dictionary keys to replace strings in pandas dataframe column with dictionary values and perform evaluate

1. We first explode your string to rows seperated by a comma, using this function.

2. Then we split the values by a whitespace (' ') to seperate columns.

3. Finally we map your dictionary to the letters and do a groupby.sum:

new  = explode_str(df.dropna(), 'col1', ',')['col1'].str.strip().str.split(' ', expand=True).append(df[df['col1'].isna()])

s = new[1].map(di) * pd.to_numeric(new[0])

df['result'] = s.groupby(s.index).sum()

Output

                  col1  result
0       3 a, 3 ab, 1 b    37.5
1  4 a, 4 ab, 1 b, 1 d    50.5
2                  NaN     0.0

---

Function used from linked answer:

def explode_str(df, col, sep):
    s = df[col]
    i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
    return df.iloc[i].assign(**{col: sep.join(s).split(sep)})

Here is on way seem over kill

df['col1'].str.split(', ',expand=True).replace({' ':'*','np.nan':'0'},regex=True).\
     stack().apply(lambda x : eval(x,di)).sum(level=0)
Out[884]: 
0    37.5
1    50.5
2     0.0
dtype: float64

comprehension

from functools import reduce
from operator import mul

def m(x): return di.get(x, x)

df.assign(result=[
    sum(
        reduce(mul, map(float, map(m, s.split())))
        for s in row.split(', ')
    ) for row in df.col1
])

                  col1  result
0       3 a, 3 ab, 1 b    37.5
1  4 a, 4 ab, 1 b, 1 d    50.5
2               np.nan     0.0