Replace multiple newlines with single newlines during reading file

You could use a second regex to replace multiple new lines with a single new line and use strip to get rid of the last new line.

import os
import re

files=[]
pars=[]

for i in os.listdir('path_to_dir_with_files'):
    files.append(i)

for f in files:
    with open('path_to_dir_with_files/'+str(f), 'r') as a:
        word = re.sub(r'someword=|\,.*|\#.*','', a.read())
        word = re.sub(r'\n+', '\n', word).strip()
        pars.append(word)

for k in pars:
   print k

Without changing your code much, one easy way would just be to check if the line is empty before you print it, e.g.:

import os
import re

files=[]
pars=[]

for i in os.listdir('path_to_dir_with_files'):
    files.append(i)

for f in files:
    with open('path_to_dir_with_files'+str(f), 'r') as a:
        pars.append(re.sub('someword=|\,.*|\#.*','',a.read()))

for k in pars:
    if not k.strip() == "":
        print k

*** EDIT Since each element in pars is actually the entire content of the file (not just a line), you need to go through an replace any double end lines, easiest to do with re

import os
import re

files=[]
pars=[]

for i in os.listdir('path_to_dir_with_files'):
    files.append(i)

for f in files:
    with open('path_to_dir_with_files'+str(f), 'r') as a:
        pars.append(re.sub('someword=|\,.*|\#.*','',a.read()))

for k in pars:
    k = re.sub(r"\n+", "\n", k)
    if not k.strip() == "":
        print k

Note that this doesn't take care of the case where a file ends with a newline and the next one begins with one - if that's a case you are worried about you need to either add extra logic to deal with it or change the way you're reading the data in