I'm attempting to construct a regex string in Swift 4 that gets characters at the start of a line where some are known and others aren't.
Let's say I've got a text file with line breaks for each word that reads as follows:
pucker
tuckered
duckerdinger
sucker punch
I'd like to get every word that contains "cker" in it that's 1 to 8 characters long.
I'm attempting to use this statement ^..cker..{1,8} as my RegEx string. All I'm getting is a partial match in Patterns (a Mac App), but Regex101.com's saying no match, and most importantly, Xcode says I'm using an invalid regex. I've also tried ^(..cker..) and a bazillion other variations.
What am I screwing up and how do I fix it? What I'm trying to do seems like it would be super simple, but I've wasted more time than I care to admit fiddling with it.
Update:
This has been the best I've been able to get so far...
"\\b..cker..", but I'm only able to get words that are exactly 8 characters long. I'd like to capture words that contain "cker" that are the 3rd, 4th, 5th, and 6th letters while capturing words up to 8 characters long.
Try this regex:
\b(?=.*cker)[a-zA-Z]{1,8}\b
Click for Demo
Explanation:
\b - matches a word boundary
(?=.*cker) - Positive Lookahead to make sure our string should contain the character sequence cker
[a-zA-Z]{1,8} - Matches 1 to 8 occurrences of a letter
\b - matches a word boundary
Related
I've been looking for a good way to see if a string of items are all numbers, and thought there might be a way of specifying a range from 0 to 9 and seeing if they're included in the string, but all that I've looked up online has really confused me.
def validate_pin(pin)
(pin.length == 4 || pin.length == 6) && pin.count("0-9") == pin.length
end
The code above is someone else's work and I've been trying to identify how it works. It's a pin checker - takes in a set of characters and ensures the string is either 4 or 6 digits and all numbers - but how does the range work?
When I did this problem I tried to use to_a? Integer and a bunch of other things including ranges such as (0..9) and ("0..9) and ("0".."9") to validate a character is an integer. When I saw ("0-9) it confused the heck out of me, and half an hour of googling and youtube has only left me with regex tutorials (which I'm interested in, but currently just trying to get the basics down)
So to sum this up, my goal is to understand a more semantic/concise way to identify if a character is an integer. Whatever is the simplest way. All and any feedback is welcome. I am a new rubyist and trying to get down my fundamentals. Thank You.
Regex really is the right way to do this. It's specifically for testing patterns in strings. This is how you'd test "do all characters in this string fall in the range of characters 0-9?":
pin.match(/\A[0-9]+\z/)
This regex says "Does this string start and end with at least one of the characters 0-9, with nothing else in between?" - the \A and \z are start-of-string and end-of-string matchers, and the [0-9]+ matches any one or more of any character in that range.
You could even do your entire check in one line of regex:
pin.match(/\A([0-9]{4}|[0-9]{6})\z/)
Which says "Does this string consist of the characters 0-9 repeated exactly 4 times, or the characters 0-9, repeated exactly 6 times?"
Ruby's String#count method does something similar to this, though it just counts the number of occurrences of the characters passed, and it uses something similar to regex ranges to allow you to specify character ranges.
The sequence c1-c2 means all characters between c1 and c2.
Thus, it expands the parameter "0-9" into the list of characters "0123456789", and then it tests how many of the characters in the string match that list of characters.
This will work to verify that a certain number of numbers exist in the string, and the length checks let you implicitly test that no other characters exist in the string. However, regexes let you assert that directly, by ensuring that the whole string matches a given pattern, including length constraints.
Count everything non-digit in pin and check if this count is zero:
pin.count("^0-9").zero?
Since you seem to be looking for answers outside regex and since Chris already spelled out how the count method was being implemented in the example above, I'll try to add one more idea for testing whether a string is an Integer or not:
pin.to_i.to_s == pin
What we're doing is converting the string to an integer, converting that result back to a string, and then testing to see if anything changed during the process. If the result is =>true, then you know nothing changed during the conversion to an integer and therefore the string is only an Integer.
