Transform values in column using python

I would suggest using df.apply to adjust the values. df["Date"].apply(transform) iterates over every value in the column Date, and will split the string on each space, keeping only the last 2 characters in the second string, and then join them using a period.

def transform(string):
    lst = string.split(" ")
    lst[1] = lst[1][-2:]
    return ".".join(lst)

df["Date"] = df["Date"].apply(transform)

Suppose that our string is as follows:

s = "Q1 2022"

We can remove the white-space characters and split the string into pieces as follows:

pieces = s.split()
print(pieces)

The result is:

pieces = ['Q1', '2022']

You can change the year '2022' to '22' as follows:

pieces = ['Q1', '2022']
# pieces[0] == "Q1"
# pieces[1] == '2022'
pieces[1] = pieces[1][-2:]

In the end, we have something like this:

def foobar(old_string:str) -> str:
    """
        EXAMPLE
            INPUT:  "Q1 2022"
            OUTPUT: "Q1.22"
    """
    old_string = "".join(str(ch) for ch in old_string)
    pieces = old_string.split()
    pieces[1] = pieces[1][-2:]
    new_string = ".".join(pieces)
    return new_string
    
result = foobar("Q1 2022")
print(result)
# prints "Q1.22"

I think the most straightforward way would be to take the first two characters, a period, and the last two characters, and concatenate them.
Something like this:

df['Date'] = df['Date'].str[:2] + "." + df['Date'].str[-2:]

We can do

df['Date'] = pd.to_datetime(df['Date'],format = 'Q%d %Y').dt.strftime('Q%d.%y')
Out[624]: 
0    Q01.22
1    Q02.22
Name: Date, dtype: object