Question

I have searched high and low for a resolution to this situation, and tested a few different methods, but I haven’t had any luck thus far. Basically, I have a file with data in the following format that I need to convert into a CSV:

(previously known as CyberWay Pte Ltd)
0 2019
01.com
0 1975
1 TRAVEL.COM
0 228
1&1 Internet
97 606
1&1 Internet AG
0 1347
1-800-HOSTING
0 8
1Velocity
0 28
1st Class Internet Solutions
0 375
2iC Systems
0 192

I’ve tried using re.sub and replacing the whitespace between the numbers on every other line with a comma, but haven’t had any success so far. I admit that I normally parse from CSVs, so raw text has been a bit of a challenge for me. I would need to maintain the string formats that are above each respective set of numbers.

I’d prefer the CSV to be formatted as such:

foo bar
0,8
foo bar
0,9
foo bar
0,10
foo bar
0,11

There’s about 50,000 entries, so manually editing this would take an obscene amount of time.

If anyone has any suggestions, I’d be most grateful.

Thank you very much.

Answer 1

If I have correctly understood your requirement, you need a strip() on all lines and a split based on whitespace on even lines (lines starting from 1):

import re

fp = open("csv.txt", "r")
while True:
    line = fp.readline()
    if '' == line:
            break
    line    = line.strip()
    fields  = re.split("\s+", fp.readline().strip())
    print "\"%s\",%s,%s" % ( line, fields[0], fields[1] )
fp.close()

The output is a CSV (you might need to escape quotes if they occur in your input):

"Content of odd line",Number1,Number2

I do not understand the ‘foo,bar’ you place as header on your example’s odd lines, though.