Reputation: 49
I have a pandas dataframe looks like as below:
date | location | occurance <br>
------------------------------------------------------
somedate |united_kingdom_london | 5
somedate |united_state_newyork | 5
I want it to transform into
date | country | city | occurance <br>
---------------------------------------------------
somedate | united kingdom | london | 5
---------------------------------------------------
somedate | united state | newyork | 5
I am new to Python and after some research I have written following code, but seems to unable to extract country and city:
df.location= df.location.replace({'-': ' '}, regex=True)
df.location= df.location.replace({'_': ' '}, regex=True)
temp_location = df['location'].str.split(' ').tolist()
location_data = pd.DataFrame(temp_location, columns=['country', 'city'])
I appreciate your response.
Upvotes: 2
Views: 9674
Reputation: 25639
Starting with this:
df = pd.DataFrame({'Date': ['somedate', 'somedate'],
'location': ['united_kingdom_london', 'united_state_newyork'],
'occurence': [5, 5]})
Try this:
df['Country'] = df['location'].str.rpartition('_')[0].str.replace("_", " ")
df['City'] = df['location'].str.rpartition('_')[2]
df[['Date','Country', 'City', 'occurence']]
Date Country City occurence
0 somedate united kingdom london 5
1 somedate united state newyork 5
Borrowing idea from @MaxU
df[['Country'," " , 'City']] = (df.location.str.replace('_',' ').str.rpartition(' ', expand= True ))
df[['Date','Country', 'City','occurence' ]]
Date Country City occurence
0 somedate united kingdom london 5
1 somedate united state newyork 5
Upvotes: 3
Reputation: 210842
I would use .str.extract() method:
In [107]: df
Out[107]:
Date location occurence
0 somedate united_kingdom_london 5
1 somedate united_state_newyork 5
2 somedate germany_munich 5
In [108]: df[['country','city']] = (df.location.str.replace('_',' ')
.....: .str.extract(r'(.*)\s+([^\s]*)', expand=True))
In [109]: df
Out[109]:
Date location occurence country city
0 somedate united_kingdom_london 5 united kingdom london
1 somedate united_state_newyork 5 united state newyork
2 somedate germany_munich 5 germany munich
In [110]: df = df.drop('location', 1)
In [111]: df
Out[111]:
Date occurence country city
0 somedate 5 united kingdom london
1 somedate 5 united state newyork
2 somedate 5 germany munich
PS please be aware that it's not possible to parse properly (to distinguish) between rows containing two-words country + one-word city and rows containing one-word country + two-words city (unless you have a full list of countries so you check it against this list)...
Upvotes: 0
Reputation: 3655
Something like this works
import pandas as pd
df = pd.DataFrame({'Date': ['somedate', 'somedate'],
'location': ['united_kingdom_london', 'united_state_newyork'],
'occurence': [5, 5]})
df.location = df.location.str[::-1].str.replace("_", " ", 1).str[::-1]
newcols = df.location.str.split(" ")
newcols = pd.DataFrame(df.location.str.split(" ").tolist(),
columns=["country", "city"])
newcols.country = newcols.country.str.replace("_", " ")
df = pd.concat([df, newcols], axis=1)
df.drop("location", axis=1, inplace=True)
print(df)
Date occurence country city
0 somedate 5 united kingdom london
1 somedate 5 united state newyork
You could use regex in the replace for a more complicated pattern but if it's just the word after the last _
I find it easier to just reverse the str twice as a hack rather than fiddling around with regular expressions
Upvotes: 0
Reputation: 8683
Try this:
temp_location = {}
splits = df['location'].str.split(' ')
temp_location['country'] = splits[0:-1].tolist()
temp_location['city'] = splits[-1].tolist()
location_data = pd.DataFrame(temp_location)
If you want it back in the original df:
df['country'] = splits[0:-1].tolist()
df['city'] = splits[-1].tolist()
Upvotes: 0
Reputation: 107587
Consider splitting the column's string value using rfind()
import pandas as pd
df = pd.DataFrame({'Date': ['somedate', 'somedate'],
'location': ['united_kingdom_london', 'united_state_newyork'],
'occurence': [5, 5]})
df['country'] = df['location'].apply(lambda x: x[0:x.rfind('_')])
df['city'] = df['location'].apply(lambda x: x[x.rfind('_')+1:])
df = df[['Date', 'country', 'city', 'occurence']]
print(df)
# Date country city occurence
# 0 somedate united_kingdom london 5
# 1 somedate united_state newyork 5
Upvotes: 0