row = [x.strip() for x in row]

(if you just want to get spaces at the end, use rstrip)

You can use the strip() method to remove trailing and leading spaces:

>>> s = '   abd cde   '
>>> s.strip()
'abd cde'

Note: the internal spaces are preserved.

Strings in Python are immutable (meaning that their data cannot be modified) so the replace method doesn't modify the string - it returns a new string. You could fix your code as follows:

for i in hello:
    j = i.replace(' ','')
    k.append(j)

However a better way to achieve your aim is to use a list comprehension. For example the following code removes leading and trailing spaces from every string in the list using strip:

hello = [x.strip(' ') for x in hello]

Looks like you just need this:

>>> myvar = [1, 2, '3']
>>> ",".join(map(str, myvar))
'1,2,3'

But, if you want to be more robust, then apply strip function to every element:

>>> myvar = [1, 2, '3']
>>> ",".join(map(lambda x: str(x).strip(), myvar))
'1,2,3'

>>> myvar = [1, 2, 3, ' 4 ']
>>> ",".join(map(lambda x: str(x).strip(), myvar))
'1,2,3,4'

For your case, with a list, you can use str.strip()

l = [x.strip() for x in List]

This will strip both trailing and leading spaces. If you only need to remove leading spaces, go with Alex' solution.

You're forgetting to reset j to zero after iterating through the first list.

Which is one reason why you usually don't use explicit iteration in Python - let Python handle the iterating for you:

>>> networks = [["  kjhk  ", "kjhk  "], ["kjhkj   ", "   jkh"]]
>>> result = [[s.strip() for s in inner] for inner in networks]
>>> result
[['kjhk', 'kjhk'], ['kjhkj', 'jkh']]