Return all alphabetic words from a list

Question

Hi, I have a list of words and I want return all alphabetic words from this list. It must look like this, but it doesn’t work:
import re
words = [ “I”, “want”, “,”,”:”, “/“, “Sleep”]
for word in words:
if word not re.match(r’^[0-9\-\.,]+$’, words):
print(word)

The out must be:
words= [“i”, “want”, “sleep”]

Answer 1

Katja, isalpha will give True for everything that only consists of letters.

Answer 2

Mirielle , it retuns me “ str” object has no attribute isAlpha

Answer 3

Ah, so you want all the words lowered?

Yeah, then just write

words2 = [w.lower() for w in words if w.isalpha()]

Answer 4

Do you need re for that? Why not just:

for word in words:
if word.isalpha():
print(word)

Answer 5

If you want to create a new list only with pure words, you can write:

words2 = [w for w in words if w.isalpha()]

Answer 6

HonFu , it looks like this:
words=[word for word in words if word.isalpha()]
words = [word.lower() for word in words]
it return exactly what i want✨

Answer 7

HonFu, yes, but it must also return lower case words

Answer 8

U can write the code using isalpha().. the code is ...
for word in words:
If word.isalpha():
print(word)

Answer 9

arr = ["i","want",",",":","/","sleep"]

words = []

for x in arr:
if x.isalpha():
words.append(x)

print(words)

####___________________________________
outputs ['i','want','sleep']

Try it and let me know whether it was perfect for you or not. If not i will try something else for you then.
but I guess this is the solution