EDIT:
The example above only works if the entire string is an Integer and won’t properly deal with leading zeros. If you want to check to make sure each and every character is an Integer then do something like this instead:
pin.prepend(“1”).to_i.to_s(1..-1) == pin
Part of the question seems to be exactly HOW the following portion of code is doing its job:
pin.count("0-9")
This piece of the code is simply returning a count of how many instances of the numbers 0 through 9 exist in the string. That's only one piece of the relevant section of code though. You need to look at the rest of the line to make sense of it:
pin.count("0-9") == pin.length
The first part counts how many instances then the second part compares that to the length of the string. If they are equal (==) then that means every character in the string is an Integer.
Sometimes negation can be used to advantage:
!pin.match?(/\D/) && [4,6].include?(pin.length)
pin.match?(/\D/) returns true if the string contains a character other than a digit (matching /\D/), in which case it it would be negated to false.
One advantage of using negation here is that if the string contains a character other than a digit pin.match?(/\D/) would return true as soon as a non-digit is found, as opposed to methods that examine all the characters in the string.
I have a word list of over 10,000 words, but this is just a sample:
'Tis midnight
sev'n words spoke
th'Immortal night
A wonder-working pow'r
Wondrous deliv'rer to me
I want to delete all words that contain apostrophes so the list should look like this:
midnight
words spoke
night
A wonder-working
Wondrous to me
How can I do this using Sublime Text so it finds apostrophes and smart apostrophes (’)?
You could use a character class['’] to match both variations of the apostrophes and match zero or more times a non-whitespace character \S* before or after the matched apostrophe followed by optional horizontal white-space chars.
\S*['’]\S*\h*
Regex demo
A slightly more optimized version without preventing the first \S* causing backtracking could be using a negated character class [^\s'’]* to match until the first apostrophe.
[^\s'’]*['’]\S*\h*
Regex demo
I have some strings with a sentence and i need to subdivise it into a substring of maximum 40 characters.
But i don't want to split the sentence in the middle of a word.
I tried with .gsub function but it's return 40 characters maximum and avoid to cut the string in the middle of a word. But it's return only the first occurence.
sentence[0..40].gsub(/\s\w+$/,'')
I tried with split but i can select only the fist 40 characters and split in the middle of a word...
sentence.split(...){40}
My string is "Sure, we will show ourselves only when we know the east door has been opened.".
The string output i want is
["Sure, we will show ourselves only when we","know the east door has
been opened."]
Do you have a solution ? Thanks
Your first attempt:
sentence[0..40].gsub(/\s\w+$/,'')
almost works, but it has one fatal flaw. You are splitting on the number of characters before cutting off the last word. This means you have no way of knowing whether the bit being trimmed off was a whole word, or a partial word.
Because of this, your code will always cut off the last word.
I would solve the problem as follows:
sentence[/\A.{0,39}[a-z]\b/mi]
\A is an anchor to fix the regex to the start of the string.
.{0,39}[a-z] matches on 1 to 40 characters, where the last character must be a letter. This is to prevent the last selected character from being punctuation or space. (Is that desired behaviour? Your question didn't really specify. Feel free to tweak/remove that [a-z] part, e.g. [a-z.] to match a full stop, if desired.)
\b is a word boundary look-around. It is a zero-width matcher, on beginning/end of words.
/mi modifiers will include case insensitive (i.e. A-Z) and multi-line matches.
One very minor note is that because this regex is matching 1 to 40 characters (rather than zero), it is possible to get a null result. (Although this is seemingly very unlikely, since you'd need a 1-word, 41+ letter string!!) To account for this edge case, call .to_s on the result if needed.
Update: Thank you for the improved edit to your question, providing a concrete example of an input/result. This makes it much clearer what you are asking for, as the original post was somewhat ambiguous.
You could solve this with something like the following:
sentence.scan(/.{0,39}[a-z.!?,;](?:\b|$)/mi)
String#scan returns an array of strings that match the pattern - so you can then re-join these strings to reconstruct the original.
Again, I have added a few more characters (!?,;) to the list of "final characters in the substring". Feel free to tweak this as desired.
(?:\b|$) means "either a word boundary, or the end of the line". This fixes the issue of the result not including the final . in the substrings. Note that I have used a non-capture group (?:) to prevent the result of scan from changing.
