Would like to find the following pattern in a string:
word-word-word++ or -word-word-word++
So that it iterates the -word or word- pattern until the end of the substring.
the string is quite large and contains many words with those^ patterns.
The following has been tried:
p = re.compile('(?:\w+\-)*\w+\s+=', re.IGNORECASE)
result = p.match(data)
but it returns NONE. Does anyone know the answer?
Your regex will only match the first pattern, match() will only find one occurrence, and that only if it is immediately followed by some whitespace and an equals sign.
Also, in your example you implied you wanted three or more words, so here's a version that was changed in the following ways:
match both patterns (note the leading -?)
match only if there are at least three words to the pattern ({2,} instead of +)
match even if there's nothing after the pattern (the \b matches a word boundary. It is not really necessary here, since the preceding \w+ guarantees we are at a word boundary anyway)
returns all matches instead of only the first one.
Here's the code:
#!/usr/bin/python
import re
data=r"foo-bar-baz not-this -this-neither nope double-dash--so-nope -yeah-this-even-at-end-of-string"
p = re.compile(r'-?(?:\w+-){2,}\w+\b', re.IGNORECASE)
print p.findall(data)
# prints ['foo-bar-baz', '-yeah-this-even-at-end-of-string']
Related
I am using Python and would like to match all the words after test till a period (full-stop) or space is encountered.
text = "test : match this."
At the moment, I am using :
import re
re.match('(?<=test :).*',text)
The above code doesn't match anything. I need match this as my output.
Everything after test, including test
test.*
Everything after test, without test
(?<=test).*
Example here on regexr.com
You need to use re.search since re.match tries to match from the beging of the string. To match until a space or period is encountered.
re.search(r'(?<=test :)[^.\s]*',text)
To match all the chars until a period is encountered,
re.search(r'(?<=test :)[^.]*',text)
In a general case, as the title mentions, you may capture with (.*) pattern any 0 or more chars other than newline after any pattern(s) you want:
import re
p = re.compile(r'test\s*:\s*(.*)')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want . to match across multiple lines, compile the regex with re.DOTALL or re.S flag (or add (?s) before the pattern):
p = re.compile(r'test\s*:\s*(.*)', re.DOTALL)
p = re.compile(r'(?s)test\s*:\s*(.*)')
However, it will retrun match this.. See also a regex demo.
You can add \. pattern after (.*) to make the regex engine stop before the last . on that line:
test\s*:\s*(.*)\.
Watch out for re.match() since it will only look for a match at the beginning of the string (Avinash aleady pointed that out, but it is a very important note!)
See the regex demo and a sample Python code snippet:
import re
p = re.compile(r'test\s*:\s*(.*)\.')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want to make sure test is matched as a whole word, add \b before it (do not remove the r prefix from the string literal, or '\b' will match a BACKSPACE char!) - r'\btest\s*:\s*(.*)\.'.
I don't see why you want to use regex if you're just getting a subset from a string.
This works the same way:
if line.startswith('test:'):
print(line[5:line.find('.')])
example:
>>> line = "test: match this."
>>> print(line[5:line.find('.')])
match this
Regex is slow, it is awkward to design, and difficult to debug. There are definitely occassions to use it, but if you just want to extract the text between test: and ., then I don't think is one of those occasions.
See: https://softwareengineering.stackexchange.com/questions/113237/when-you-should-not-use-regular-expressions
For more flexibility (for example if you are looping through a list of strings you want to find at the beginning of a string and then index out) replace 5 (the length of 'test:') in the index with len(str_you_looked_for).
Let's say that I have a string that looks like this:
a = '1253abcd4567efgh8910ijkl'
I want to find all substrings that starts with a digit, and ends with an alphabet.
I tried,
b = re.findall('\d.*\w',a)
but this gives me,
['1253abcd4567efgh8910ijkl']
I want to have something like,
['1234abcd','4567efgh','8910ijkl']
How can I do this? I'm pretty new to regex method, and would really appreciate it if anyone can show how to do this in different method within regex, and explain what's going on.
