How to create a ranking column based on date value range in another column?

We can use groupby apply to compute the difference per group between days, then use cumsum on the to "count" how many differences are greater than (`gt) 3 days:

# Convert to datetime (if not already)
df['Dates'] = pd.to_datetime(df['Dates'])
# Sort in correct order
df = df.sort_values(['Item_ID', 'Dates'], ignore_index=True)
# Calculate Ranking per Group
df['Date Order'] = (
    df.groupby('Item_ID')['Dates'].apply(
        lambda s: s.diff().gt(pd.Timedelta(days=3)).cumsum() + 1
    )
)

Could also groupby twice and use groupby diff and groupby cumsum:

# Convert to datetime (if not already)
df['Dates'] = pd.to_datetime(df['Dates'])
# Sort in correct order
df = df.sort_values(['Item_ID', 'Dates'], ignore_index=True)

# Reuse same Grouper
g = df.groupby('Item_ID') 
# Calculate Difference per group and compare (whole Series)
df['Date Order'] = g['Dates'].diff().gt(pd.Timedelta(days=3))
# Calculate cumsum per group
df['Date Order'] = g['Date Order'].cumsum() + 1

Both produce df:

  Item_ID      Dates  Date Order
0  Item_1 2020-05-27           1
1  Item_1 2020-06-01           2
2  Item_1 2020-06-02           2
3  Item_2 2018-04-15           1
4  Item_2 2018-04-18           1
5  Item_2 2018-04-22           2
6  Item_2 2018-04-28           3

Here is a breakdown of the steps per group as a DataFrame:

s = pd.Series([pd.Timestamp('2020-05-27 00:00:00'),
               pd.Timestamp('2020-06-01 00:00:00'),
               pd.Timestamp('2020-06-02 00:00:00')],
              name='Dates',
              index=pd.Series({0: 'Item_1', 1: 'Item_1', 2: 'Item_1'},
                              name='Item_ID'))
steps_per_group = pd.DataFrame({
    'diff': s.diff(),
    'gt': s.diff().gt(pd.Timedelta(days=3)),
    'cumsum': s.diff().gt(pd.Timedelta(days=3)).cumsum(),
    'cumsum 1 start': s.diff().gt(pd.Timedelta(days=3)).cumsum() + 1
})

          diff     gt  cumsum  cumsum 1 start
Item_ID                                      
Item_1     NaT  False       0               1
Item_1  5 days   True       1               2
Item_1  1 days  False       1               2

From your DataFrame :

>>> import pandas as pd

>>> data = [
...     ["Item_1", "2020-05-27"],
...     ["Item_1", "2020-06-01"],
...     ["Item_1", "2020-06-02"],
...     ["Item_2", "2018-04-15"],
...     ["Item_2", "2018-04-18"],
...     ["Item_2", "2018-04-22"],
...     ["Item_2", "2018-04-28"],
... ]
>>> df = pd.DataFrame(data, columns=["Item_ID", "Dates"])
>>> df['Dates'] = pd.to_datetime(df['Dates'], format="%Y-%m-%d")
>>> df
    Item_ID     Dates
0   Item_1  2020-05-27
1   Item_1  2020-06-01
2   Item_1  2020-06-02
3   Item_2  2018-04-15
4   Item_2  2018-04-18
5   Item_2  2018-04-22
6   Item_2  2018-04-28

We can get the date diff grouped by Item_ID like so :

>>> window_size = 3
>>> df['diff'] = df.groupby('Item_ID')["Dates"].diff().dt.days.gt(window_size)
>>> df
    Item_ID     Dates   diff
0   Item_1  2020-05-27  False
1   Item_1  2020-06-01  True
2   Item_1  2020-06-02  False
3   Item_2  2018-04-15  False
4   Item_2  2018-04-18  False
5   Item_2  2018-04-22  True
6   Item_2  2018-04-28  True

Then, by grouping again by Item_ID and applying a cumsum, we get the expected result :

>>> df['Date Order'] = df.groupby('Item_ID')["diff"].cumsum()+1
>>> df
    Item_ID     Dates   diff    Date Order
0   Item_1  2020-05-27  False   1
1   Item_1  2020-06-01  True    2
2   Item_1  2020-06-02  False   2
3   Item_2  2018-04-15  False   1
4   Item_2  2018-04-18  False   1
5   Item_2  2018-04-22  True    2
6   Item_2  2018-04-28  True    3