Background
import pandas as pd
Names = [list(['Jon', 'Smith', 'jon', 'John']),
list([]),
list(['Bob', 'bobby', 'Bobs'])]
df = pd.DataFrame({'Text' : ['Jon J Smith is Here and jon John from ',
'',
'I like Bob and bobby and also Bobs diner '],
'P_ID': [1,2,3],
'P_Name' : Names
})
#rearrange columns
df = df[['Text', 'P_ID', 'P_Name']]
df
Text P_ID P_Name
0 Jon J Smith is Here and jon John from 1 [Jon, Smith, jon, John]
1 2 []
2 I like Bob and bobby and also Bobs diner 3 [Bob, bobby, Bobs]
Goal
I would like to use the following function
df['new']=df.Text.replace(df.P_Name,'**BLOCK**',regex=True)
but skip row 2, since it has an empty list []
Tried
I have tried the following
try:
df['new']=df.Text.replace(df.P_Name,'**BLOCK**',regex=True)
except ValueError:
pass
But I get the following output
Text P_ID P_Name
0 Jon J Smith is Here and jon John from 1 [Jon, Smith, jon, John]
1 2 []
2 I like Bob and bobby and also Bobs diner 3 [Bob, bobby, Bobs]
Desired Output
Text P_ID P_Name new
0 `**BLOCK**` J `**BLOCK**` is Here and `**BLOCK**` `**BLOCK**` from
1 []
2 I like `**BLOCK**` and `**BLOCK**` and also `**BLOCK**` diner
Question
How do I get my desired output by skipping row 2 and continuing with my function?
Locate the rows which do not have an empty list and use your replace
method only on those rows:
# Boolean indexing the rows which do not have an empty list
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)
Output
Text P_ID P_Name New
0 Jon J Smith is Here and jon John from 1 [Jon, Smith, jon, John] **BLOCK** J **BLOCK** is Here and **BLOCK** **BLOCK** from
1 Test 2 [] NaN
2 I like Bob and bobby and also Bobs diner 3 [Bob, bobby, Bobs] I like **BLOCK** and **BLOCK** and also **BLOCK**s diner