps. They may be timer values giving minutes:seconds, but in that case they’re durations. Why are you trying to interpret them as times of day? What time does 59:23.4 represent?