How to count consecutive repetitions in a pandas series

The cumsum trick is useful here, it's a little tricky with the NaNs though, so I think you need to handle these separately:

In [11]: df.id.isnull() & df.id.shift(-1).isnull()
Out[11]:
0      True
1     False
2     False
3     False
4     False
5     False
6     False
7      True
8     False
9     False
10    False
11    False
12     True
Name: id, dtype: bool

In [12]: df.id.eq(df.id.shift(-1))
Out[12]:
0     False
1     False
2      True
3     False
4      True
5      True
6     False
7     False
8     False
9      True
10     True
11    False
12    False
Name: id, dtype: bool

In [13]: (df.id.isnull() & df.id.shift(-1).isnull()) | (df.id.eq(df.id.shift(-1)))
Out[13]:
0      True
1     False
2      True
3     False
4      True
5      True
6     False
7      True
8     False
9      True
10     True
11    False
12     True
Name: id, dtype: bool

In [14]: ((df.id.isnull() & df.id.shift(-1).isnull()) | (df.id.eq(df.id.shift(-1)))).cumsum()
Out[14]:
0     1
1     1
2     2
3     2
4     3
5     4
6     4
7     5
8     5
9     6
10    7
11    7
12    8
Name: id, dtype: int64

Now you can use this labeling in your groupby:

In [15]: g = df.groupby(((df.id.isnull() & df.id.shift(-1).isnull()) | (df.id.eq(df.id.shift(-1)))).cumsum())

In [16]: pd.DataFrame({"count": g.id.size(), "id": g.id.nth(0)})
Out[16]:
    count   id
id
1       2  NaN
2       2  1.0
3       1  2.0
4       2  2.0
5       2  NaN
6       1  1.0
7       2  1.0
8       1  NaN

Here is another approach using fillna to handle NaN values:

s = df.id.fillna('nan')
mask = s.ne(s.shift())

ids = s[mask].to_numpy()
counts = s.groupby(mask.cumsum()).cumcount().add(1).groupby(mask.cumsum()).max().to_numpy()

# Convert 'nan' string back to `NaN`
ids[ids == 'nan'] = np.nan
ser_out = pd.Series(counts, index=ids, name='counts')

[out]

nan    2
1.0    2
2.0    3
nan    2
1.0    3
nan    1
Name: counts, dtype: int64