在第二次出现''（空格）后如何删除所有字符

Question

My name regex has been proven faulty on a couple entries:

find_name = re.search(r'^[^\d]*', clean_content)

The above would output something like this on a few entries:

TERRI BROWSING APT A # current output

So, I need a way to trim that out; it's tripping the rest of my program. The only identifier I can think of is if I can somehow detect the second space; and remove all characters after it.

I only need the first and last name; ie

TERRI BROWSING # desired

After I remove those characters I could just .strip() out the trailing space, just need a way to remove all after second space.... or maybe detect only to get two words, nothing more.

Answer 1

Maybe you don't need regex but can use simple splits and joins:

text = 'TERRI BROWSING APT A'
' '.join(text.split(' ')[0:2])

# 'TERRI BROWSING'

Answer 2

You can do:

^\S+\s+\S+

• ^ matches the start of the string

• \\S+ matches one or more non-whitespaces

• \\s+ matches one or more whitespaces

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Also, assuming the whitespace is actually a space character, you can find the index of the second space using str.find and slice the string upto that point:

text[:text.find(' ', text.find(' ') + 1)] 

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Example:

In [326]: text = 'TERRI BROWSING APT A'                                                                                                                                                                     

In [327]: re.search(r'^\S+\s+\S+', text).group()                                                                                                                                                            
Out[327]: 'TERRI BROWSING'

In [338]: text[:text.find(' ', text.find(' ') + 1)]                                                                                                                                                         
Out[338]: 'TERRI BROWSING'

Answer 3

If you want to remove the rest, you could match 2 times a non whitespace char \\S* followed by a space and capture that in a group. Then match any char 0+ times and replace with the first capturing group using re.sub

^(\S* \S* ).*

Regex demo | Python demo

import re

print(re.sub(r"^(\S* \S* ).*", r"\1", "TERRI BROWSING APT A"))

Result

TERRI BROWSING