I am looking for a pattern that matches everything until the first occurrence of a specific character, say a ";" - a semicolon.
I wrote this:
/^(.*);/
But it actually matches everything (including the semicolon) until the last occurrence of a semicolon.
You need
/^[^;]*/
The [^;] is a character class, it matches everything but a semicolon.
^ (start of line anchor) is added to the beginning of the regex so only the first match on each line is captured. This may or may not be required, depending on whether possible subsequent matches are desired.
To cite the perlre manpage:
You can specify a character class, by enclosing a list of characters in [] , which will match any character from the list. If the first character after the "[" is "^", the class matches any character not in the list.
This should work in most regex dialects.
Would;
/^(.*?);/
work?
The ? is a lazy operator, so the regex grabs as little as possible before matching the ;.
/^[^;]*/
The [^;] says match anything except a semicolon. The square brackets are a set matching operator, it's essentially, match any character in this set of characters, the ^ at the start makes it an inverse match, so match anything not in this set.
None of the proposed answers did work for me. (e.g. in notepad++)
But
^.*?(?=\;)
did.
Try /[^;]*/
Google regex character classes for details.
sample text:
"this is a test sentence; to prove this regex; that is g;iven below"
If for example we have the sample text above, the regex /(.*?\;)/ will give you everything until the first occurence of semicolon (;), including the semicolon: "this is a test sentence;"
Try /[^;]*/
That's a negating character class.
This was very helpful for me as I was trying to figure out how to match all the characters in an xml tag including attributes. I was running into the "matches everything to the end" problem with:
/<simpleChoice.*>/
but was able to resolve the issue with:
/<simpleChoice[^>]*>/
after reading this post. Thanks all.
this is not a regex solution, but something simple enough for your problem description. Just split your string and get the first item from your array.
$str = "match everything until first ; blah ; blah end ";
$s = explode(";",$str,2);
print $s[0];
output
$ php test.php
match everything until first
This will match up to the first occurrence only in each string and will ignore subsequent occurrences.
/^([^;]*);*/
"/^([^\/]*)\/$/" worked for me, to get only top "folders" from an array like:
a/ <- this
a/b/
c/ <- this
c/d/
/d/e/
f/ <- this
Really kinda sad that no one has given you the correct answer....
In regex, ? makes it non greedy. By default regex will match as much as it can (greedy)
Simply add a ? and it will be non-greedy and match as little as possible!
Good luck, hope that helps.
This works for getting the content from the beginning of a line till the first word,
/^.*?([^\s]+)/gm
I faced a similar problem including all the characters until the first comma after the word entity_id. The solution that worked was this in Bigquery:
SELECT regexp_extract(line_items,r'entity_id*[^,]*')
Related
I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!
Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet
If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.
I'm currently tryng to split text especially where there is no space after the '.'(dot). The df is a csv file.
My current code (don't mind the spaces please) :
for i in df['blurb']:
try:
df.loc[i,'blurb'] = df.loc[i,'blurb'].replace('.A', '.\nA')
except:
pass
...
try:
df.loc[i,'blurb'] = df.loc[i,'blurb'].replace('.Z', '.\nZ')
except:
pass
and this for every letter from the alphabet, since i' looking to put a /n [nem line] after every such dots.
the result is the same as the original (it does not want to save over the original). If i create another column, [blurb2], it gives the same outcome as the original blurb column. I've already looked for a few hours on this site for answers , but nothing seems to work [no error messages though] ... This is driving me crazy...
Anyone have any tips ? Thanks a mill in advance !
Cheers
To insert a newline after a dot that has a non-whitespace after it you can use
df['blurb'] = df['blurb'].str.replace(r'\.(?=\S)', '\\g<0>\n')
Note here
\. - matches a dot char that is followed with
(?=\S) - any char other than whitespace. Since it is a regex lookahead, its pattern is only checked for, but does not get consumed.
The \g<0> is the whole match value matched by the regex.
I'm trying to catch the last part after the last backslash
I need the \Web_ERP_Assistant (with the \)
My idea was :
C:\Projects\Ensure_Solution\Assistance\App_WebReferences\Web_ERP_WebService\Web_ERP_Assistant
\\.+?(?!\\) // I know there is something with negative look -ahead `(?!\\)`
But I can't find it.
