在pandas数据框中切片字符串并分配一个新列

Question

Hi I have the following data frame:

df = pd.DataFrame()
df['Name'] = ['P. John','Merry','P. John travis']
df['First_Name'] = df.Name.str.split('.', expand = True)[0]
df['Last_Name'] = df.Name.str.split('.', expand = True)[1]

I want to slice the column base on period "." and used as last name. I could do for all but for "merry" it shows None as follow:

0            John
1            None
2     John travis

How could I replace with all None in last name with First Name ? A searched in the forum and could not find one.

And second question is that I have another data frame as follow:

df1 = pd.DataFrame({'Name':['John','Merry','John travis'],"Position":['CEO','CTO','Engr']})

I am creating a new column ** Position** for df by using map function.

 df ['Position'] = df.Last_Name.map (df1.set_index('Name').Position)

but the new column in df show me some Nan value as follow:

The data frame I shown in this post replicates the real problem that I am solving. However, using the map fucntion in real problem give me the follwoing error code:

Reindexing only valid with uniquely valued Index objects.

Can anyone advise me on that.?

Thanks.

Answer 1

You can simplify your code by one split with parameter n=1 for split by first . if possible multiple one and then replace None by fillna :

df = pd.DataFrame({'Name':['P. John','Merry','P. John travis']})

df[['First_Name', 'Last_Name']] = df.Name.str.split('.\s+', expand = True, n=1)
#if always only one .
#df[['First_Name', 'Last_Name']] = df.Name.str.split('.\s+', expand = True, n=1)
df['Last_Name'] = df['Last_Name'].fillna(df['First_Name'])
print (df)
             Name First_Name     Last_Name
0         P. John          P          John
1           Merry      Merry         Merry
2  P. John travis          P   John travis

Or remove expand=True for Series of list s and select first and last values:

splitted = df.Name.str.split('.\s+', n=1)
df['first_Name'] = splitted.str[0]
df['Last_Name'] = splitted.str[-1]
print (df)
             Name first_Name     Last_Name
0         P. John          P          John
1           Merry      Merry         Merry
2  P. John travis          P   John travis

Answer 2

Using fillna

Ex:

import pandas as pd
df = pd.DataFrame()
df['Name'] = ['P. John','Merry','P. John travis']
df['First_Name'] = df.Name.str.split('.', expand = True)[0]
df['Last_Name'] = (df.Name.str.split('.', expand = True)[1]).fillna(df["First_Name"])
print(df)

Output:

             Name First_Name     Last_Name
0         P. John          P          John
1           Merry      Merry         Merry
2  P. John travis          P   John travis

Answer 3

you could use a list comprehension and negative indexing

df['Last_Name'] = [x.split('.')[-1] for x in df.Name]

             Name     Last_Name
0         P. John          John
1           Merry         Merry
2  P. John travis   John travis

here's an extension of the above technique that returns a whole new dataframe with the name split as desired, in a single statement

pd.DataFrame([(lambda x: (y, x[0], x[-1]))(y.split('.')) 
              for y in df.Name], 
             columns=['Name', 'First_Name', 'Last_Name'])

             Name First_Name     Last_Name
0         P. John          P          John
1           Merry      Merry         Merry
2  P. John travis          P   John travis