You are performing the rounding operation passing a series as an argument. Instead you need to fix this to perform the rounding up for each value in the series. I suggest you use map with a lambda in the function to do it:

Data['Numerator'] = Data['Numerator'].map(lambda x: Decimal(x).quantize(Decimal('.1'), rounding=ROUND_HALF_UP))

The output we get is as expected:

  Code Disaggregation Numerator
0    x              a      19.3
1    x              b      82.1
2    x          Total     101.2

timestamps

You need to use dt.round. This is however a bit as the previous/next hour behavior depends on the hour itself. You can force it by adding or subtracting a small amount of time (here 1ns):

s = pd.to_datetime(pd.Series(['1/2/2021 3:45', '25/4/2021 12:30', 
                              '25/4/2021 13:30', '12/4/2022 23:45']))

# xx:30 -> rounding depending on the hour parity (default)
s.dt.round(freq='1h')

0   2021-01-02 04:00:00
1   2021-04-25 12:00:00    <- -30min
2   2021-04-25 14:00:00    <- +30min
3   2022-12-05 00:00:00
dtype: datetime64[ns]


# 00:30 -> 00:00 (force down)
s.sub(pd.Timedelta('1ns')).dt.round(freq='1h')

0   2021-01-02 04:00:00
1   2021-04-25 12:00:00
2   2021-04-25 13:00:00
3   2022-12-05 00:00:00
dtype: datetime64[ns]


# 00:30 -> 01:00 (force up)
s.add(pd.Timedelta('1ns')).dt.round(freq='1h')

0   2021-01-02 04:00:00
1   2021-04-25 12:00:00
2   2021-04-25 13:00:00
3   2022-12-05 00:00:00
dtype: datetime64[ns]

floats

IIUC, you can use divmod (or numpy.modf) to get the integer and decimal part, then perform simple boolean arithmetic:

s = pd.Series([7.15, 5.25, 22.30, 18.45])

s2, r = s.divmod(1)  # or np.modf(s)

s2[r.ge(0.3)] += 1

s2 = s2.astype(int)

Alternative: using mod and boolean to int equivalence:

s2 = s.astype(int)+s.mod(1).ge(0.3)

output:

0     7
1     5
2    23
3    19
dtype: int64

Note on precision. It is not always easy to compare floats due to floating point arithmetics. For instance using gt would fail on the 22.30 here. To ensure precision round to 2 digits first.

s.mod(1).round(2).ge(0.3)

or use integers:

s.mod(1).mul(100).astype(int).ge(30)

You can use round and a little arithmetic to get your desired result:

round(num*2)/2 will round a number to the nearest 0.5.

Here is how you would apply this to a dataframe:

>>> df = pd.DataFrame(data={'col1': [3.6, 2.1, 4.7, 3.8, 1.9]})
>>> df
   col1
0   3.6
1   2.1
2   4.7
3   3.8
4   1.9
>>> df['col2'] = df.col1.mul(2).round().div(2)
>>> df
   col1  col2
0   3.6   3.5
1   2.1   2.0
2   4.7   4.5
3   3.8   4.0
4   1.9   2.0

You can add some tiny value to orig when the decimal is 0.5. That guarantees that any integer + 0.5 will always round up to the next integer.

import numpy as np
df['round_up'] = np.round(np.where(df['orig'] % 1 == 0.5,
                                   df['orig'] + 0.1,
                                   df['orig']))
print(df)
       orig  round_up
0  0.500000       1.0
1  1.499999       2.0
2  1.500000       2.0
3  2.500000       3.0
4  3.500000       4.0
5  4.500000       5.0
6  5.500000       6.0
7  6.500000       7.0