Python function not accepting the second argument

Proloy Chatterjee 0
Viewed 57

Sample Data frame -

df = pd.DataFrame({'City':['York New', 'Parague', 'New Delhi', 'Venice', 'new Orleans'],
                    'Event':['Music', 'Poetry', 'Theatre', 'Comedy', 'Tech_Summit'],
                    'Cost':[10000, 5000, 15000, 2000, 12000]})
  
index_ = [pd.Period('02-2018'), pd.Period('04-2018'),
          pd.Period('06-2018'), pd.Period('10-2018'), pd.Period('12-2018')]
  
df.index = index_
print(df)

Problem Statement - For those cities which starts with the keyword ‘New’ or ‘new’, change it to ‘New_’

First I have created a new column in the dataframe to find if the City has "new" in the name and if yes then at what position

df["pos"] = df["City"].apply(lambda x: x.lower().find("new"))

Then I have created a function to replace "New" or "new" by "New_" if they are present in the starting of the city name -

def replace_new(city,pos):
    if pos==0:
        return city.replace("[Nn]ew", "New_", regex = True)
    else:
        return city

df = df[["City","pos"]].apply(replace_new, axis = 1)

When I execute the above code line I am getting this error - "("replace_new() missing 1 required positional argument: 'pos'", 'occurred at index 2018-02')"

What am I doing wrong here? Please help

data-cleaning dataframe pandas python
2 Answers

Call the method this way

df = df.apply(lambda x: replace_new(x.City, x.pos), axis=1)

Also the replace method is not working properly use Re

import re
def replace_new(city,pos):
    if pos==0:
        return re.sub('[Nn]ew', 'New_', city)
    else:
        return city
khizer0

Use str.replace with a regex:

df['City'] = df['City'].str.replace(r'^new\s*', 'New_', case=False, regex=True)

output:

                City        Event   Cost
2018-02     York New        Music  10000
2018-04      Parague       Poetry   5000
2018-06    New_Delhi      Theatre  15000
2018-10       Venice       Comedy   2000
2018-12  New_Orleans  Tech_Summit  12000

regex:

^       # match start of line
new     # match "new"
\s*     # match zero or more spaces
mozway0
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