inserting a value into a dataframe based on a comparison condition

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I am trying to insert a value into a dataframe based on a comparison with another dataframe. Here is an example:

>>> import pandas as pd
>>> import numpy as np
>>> print(df)
>>> df
      name                          
  0  richard Finn, Tim Maltby       
  1  Fernando Lebrija                          

>>> df2

       Fullname             id
  0   richard Finn          500
  1   Tim Maltby            699
  2   Fernando Lebrija      300

The desired output is :

 >>> df
      name                            id              
  0  richard Finn, Tim Maltby        500,699
  1  Fernando Lebrija                300

I tried using :

df['id'] = np.where((df['name']==df2['Fullname']), df2['id]', df['id'])

but it gives me the following error: `SyntaxError: invalid syntax

4 Answers

You can do a split, explode, then map and groupby:

df['id'] = (df['name'].str.split(',\s*')
    .explode()
    .map(df2.set_index('Fullname')['id'])
    .groupby(level=0).agg(list)
)

Output:

                       name          id
0  richard Finn, Tim Maltby  [500, 699]
1          Fernando Lebrija       [300]

Another way, using List comprehension

mapper = df2.set_index('Fullname')['id'].to_dict()
df['id'] = df['name'].apply(lambda x: ','.join([str(mapper.get(i.strip(), '')) for i in x.split(',')]))



    name                        id
0   richard Finn, Tim Maltby    500,699
1   Fernando Lebrija            300

We can also explore series.replace:

s = dict(df2[['Fullname','id']].astype(str).to_numpy())
df1['id'] = df1['name'].replace(s,regex=True)

print(df1)

                       name        id
0  richard Finn, Tim Maltby  500, 699
1          Fernando Lebrija       300

we can use use str.split stack and merge

final = pd.merge(
    df1["name"]
    .str.split(",", expand=True)
    .stack()
    .str.strip()
    .to_frame("Fullname")
    .reset_index(level=0),
    df2,
    on="Fullname",
).astype(str).groupby("level_0").agg(",".join).rename_axis("", axis=0)

print(final)

                  Fullname       id

0  richard Finn,Tim Maltby  500,699
1         Fernando Lebrija      300
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