[Regexer Demo]
Your negative lookahead solution would e.g. be this:
\\(?:.(?!\\))+$
See it here on Regexr
One that worked for me was:
.+(\\.+)$
Try it online!
Explanation:
.+ - any character except newline
( - create a group
\\.+ - match a backslash, and any characters after it
) - end group
$ - this all has to happen at the end of the string
A negative look ahead is a correct answer, but it can be written more cleanly like:
(\\)(?!.*\\)
This looks for an occurrence of \ and then in a check that does not get matched, it looks for any number of characters followed by the character you don't want to see after it. Because it's negative, it only matches if it does not find a match.
You can try anchoring it to the end of the string, something like \\[^\\]*$. Though I'm not sure if one absolutely has to use regexp for the task.
What about this regex: \\[^\\]+$
If you don't want to include the backslash, but only the text after it, try this: ([^\\]+)$ or for unix: ([^\/]+)$
I used below regex to get that result also when its finished by a \
(\\[^\\]+)\\?$
[Regex Demo]
Could someone help me on regex to match German words/sentences in
python? It does not work on jupyter notebook. I tried same in jsfiddle
it works fine. I tried using this below script but does not work
import re
pattern = re.compile(r'\[^a-zA-Z0-9äöüÄÖÜß]\\', re.UNICODE)
print(pattern.search(text))
Your expression will always fail:
\[^a-zA-Z0-9äöüÄÖÜß]\\
Broken down, you require
[ # literally
^ # start of the line / text
a-z # literally, etc.
The problem is that you require a [ literally right before the start of a line which can never be true (either there's nothing or a newline). So in the end, either remove the backslash to get a proper character class as in:
[^a-zA-Z0-9äöüÄÖÜß]+
But this will surely not match the words you're looking for (quite the opposite). So either use something as simple as \w+ or the solution proposed by #Wiktor in the comments section.
The square brackets define a range of characters you want to look for, however the '^' negates these characters if it appears within the character class.
If you want to specify the beginning of the line you need to put the '^' before the brackets.
Also you need to add a multiplier behind the class to search for more than just one character in this case:
r'^[a-zA-Z0-9äöüÄÖÜß]+'
One ore more characters contained in the brackets are matched as long as they are not seperated by any other character not listed between '[]'
Here's the link to the official documentation
I'm developing a calculator program in Python, and need to remove leading zeros from numbers so that calculations work as expected. For example, if the user enters "02+03" into the calculator, the result should return 5. In order to remove these leading zeroes in-front of digits, I asked a question on here and got the following answer.
self.answer = eval(re.sub(r"((?<=^)|(?<=[^\.\d]))0+(\d+)", r"\1\2", self.equation.get()))
I fully understand how the positive lookbehind to the beginning of the string and lookbehind to the non digit, non period character works. What I'm confused about is where in this regex code can I find the replacement for the matched patterns?
I found this online when researching regex expressions.
result = re.sub(pattern, repl, string, count=0, flags=0)
Where is the "repl" in the regex code above? If possible, could somebody please help to explain what the r"\1\2" is used for in this regex also?
Thanks for your help! :)
The "repl" part of the regex is this component:
r"\1\2"
In the "find" part of the regex, group capturing is taking place (ordinarily indicated by "()" characters around content, although this can be overridden by specific arguments).
In python regex, the syntax used to indicate a reference to a positional captured group (sometimes called a "backreference") is "\n" (where "n" is a digit refering to the position of the group in the "find" part of the regex).
So, this regex is returning a string in which the overall content is being replaced specifically by parts of the input string matched by numbered groups.
Note: I don't believe the "\1" part of the "repl" is actually required. I think:
r"\2"
...would work just as well.
Further reading: https://www.regular-expressions.info/brackets.html
Firstly, repl includes what you are about to replace.
To understand \1\2 you need to know what capture grouping is.
Check this video out for basics of Group capturing.
Here , since your regex splits every match it finds into groups which are 1,2... so on. This is so because of the parenthesis () you have placed in the regex.
$1 , $2 or \1,\2 can be used to refer to them.
In this case: The regex is replacing all numbers after the leading 0 (which is caught by group 2) with itself.
Note: \1 is not necessary. works fine without it.
See example:
>>> import re
>>> s='awd232frr2cr23'
>>> re.sub('\d',' ',s)
'awd frr cr '
>>>
Explanation:
As it is, '\d' is for integer so removes them and replaces with repl (in this case ' ').