I need to split a string that has a character, that is repeated but only split the last 3 instances and keep the 0 to any number of instances before those 3 instances of the character.
ex1: "Hi####there" after regex (and splitting) "Hi#","there"
ex2: "Hi###there" after regex (and splitting) "Hi","there"
Splitting on #{3} does not give the result I want: "Hi","#there"
edit:
I simplified my quesiton a bit; serialization and the character I am using is not relevant. I'm just interested in the regex that will get me the last x number in a list of y number of a repetitive list of characters. (#{3})(\w+) results in only returning hi# and the rest of the string is lost.
This is done in Javascript, but works
"Hi####there".split(/###(?=[^#])/); // Output = > ["Hi#", "there"]
"Hi###there".split(/###(?=[^#])/); // Output = > ["Hi", "there"]
// Tested on win7 with Chrome 45+
(?=[^#]) ... Lookahead to check if there is no # after the final/splitter ###
Related
Here is a sample string.
"BLAH, blah, going to the store &^5, light Version 12.7(2)L6, anyway
plus other stuff Version 3.3.4.6. Then goes on an on for several lines..."
I want to capture only the first version number without including the word version if possible but not include the periods and parenthesis. The result would stop when it encounters a comma. The result would be:
"1272L6"
I don't want it to include other instances of version in the text. Can this be done?
I've tried (?<=version)[^,]* I know it does not address removing the periods and parens and does not address the subsequent versions.
This exact RegEx, maybe not the best solution, but it might help you to get 1272L6:
([0-9]{2})\.([0-9]{1})\(([0-9]{1})\)([A-Z]{1}[0-9]{1})
It creates four groups (where $1$2$3$4 is your target 1272L6) and passes ., ) and (.
You might change {1} to other numbers of repetitions, such as {1,2}.
Assuming the version number is fixed on format but not on the specific digits or letters, you could do this.
String s = "this is a test 12.7(2)L6, 13.7(2)L6, 14.7(2)L6";
String reg = "(\\d\\d\\.\\d\\(\\d\\)[A-Z]\\d),";
Matcher m = Pattern.compile(reg).matcher(s);
if (m.find()) { // should only find first one
System.out.println(m.group(1).replaceAll("[.()]", ""));
}
I'm using Ruby 2.1. I have this logic that looks for consecutive pairs of strings in a bigger string
results = line.scan(/\b((\S+?)\b.*?\b(\S+?))\b/)
My question is, how do I iterate over the list of results and print out whether there are three or more characters between the two strings? For instance if my string were
"abc def"
The above would produce
[["abc def", "abc", "def"]]
and I'd like to know whether there are three or more characters between "abc" and "def."
Use a quantifier for the spaces inbetween: \b((\S+?)\b\s{3,}\b(\S+?))\b
Also, the inner boundries are not really needed:
\b((\S+?)\s{3,}(\S+?))\b
A straightforward way to check this is by running a separate regex:
results.select!{|x|p x[/\S+?\b(.*?)\b\S+?/,1].size}
will print the size for every of the bunch.
Another way is to take the size of the captured groups and subtract them:
results = []
line.scan(/\b((\S+?)\b.*?\b(\S+?))\b/) do |s, group1, group2|
results << $~ if s.size - group1.size - group2.size >= 3
end
I have a string, and I want to extract, using regular expressions, groups of characters that are between the character : and the other character /.
typically, here is a string example I'm getting:
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
and so, I want to retrieved, 45.72643,4.91203 and also hereanotherdata
As they are both between characters : and /.
I tried with this syntax in a easier string where there is only 1 time the pattern,
[tt]=regexp(str,':(\w.*)/','match')
tt = ':45.72643,4.91203/'
but it works only if the pattern happens once. If I use it in string containing multiples times the pattern, I get all the string between the first : and the last /.
How can I mention that the pattern will occur multiple time, and how can I retrieve it?
Use lookaround and a lazy quantifier:
regexp(str, '(?<=:).+?(?=/)', 'match')
Example (Matlab R2016b):
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = regexp(str, '(?<=:).+?(?=/)', 'match')
result =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
In most languages this is hard to do with a single regexp. Ultimately you'll only ever get back the one string, and you want to get back multiple strings.
