I am trying to extract words after the first space using
species<-gsub(".* ([A-Za-z]+)", "\1", x=genus)
This works fine for the other rows that have two words, however row [9] "Eulamprus tympanum marnieae" has 3 words and my code is only returning the last word in the string "marnieae". How can I extract the words after the first space so I can retrieve "tympanum marnieae" instead of "marnieae" but have the answers stored in one variable called >species.
genus
[9] "Eulamprus tympanum marnieae"
Your original pattern didn't work because the subpattern [A-Za-z]+ doesn't match spaces, and therefore will only match a single word.
You can use the following pattern to match any number of words (other than 0) after the first, within double quotes:
"[A-Za-z]+ ([A-Za-z ]+)" https://regex101.com/r/p6ET3I/1
https://regex101.com/r/p6ET3I/2
This is a relatively simple, but imperfect, solution. It will also match trailing spaces, or just one or more spaces after the first word even if a second word doesn't exist. "Eulamprus " for example will successfully match the pattern, and return 5 spaces. You should only use this pattern if you trust your data to be properly formatted.
A more reliable approach would be the following:
"[A-Za-z]+ ([A-Za-z]+(?: [A-Za-z]+)*)"
https://regex101.com/r/p6ET3I/3
This pattern will capture one word (following the first), followed by any number of addition words (including 0), separated by spaces.
However, from what I remember from biology class, species are only ever comprised of one or two names, and never capitalized. The following pattern will reflect this format:
"[A-Za-z]+ ([a-z]+(?: [a-z]+)?)"
https://regex101.com/r/p6ET3I/4
I would like to construct regular expression which will match password if there is no character repeating 4 or more times.
I have come up with regex which will match if there is character or group of characters repeating 4 times:
(?:([a-zA-Z\d]{1,})\1\1\1)
Is there any way how to match only if the string doesn't contain the repetitions? I tried the approach suggested in Regular expression to match a line that doesn't contain a word? as I thought some combination of positive/negative lookaheads will make it. But I haven't found working example yet.
By repetition I mean any number of characters anywhere in the string
Example - should not match
aaaaxbc
abababab
x14aaaabc
Example - should match
abcaxaxaz
(a is here 4 times but it is not problem, I want to filter out repeating patterns)
That link was very helpful, and I was able to use it to create the regular expression from your original expression.
^(?:(?!(?<char>[a-zA-Z\d]+)\k<char>{3,}).)+$
or
^(?:(?!([a-zA-Z\d]+)\1{3,}).)+$
Nota Bene: this solution doesn't answer exaactly to the question, it does too much relatively to the expressed need.
-----
In Python language:
import re
pat = '(?:(.)(?!.*?\\1.*?\\1.*?\\1.*\Z))+\Z'
regx = re.compile(pat)
for s in (':1*2-3=4#',
':1*1-3=4#5',
':1*1-1=4#5!6',
':1*1-1=1#',
':1*2-a=14#a~7&1{g}1'):
m = regx.match(s)
if m:
print m.group()
else:
print '--No match--'
result
:1*2-3=4#
:1*1-3=4#5
:1*1-1=4#5!6
--No match--
--No match--
It will give a lot of work to the regex motor because the principle of the pattern is that for each character of the string it runs through, it must verify that the current character isn't found three other times in the remaining sequence of characters that follow the current character.
But it works, apparently.
Beginner RegExp question. I have lines of JSON in a textfile, each with slightly different Fields, but there are 3 fields I want to extract for each line if it has it, ignoring everything else. How would I use a regex (in editpad or anywhere else) to do this?
Example:
"url":"http://www.netcharles.com/orwell/essays.htm",
"domain":"netcharles.com",
"title":"Orwell Essays & Journalism Section - Charles' George Orwell Links",
"tags":["orwell","writing","literature","journalism","essays","politics","essay","reference","language","toread"],
"index":2931,
"time_created":1345419323,
"num_saves":24
I want to extract URL,TITLE,TAGS,
/"(url|title|tags)":"((\\"|[^"])*)"/i
I think this is what you're asking for. I'll provide an explanation momentarily. This regular expression (delimited by / - you probably won't have to put those in editpad) matches:
"
A literal ".
(url|title|tags)
Any of the three literal strings "url", "title" or "tags" - in Regular Expressions, by default Parentheses are used to create groups, and the pipe character is used to alternate - like a logical 'or'. To match these literal characters, you'd have to escape them.
