If statement that keeps characters only

Question

Let's say we have a list list_a = [a,b,C,.,/,!,d,E,f,]

i want to append to a new list only the letters of the alphabet.

So the new list will be list_b = [a,b,C,d,E,f]. So far i have tried doing it like that way:

list_b = []
for elements in list_a:
    try:
        if elements == str(elements):
            list_b.append(elements)
    except ValueError: #Catches the Error when the element is not a letter
        continue

However, when i print the list_b it has all the elements of list_a , it doesn't do the job i expected. Any ideas ?

PS: comma in the specific example brings Error too.

Does [this answer](https://stackoverflow.com/questions/3640359/regular-expressions-search-in-list) help? — tadman, Feb 28 '18 at 19:06
Can you explain why you think this code would work? I'm not understanding your logic. — Daniel Roseman, Feb 28 '18 at 19:07
Simple.If the element of the list is a letter/character it gets appended in a new list else the for loop goes on.What do you not understand ? — Vaggelis Manousakis, Feb 28 '18 at 19:08
I can't understand why you would think that `str('.')` does not equal `'.'`, or what would raise a ValueError. — Daniel Roseman, Feb 28 '18 at 19:16
You are right.str( 3 ) is not equal ' 3 '.Allthough this works for numbers it doesn't works for commas , questionmarks etc.I am new to Pytho sorry.. — Vaggelis Manousakis, Feb 28 '18 at 19:22

wpercy · Answer 1 · 2018-02-28T19:14:36.693

3

You can use the .isalpha() method of the string type.

In [1]: list_a = ['a','b','C','.','/','!','d','E','f']

In [2]: list_b = [i for i in list_a if i.isalpha()]

In [3]: list_b
Out[3]: ['a', 'b', 'C', 'd', 'E', 'f']

edited Feb 28 '18 at 19:14

answered Feb 28 '18 at 19:09

wpercy

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score 2 · Answer 2 · edited Feb 28 '18 at 19:12

2

Try checking if the character is an alphabet by using the .isalpha() function.

list_b = []
for elements in list_a:
   if elements.isalpha():
        list_b.append(elements)

edited Feb 28 '18 at 19:12

wpercy

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answered Feb 28 '18 at 19:08

Tim Lee

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8

Sorry. Got confused with Java – Tim Lee Feb 28 '18 at 19:12

score 2 · Answer 3 · answered Feb 28 '18 at 19:14

2

You are missing the fact that the str() function does not return the "str" elements you think it does, just an str representation of them. Try creating a list with you dictionary [a-zA-Z] (not very pythonic but simple to grasp) and check if your character exists in it. I suggest writing your own code from scratch instead of copy/pasting, that is the only way to really understand the problem....

answered Feb 28 '18 at 19:14

Theo

21
1

Thanks for letting me know my mistake.I will keep that in mind! – Vaggelis Manousakis Feb 28 '18 at 20:20

kerk12 · Accepted Answer · 2018-02-28T19:24:03.970

You can try this:

import string
for item in list_a:
    if item in string.ascii_letters:
        list_b.append(item)

Also, check out the string module. It has a lot of additional methods that can help you, should you wish to work with strings. Do note that this works only with ascii characters. If you want to check for every character in an alphabet, then you can via the isalpha() method, as the others have noted above.

score 0 · Answer 5 · answered Feb 28 '18 at 19:33

Well, it is basically the same logic of using isalpha() method, but you can do this by using filter:

list_a = ['a','b','C','.','/','!','d','E','f']

list_b = list(filter(lambda i: i.isalpha(), list_a))

print(list_b)

Output:

['a', 'b', 'C', 'd', 'E', 'f']

Brendan Abel · Answer 6 · 2018-02-28T19:09:02.950

-2

You can use the string or re package to do this

import re
new_list = [c for c in old_list if re.match(r'[a-zA-Z]', c)]

Or with string

import string
new_list = [c for c in old_list if c in string.ascii_letters]

edited Feb 28 '18 at 19:09

answered Feb 28 '18 at 19:07

Brendan Abel

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This seems unnecessarily complex. – Daniel Roseman Feb 28 '18 at 19:08
@DanielRoseman They're single line statements with a single if test, I'm not sure how much simpler they can get. – Brendan Abel Feb 28 '18 at 19:09
Why re? for such a small thing. – Reck Feb 28 '18 at 19:10
stick to KISS rule – Reck Feb 28 '18 at 19:11
Works fine.However kerk12's answer is much simpler and faster.Thanks anyway ! – Vaggelis Manousakis Feb 28 '18 at 19:19

If statement that keeps characters only

6 Answers6