I've never used Matlab, so it may be possible in that language, but based on other languages, this is how I'd approach it...
I can't give you the exact code, but a search indicates that in Matlab there is a function called strsplit, example...
C = strsplit(data,':')
That should will break your original string up into an array of strings, using the ":" as the break point. You can then ignore the first array index (as it contains text before a ":"), loop the rest of the array and regexp to extract everything that comes before a "/".
So for instance...
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
Breaks down into an array with parts...
1 - 'abcd'
2 - '45.72643,4.91203/Rou'
3 - 'hereanotherdata/defgh'
Then Ignore 1, and extract everything before the "/" in 2 and 3.
As John Mawer and Adriaan mentioned, strsplit is a good place to start with. You can use it for both ':' and '/', but then you will not be able to determine where each of them started. If you do it with strsplit twice, you can know where the ':' starts :
A='abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
B=cellfun(#(x) strsplit(x,'/'),strsplit(A,':'),'uniformoutput',0);
Now B has cells that start with ':', and has two cells in each cell that contain '/' also. You can extract it with checking where B has more than one cell, and take the first of each of them:
C=cellfun(#(x) x{1},B(cellfun('length',B)>1),'uniformoutput',0)
C =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
Starting in 16b you can use extractBetween:
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = extractBetween(str,':','/')
result =
2×1 cell array
{'45.72643,4.91203'}
{'hereanotherdata' }
If all your text elements have the same number of delimiters this can be vectorized too.
This is a follow up to A regex to detect periodic strings .
A period p of a string w is any positive integer p such that w[i]=w[i+p]
whenever both sides of this equation are defined. Let per(w) denote
the size of the smallest period of w . We say that a string w is
periodic iff per(w) <= |w|/2.
So informally a periodic string is just a string that is made up from a another string repeated at least once. The only complication is that at the end of the string we don't require a full copy of the repeated string as long as it is repeated in its entirety at least once.
For, example consider the string x = abcab. per(abcab) = 3 as x[1] = x[1+3] = a, x[2]=x[2+3] = b and there is no smaller period. The string abcab is therefore not periodic. However, the string ababa is periodic as per(ababa) = 2.
As more examples, abcabca, ababababa and abcabcabc are also periodic.
#horcruz, amongst others, gave a very nice regex to recognize a periodic string. It is
\b(\w*)(\w+\1)\2+\b
I would like to find all maximal periodic substrings in a longer string. These are sometimes called runs in the literature.
Formally a substring w is a maximal periodic substring if it is periodic and neither w[i-1] = w[i-1+p] nor w[j+1] = w[j+1-p]. Informally, the "run" cannot be contained in a larger "run"
with the same period.
The four maximal periodic substrings (runs) of string T = atattatt are T[4,5] = tt, T[7,8] = tt, T[1,4] = atat, T[2,8] = tattatt.
The string T = aabaabaaaacaacac contains the following 7 maximal periodic substrings (runs):
T[1,2] = aa, T[4,5] = aa, T[7,10] = aaaa, T[12,13] = aa, T[13,16] = acac, T[1,8] = aabaabaa, T[9,15] = aacaaca.
The string T = atatbatatb contains the following three runs. They are:
T[1, 4] = atat, T[6, 9] = atat and T[1, 10] = atatbatatb.
Is there a regex (with backreferences) that will capture all maximal
periodic substrings?
I don't really mind which flavor of regex but if it makes a difference, anything that the Python module re supports. However I would even be happy with PCRE if that makes the problem solvable.
(This question is partly copied from https://codegolf.stackexchange.com/questions/84592/compute-the-maximum-number-of-runs-possible-for-as-large-a-string-as-possible . )
Let's extend the regex version to the very powerful https://pypi.python.org/pypi/regex . This supports variable length lookbehinds for example.
This should do it, using Python's re module:
(?<=(.))(?=((\w*)(\w*(?!\1)\w\3)\4+))
Fiddle: https://regex101.com/r/aA9uJ0/2
Notes:
You must precede the string being scanned by a dummy character; the # in the fiddle. If that is a problem, it should be possible to work around it in the regex.
