Mapping alphabets to its numeric equivalent?

Please see my code below. I'm iterating through strings like '1A', '4D', etc, and I want the output to instead be 1.1, 4.4, and so on..see below.

Instead of 1A I want 1.1, 1B= 1.2, 4A = 4.1, 5D = 5.4, etc...

Convert alphabet letters to number in Python

data = ['1A','1B','4A', '5D','']
df = pd.DataFrame(data, columns = ['Score'])

newcol = []

for col, row in df['Score'].iteritems()
    if pd.isnull(row):
        newcol.append(row)       
    elif pd.notnull(row): 
        newcol.append(#FIRST ELEMENT OF ROW, 1-5,'.', 
                      #NUMERIC EQUIVALENT OF ALPHA, IE, A=1, B=2, C=3, D=4, etc)

You can use str.replace:

df['Score'] = df['Score'].str.replace('\D',
              lambda x: f'.{ord(x.group(0).upper())-64}', regex=True)

output:

  Score
0   1.1
1   1.2
2   4.1
3   5.4
4      

Use (with @Ch3steR's comment)-

from string import ascii_uppercase
dic = {j:str(i) for i,j in enumerate(ascii_uppercase, 1)}
df['Score'].str[:-1] + '.' + df['Score'].str[-1].map(dic)

Output

0    1.1
1    1.2
2    4.1
3    5.4
4    NaN
Name: Score, dtype: object