Adding a blank and fixed that problem from your own code

df['Text'].str.replace('eng ', 'engine ')
Out[736]: 
0            engine is here
1      engine needs washing
2    engine is overheating 
Name: Text, dtype: object

Update

df.Text.str.split(' ',expand=True).replace('eng','engine').fillna('').apply(' '.join,1)
Out[752]: 
0           engine is here 
1     engine needs washing 
2    engine is overheating 
dtype: object

You could use str.rpartition to split the ids on the last space.

In [169]: parts = df['id'].str.rpartition(' ')[[0,2]]; parts
Out[169]: 
                0   2
0      abarth 1.4   a
1        abarth 1   a
2  land rover 1.3   r
3      land rover   2
4    land rover 5   g
5      mazda 4.55  bl

Then you could use == to compare parts[2] to df['series']:

In [170]: mask = (parts[2] == df['series']); mask
Out[170]: 
0     True
1     True
2     True
3    False
4     True
5     True
dtype: bool

And finally, use df['id'].where to replace df['id] with parts[0] where mask is True:

import pandas as pd
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'] })
parts = df['id'].str.rpartition(' ')[[0,2]]
mask = (parts[2] == df['series'])
df['id'] = df['id'].where(~mask, parts[0], axis=0)
print(df)

yields

               id series
0      abarth 1.4      a
1        abarth 1      a
2  land rover 1.3      r
3    land rover 2       
4    land rover 5      g
5      mazda 4.55     bl

Alternatively, you could use

import re
def remove_series(x):
    pat = r'{}$'.format(x['series'])
    return re.sub(pat, '', x['id'])
df['id'] = df.apply(remove_series, axis=1)

But calling df.apply with a custom function tends to be much slower than using built-in vectorize methods such as those used in the first method.

Considering data is your pandas DataFrame you can also use:

data.replace({'very bad': 1, 'bad': 2, 'poor': 3, 'good': 4, 'very good': 5}, inplace=True)