str.split actually returns a list.

Return a list of the words in the string, using sep as the delimiter string.

Since you are appending the returned list to newlist, you are getting a list of lists. Instead use list.extend method, like this

for word in lines:
    newlist.extend(word.split(","))

But you can simply use nested list comprehension like this

>>> data = ['60,78', '70,77', '80,74', '90,75', '100,74', '110,75']
>>> [item for items in data for item in items.split(",")]
['60', '78', '70', '77', '80', '74', '90', '75', '100', '74', '110', '75']

I came to add:

map(str.strip, string.split(','))

but saw it had already been mentioned by Jason Orendorff in a comment.

Reading Glenn Maynard's comment on the same answer suggesting list comprehensions over map I started to wonder why. I assumed he meant for performance reasons, but of course he might have meant for stylistic reasons, or something else (Glenn?).

So a quick (possibly flawed?) test on my box (Python 2.6.5 on Ubuntu 10.04) applying the three methods in a loop revealed:

$ time ./list_comprehension.py  # [word.strip() for word in string.split(',')]
real    0m22.876s

$ time ./map_with_lambda.py     # map(lambda s: s.strip(), string.split(','))
real    0m25.736s

$ time ./map_with_str.strip.py  # map(str.strip, string.split(','))
real    0m19.428s

making map(str.strip, string.split(',')) the winner, although it seems they are all in the same ballpark.

Certainly though map (with or without a lambda) should not necessarily be ruled out for performance reasons, and for me it is at least as clear as a list comprehension.

Use .split on the string class:

>>> industries_list=["Computers, Internet","Photography, Tourism","Motoring, Manufacturing"]
>>> [var.split(',') for var in industries_list]
[['Computers', ' Internet'], ['Photography', ' Tourism'], ['Motoring', ' Manufacturing']]

If you'd like no spaces:

>>> [[s.strip() for s in var.split(',')] for var in industries_list]
[['Computers', 'Internet'], ['Photography', 'Tourism'], ['Motoring', 'Manufacturing']]

Live demo.

I can recomend you to use list comprehension:

newList = [element for item in myList for element in item.split(',')]

How about a list?

mystring.split(",")

It might help if you could explain what kind of info we are looking at. Maybe some background info also?

EDIT:

I had a thought you might want the info in groups of two?

then try:

re.split(r"\d*,\d*", mystring)

and also if you want them into tuples

[(pair[0], pair[1]) for match in re.split(r"\d*,\d*", mystring) for pair in match.split(",")]

in a more readable form:

mylist = []
for match in re.split(r"\d*,\d*", mystring):
    for pair in match.split(",")
        mylist.append((pair[0], pair[1]))