":"
Another literal string.
(
The beginning of another group. (Group 2)
(
Another group (3)
\\"
The literal string \" - you have to escape the backslash because otherwise it will be interpreted as escaping the next character, and you never know what that'll do.
|
or...
[^"]
Any single character except a double quote The brackets denote a Character Class/Set, or a list of characters to match. Any given class matches exactly one character in the string. Using a carat (^) at the beginning of a class negates it, causing the matcher to match anything that's not contained in the class.
)
End of group 3...
*
The asterisk causes the previous regular expression (in this case, group 3), to be repeated zero or more times, In this case causing the matcher to match anything that could be inside the double quotes of a JSON string.
)"
The end of group 2, and a literal ".
I've done a few non-obvious things here, that may come in handy:
Group 2 - when dereferenced using Backreferences - will be the actual string assigned to the field. This is useful when getting the actual value.
The i at the end of the expression makes it case insensitive.
Group 1 contains the name of the captured field.
EDIT: So I see that the tags are an array. I'll update the regular expression here in a second when I've had a chance to think about it.
Your new Regex is:
/"(url|title|tags)":("(\\"|[^"])*"|\[("(\\"|[^"])*"(,"(\\"|[^"])*")*)?\])/i
All I've done here is alternate the string regular expression I had been using ("((\\"|[^"])*)"), with a regular expression for finding arrays (\[("(\\"|[^"])*"(,"(\\"|[^"])*")*)?\]). No so easy to Read, is it? Well, substituting our String Regex out for the letter S, we can rewrite it as:
\[(S(,S)*)?\]
Which matches a literal opening bracket (hence the backslashes), optionally followed by a comma separated list of strings, and a closing bracket. The only new concept I've introduced here is the question mark (?), which is itself a type of repetition. Commonly referred to as 'making the previous expression optional', it can also be thought of as exactly 0 or 1 matches.
With our same S Notation, here's the whole dirty Regular Expression:
/"(url|title|tags)":(S|\[(S(,S)*)?\])/i
If it helps to see it in action, here's a view of it in action.
This question is a bit older, but I have had browsed a bit on my PC and found that expression. I passed him as GIST, could be useful to others.
EDIT:
# Expression was tested with PHP and Ruby
# This regular expression finds a key-value pair in JSON formatted strings
# Match 1: Key
# Match 2: Value
# https://regex101.com/r/zR2vU9/4
# http://rubular.com/r/KpF3suIL10
(?:\"|\')(?<key>[^"]*)(?:\"|\')(?=:)(?:\:\s*)(?:\"|\')?(?<value>true|false|[0-9a-zA-Z\+\-\,\.\$]*)
# test document
[
{
"_id": "56af331efbeca6240c61b2ca",
"index": 120000,
"guid": "bedb2018-c017-429E-b520-696ea3666692",
"isActive": false,
"balance": "$2,202,350",
"object": {
"name": "am",
"lastname": "lang"
}
}
]
the json string you'd like to extract field value from
{"fid":"321","otherAttribute":"value"}
the following regex expression extract exactly the "fid" field value "321"
(?<=\"fid\":\")[^\"]*
Please try below expression:
/"(url|title|tags)":("([^""]+)"|\[[^[]+])/gm
Explanation:
1st Capturing Group (url|title|tags): This is alternatively capturing the characters 'url','title' and 'tags' literally (case sensitive).
2nd Capturing Group ("([^""]+)"|[[^[]+]):
1st Alternative "([^""]+)" is matches all words within " and " including " and "
2nd Alternative [[^[]+] is matches all words within [ and ] including [ and ]
I have tested here
I adapted regex to work with JSON in my own library. I've detailed algorithm behavior below.
First, stringify the JSON object. Then, you need to store the starts and lengths of the matched substrings. For example:
"matched".search("ch") // yields 3
For a JSON string, this works exactly the same (unless you are searching explicitly for commas and curly brackets in which case I'd recommend some prior transform of your JSON object before performing regex (i.e. think :, {, }).
