I'm looking to find the first whole occurance of a number within a string. I'm not looking for the first digit, rather the whole first number. So, for example, the first number in: w134fklj342 is 134, while the first number in 1235alkj9342klja9034 is 1235.
I have attempted to use \d but I'm unsure how to expand that to include multiple digits (without specifying how long the number is).
I think, you're looking for this regex
\d+
"Plus" means "one or more". This regex will match all numbers within a string, so pick first one.
strings = ['w134fklj342', '1235alkj9342klja9034']
strings.each do |s|
puts s[/\d+/]
end
# >> 134
# >> 1235
Demo: http://rubular.com/r/YE8kPE2SyW
The easiest way to understand regexes is to think of eachbit is one character; e.g: \d or [1234567890] or [0-9] will match one digit.
To expand this one character you have 2 basic options: * and +
* will match the character 0 or more times
+ will match it one or more times
Like Sergio said you should use \d+ to match many digits.
Excellent tutorial for regexes in general: http://www.regular-expressions.info/tutorial.html
Related
I have a special use case where I want to discard all the contractions from the string and select only words followed by alphabets which do not contain any special character.
For eg:
string = "~ ASAP ASCII Achilles Ada Stackoverflow James I'd I'll I'm I've"
string.scan(/\b[A-z][a-z]+\b/)
#=> ["Achilles", "Ada", "Stackoverflow", "James", "ll", "ve"]
Note: It's not discarding the whole word I'll and I've
Can someone please help how to discard the whole word which contains contractions?
Try this Regex:
(?:(?<=\s)|(?<=^))[a-zA-Z]+(?=\s|$)
Explanation:
(?:(?<=\s)|(?<=^)) - finds the position immediately preceded by either start of the line or by a white-space
[a-zA-Z]+ - matches 1+ occurrences of a letter
(?=\s|$) - The substring matched above must be followed by either a whitespace or end of the line
Click for Demo
Update:
To make sure that not all the letters are in upper case, use the following regex:
(?:(?<=\s)|(?<=^))(?=\S*[a-z])[a-zA-Z]+(?=\s|$)
Click for Demo
The only thing added here is (?=\S*[a-z]) which means that there must be atleast one lowercase letter
I know that there's an accepted answer already, but I'd like to give my own shot:
(?<=\s|^)\w+[a-z]\w*
You can test it here. This regex is shorter and more efficient (157 steps against 315 from the accepted answer).
The explanation is rather simple:
(?<=\s|^)- This is a positive look behind. It means that we want strings preceded by a whitespace character or the start of the string.
\w+[a-z]\w* - This one means that we want strings composed by letters only (word characters) containing least one lowercase letter, thus discarding words which are whole uppercase. Along with the positive look behind, the whole regex ends up discarding words containing special characters.
NOTE: this regex won't take into account one-letter words. If you want to accomplish that, then you should use \w*[a-z]\w* instead, with a little efficiency cost.
I have the following text:
"Showing1-30\nof 1404results"
What I want to pull out is the 1404.
How do I do that?
I was thinking I would use a regexp to match just the string between the words of and results, but can't quite figure out how to do that.
Or is there another way, say a built-in Ruby method I could use that is efficient?
I was also considering using split, but the spacing is off so it looks like this:
=> ["Showing1-30", "of", "1404results"]
How do I do what I want?
You could just do
["Showing1-30", "of", "1404results"].last.to_i
Or use a regex like
/of (\d+)results/
Match "of" followed by one or more spaces, followed by one or more digits in capture group 1, followed by "results", then retrieve the contents of capture group 1.
"Showing1-30\nof 1404results"[/of\s+(\d+)results/,1]
#=> "1404"
or
Match the string that is preceded by "of" followed by one1 space (positive lookbehind) and is followed by "results" (positive lookahead)
"Showing1-30\nof 1404results"[/(?<=of\s)\d+(?=results)/]
#=> "1404"
or
Match "of" followed by one or more spaces, forget everything matched so far (\K), match one or more digits followed by "results" (positive lookahead)
"Showing1-30\nof 1404results"[/of\s+\K\d+(?=results)/]
#=> "1404"
It may be desirable to change the first regex to
/(?<=of\s)\d+\s*(?=results)/
in case someone decides to "correct" the string to read "Showing 1-30\nof 1404 results"[/(?<=of\s)\d+(?=results)/]. (Same for the other two.)
1 Ruby's positive lookbehinds cannot be variable length; hence, \s+ is not permitted here.
I'd use:
"Showing1-30\nof 1404results"[/(\d+)results/, 1] # => "1404"
"Showing1-30\nof 1404results" is not overly readable. If you are in charge of generating it, or if it is likely to change to something more readable, such as "Showing 1-30\nof 1404 results", then a simple tweak will help:
"Showing1-30\nof 1404results"[/(\d+)\s*results/, 1] # => "1404"
where \s* will allow 0, 1 or multiple whitespace characters.