I have been coding a program in Lua that automatically formats IRC logs from a roleplay. In the roleplay logs there is a specific guideline for "Out of character" conversation, which we use double parentheses for. For example: ((<Things unrelated to roleplay go here>)). I have been trying to have my program remove text between double brackets (and including both brackets). The code is:
ofile = io.open("Output.txt", "w")
rfile = io.open("Input.txt", "r")
p = rfile:read("*all")
w = string.gsub(p, "%(%(.*?%)%)", "")
ofile:write(w)
The pattern here is > "%(%(.*?%)%)" I've tried multiple variations of the pattern. All resulted in fruitless results:
1. %(%(.*?%)%) --Wouldn't do anything.
2. %(%(.*%)%) --Would remove *everything* after the first OOC message.
Then, my friend told me that prepending the brackets with percentages wouldn't work, and that I had to use backslashes to 'escape' the parentheses.
3. \(\(.*\)\) --resulted in the output file being completely empty.
4. (\(\(.*\)\)) --Same result as above.
5. (\(\(.*?\)\) --would for some reason, remove large parts of the text for no apparent reason.
6. \(\(.*?\)\) --would just remove all the text except for the last line.
The short, absolute question:
What pattern would I need to use to remove all text between double parentheses, and remove the double parentheses themselves too?
You're friend is thinking of regular expressions. Lua patterns are similar, but different. % is the correct escape character.
Your pattern should be %(%(.-%)%). The - is similar to * in that it matches any number of the preceding sequence, but while * tries to match as many characters as it can (it's greedy), - matches the least amount of characters possible (it's non-greedy). It won't go overboard and match extra double-close-parenthesis.
I can't figure out what does this regex match:
A: "\\/\\/c\\/(\\d*)"
B: "\\/\\/(\\d*)"
I suppose they are matching some kind of number sequence since \d matches any digit but I'd like to know an example of a string that would be a match for this regex.
The pattern syntax is that specified by ICU. Expressions are created with NSRegularExpression in an iOS app and are correct.
The first matches //c/ + 0 or more digits. The second matches // + 0 or more digits. In both the digits are captured.
An example of a match for A) is //c/123
An example of a match for B) is //12345
When I use Cygwin which emulates Bash on Windows, I sometimes run into situations where I have to escape my escape characters which is what I think is making this expression look so weird. For instance, when I use sed to look for a single '\' I sometimes have to write it as '\\\\'. (Funny, StackOverflow proved my point. If you write 4 backslashes in the comment, it only shows two. So if you process it again, they might all disappear depending on your situation).
Considering this, it might be helpful to think of pairs of backslashes as representing only one if you're coming from a similar situation. My guess would be you are. Because of this I would say Erik Duymelinck is probably spot on. This will capture a sequence of digits that may or may not follow a couple slashes and a c:
//c/000
//00000
This regex matches an odd sequence of characters, which, at first glance, almost seem like a regex, since \d is a digit, and followed by an asterisk (\d*) would mean zero-or-more digits. But it's not a digit, because the escape-slash is escaped.
\\/\\/c\\/(\\d*)
So, for instance, this one matches the following text:
\/\/c\/\
\/\/c\/\d
\/\/c\/\dd
\/\/c\/\ddd
\/\/c\/\dddd
\/\/c\/\ddddd
\/\/c\/\dddddd
...
This one is almost the same
\\/\\/(\\d*)
except you just delete the c\/ from the above results:
\/\/\
\/\/\d
\/\/\dd
\/\/\ddd
\/\/\dddd
\/\/\ddddd
\/\/\dddddd
...
In both cases, the final \ and optional d is [capture group][1] one.
My first impression was that these regexes were intended for escaping in Java strings, meaning they would be completely invalid. If the were escaped for Java strings, such as
Pattern p = Pattern.compile("\\/\\/c\\/(\\d*)");
It would be invalid, because after un-escaping, it would result in this invalid regex:
\/\/c\/(\d*)
The single escape-slashes (\) are invalid. But the \d is valid, as it would mean any digit.
But again, I don't think they're invalid, and they're not escaped for a Java string. They're just odd.