Next, you need to reconstruct the JSON object. The algorithm I authored does this by detecting JSON syntax by recursively going backwards from the match index. For instance, the pseudo code might look as follows:
find the next key preceding the match index, call this theKey
then find the number of all occurrences of this key preceding theKey, call this theNumber
using the number of occurrences of all keys with same name as theKey up to position of theKey, traverse the object until keys named theKey has been discovered theNumber times
return this object called parentChain
With this information, it is possible to use regex to filter a JSON object to return the key, the value, and the parent object chain.
You can see the library and code I authored at http://json.spiritway.co/
if your json is
{"key1":"abc","key2":"xyz"}
then below regex will extract key1 or key2 based on a key that you pass in regex
"key2(.*?)(?=,|}|$)
you can verify it here - regex101.com
Why does it have to be a Regular Expression object?
Here we can just use a Hash object first and then go search it.
mh = {"url":"http://www.netcharles.com/orwell/essays.htm","domain":"netcharles.com","title":"Orwell Essays & Journalism Section - Charles' George Orwell Links","tags":["orwell","writing","literature","journalism","essays","politics","essay","reference","language","toread"],"index":2931,"time_created":1345419323,"num_saves":24}
The output of which would be
=> {:url=>"http://www.netcharles.com/orwell/essays.htm", :domain=>"netcharles.com", :title=>"Orwell Essays & Journalism Section - Charles' George Orwell Links", :tags=>["orwell", "writing", "literature", "journalism", "essays", "politics", "essay", "reference", "language", "toread"], :index=>2931, :time_created=>1345419323, :num_saves=>24}
Not that I want to avoid using Regexp but don't you think it would be easier to take it a step at a time until your getting the data you want to further search through? Just MHO.
mh.values_at(:url, :title, :tags)
The output:
["http://www.netcharles.com/orwell/essays.htm", "Orwell Essays & Journalism Section - Charles' George Orwell Links", ["orwell", "writing", "literature", "journalism", "essays", "politics", "essay", "reference", "language", "toread"]]
Taking the pattern that FrankieTheKneeman gave you:
pattern = /"(url|title|tags)":"((\\"|[^"])*)"/i
we can search the mh hash by converting it to a json object.
/#{pattern}/.match(mh.to_json)
The output:
=> #<MatchData "\"url\":\"http://www.netcharles.com/orwell/essays.htm\"" 1:"url" 2:"http://www.netcharles.com/orwell/essays.htm" 3:"m">
Of course this is all done in Ruby which is not a tag that you have but relates I hope.
But oops! Looks like we can't do all three at once with that pattern so I will do them one at a time just for sake.
pattern = /"(title)":"((\\"|[^"])*)"/i
/#{pattern}/.match(mh.to_json)
#<MatchData "\"title\":\"Orwell Essays & Journalism Section - Charles' George Orwell Links\"" 1:"title" 2:"Orwell Essays & Journalism Section - Charles' George Orwell Links" 3:"s">
pattern = /"(tags)":"((\\"|[^"])*)"/i
/#{pattern}/.match(mh.to_json)
=> nil
Sorry about that last one. It will have to be handled differently.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Extract info inside all parenthesis in R (regex)
I inported data from excel and one cell consists of these long strings that contain number and letters, is there a way to extract only the numbers from that string and store it in a new variable? Unfortunately, some of the entries have two sets of brackets and I would only want the second one? Could I use grep for that?
the strings look more or less like this, the length of the strings vary however:
"East Kootenay C (5901035) RDA 01011"
or like this:
"Thompson-Nicola J (Copper Desert Country) (5933039) RDA 02020"
All I want from this is 5901035 and 5933039
Any hints and help would be greatly appreciated.
There are many possible regular expressions to do this. Here is one:
x=c("East Kootenay C (5901035) RDA 01011","Thompson-Nicola J (Copper Desert Country) (5933039) RDA 02020")
> gsub('.+\\(([0-9]+)\\).+?$', '\\1', x)
[1] "5901035" "5933039"
Lets break down the syntax of that first expression '.+\\(([0-9]+)\\).+'
.+ one or more of anything
\\( parentheses are special characters in a regular expression, so if I want to represent the actual thing ( I need to escape it with a \. I have to escape it again for R (hence the two \s).
([0-9]+) I mentioned special characters, here I use two. the first is the parentheses which indicate a group I want to keep. The second [ and ] surround groups of things. see ?regex for more information.
?$ The final piece assures that I am grabbing the LAST set of numbers in parens as noted in the comments.
I could also use * instead of . which would mean 0 or more rather than one or more i in case your paren string comes at the beginning or end of a string.
