I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?
Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ from the start till the end of the string
[...] one of the given character
...{3} three time of the phrase before
(...)* 0 till n times of the characters in the brackets
What you're asking it to find with your regex is "at least one triple of letters a, b, c" - that's what "+" gives you. Whatever follows after that doesn't really matter to the regex. You might want to include "$", which means "end of the line", to be sure that the line must all consist of allowed triples. However in the current form your regex would also demand that the last triple ends in a comma, so you should explicitly code that it's not so.
Try this:
re.match('([abc][abc][abc],)*([abc][abc][abc])$'
This finds any number of allowed triples followed by a comma (maybe zero), then a triple without a comma, then the end of the line.
Edit: including the "^" (start of string) symbol is not necessary, because the match method already checks for a match only at the beginning of the string.
The obligatory "you don't need a regex" solution:
all(letter in 'abc,' for letter in data) and all(len(item) == 3 for item in data.split(','))
You need to iterate over sequence of found values.
data_string = "abc,bca,df"
imatch = re.finditer(r'(?P<value>[abc]{3})(,|$)', data_string)
for match in imatch:
print match.group('value')
So the regex to check if the string matches pattern will be
data_string = "abc,bca,df"
match = re.match(r'^([abc]{3}(,|$))+', data_string)
if match:
print "data string is correct"
Your result is not surprising since the regular expression
([abc][abc][abc],)+
tries to match a string containing three characters of [abc] followed by a comma one ore more times anywhere in the string. So the most important part is to make sure that there is nothing more in the string - as scessor suggests with adding ^ (start of string) and $ (end of string) to the regular expression.
An alternative without using regex (albeit a brute force way):
>>> def matcher(x):
total = ["".join(p) for p in itertools.product(('a','b','c'),repeat=3)]
for i in x.split(','):
if i not in total:
return False
return True
>>> matcher("abc,bca,aaa")
True
>>> matcher("abc,bca,xyz")
False
>>> matcher("abc,aaa,bb")
False
If your aim is to validate a string as being composed of triplet of letters a,b,and c:
for ss in ("abc,bbc,abb,baa,bbb",
"acc",
"abc,bbc,abb,bXa,bbb",
"abc,bbc,ab,baa,bbb"):
print ss,' ',bool(re.match('([abc]{3},?)+\Z',ss))
result
abc,bbc,abb,baa,bbb True
acc True
abc,bbc,abb,bXa,bbb False
abc,bbc,ab,baa,bbb False
\Z means: the end of the string. Its presence obliges the match to be until the very end of the string
By the way, I like the form of Sonya too, in a way it is clearer:
bool(re.match('([abc]{3},)*[abc]{3}\Z',ss))
To just repeat a sequence of patterns, you need to use a non-capturing group, a (?:...) like contruct, and apply a quantifier right after the closing parenthesis. The question mark and the colon after the opening parenthesis are the syntax that creates a non-capturing group (SO post).
For example:
(?:abc)+ matches strings like abc, abcabc, abcabcabc, etc.
(?:\d+\.){3} matches strings like 1.12.2., 000.00000.0., etc.
Here, you can use
^[abc]{3}(?:,[abc]{3})*$
^^
Note that using a capturing group is fraught with unwelcome effects in a lot of Python regex methods. See a classical issue described at re.findall behaves weird post, for example, where re.findall and all other regex methods using this function behind the scenes only return captured substrings if there is a capturing group in the pattern.
In Pandas, it is also important to use non-capturing groups when you just need to group a pattern sequence: Series.str.contains will complain that this pattern has match groups. To actually get the groups, use str.extract. and
the Series.str.extract, Series.str.extractall and Series.str.findall will behave as re.findall.
I am trying to capture parts of an equation using a regular expression.
Equation: 1×2÷3×4
Regular expression: \d+(×|÷)\d+
I expect this to result in:
1×2
2÷3
3×4
But it only returns:
1×2
3×4
I assume this has something to do with the structure, but I'm not even sure where to start or what to google to find the answer.
If your regex matches something then it will continue after that match so that's why you are getting only two matches. You can use (?=abc) positive lookahead to just see that if there is ([×÷]) and capture it and (\d) after the match.
You can use
/\d(?=([×÷])(\d))/g
The below code is specifically in Javascript
const regex = /\d(?=([×÷])(\d))/g;
const str = "1×2÷3×4";
const results = [...str.matchAll(regex)].map((arr) => {
return `${arr[0]}${arr[1]}${arr[2]}`;
});
console.log(results);
Each part of the string will only be matched to the pattern one time - once the substring "1x2" has matched the regular expression, the '2' won't be re-used in subsequent matches. Consider the string "×2÷3×4" (i.e. drop the first '1') - in this case the first (and only) match is "2÷3".
I want to find the pattern of a regular expression from a character string. My goal is to be able to reuse this pattern to find a string in another context but checking the pattern.
from sting "1example4whatitry2do",
I want to find pattern like: [0-9]{1}[a-z]{7}[0-9]{1}[a-z]{8}[0-9]{1}[a-z]{2}
So I can reuse this pattern to find this other example of sting 2eytmpxe8wsdtmdry1uo
I can do a loop on each caracter, but I hope there is a fast way
Thanks for your help !
You can puzzle this out:
go over your strings characterwise
if the character is a text character add a 't' to a list
if the character is a number add a 'd' to a list
if the character is something else, add itself to the list
Use itertools.groupby to group consecutive identical letters into groups.
Create a pattern from the group-key and the length of the group using some string literal formatting.
