How to write a function to remove digits from a str

[v13tn1g]

Given a string consisting of letters and digits only, return a list whose elements are those substrings of the given string that consist of consecutive letters only. For example, alpha("1up2down 33sideways77") will return the list ["up", "down", "sideways"].

All i know is that i would have to use s.split()..i think?

any help would be appreciated. thanks.

[boxfish]

I haven't seen any built in string functions that look like they could do this for you, although isalpha() could be useful. myString.isalph a() returns true if all the characters in myString are letters. You'll probably have to loop through each character in the string, put it in your list if it's a letter, and append a new string to your list if you find the beginning of a new word.
I hope this gives you some idea of how to solve the problem.

[bvdet]

re.split() should work well for this task.

>>> import re
>>> re.split('[0-9]', '1up2down33sideways77')
['', 'up', 'down', '', 'sideways', '', '']
# eliminate null strings with a list comprehension
>>> [item for item in re.split('[0-9]', '1up2down33sideways77') if item]
['up', 'down', 'sideways']
>>>

[boxfish]

Wow, bvdet. Looks like I'd better read up on the re module.

[bvdet]

The re module is very powerful, but it can be cryptic and difficult. Here's a good tutorial: click here

Here's another, not so succinct method of achieving the same result:

aStr = "1up2down33sideways77"

# create a list of indices of the digits in aStr
idxList = [i for i, s in enumerate(aStr) if s.isdigit()]
print idxList
strList = [aStr[idxList[i]+1:idxList[i+1]] for i in range(len(idxList)-1)]
# this list will contain null strings
print strList

# similar to the list comprehension above, eliminates null strings
strList2 = []
for i in range(len(idxList)-1):
    s = aStr[idxList[i]+1:idxList[i+1]]
    if s:
        strList2.append(s)

print strList2