The second piece of the gsub is what I am replacing the first portion with. I used: \\1. This says use group 1 (the stuff inside the ( ) from above. I need to escape it twice again, once for the regex and once for R.
Clear as mud to be sure! Enjoy your data munging project!
Here is a gsubfn solution:
library(gsubfn)
strapplyc(x, "[(](\\d+)[)]", simplify = TRUE)
[(] matches an open paren, (\\d+) matches a string of digits creating a back-reference owing to the parens around it and finally [)] matches a close paren. The back-reference is returned.
I saw other questions dealing with the finding the n-th occurrence of a word/pattern, but I couldn't find how you would actually substitute the n-th occurrence of a pattern in vim. There's the obvious way of hard coding all the occurrences like
:s/.*\(word\).*\(word\).*\(word\).*/.*\1.*\2.*newWord.*/g
Is there a better way of doing this?
For information,
s/\%(\(pattern\).\{-}\)\{41}\zs\1/2/
also works to replace 42th occurrence. However, I prefer the solution given by John Kugelman which is more simple -- even if it will not limit itself to the current line.
You can do this a little more simply by using multiple searches. The empty pattern in the :s/pattern/repl/ command means replace the most recent search result.
:/word//word//word/ s//newWord/
or
:/word//word/ s/word/newWord/
You could then repeat this multiple times by doing #:, or even 10#: to repeat the command 10 more times.
Alternatively, if I were doing this interactively I would do something like:
3/word
:s//newWord/r
That would find the third occurrence of word starting at the cursor and then perform a substitution.
Replace each Nth occurrence of PATTERN in a line with REPLACE.
:%s/\(\zsPATTERN.\{-}\)\{N}/REPLACE/
To replace the nth occurrence of PATTERN in a line in vim, in addtion to the above answer I just wanted to explain the pattern matching i.e how it is actually working for easy understanding.
So I will be discussing the \(.\{-}\zsPATTERN\)\{N} solution,
The example I will be using is replacing the second occurrence of more than 1 space in a sentence(string).
According to the pattern match code->
According to the zs doc,
\zs - Scroll the text horizontally to position the cursor at the start (left
side) of the screen.
.\{-} 0 or more as few as possible (*)
Here . is matching any character and {} the number of times.
e.g ab{2,3}c here it will match where b comes either 2 or 3 times.
In this case, we can also use .* which is 0 or many as many possible.
According to vim non-greedy docs, "{-}" is the same as "*" but uses the shortest match first algorithm.
\{N} -> Matches n of the preceding atom
/\<\d\{4}\> search for exactly 4 digits, same as /\<\d\d\d\d>
**ignore these \<\> they are for exact searching, like search for fred -> \<fred\> will only search fred not alfred.
\( \) combining the whole pattern.
PATTERN here is your pattern you are matching -> \s\{1,} (\s - space and {1,} as explained just above, search for 1 or more space)
"abc subtring def"
:%s/\(.\{-}\zs\s\{1,}\)\{2}/,/
OUTPUT -> "abc subtring,def"
# explanation: first space would be between abc and substring and second
# occurence of the pattern would be between substring and def, hence that
# will be replaced by the "," as specified in replace command above.
This answers your actual question, but not your intent.
You asked about replacing the nth occurrence of a word (but seemed to mean "within a line"). Here's an answer for the question as asked, in case someone finds it like I did =)
For weird tasks (like needing to replace every 12th occurrence of "dog" with "parrot"), I like to use recursive recordings.
First blank the recording in #q
qqq
Now start a new recording in q
qq
Next, manually do the thing you want to do (using the example above, replace the 12th occurrence of "dog" with "parrot"):
/dog
nnnnnnnnnnn
delete "dog" and get into insert
diwi
type parrot
parrot
Now play your currently empty "#q" recording
#q
which does nothing.
Finally, stop recording:
q
Now your recording in #q calls itself at the end. But because it calls the recording by name, it won't be empty anymore. So, call the recording:
#q
It will replay the recording, then at the end, as the last step, replay itself again. It will repeat this until the end of the file.
TLDR;
qq
q
/dog
nnnnnnnnnnndiwiparrot<esc>
#q
q
#q
Well, if you do /gc then you can count the number of times it asks you for confirmation, and go ahead with the replacement when you get to the nth :D