Автоматизированная среда тестирования - Windows CE

Вы можете использовать str.extract с contains и loc с помощью boolean indexing :

df1 = (df.AgeuponOutcome.str.extract('(\d+) (\w+)', expand=True))
df1.columns = ['a','b']
print (df1)
    a       b
0   1    year
1   1    year
2   2   years
3   3   weeks
4   2   years
5   1   month
6   3   weeks
7   3   weeks
8   5  months
9   1    year
10  2   years
11  2   years
12  4   years

print (df1.loc[df1.b.str.contains('month'), 'a'])
5    1
8    5
Name: a, dtype: object

print (df1.loc[df1.b.str.contains('year'), 'a'])
0     1
1     1
2     2
4     2
9     1
10    2
11    2
12    4
Name: a, dtype: object

Если вам нужны выходные данные в качестве новых столбцов:

df1['month'] = (df1.loc[df1.b.str.contains('month'), 'a'])
df1['year'] = (df1.loc[df1.b.str.contains('year'), 'a'])
df1['week'] = (df1.loc[df1.b.str.contains('week'), 'a'])
print (df1)
    a       b month year week
0   1    year   NaN    1  NaN
1   1    year   NaN    1  NaN
2   2   years   NaN    2  NaN
3   3   weeks   NaN  NaN    3
4   2   years   NaN    2  NaN
5   1   month     1  NaN  NaN
6   3   weeks   NaN  NaN    3
7   3   weeks   NaN  NaN    3
8   5  months     5  NaN  NaN
9   1    year   NaN    1  NaN
10  2   years   NaN    2  NaN
11  2   years   NaN    2  NaN
12  4   years   NaN    4  NaN

ИЗМЕНИТЬ по комментарию:

Вы можете использовать:

#convert to int
df1['a'] = df1.a.astype(int)

#divide by constant to column a
df1.loc[df1.b.str.contains('month'), 'a'] = df1.loc[df1.b.str.contains('month'), 'a'] / 12
df1.loc[df1.b.str.contains('week'), 'a'] = df1.loc[df1.b.str.contains('week'), 'a']  /52.1429
print (df1)
           a       b
0   1.000000    year
1   1.000000    year
2   2.000000   years
3   0.057534   weeks
4   2.000000   years
5   0.083333   month
6   0.057534   weeks
7   0.057534   weeks
8   0.416667  months
9   1.000000    year
10  2.000000   years
11  2.000000   years
12  4.000000   years