A one liner without map is:

df['E'] = df['B'].str.replace('\W', '')

The regex groups the digits on either side of the '.' ignoring all non-digits. The code uses these groups to create the required output. Regex101

import pandas as pd

def clean_input(m):
    print(m.group(0))
    if m:
        val = m.group(1)
        if m.group(2):
            val = val + '.' +m.group(2)
    return val

a = pd.DataFrame({'colA':
   ['7.8.',
    '5..3',
    '%3.2',
    '   ',
    '3.*8',
    '3.8*',
    '140',
    '5.5.',
    '14.5 of HGB',
    '>14.5',
    '<14.5',
    '14,5',
   '14. 5']})
a['colA'].str.replace('[^\d]*(\d+)[^\d]*(?:\.)?[^\d]*(\d)*[^\d]*', clean_input)

Output:

0      7.8
1      5.3
2      3.2
3         
4      3.8
5      3.8
6      140
7      5.5
8     14.5
9     14.5
10    14.5
11    14.5
12    14.5

Regex explanation:

• \d - matches a digit
• [^<pattern>] - matches any character except the
• [^\d] - matches any character except for digits.
• [^\d]+- matches one or more of the above.
• (?:) - is non-capturing group where the matched characters are not captured.
• <pattern>? - zero or one occurance of the pattern.
• \. - since . is a meta character, it has to be escaped with \

Is that what you want?

In [71]: df.nonhashtag.apply(' '.join).str.replace('[^A-Za-z\s]+', '') \
           .str.split(expand=False)
Out[71]:
0    [want, better, than, Dhabi, United, Arab, Emir...
1                  [Just, posted, photo, Rasim, Villa]
2          [Dhabi, International, Airport, AUH, Dhabi]
3       [just, shrug, off, Dubai, Mall, Burj, Khalifa]
4    [out, Cowboy, steppin, Notorious, going, sleep...
5                  [Buona, notte, Viceroy, Hotel, Yas]
Name: nonhashtag, dtype: object

'[^A-Za-z\s]+' - is a RegEx meaning take all characters except those:

• with ASCII codes from A to Z
• from a to z
• spaces and tabs

So .str.replace('[^A-Za-z\s]+', '') will remove all characters except letters belonging to english alphabet, spaces and tabs