'negative' pattern matching in python

Question

I have the following input,

OK SYS 10 LEN 20 12 43
1233a.fdads.txt,23 /data/a11134/a.txt
3232b.ddsss.txt,32 /data/d13f11/b.txt
3452d.dsasa.txt,1234 /data/c13af4/f.txt
.

And I'd like to extract all of the input except the line containing "OK SYS 10 LEN 20" and the last line which contains a single "." (dot). That is, I want to extract the following

1233a.fdads.txt,23 /data/a11134/a.txt
3232b.ddsss.txt,32 /data/d13f11/b.txt
3452d.dsasa.txt.1234 /data/c13af4/f.txt

I tried the following,

for item in output:
    match_obj = re.search("^(?!OK) | ^(?!\\.)", item)
    if match_obj :
        print("got item " + item)

but it does not work, as it does not produce any output.

@alannaC it's used to escape the special character `.` which stands for any character in many regex implementations. — kaiya, May 10 '22 at 15:37

score 59 · Accepted Answer · edited Apr 11 '23 at 10:27

59

See it in action:

match_obj = re.search("^(?!OK|\\.).*", item)

Don't forget to put .* after negative look-ahead, otherwise you couldn't get any match

edited Apr 11 '23 at 10:27

Neuron

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answered Aug 23 '12 at 12:25

mmdemirbas

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The negative look-ahead doesn't match anything, it just looks ahead *before* matching, and applies to the next pattern that follows it (eg. `.*`) Took me a while to figure that out. – Hippo Jun 26 '18 at 12:37

score 8 · Answer 2 · edited Apr 11 '23 at 10:27

8

Use a negative match. (Also note that whitespace is significant, by default, inside a regex so don't space things out. Alternatively, use re.VERBOSE.)

for item in output:
    match_obj = re.search("^(OK|\\.)", item)
    if not match_obj:
        print("got item " + item)

edited Apr 11 '23 at 10:27

Neuron

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answered Aug 23 '12 at 12:15

Marcelo Cantos

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score 6 · Answer 3 · edited Apr 11 '23 at 10:28

6

if not (line.startswith("OK ") or line.strip() == "."):
    print(line)

edited Apr 11 '23 at 10:28

Neuron

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answered Aug 23 '12 at 12:08

Jochen Ritzel

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1

You'd need to handle the `.` too, but that's easy. – DSM Aug 23 '12 at 12:13
Thanks, added for completeness' sake. – Jochen Ritzel Aug 23 '12 at 12:21

score 4 · Answer 4 · edited Apr 11 '23 at 10:28

4

Why don't you match the OK SYS row and not return it.

for item in output:
    match_obj = re.search("(OK SYS|\\.).*", item)
    if not match_obj :
        print("got item " + item)

edited Apr 11 '23 at 10:28

Neuron

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answered Aug 23 '12 at 12:08

Pablo Jomer

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score 1 · Answer 5 · answered Aug 23 '12 at 12:09

If this is a file, you can simply skip the first and last lines and read the rest with csv:

>>> s = """OK SYS 10 LEN 20 12 43
... 1233a.fdads.txt,23 /data/a11134/a.txt
... 3232b.ddsss.txt,32 /data/d13f11/b.txt
... 3452d.dsasa.txt,1234 /data/c13af4/f.txt
... ."""
>>> stream = StringIO.StringIO(s)
>>> rows = [row for row in csv.reader(stream,delimiter=',') if len(row) == 2]
>>> rows
[['1233a.fdads.txt', '23 /data/a11134/a.txt'], ['3232b.ddsss.txt', '32 /data/d13f11/b.txt'], ['3452d.dsasa.txt', '1234 /data/c13af4/f.txt']]

If its a file, then you can do this:

with open('myfile.txt','r') as f:
   rows = [row for row in csv.reader(f,delimiter=',') if len(row) == 2]

score 1 · Answer 6 · edited May 01 '15 at 16:17

1

and(re.search("bla_bla_pattern", str_item, re.IGNORECASE) == None)

is working.

edited May 01 '15 at 16:17

tbraun89

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answered May 01 '15 at 15:38

hkn06tr

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score 0 · Answer 7 · answered Nov 28 '18 at 09:58

You can also do it without negative look ahead. You just need to add parentheses to that part of expression which you want to extract. This construction with parentheses is named group.

Let's write python code:

string = """OK SYS 10 LEN 20 12 43
1233a.fdads.txt,23 /data/a11134/a.txt
3232b.ddsss.txt,32 /data/d13f11/b.txt
3452d.dsasa.txt,1234 /data/c13af4/f.txt
.
"""

search_result = re.search(r"^OK.*\n((.|\s)*).", string)

if search_result:
    print(search_result.group(1))

Output is:

1233a.fdads.txt,23 /data/a11134/a.txt
3232b.ddsss.txt,32 /data/d13f11/b.txt
3452d.dsasa.txt,1234 /data/c13af4/f.txt

^OK.*\n will find first line with OK statement, but we don't want to extract it so leave it without parentheses. Next is part which we want to capture: ((.|\s)*), so put it inside parentheses. And in the end of regexp we look for a dot ., but we also don't want to capture it.

P.S: I find this answer is super helpful to understand power of groups. https://stackoverflow.com/a/3513858/4333811

vt220 · Answer 8 · 2020-05-06T08:41:07.417

0

If the OK line is the first line and the last line is the dot you could consider slice them off like this:

TestString = '''OK SYS 10 LEN 20 12 43
1233a.fdads.txt,23 /data/a11134/a.txt
3232b.ddsss.txt,32 /data/d13f11/b.txt
3452d.dsasa.txt,1234 /data/c13af4/f.txt
.
'''
print('\n'.join(TestString.split()[1:-1]))

However if this is a very large string you may run into memory problems.

edited May 06 '20 at 08:41

answered May 06 '20 at 08:23

vt220

1
1

'negative' pattern matching in python

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