\w will match any wordcharacter which consists of numbers, alphabets and the underscore sign. You need to use [a-zA-Z] to capture letters only. See this example.
import re
a = '1253abcd4567efgh8910ijkl'
b = re.findall('(\d+[A-Za-z]+)',a)
Output:
['1253abcd', '4567efgh', '8910ijkl']
\d will match digits. \d+ will match one or more consecutive digits. For e.g.
>>> re.findall('(\d+)',a)
['1253', '4567', '8910']
Similarly [a-zA-Z]+ will match one or more alphabets.
>>> re.findall('([a-zA-Z]+)',a)
['abcd', 'efgh', 'ijkl']
Now put them together to match what you exactly want.
From the Python manual on regular expressions, it tells us that \w:
matches any alphanumeric character and the underscore; this is equivalent to the set [a-zA-Z0-9_]
So you are actually over capturing what you need. Refine your regular expression a bit:
>>> re.findall(r'(\d+[a-z]+)', a, re.I)
['1253abcd', '4567efgh', '8910ijkl']
The re.I makes your expression case insensitive, so it will match upper and lower case letters as well:
>>> re.findall(r'(\d+[a-z]+)', '12124adbad13434AGDFDF434348888AAA')
['12124adbad']
>>> re.findall(r'(\d+[a-z]+)', '12124adbad13434AGDFDF434348888AAA', re.I)
['12124adbad', '13434AGDFDF', '434348888AAA']
\w matches string with any alphanumeric character. And you have used \w with *. So your code will provide a string which is starting with a digit and contains alphanumeric characters of any length.
Solution:
>>>b=re.findall('\d*[A-Za-z]*', a)
>>>b
['1253abcd', '4567efgh', '8910ijkl', '']
you will get '' (an empty string) at the end of the list to display no match. You can remove it using
b.pop(-1)
I'm looking to find words in a string that match a specific pattern.
Problem is, if the words are part of an email address, they should be ignored.
To simplify, the pattern of the "proper words" \w+\.\w+ - one or more characters, an actual period, and another series of characters.
The sentence that causes problem, for example, is a.a b.b:c.c d.d#e.e.e.
The goal is to match only [a.a, b.b, c.c] . With most Regexes I build, e.e returns as well (because I use some word boundary match).
For example:
>>> re.findall(r"(?:^|\s|\W)(?<!#)(\w+\.\w+)(?!#)\b", "a.a b.b:c.c d.d#e.e.e")
['a.a', 'b.b', 'c.c', 'e.e']
How can I match only among words that do not contain "#"?
I would definitely clean it up first and simplify the regex.
first we have
words = re.split(r':|\s', "a.a b.b:c.c d.d#e.e.e")
then filter out the words that have an # in them.
words = [re.search(r'^((?!#).)*$', word) for word in words]
Properly parsing email addresses with a regex is extremely hard, but for your simplified case, with a simple definition of word ~ \w\.\w and the email ~ any sequence that contains #, you might find this regex to do what you need:
>>> re.findall(r"(?:^|[:\s]+)(\w+\.\w+)(?=[:\s]+|$)", "a.a b.b:c.c d.d#e.e.e")
['a.a', 'b.b', 'c.c']
The trick here is not to focus on what comes in the next or previous word, but on what the word currently captured has to look like.
Another trick is in properly defining word separators. Before the word we'll allow multiple whitespaces, : and string start, consuming those characters, but not capturing them. After the word we require almost the same (except string end, instead of start), but we do not consume those characters - we use a lookahead assertion.
You may match the email-like substrings with \S+#\S+\.\S+ and match and capture your pattern with (\w+\.\w+) in all other contexts. Use re.findall to only return captured values and filter out empty items (they will be in re.findall results when there is an email match):
import re
rx = r"\S+#\S+\.\S+|(\w+\.\w+)"
s = "a.a b.b:c.c d.d#e.e.e"
res = filter(None, re.findall(rx, s))
print(res)
# => ['a.a', 'b.b', 'c.c']
See the Python demo.
See the regex demo.
I have about 15,000 files I need to parse which could contain one or more strings/numbers from a list I have. I need to separate the files with matching strings.
