[FIXED] Extracting missing datetimeindex date/hour from dataframe

Issue

I have a dataframe with a datetimeindex as shown with the format, the raw data is supposed to contains record every hourly for a year(each day having 24 record). Some hours/days are missing and not recorded in the data.

How can i get a list of all the missing datetimeindex hour.

Example: 01 hour is missing, how can i find and print out 2012-10-02 01:00:00

I'm currently able to get the missing days but unable to do so for the hour.

    missing_day = pd.date_range(start = mdf.index[0], end = mdf.index[-1]).difference(mdf.index)
missing_day = missing_day.strftime('%Y%m%d')
missing = pd.Series(missing_day).array
for i in missing:
    print(i)
    for x in range(24):
        x = str(x)
        m = i + x
        m = datetime.strptime(m,'%Y%m%d%H')
        print(m)

Output(printing 24 hour for each missing days)

What would be the best way to list out all of the missing datetime.

Solution

Use set predicates to find missing index:

out = pd.date_range(df.index.min(), df.index.max(), freq='H').difference(df.index)
print(out)

# Output
DatetimeIndex(['2022-01-01 06:00:00', '2022-01-01 12:00:00',
               '2022-01-01 14:00:00', '2022-01-01 16:00:00'],
              dtype='datetime64[ns]', freq=None)

Setup:

df = pd.DataFrame({'A':[0]}, index=pd.date_range('2022-01-01', freq='H', periods=24))
df = df.sample(n=20).sort_index()
print(df)

# Output
                     A
2022-01-01 00:00:00  0
2022-01-01 01:00:00  0
2022-01-01 02:00:00  0
2022-01-01 03:00:00  0
2022-01-01 04:00:00  0
2022-01-01 05:00:00  0
2022-01-01 07:00:00  0
2022-01-01 08:00:00  0
2022-01-01 09:00:00  0
2022-01-01 10:00:00  0
2022-01-01 11:00:00  0
2022-01-01 13:00:00  0
2022-01-01 15:00:00  0
2022-01-01 17:00:00  0
2022-01-01 18:00:00  0
2022-01-01 19:00:00  0
2022-01-01 20:00:00  0
2022-01-01 21:00:00  0
2022-01-01 22:00:00  0
2022-01-01 23:00:00  0

Answered By - Corralien