Keep regular expressions as simple as possible until it's proven they need to be more complex. As complexity increases the odds of slowing the match increases which, in a loop, can be drastic with long strings. Also, the odds of adding a hole that leads to false positives goes up too, which can be hard to debug.
If the position of this number is fixed, the the following is the fastest
"Showing1-30\nof 1404results"[-12..-8]
The [-12..-8] is a range, you can see the string as an array of characters and specify the characters between the 8th and the 12th position counting from the right, -1 is the end of the line, -2 the last character etc..
In not, then a regular expression like
"Showing1-30\nof 14results"[/ \d+/].strip
You look for a space followed by a number, then you remove the leading space.
This is simpler than having to use a capture group.
I need to match a string of variable length(between 5 and 12), composed of uppercase letters and one or more digits between 1 and 8.
How can I specify that I need the whole captured group's length to be between 5 and 12?
I have tried with parenthesis but with no luck.
I have tried this
\s([A-Z]+[1-8]+[A-Z]+){5,12}\s
My idea was to use the quantifier {5,12} to limit the length of the captured group between parenthesis, but clearly it doesn't work like that.
The string needs to be identified inside a normal text just like
"THE STRING I NEED TO DECODE IS SOMETHING LIKE FD1531FHHKWF BUT NOT LIKE g4G58234JJ"
You actually have two conditions to met:
The length of the match is to be specified with curly brackets {5,12}, and before and after there should be not letters/digits. So:
/(?!\b[A-Z]+\b)\b[A-Z1-8]{5,12}\b/
First, we assure that the lookahead for letters only is negative, then we look for the pattern.
Use positive look-ahead on total size of regex
\s(?=^.{5,12}$)([A-Z]+[1-8]+[A-Z]+)\s
Explanation
(?= # look-ahead match start
^.{5,12}$ # 3 to 15 characters from start to end
) # look-ahead match end
I'm working to create a ruby regex that meets the following conditions:
Supported:
A-Z, a-z, 0-9, dashes in the middle but never starting or ending in a dash.
At least 5, no more than 500 characters
So far I have:
[0-9a-z]{5,500}
Any suggestions on how to update to meet the criteria above?
Thanks
[A-Za-z\d][-A-Za-z\d]{3,498}[A-Za-z\d]
If you are willing to treat _ as a letter also, it's even simpler:
\w[-\w]{3,498}\w
This should work:
[0-9A-Za-z][0-9A-Za-z-]{3,498}[0-9A-Za-z]
Here you go:
/^[0-9A-Za-z][0-9A-Za-z\-]{3,498}[0-9A-Za-z]$/
or if you want the beginning and end to be only 0-9,A-Z,a-z (instead of non dash) then:
Explanation:
The first ^ matches beginning of string.
The next [] matches a A-Z,a-z,0-9
The next [] matches 3 to 498 chars of A-Z,a-z,0-9,dash. Note that we match 3 to 498 chars because we match one char in the beginning and one in the end.
The next [^] is again a A-Z,a-z,0-9.
And lastly we match $ for the end of the string.
This assumes that there are either always dashes or never dashes. It also assumes only one dash is allowed between alphanumeric characters. It's the only way I can think off hand to limit characters instead of number of instances of the string.
(([0-9a-zA-Z]{4,499})|([0-9a-zA-Z][\d]?){2,249})[0-9a-zA-Z]
Assuming there's no limit to the number of adjacent dashes allowed, this would work:
[0-9a-zA-Z][0-9a-zA-Z\d]{3,498}[0-9a-zA-Z]
I need to filter all lines with words starting with a letter followed by zero or more letters or numbers, but no special characters (basically names which could be used for c++ variable).
egrep '^[a-zA-Z][a-zA-Z0-9]*'
This works fine for words such as "a", "ab10", but it also includes words like "b.b". I understand that * at the end of expression is problem. If I replace * with + (one or more) it skips the words which contain one letter only, so it doesn't help.
EDIT:
I should be more precise. I want to find lines with any number of possible words as described above. Here is an example:
int = 5;
cout << "hello";
//some comments
In that case it should print all of the lines above as they all include at least one word which fits the described conditions, and line does not have to began with letter.
Your solution will look roughly like this example. In this case, the regex requires that the "word" be preceded by space or start-of-line and then followed by space or end-of-line. You will need to modify the boundary requirements (the parenthesized stuff) as needed.
'(^| )[a-zA-Z][a-zA-Z0-9]*( |$)'
Assuming the line ends after the word:
'^[a-zA-Z][a-zA-Z0-9]+|^[a-zA-Z]$'
You have to add something to it. It might be that the rest of it can be white spaces or you can just append the end of line.(AFAIR it was $ )
Your problem lies in the ^ and $ anchors that match the start and end of the line respectively. You want the line to match if it does contain a word, getting rid of the anchors does what you want:
egrep '[a-zA-Z][a-zA-Z0-9]+'
Note the + matches words of length 2 and higher, a * in that place would signel chars too.