Given a string: 3423423987, it could appear independently as "3423423987", or as "3423423987_1" or "3423423987_1a", "3423423987-1a", but it could also be "2133423423987". However, I only want to detect the matching sequence where it is not a part of another number, only when it has a suffix of some sort.
So 3423423987_1 is acceptable, but 13423423987 is not.
I'm having trouble with regex, haven't used it much to be honest.
Simply speaking, if I simulate this with a list of possible positives and negatives, I should get 7 hits, for the given list. I would like to extract the text till the end of the word, so that I can record that later.
Here's my code:
def check_text_for_string(text_to_parse, string_to_find):
import re
matches = []
pattern = r"%s_?[^0-9,a-z,A-Z]\W"%string_to_find
return re.findall(pattern, text_to_parse)
if __name__ =="__main__":
import re
word_to_match = "3423423987"
possible_word_list = [
"3423423987_1 the cake is a lie", #Match
"3423423987sdgg call me Ishmael", #Not a match
"3423423987 please sir, can I have some more?", #Match
"3423423987", #Match
"3423423987 ", #Match
"3423423987\t", #Match
"adsgsdzgxdzg adsgsdag\t3423423987\t", #Match
"1233423423987", #Not a match
"A3423423987", #Not a match
"3423423987-1a\t", #Match
"3423423987.0", #Not a match
"342342398743635645" #Not a match
]
print("%d words in sample list."%len(possible_word_list))
print("Only 7 should match.")
matches = check_text_for_string("\n".join(possible_word_list), word_to_match)
print("%d matched."%len(matches))
print(matches)
But clearly, this is wrong. Could someone help me out here?
It seems you just want to make sure the number is not matched as part of a, say, float number. You then need to use lookarounds, a lookbehind and a lookahead to disallow dots with digits before and after.
(?<!\d\.)(?:\b|_)3423423987(?:\b|_)(?!\.\d)
See the regex demo
To also match the "prefixes" (or, better call them "suffixes" here), you need to add something like \S* (zero or more non-whitespaces) or (?:[_-]\w+)? (an optional sequence of a - or _ followed with 1+ word chars) at the end of the pattern.
Details:
(?<!\d\.) - fail the match if we have a digit and a dot before the current position
(?:\b|_) - either a word boundary or a _ (we need it as _ is a word char)
3423423987 - the search string
(?:\b|_) - ibid
(?!\.\d) - fail the match if a dot + digit is right after the current position.
So, use
pattern = r"(?<!\d\.)(?:\b|_)%s(?:\b|_)(?!\.\d)"%string_to_find
See the Python demo
If there can be floats like Text with .3423423987 float value, you will need to also add another lookbehind (?<!\.) after the first one: (?<!\d\.)(?<!\.)(?:\b|_)3423423987(?:\b|_)(?!\.\d)
You can use this pattern:
(?:\b|^)3423423987(?!\.)(?=\b|_|$)
(?:\b|^) asserts that there are no other numbers to the left
(?!\.) asserts the number isn't followed by a dot
(?=\b|_|$) asserts the number is followed by a non word character, an underscore or nothing
I have a multiple regex which combines thousands of different regexes e.g r"reg1|reg2|...".
I'd like to know which one of the regexes gave a match in re.search(r"reg1|reg2|...", text), and I cannot figure how to do it since `re.search(r"reg1|reg2|...", text).re.pattern gives the whole regex.
For example, if my regex is r"foo[0-9]|bar", my pattern "foo1", I'd like to get as an answer "foo[0-9].
Is there any way to do this ?
Wrap each sub-regexp in (). After the match, you can go through all the groups in the matcher (match.group(index)). The non-empty group will be the one that matched.
You could put each possible regex into a list, then checking them in series, as this would be faster than one very large regex, and allow you to figure out which matched as you need to:
mystring = "Some string you're searching in."
regs = ['reg1', 'reg2', 'reg3', ...]
matching_reg = None
for reg in regs:
match = re.search(reg, mystring)
if match:
matching_reg = reg
break
After that, match and matching_reg will both be None if no match was found. If a match was found, match will contain the regex result and matching_reg will contain the regex search string from regs that matched.
Note that break is used to stop attempting to match as soon as a match is found.