Code:
from itertools import groupby
from string import ascii_lowercase
lower_case = set(ascii_lowercase) # set for faster lookup
def find_regex(p):
cum = []
for c in p:
if c.isdigit():
cum.append("d")
elif c in lower_case:
cum.append("t")
else:
cum.append(c)
grp = groupby(cum)
return ''.join(f'\\{what}{{{how_many}}}'
if how_many>1 else f'\\{what}'
for what,how_many in ( (g[0],len(list(g[1]))) for g in grp))
pattern = "1example4...whatit.ry2do"
print(find_regex(pattern))
Output:
\d\t{7}\d\.{3}\t{6}\.\t{2}\d\t{2}
The ternary in the formatting removes not needed {1} from the pattern.
See:
str.isdigit()
If you now replace '\t'with '[a-z]' your regex should fit. You could also replace isdigit check using a regex r'\d' or a in set(string.digits) instead.
pattern = "1example4...whatit.ry2do"
pat = find_regex(pattern).replace(r"\t","[a-z]")
print(pat) # \d[a-z]{7}\d\.{3}[a-z]{6}\.[a-z]{2}\d[a-z]{2}
See
string module for ascii_lowercase and digits
I am trying to pull a certain number from various strings. The number has to be standalone, before ', or before (. The regex I came up with was:
\b(?<!\()(x)\b(,|\(|'|$) <- x is the numeric number.
If x is 2, this pulls the following string (almost) fine, except it also pulls 2'abd'. Any advice what I did wrong here?
2(2'Abf',3),212,2'abc',2(1,2'abd',3)
Your actual question is, as I understand it, get these specific number except those in parenthesis.
To do so I suggest using the skip_what_to_avoid|what_i_want pattern like this:
(\((?>[^()\\]++|\\.|(?1))*+\))
|\b(2)(?=\b(?:,|\(|'|$))
The idea here is to completely disregard the overall matches (and there first group use for the recursive pattern to capture everything between parenthesis: (\((?>[^()\\]++|\\.|(?1))*+\))): that's the trash bin. Instead, we only need to check capture group $2, which, when set, contains the asterisks outside of comments.
Demo
Sample Code:
import regex as re
regex = r"(\((?>[^()\\]++|\\.|(?1))*+\))|\b(2)(?=\b(?:,|\(|'|$))"
test_str = "2(2'Abf',3),212,2'abc',2(1,2'abd',3)"
matches = re.finditer(regex, test_str, re.MULTILINE)
for matchNum, match in enumerate(matches):
matchNum = matchNum + 1
if match.groups()[1] is not None:
print ("Found at {start}-{end}: {group}".format(start = match.start(2), end = match.end(2), group = match.group(2)))
Output:
Found at 0-1: 2
Found at 16-17: 2
Found at 23-24: 2
This solution requires the alternative Python regex package.
I've got two files:
1st: Entries.txt
confirmation.resend
send
confirmation.showResendForm
login.header
login.loginBtn
2nd: Used_Entries.txt
confirmation.showResendForm = some value
login.header = some other value
I want to find all entries from the first file (Entries.txt) that have not been asigned a value in the 2nd file (Used_Entries.txt)
In this example I'd like the following result:
confirmation.resend
send
login.loginBtn
In the result confirmation.showResendForm and login.header do not show up because these exist in the Used_Entries.txt
How do I do this? I've been playing around with regular expressions but haven't been able to solve it. A bash script or sth would be much appreciated!
You can do this with regex. But get your code mood ready, because you can't match both files with regex at once, and we do want to match both contents with regex at once. Well, that means you must have at least some understanding of your language, I would like you to concatenate the contents from the two files with at least a new line in between.
This regex solution expects your string to be matched to be in this format:
text (no equals sign)
text
text
...
key (no equals sign) ␣ (optional whitespace) = (literal equal) whatever (our regex will skip this part.)
key=whatever
key=whatever
Do I have your attention? Yes? Please see the following regex (using techniques accessible to most regex engines):
/(^[^=\n]+$)(?!(?s).*^\1\s*=)/m
Inspired from a recent answer I saw from zx81, you can switch to (?s) flag in the middle to switch to DOTALL mode suddenly, allowing you to start multiline matching with . in the middle of a RegExp. Using this technique and the set syntax above, here's what the regex does, as an explanation:
(^[^=\n]+$) Goes through all the text (no equals sign) elements. Enforces no equals signs or newlines in the capture. This means our regex hits every text element as a line, and tries to match it appropriately.
(?! Opens a negative lookahead group. Asserts that this match will not locate the following:
(?s).* Any number of characters or new lines - As this is a greedy match, will throw our matcher pointer to the very end of the string, skipping to the last parts of the document to backtrack and scoop up quickly.
^\1\s*= The captured key, followed by an equals sign after some optional whitespaces, in its own line.
) Ends our group.
View a Regex Demo!
A regex demo with more test cases
I'm stupid. I could had just put this:
/(^[^=\n]+$)(?!.*^\1\s*=)/sm
I've been going at this a little bit to complex and just solved it with a small script in scala:
import scala.io.Source
object HelloWorld {
def main(args: Array[String]) {
val entries = (for(line <- Source.fromFile("Entries.txt").getLines()) yield {
line
}).toList
val usedEntries = (for(line <- Source.fromFile("Used_Entries.txt").getLines()) yield {
line.dropRight(line.length - line.indexOf(' '))
}).toList
println(entries)
println(usedEntries)
val missingEntries = (for {
entry <- entries
if !usedEntries.exists(_ == entry)
} yield {
entry
}).toList
println(missingEntries)
println("Missing Entries: ")
println()
for {
missingEntry <- missingEntries
} yield {
println(missingEntry)
}
}
}
import re
e=open("Entries.txt",'r')
m=e.readlines()
u=open("Used_Entries.txt",'r')
s=u.read()
y=re.sub(r"= .*","",s)
for i in m:
if i.strip() in [k.strip() for k in y.split("\n")] :
pass
else:
print i.strip()