In Google Sheets, I'd like to replace the following text:
The ANS consists of fibers that stimulate ((1smooth muscle)), ((2cardiac muscle)), and ((3glandular cells)).
with this text below:
The ANS consists of fibers that stimulate {{c1::smooth muscle}}, {{c2::cardiac muscle}}, and {{c3::glandular cells}}.
I know if I use =REGEXREPLACE(E3, "\(\([0-9]*", "{{c::") I can get here:
The ANS consists of fibers that stimulate {{c::smooth muscle)), {{c::cardiac muscle)), and {{c::glandular cells)).
BUT I don't know how to keep the original numbers
Nvm, figured it out.
Putting parentheses around the term allows you to reference it again in your replacement string.
For example, my solution for my problem was this:
=REGEXREPLACE(E3, "\(\(([0-9]*)", "{{c$1::")
This works because putting [0-9]* in parentheses like so: ([0-9]*) allowed it to be referenced as $1 in my substitution string.
I assume that if I had another phrase enclosed in parentheses after that it would be able to be referenced with $2.
Hope this helps someone in the future.
Pass 1:
Search Pattern:(((\d)(.+)
Replacement:{{C$1::$2
Pass 2:
Search Pattern: ))
Replacment: }}
I played around with it some more and this will do it in one pass:
Search Pattern: \(\((\d)(.+?)\b\)\)
Replacements: {{C$1::$2}}
Related
I'm using the following regex to find URLs in a text file:
/http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+/
It outputs the following:
http://rda.ucar.edu/datasets/ds117.0/.
http://rda.ucar.edu/datasets/ds111.1/.
http://www.discover-earth.org/index.html).
http://community.eosdis.nasa.gov/measures/).
Ideally they would print out this:
http://rda.ucar.edu/datasets/ds117.0/
http://rda.ucar.edu/datasets/ds111.1/
http://www.discover-earth.org/index.html
http://community.eosdis.nasa.gov/measures/
Any ideas on how I should tweak my regex?
Thank you in advance!
UPDATE - Example of the text would be:
this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/).
This will trim your output containing trail characters, ) .
import re
regx= re.compile(r'(?m)[\.\)]+$')
print(regx.sub('', your_output))
And this regex seems workable to extract URL from your original sample text.
https?:[\S]*\/(?:\w+(?:\.\w+)?)?
Demo,,, ( edited from https?:[\S]*\/)
Python script may be something like this
ss=""" this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/). """
regx= re.compile(r'https?:[\S]*\/(?:\w+(?:\.\w+)?)?')
for m in regx.findall(ss):
print(m)
So for the urls you have here:
https://regex101.com/r/uSlkcQ/4
Pattern explanation:
Protocols (e.g. https://)
^[A-Za-z]{3,9}:(?://)
Look for recurring .[-;:&=+\$,\w]+-class (www.sub.domain.com)
(?:[\-;:&=\+\$,\w]+\.?)+`
Look for recurring /[\-;:&=\+\$,\w\.]+ (/some.path/to/somewhere)
(?:\/[\-;:&=\+\$,\w\.]+)+
Now, for your special case: ensure that the last character is not a dot or a parenthesis, using negative lookahead
(?!\.|\)).
The full pattern is then
^[A-Za-z]{3,9}:(?://)(?:[\-;:&=\+\$,\w]+\.?)+(?:\/[\-;:&=\+\$,\w\.]+)+(?!\.|\)).
There are a few things to improve or change in your existing regex to allow this to work:
http[s]? can be changed to https?. They're identical. No use putting s in its own character class
[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),] You can shorten this entire thing and combine character classes instead of using | between them. This not only improves performance, but also allows you to combine certain ranges into existing character class tokens. Simplifying this, we get [a-zA-Z0-9$-_#.&+!*\(\),]
We can go one step further: a-zA-Z0-9_ is the same as \w. So we can replace those in the character class to get [\w$-#.&+!*\(\),]
In the original regex we have $-_. This creates a range so it actually inclues everything between $ and _ on the ASCII table. This will cause unwanted characters to be matched: $%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_. There are a few options to fix this:
[-\w$#.&+!*\(\),] Place - at the start of the character class
[\w$#.&+!*\(\),-] Place - at the end of the character class
[\w$\-#.&+!*\(\),] Escape - such that you have \- instead
You don't need to escape ( and ) in the character class: [\w$#.&+!*(),-]
[0-9a-fA-F][0-9a-fA-F] You don't need to specify [0-9a-fA-F] twice. Just use a quantifier like so: [0-9a-fA-F]{2}
(?:%[0-9a-fA-F][0-9a-fA-F]) The non-capture group isn't actually needed here, so we can drop it (it adds another step that the regex engine needs to perform, which is unnecessary)
So the result of just simplifying your existing regex is the following:
https?://(?:[$\w#.&+!*(),-]|%[0-9a-fA-F]{2})+
Now you'll notice it doesn't match / so we need to add that to the character class. Your regex was matching this originally because it has an improper range $-_.