Get captured group 2 from each match to get the collection of maximal periodic substrings.
Haven't tried it with longer strings; performance may be an issue.
Explanation:
(?<=(.)) - look-behind to the character preceding the maximal periodic substring; captured as group 1
(?=...) - look-ahead, to ensure overlapping patterns are matched; see How to find overlapping matches with a regexp?
(...) - captures the maximal periodic substring (group 2)
(\w*)(\w*...\w\3)\4+ - #horcruz's regex, as proposed by OP
(?!\1) - negative look-ahead to group 1 to ensure the periodic substring is maximal
As pointed out by #ClasG, the result of my regex may be incomplete. This happens when two runs start at the same offset. Examples:
aabaab has 3 runs: aabaab, aa and aa. The first two runs start at the same offset. My regex will fail to return the shortest one.
atatbatatb has 3 runs: atatbatatb, atat, atat. Same problem here; my regex will only return the first and third run.
This may well be impossible to solve within the regex. As far as I know, there is no regex engine that is capable of returning two different matches that start at the same offset.
I see two possible solutions:
Ignore the missing runs. If I am not mistaken, then they are always duplicates; an identical run will follow within the same encapsulating run.
Do some postprocessing on the result. For every run found (let's call this X), scan earlier runs trying to find one that starts with the same character sequence (let's call this Y). When found (and not already 'used'), add an entry with the same character sequence as X, but the offset of Y.
I think it is not possible. Regular expressions cannot do complex nondeterministic jobs, even with backreferences. You need an algorithm for this.
This kind of depends on your input criteria... There is no infinite string of characters.. using back references you will be able to create a suitable representation of the last amount of occurrences of the pattern you wish to match.
\
Personally I would define buckets of length of input and then fill them.
I would then use automata to find patterns in the buckets and then finally coalesce them into larger patterns.
It's not how fast the RegEx is going to be in this case it's how fast you are going to be able to recognize a pattern and eliminate the invalid criterion.
I'm trying to produce a regular expression that can identify a number within an interval in a string in VBA. Sometimes this number has characters around it, other times not (non-consistent notation from a supplier). The expression should identify that 1413 in the three examples below are within the number range 500-2000 (or alternatively that it's not in the number range 0-50 or 51-499).
Example:
Test 12/2014. Tot.flow:1413 m3 or
Test 12/2014. Tot.flow:1413m3 or
Test 12/2014. Tot.flow: 1413
These strings have some identifiers:
there will always be a colon before the number
there may be a white space between the colon and the number
there may be a white space between the number and the m3
m3 is not necessarily always present, and if not, the number is at the end of the string
So far what I have in my attempt to make an regex that find the number range is ([5-9][0-9][0-9]|[1]\d{3}|2000), but this matches all three digit numbers as well (2001 gives a match on 200). However, I understand that I'm missing out on a couple of concepts to achieve the ultimate goal here. I guess my problems are as following:
How to start the interval at something not being zero (found lots of questions on intervals starting on zero)
How to take into account the variations in notation both for flow: and m3?
I'm only interested in checking that the number lies within the number range. This is driving me bonkers, all help is highly appreciated!
You can just extract the number with regExp.Replace() using the following regex:
^.*:\s*(\d+).*$
The replacement part is $1.
Then, use usual number comparison to check whether the value is in the expected range (e.g. If CLng(result) > 499 And If CLng(result) < 2001 Then ...).
Test macro:
Dim re As RegExp, tgt As String, src As String
Set re = New RegExp
With re
.pattern = "^.*:\s*(\d+).*$"
.Global = False
End With
src = "Test 12/2014. Tot.flow: 1413"
tgt = re.Replace(src, "$1")
MsgBox (CLng(tgt) > 499 And CLng(tgt) < 2001)
You can try with:
:\s?([5-9]\d\d|1\d{3}|2000)\s?(m3|\n)
also, your regex ([5-9][0-9][0-9]|[1]\d{3}|2000) in my opinion is fine, it should not match numbers >500 and 2000<.