If I have a list of strings such as:


What should I do in order to get rid of all the 8s in each string? I tried using strip or replace in a for loop but it doesn’t work like it would in a normal string (that not in a list). Does anyone have a suggestion?

Try this:

lst = [("aaaa8"),("bb8"),("ccc8"),("dddddd8")]
print([s.strip('8') for s in lst]) # remove the 8 from the string borders
print([s.replace('8', '') for s in lst]) # remove all the 8s 

Beside using loop and for comprehension, you could also use map

lst = [("aaaa8"),("bb8"),("ccc8"),("dddddd8")]
mylst = map(lambda each:each.strip("8"), lst)
print mylst

A faster way is to join the list, replace 8 and split the new string:

mylist = [("aaaa8"),("bb8"),("ccc8"),("dddddd8")]
mylist=" ".join(mylist).replace('8','').split()
print mylist

mylist = [("aaaa8"),("bb8"),("ccc8"),("dddddd8")]
print mylist
for i in mylist:
print mylist

Here’s a short one-liner using regular expressions:

print [re.compile(r"8").sub("", m) for m in mylist]

If we separate the regex operations and improve the namings:

pattern = re.compile(r"8") # Create the regular expression to match
res = [pattern.sub("", match) for match in mylist] # Remove match on each element
print res

lst = [("aaaa8"),("bb8"),("ccc8"),("dddddd8")...]

msg = filter(lambda x : x != "8", lst)

print msg

For anyone who came across this post, just for understanding the above removes any elements from the list which are equal to 8.

Supposing we use the above example the first element (“aaaaa8”) would not be equal to 8 and so it would be dropped.

To make this (kinda work?) with how the intent of the question was we could perform something similar to this

msg = filter(lambda x: x != "8", map(lambda y: list(y), lst))
  • I am not in an interpreter at the moment so of course mileage may vary, we may have to index so we do list(y[0]) would be the only modification to the above for this explanation purposes.

What this does is split each element of list up into an array of characters so (“aaaa8”) would become [“a”, “a”, “a”, “a”, “8”].

This would result in a data type that looks like this

msg = [[“a”, “a”, “a”, “a”], [“b”, “b”]…]

So finally to wrap that up we would have to map it to bring them all back into the same type roughly

msg = list(map(lambda q: ''.join(q), filter(lambda x: x != "8", map(lambda y: list(y[0]), lst))))

I would absolutely not recommend it, but if you were really wanting to play with map and filter, that would be how I think you could do it with a single line.