https?://(?:[$\w#.&+!*(),/-]|%[0-9a-fA-F]{2})+
Unfortunately, even with this change, it'll still match ). at the end. That's because your regex isn't told to stop matching after /. Even implementing this will now cause it to not match file names like index.html. So a better solution is needed. If you give me a couple of days, I'm working on a fully functional RFC-compliant regex that matches URLs. I figured, in the meantime, I would at least explain why your regex isn't working as you'd expect it to.
Thanks all for the responses. A coworker ended up helping me with it. Here is the solution:
des_links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', des)
for i in des_links:
tmps = "/".join(i.split('/')[0:-1])
print(tmps)
I need to put together a regex that matches a patter only if string does not begin with 'N'.
Here is my pattern so far [A-E]+[-+]?.
Now I want to make sure that it does not match something like:
N\A
NA
NB+
NB-
NCAB
This is for REGEXP_SUBSTR command in Oracle SQL DB
UPDATE
It looks like I should have been more specific, sorry
I want to extract from a string [A-E]+[-+]? but if the string also matches ^(N|n) then I want my regex to return nothing.
See examples below:
String Returns
N/A
F1/AAA AAA
NABC
FABC ABC
To match a character between A and E not preceded by N, you can use:
([^N]|^)[A-E]+
If you want to avoid fields that contains N[A-E] use a negation in your query using the pattern N[A-E] (in other words, use two predicates, this one to exclude NA and the first to find A)
To be more clear:
WHERE NOT REGEXP_LIKE(coln, 'N[A-E]') AND REGEXP_LIKE(coln, '[A-E]')
Ok I figured it out, I broadened the scope of the problem a little, I realized that I can also play with other parameters of REGEXP_SUBSTR in this case that I can have returned only second substring.
REGEXP_SUBSTR(field1, '^([^NA-D][^A-D]*)?([A-D]+[-+]?)',1,1,'i',2)
I still have to give you guys the credit, lot of good ideas that led me to here.
Just throw a [^N]? in front. That should do it.
OOPS...
That actually needs to include an " OR ^ "...
It should look like this:
([^N]|^)[A-E]+[-+]?
Sorry about that...It looks like the right answer already got posted anyway.
I've searched around quite a bit now, but I can't get any suggestions to work in my situation. I've seen success with negative lookahead or lookaround, but I really don't understand it.
I wish to use RegExp to find URLs in blocks of text but ignore them when quoted. While not perfect yet I have the following to find URLs:
(https?\://)?(\w+\.)+\w{2,}(:[0-9])?\/?((/?\w+)+)?(\.\w+)?
I want it to match the following:
www.test.com:50/stuff
http://player.vimeo.com/video/63317960
odd.name.amazone.com/pizza
But not match:
"www.test.com:50/stuff
http://plAyerz.vimeo.com/video/63317960"
"odd.name.amazone.com/pizza"
Edit:
To clarify, I could be passing a full paragraph of text through the expression. Sample paragraph of what I'd like below:
I would like the following link to be found www.example.com. However this link should be ignored "www.example.com". It would be nice, but not required, to have "www.example.com and www.example.com" ignored as well.
A sample of a different one I have working below. language is php:
$articleEntry = "Hey guys! Check out this cool video on Vimeo: player.vimeo.com/video/63317960";
$pattern = array('/\n+/', '/(https?\:\/\/)?(player\.vimeo\.com\/video\/[0-9]+)/');
$replace = array('<br/><br/>',
'<iframe src="http://$2?color=40cc20" width="500" height="281" frameborder="0" webkitAllowFullScreen mozallowfullscreen allowFullScreen></iframe>');
$articleEntry = preg_replace($pattern,$replace,$articleEntry);
The result of the above will replace any new lines "\n" with a double break "" and will embed the Vimeo video by replacing the Vimeo address with an iframe and link.
I've found a solution!
(?=(([^"]+"){2})*[^"]*$)((https?:\/\/)?(\w+\.)+\w{2,}(:[0-9]+)?((\/\w+)+(\.\w+)?)?\/?)
The first part from (? to *$) what makes it work for me. I found this as an answer in java Regex - split but ignore text inside quotes? by https://stackoverflow.com/users/548225/anubhava
While I had read that question before, I had overlooked his answer because it wasn't the one that "solved" the question. I just changed the single quote to double quote and it works out for me.
add ^ and $ to your regex
^(https?\://)?(\w+\.)+\w{2,}(:[0-9])?\/?((/?\w+)+)?(\.\w+)?$
please notice you might need to escape the slashes after http (meaning https?\:\/\/)
update
if you want it to be case sensitive, you shouldn't use \w but [a-z]. the \w contains all letters and numbers, so you should be careful while using it.
