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Let's say I have this string

'1. A4  1... d5  2. c4  2... Yf6  3. NP3  3... dxc4  4. BO3  4... BK4  5. e3  5... Bf3  6. Q3  6... e6  7. Bc4  7... B4  8. O-O  8... B3  9. b3  9... O-O  10. B3  10... Re8  11. Q7  1-0'

I want to remove the numbers not attached to letters and if you scroll to the end I want the 1-0 removed as well, so something like this

['A4', 'd5', ..., 'O-O', ..., 'Q7']

So I tried this,

re.findall(r'(?:[^\W\d_]+\d|\d+[^\W\d_])[^\W_]*|[^\W\d_]+', text)

but got this,

['A4', 'd5', ..., 'O', 'O', ..., 'Q7']

So it is removing the - for 1-0 which I want, but also to O-O.

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  • What do you mean by "it is removing the - for 1-0"? Isn't it removing all of 1-0? Commented May 24, 2021 at 16:25
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3 Answers 3

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1

Why bother with regular expressions here? You can just split the string and take the odd numbered indices:

>>> s = '1. A4  1... d5  2. c4  2... Yf6  3. NP3  3... dxc4  4. BO3  4... BK4  5. e3  5... Bf3  6. Q3  6... e6  7. Bc4  7... B4  8. O-O  8... B3  9. b3  9... O-O  10. B3  10... Re8  11. Q7  1-0'
>>> list(itertools.islice(s.split(), 1, None, 2))
['A4',
 'd5',
 'c4',
 'Yf6',
 'NP3',
 'dxc4',
 'BO3',
 'BK4',
 'e3',
 'Bf3',
 'Q3',
 'e6',
 'Bc4',
 'B4',
 'O-O',
 'B3',
 'b3',
 'O-O',
 'B3',
 'Re8',
 'Q7']
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2 Comments

the OP's requirement is remove the numbers not attached to letters, this answer works but would fail on 2 digits numbers, or numbers with unknown lengths.
@Jared I don't believe it would fail at all. There's nothing about this solution that distinguishes 1-digit from 2- or more digit numbers, or even cares about the contents of each token. The input is a list moves in a single game of chess. It seems pretty clear the poster wants to filter out the turn numbers from the moves.
1

Find all words with at least a single letter in it :

re.findall(r'\S*[a-zA-Z]\S*', text)

['A4', 'd5', 'c4', 'Yf6', 'NP3', 'dxc4', 'BO3', 'BK4', 'e3', 'Bf3', 'Q3', 'e6', 'Bc4', 'B4', 'O-O', 'B3', 'b3', 'O-O', 'B3', 'Re8', 'Q7']
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Comments

0

You can use a list comprehension to iterate through the string and isidigt to check if each item contains a digit or not.

str1 = '1. A4  1... d5  2. c4  2... Yf6  3. NP3  3... dxc4  4. BO3  4... BK4  5. e3  5... Bf3  6. Q3  6... e6  7. Bc4  7... B4  8. O-O  8... B3  9. b3  9... O-O  10. B3  10... Re8  11. Q7  1-0'
output = list(filter(None, [i if i.replace("-","").isdigit()==False else None for i in str1.replace(".", "").split(' ')]))

print(output) gives:

['A4', 'd5', 'c4', 'Yf6', 'NP3', 'dxc4', 'BO3', 'BK4', 'e3', 'Bf3', 'Q3', 'e6', 'Bc4', 'B4', 'O-O', 'B3', 'b3', 'O-O', 'B3', 'Re8', 'Q7']
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1 Comment

but you have the '1-0' at the end and I don't want that, all that have 'number - number' are no good

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