"0000-22N-06W-01"
"0000-22N-06W-02"
"0000-22N-06W-03"
"0000-22N-06W-04"

import re
output = open("output.txt","w")
input = open("input.txt")

for line in input:
    output.write(re.sub(r'^(.{4})-(.{3})-(.{3})-(.{2})$', r'\1-\4-\2-\3', line))

input.close()
output.close()

NOTE: If you actually have " in your data then you should change your regular expression to this one:

^"(.{4})-(.{4})-(.{3})-(.{3})"$

Regex101 Demo

use "r" more to read file, "w" mode will create empty file for writing. .readline() will get and pass the string to the re.sub(). r" .*" will return a string you want to replace after the 'space' character. i assume 'hi 873840' is the only text in your file and your desired output is only 'hi 90'

echo "hi 873840" > hi.txt

python3.6

import re
fp = open("hi.txt", "r")
print(re.sub(r" .*", " 90", fp.readline()))

Fixing the regex:

>>> pattern = re.compile(r'\.CSV', re.IGNORECASE)
>>> pattern.sub(repl='.xlsx', string='test.CSV')
'test.xlsx'
>>> pattern.sub(repl='.xlsx', string='test.csv')
'test.xlsx'

Not using regex in the first place:

base, ext = os.path.splitext(fname)
if ext.lower() == '.csv':
    fname = base + '.xlsx'

Apply all your substitutions in one go on the current line:

with open('retweet.csv', 'rb') as inputfile:
    read=csv.reader(inputfile, delimiter=',')
    for row in read:
        text = row[0]
        text = re.sub(r'@\s\w+', ' ', text)
        text = re.sub(another_expression, another_replacement, text)
        # etc.
        f.write(text + '\n')

Note that opening a file with csv.reader(..., delimiter='\n') sounds awfully much as if you are treating that file as a sequence of lines; you could just loop over the file:

with open('output.csv', 'rb') as inputfile2:
    for line in inputfile2:

You could try this:

>>> import re    
>>> text = 'file1 file2 file3'
>>> x = re.sub(r'file([1-9])',r'file0\1',text)
'file01 file02 file03'

The brackets wrapped around the [1-9] captures the match, and it is the first match. You will see I used it in the replace using \1 meaning the first catch in the match.

Also, if you don't want to add the zero for files with 2 digits or more, you could add [^\d] in the regexp:

x = re.sub(r'file(1-9)',r'file0\1',text)

A bit more of a generic solution now that I'm revisiting this answer using str.format() and a lambda expression:

import re
fmt = '{:03d}'                 # Let's say we want 3 digits with leading zeroes
s = 'file1 file2 file3 text40'
result = re.sub(r"([A-Za-z_]+)([0-9]+)", \
                lambda x: x.group(1) + fmt.format(int(x.group(2))), \
                s)
print(result)
# 'file001 file002 file003 text040'

A bit of details about the lambda expression:

lambda x: x.group(1) + fmt.format(int(x.group(2)))
#         ^--------^   ^-^        ^-------------^
#          filename   format     file number ([0-9]+) converted to int
#        ([A-Za-z_]+)            so format() can work with our format

I am using the expression [A-Za-z_]+ assuming the filename contains letters and underscores only besides the training digits. Do pick a more appropriate expression if required.