including word boundary in string modification to be more specific - python-3.x

Question

Background
The following is a minor change from modification of skipping empty list and continuing with function
import pandas as pd
Names = [list(['ann']),
list([]),
list(['elisabeth', 'lis']),
list(['his','he']),
list([])]
df = pd.DataFrame({'Text' : ['ann had an anniversery today',
'nothing here',
'I like elisabeth and lis 5 lists ',
'one day he and his cheated',
'same here'
],
'P_ID': [1,2,3, 4,5],
'P_Name' : Names
})
#rearrange columns
df = df[['Text', 'P_ID', 'P_Name']]
df
Text P_ID P_Name
0 ann had an anniversery today 1 [ann]
1 nothing here 2 []
2 I like elisabeth and lis 5 lists 3 [elisabeth, lis]
3 one day he and his cheated 4 [his, he]
4 same here 5 []
The code below works
m = df['P_Name'].str.len().ne(0)
df.loc[m, 'New'] = df.loc[m, 'Text'].replace(df.loc[m].P_Name,'**BLOCK**',regex=True)
And does the following
1) uses the name in P_Name to block the corresponding text in the Text column by placing **BLOCK**
2) produces a new column New
This is shown below
Text P_ID P_Name New
0 **BLOCK** had an **BLOCK**iversery today
1 NaN
2 I like **BLOCK** and **BLOCK** 5 **BLOCK**ts
3 one day **BLOCK** and **BLOCK** c**BLOCK**ated
4 NaN
Problem
However, this code works a little "too well."
Using ['his','he'] from P_Name to block Text:
Example: one day he and his cheated becomes one day **BLOCK** and **BLOCK** c**BLOCK**ated
Desired: one day he and his cheated becomes one day **BLOCK** and **BLOCK** cheated
In this example, I would like cheated to stay as cheated and not become c**BLOCK**ated
Desired Output
Text P_ID P_Name New
0 **BLOCK** had an anniversery today
1 NaN
2 I like **BLOCK** and **BLOCK**5 lists
3 one day **BLOCK** and **BLOCK** cheated
4 NaN
Question
How do I achieve my desired output?

Answer 1

You need to add word boundary to each string in lists of df.loc[m].P_Name as follows:
s = df.loc[m].P_Name.map(lambda x: [r'\b'+item+r'\b' for item in x])
Out[71]:
0 [\bann\b]
2 [\belisabeth\b, \blis\b]
3 [\bhis\b, \bhe\b]
Name: P_Name, dtype: object
df.loc[m, 'Text'].replace(s, '**BLOCK**',regex=True)
Out[72]:
0 **BLOCK** had an anniversery today
2 I like **BLOCK** and **BLOCK** 5 lists
3 one day **BLOCK** and **BLOCK** cheated
Name: Text, dtype: object

Answer 2

Sometime for loop is good practice
df['New']=[pd.Series(x).replace(dict.fromkeys(y,'**BLOCK**') ).str.cat(sep=' ')for x , y in zip(df.Text.str.split(),df.P_Name)]
df.New.where(df.P_Name.astype(bool),inplace=True)
df
Text ... New
0 ann had an anniversery today ... **BLOCK** had an anniversery today
1 nothing here ... NaN
2 I like elisabeth and lis 5 lists ... I like **BLOCK** and **BLOCK** 5 lists
3 one day he and his cheated ... one day **BLOCK** and **BLOCK** cheated
4 same here ... NaN
[5 rows x 4 columns]