I made an article spinner that used regex to find words in this syntax:
{word1|word2}
And then split them up at the "|", but I need a way to make it support tier 2 brackets, such as:
{{word1|word2}|{word3|word4}}
What my code does when presented with such a line, is take "{{word1|word2}" and "{word3|word4}", and this is not as intended.
What I want is when presented with such a line, my code breaks it up as "{word1|word2}|{word3|word4}", so that I can use this with the original function and break it into the actual words.
I am using c#.
Here is the pseudo code of how it might look like:
Check string for regex match to "{{word1|word2}|{word3|word4}}" pattern
If found, store each one as "{word1|word2}|{word3|word4}" in MatchCollection (mc1)
Split the word at the "|" but not the one inside the brackets, and select a random one (aka, "{word1|word2}" or "{word3|word4}")
Store the new results aka "{word1|word2}" and "{word3|word4}" in a new MatchCollection (mc2)
Now search the string again, this time looking for "{word1|word2}" only and ignore the double "{{" "}}"
Store these in mc2.
I can not split these up normally
Here is the regex I use to search for "{word1|word2}":
Regex regexObj = new Regex(#"\{.*?\}", RegexOptions.Singleline);
MatchCollection m = regexObj.Matches(originalText); //How I store them
Hopefully someone can help, thanks!
Edit: I solved this using a recursive method. I was building an article spinner btw.
That is not parsable using a regular expression, instead you have to use a recursive descent parser. Map it to JSON by replacing:
{ with [
| with ,
wordX with "wordX" (regex \w+)
Then your input
{{word1|word2}|{word3|word4}}
becomes valid JSON
[["word1","word2"],["word3","word4"]]
and will map directly to PHP arrays when you call json_decode.
In C#, the same should be possible with JavaScriptSerializer.
I'm really not completely sure WHAT you're asking for, but I'll give it a go:
If you want to get {word1|word2}|{word3|word4} out of any occurrence of {{word1|word2}|{word3|word4}} but not {word1|word2} or {word3|word4}, then use this:
#"\{(\{[^}]*\}\|\{[^}]*\})\}"
...which will match {{word1|word2}|{word3|word4}}, but with {word1|word2}|{word3|word4} in the first matching group.
I'm not sure if this will be helpful or even if it's along the right track, but I'll try to check back every once in a while for more questions or clarifications.
s = "{Spinning|Re-writing|Rotating|Content spinning|Rewriting|SEO Content Machine} is {fun|enjoyable|entertaining|exciting|enjoyment}! try it {for yourself|on your own|yourself|by yourself|for you} and {see how|observe how|observe} it {works|functions|operates|performs|is effective}."
print spin(s)
If you want to use the [square|brackets|syntax] use this line in the process function:
'/[(((?>[^[]]+)|(?R))*)]/x',
I'm trying to search a field in a database to extract URLs. Sometimes there will be more than 1 URL in a field and I would like to extract those in to separate variables (or an array).
I know my regex isn't going to cover all possibilities. As long as I flag on anything that starts with http and ends with a space I'm ok.
The problem I'm having is that my efforts either seem to get only 1 URL per record or they get only 1 the last letter from each URL. I've tried a couple different techniques based on solutions other have posted but I haven't found a solution that works for me.
Sample input line:
Testing http://marko.co http://tester.net Just about anything else you'd like.
Output goal
$var[0] = http://marko.co
$var[1] = http://tester.net
First try:
if ( $status =~ m/http:(\S)+/g ) {
print "$&\n";
}
Output:
http://marko.co
Second try:
#statusurls = ($status =~ m/http:(\S)+/g);
print "#statusurls\n";
Output:
o t
I'm new to regex, but since I'm using the same regex for each attempt, I don't understand why it's returning such different results.
Thanks for any help you can offer.
I've looked at these posts and either didn't find what I was looking for or didn't understand how to implement it:
This one seemed the most promising (and it's where I got the 2nd attempt from, but it didn't return the whole URL, just the letter: How can I store regex captures in an array in Perl?
This has some great stuff in it. I'm curious if I need to look at the URL as a word since it's bookended by spaces: Regex Group in Perl: how to capture elements into array from regex group that matches unknown number of/multiple/variable occurrences from a string?
This one offers similar suggestions as the first two. How can I store captures from a Perl regular expression into separate variables?
Solution:
#statusurls = ($status =~ m/(http:\S+)/g);
print "#statusurls\n";
Thanks!
I think that you need to capture more than just one character. Try this regex instead:
m/http:(\S+)/g