I have the next code which reads from multiple files, parses obtained lines and prints the result:
import os
import re
files=[]
pars=[]
for i in os.listdir(\'pa
Just would like to point out: regexes aren't the best way to handle that. Replacing two empty lines by one in a Python str is quite simple, no need for re:
entire_file = "whatever\nmay\n\nhappen"
entire_file = entire_file.replace("\n\n", "\n")
And voila! Much faster than re and (in my opinion) much easier to read.
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