In Python, I want to extract only the characters from a string.
Consider I have the following string,
input = "{('players',): 24, ('year',): 28, ('money',): 19, ('ipod',): 36, ('case',): 23, ('mini',): 46}" I want the result as,
output = "players year money ipod case mini" I tried to split considering only the alphabets,
word1 = st.split("[a-zA-Z]+") But the split is not happening.
You could do it with re, but the string split method doesnt take a regex, it takes a string.
Heres one way to do it with re:
import re word1 = " ".join(re.findall("[a-zA-Z]+", st)) string.split() doesn't take regular expressions. You want something like:
re.split("[^a-zA-Z]*", "your string") and to get a string:
" ".join(re.split("[^a-zA-Z]*", "your string")) I think that you want all words, not characters.
result = re.findall(r"(?i)\b[a-z]+\b", subject) Explanation:
" \b # Assert position at a word boundary [a-z] # Match a single character in the range between “a” and “z” + # Between one and unlimited times, as many times as possible, giving back as needed (greedy) \b # Assert position at a word boundary " 
(?i), that's why I asked :)
What about doing this?
>>> import ast >>> " ".join([k[0] for k in ast.literal_eval("{('players',): 24, ('year',): 28, ('money',): 19, ('ipod',): 36, ('case',): 23, ('mini',): 46}").keys()]) 'case mini year money ipod players' 
ast.literal_eval() does?
You can take the approach of iterating through the string, and using the isalpha function to determine if it's a alpha character or not. If it is you can append it to the output string.
a = "Some57 996S/tr::--!!ing" q = "" for i in a: if i.isalpha(): q = "".join([q,i]) Or if you want all characters regardless of words or empty spaces
a = "Some57 996S/tr::--!!ing" q = "" for i in a: if i.isalpha(): q = "".join([q,i]) print q 'SomeString'
import re string = ''.join([i for i in re.findall('[\w +/.]', string) if i.isalpha()]) #'[\w +/.]' -> it will give characters numbers and punctuation, then 'if i.isalpha()' this condition will only get alphabets out of it and then join list to get expected result. # It will remove spaces also. 
[a-zA-Z]+as the delimiter, so it is removed.