Syntax error in Python regexp

Question

When I started writing one function, i got Syntax error. I tried execute line at REPL - and it's worked. But i want to do it at IDE. Can somebody help me?

My code:

def sentence_splitter(file_name):
    with open(file_name) as f:
        input_str = f.read()
        period_indexes = get_periods(input_str)
        for el in period_indexes:
            sub_str = input_str[el - 14:el + 14]
            if not re.search(r'\.\s+[A-Za-z]{1,3}\w+', sub_str) and # Error here
            re.search(r'\.\d+', sub_str) and
            re.search(r'\.\s+[a-z]+', sub_str) and
            re.search(r'([A-Za-z\.]+\.\w+){1,50}', sub_str) and
            re.search(r'\w+\.[\.,]+', s):
                pass

You need a colon `:` after an `if` statement. Is that missing or just an error from copying your code? — mdml, Nov 11 '13 at 17:58
it's not end of if statement. I have six re.search() lines in my condition — Absolut, Nov 11 '13 at 18:05
@Absolut Well how do you know that it's not in one of those? — Waleed Khan, Nov 11 '13 at 18:09
Because my IDE indicate this line. Alone or with other functions in condition - anyway. — Absolut, Nov 11 '13 at 18:13
Regardless of where the error is, that code has a weird structure. It would be much better form to have an array of regexps, and iterate through it as long as there is a match/not a match (you have a not in your expression, but it only applies to the first regexp- can't tell if this is a mistake or intentional on your part). — bitgarden, Nov 11 '13 at 18:29
try to add parentheses around the multiline-if-condition: `if (stuff \n more stuff):` — tobias_k, Nov 11 '13 at 18:30

Chris · Accepted Answer · 2013-11-11T18:39:49.020

3

You need parentheses around your if statement:

def sentence_splitter(file_name):
    with open(file_name) as f:
        input_str = f.read()
        period_indexes = get_periods(input_str)
        for el in period_indexes:
            sub_str = input_str[el - 14:el + 14]
            if not (re.search(r'\.\s+[A-Za-z]{1,3}\w+', sub_str) and # Error here
            re.search(r'\.\d+', sub_str) and
            re.search(r'\.\s+[a-z]+', sub_str) and
            re.search(r'([A-Za-z\.]+\.\w+){1,50}', sub_str) and
            re.search(r'\w+\.[\.,]+', s)):
                pass

Technically backslashes will work, but parentheses are more Pythonic, see PEP8: http://www.python.org/dev/peps/pep-0008/#maximum-line-length

edited Nov 11 '13 at 18:39

answered Nov 11 '13 at 18:32

Chris

1,416
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It depends if you want the `not` to apply to the whole statement or just one condition. `not (1==1 and 1==2)` is true, but `(not 1==1 and 1==2)` is false. – Chris Nov 11 '13 at 18:44
Don't tell me! ;-) Maybe you want to add this explanation to your answer, after all, the default behaviour if they were all on one line would be that the `not` applies only to the first term. I.e., you fixed the error (+1) but also changed (fixed?) the condition. – tobias_k Nov 11 '13 at 18:50

score 2 · Answer 2 · edited May 23 '17 at 12:29

Your conditional spans multiple lines. You need to add the line continuation character \.

if not re.search(r'\.\s+[A-Za-z]{1,3}\w+', sub_str) and \
            re.search(r'\.\d+', sub_str) and \
            re.search(r'\.\s+[a-z]+', sub_str) and \
            re.search(r'([A-Za-z\.]+\.\w+){1,50}', sub_str) and \
            re.search(r'\w+\.[\.,]+', s):

Further information about this is available in PEP8 and this answer.

One note specific to your code:

re.search(r'\w+\.[\.,]+', s)
                          ^---- This variable is not assigned 
                                (likely should be sub_str)

score 1 · Answer 3 · answered Nov 11 '13 at 18:33

In your last regexp:

re.search(r'\w+\.[\.,]+', s)

You perform a search on s, which is not defined. All the other regexps perform a search on substr, which is probably what you want. That would raise a NameError though, not a SyntaxError.

Additionally, you probably want to refactor your code to make it easier to read, as explained in my comment to your question.

Syntax error in Python regexp

3 Answers3