熊猫将名称列拆分为名字和姓氏（如果包含一个空格）

Question

Let's say I have a pandas DataFrame containing names like so:

name_df = pd.DataFrame({'name':['Jack Fine','Kim Q. Danger','Jane Smith', 'Juan de la Cruz']})

    name
0   Jack Fine
1   Kim Q. Danger
2   Jane Smith
3   Juan de la Cruz

and I want to split the name column into first_name and last_name IF there is one space in the name. Otherwise I want the full name to be shoved into first_name .

So the final DataFrame should look like:

  first_name     last_name
0 Jack           Fine
1 Kim Q. Danger
2 Jane           Smith
3 Juan de la Cruz

I've tried to accomplish this by first applying the following function to return names that can be split into first and last name:

def validate_single_space_name(name: str) -> str:
    pattern = re.compile(r'^.*( ){1}.*$')
    match_obj = re.match(pattern, name)
    if match_obj:
        return name
    else:
        return None

However applying this function to my original name_df, leads to an empty DataFrame, not one populated by names that can be split and Nones.

Help getting my current approach to work, or solutions invovling a different approach would be appreciated!

Answer 1

You can use str.split to split the strings, then test the number of splits using str.len and use this as a boolean mask to assign just those rows with the last component of the split:

In [33]:
df.loc[df['name'].str.split().str.len() == 2, 'last name'] = df['name'].str.split().str[-1]
df

Out[33]:
              name last name
0        Jack Fine      Fine
1    Kim Q. Danger       NaN
2       Jane Smith     Smith
3  Juan de la Cruz       NaN

EDIT

You can call split with param expand=True this will only populate where the name lengths are exactly 2 names:

In [16]:
name_df[['first_name','last_name']] = name_df['name'].loc[name_df['name'].str.split().str.len() == 2].str.split(expand=True)
name_df

Out[16]:
              name first_name last_name
0        Jack Fine       Jack      Fine
1    Kim Q. Danger        NaN       NaN
2       Jane Smith       Jane     Smith
3  Juan de la Cruz        NaN       NaN

You can then replace the missing first names using fillna :

In [17]:
name_df['first_name'].fillna(name_df['name'],inplace=True)
name_df
​
Out[17]:
              name       first_name last_name
0        Jack Fine             Jack      Fine
1    Kim Q. Danger    Kim Q. Danger       NaN
2       Jane Smith             Jane     Smith
3  Juan de la Cruz  Juan de la Cruz       NaN

Answer 2

I was having some issues with IndexError: list index out of range because the names could be test , kk and other weird user input. So ended up with something like this:

items['fullNameSplitLength'] = items['fullName'].str.split().str.len()
items['firstName'] = items['lastName'] = ''
items.loc[
  items['fullNameSplitLength'] >= 1,
  'firstName'
] = items.loc[items['fullNameSplitLength'] >= 1]['fullName'].str.split().str[0]
items.loc[
  items['fullNameSplitLength'] >= 2,
  'lastName'
] = items.loc[items['fullNameSplitLength'] >= 2]['fullName'].str.split().str[-1]