pandas dataframe return first word in string for column

Question

I have a dataframe:

df = pd.DataFrame({'id' : ['abarth 1.4 a','abarth 1 a','land rover 1.3 r','land rover 2',
                           'land rover 5 g','mazda 4.55 bl'], 
                   'series': ['a','a','r','','g', 'bl'] })

I would like to remove the 'series' string from the corresponding id, so the end result should be:

Final result should be 'id': ['abarth 1.4','abarth 1','land rover 1.3','land rover 2','land rover 5', 'mazda 4.55']

Currently I am using df.apply:

df.id = df.apply(lambda x: x['id'].replace(x['series'], ''), axis =1)

But this removes all instances of the strings, even in other words, like so: 'id': ['brth 1.4','brth 1','land ove 1.3','land rover 2','land rover 5', 'mazda 4.55']

Should I somehow mix and match regex with the variable inside df.apply, like so?

df.id = df.apply(lambda x: x['id'].replace(r'\b' + x['series'], ''), axis =1)

Answer 1

Use str.split and str.get and assign using loc only where df.make == ''

df.loc[df.make == '', 'make'] = df.id.str.split().str.get(0)

print df

               id    make
0      abarth 1.4  abarth
1        abarth 1  abarth
2  land rover 1.3   rover
3    land rover 2   rover
4    land rover 5   rover
5      mazda 4.55   mazda

Answer 2

It's simple. Use as follows:

df['make'] = df['id'].str.split(' ').str[0]

Answer 3

IDK why but with the part below

df.loc[df.make == '', 'make']

OR

df.loc[df['make'] == '', 'make']

I get the error - KeyError: 'make'

So instead I did (in case someone sees the same error):

df['make'] = df['id']
df['make'] = df.id.str.split().str.get(0)

Worked for me.

Answer 4

Consider a regex solution with loc where it extracts everything before first space:

df.loc[df['make']=='', 'make'] = df['id'].str.extract('(.*) ', expand=False)

Alternatively, use numpy's where which allows the if/then/else conditional logic:

df['make'] = np.where(df['make']=='', 
                      df['id'].str.extract('(.*) ', expand=False), 
                      df['make'])

Answer 5

如果我正确回答了您的问题，您可以使用replace功能：

df.make = df.make.replace("", test.id)