I have a bunch of input files in a loop and I am extracting tag from them. However, I want to separate some of the words. The incoming strings are in the form cs### where ### => is any number from 0-9. I want the result to be cs ###. The closest answer I found was this, Regex to separate Numeric from Alpha . But I cannot get this to work, as the string is being predefined (Static) and mine changes.
Found answer:
Nevermind, I found the answer the following sperates alpha-numeric characters and removes any unwanted non-alphanumeric characters so anything like ab5#6$% =>ab 56
gsub(/(?<=[0-9])(?=[a-z])|(?<=[a-z])(?=[0-9])/i, ' ').gsub(/[^0-9a-z ]/i, ' ')
If your string is something like
str = "cs3232
cs23
cs423"
Then you can do something like
str.scan(/((cs)(\d{1,10}))/m).collect{|e| e.shift; e }
# [["cs", "3232"], ["cs", "23"], ["cs", "423"]]
Related
Using a regex, how can I match strings that end with exactly one . as:
This is a string.
but not those that end with more than one . as:
This is a string...
I have a regex that detects a single .:
/[\.]{1}\z/
but I do not want it to match strings that end in ....
What you want is a 'negative lookbehind' assertion:
(?<!\.)\.\z
This looks for a period at the end of a string that isn't preceded by a period. The other answers won't match the following string: "."
Also, you may need to look out for unicode ellipsis characters…
You can detect this like so: str =~ /\u{2026}/
You can use:
[^\.][\.]\z
You are looking for a string that before the last dot there is a char that is not a dot.
I like Regexr a lot!
Solution similar to Dekel:
[^.]+[.]
Live demo
I need to replace all special characters within a string with their index.
For example,
"I-need_to#change$all%special^characters^"
should become:
"I1need6to9change16all20special28characters39"
The index of all special character differs.
I have checked many links replacing all with single character, occurances of a character.
I found very similar link but it I do not want to adopt these replace its index number as I need to replace all of the special characters.
I have also tried to do something like this:
str.gsub!(/[^0-9A-Za-z]/, '')
Here str is my example string.
As this replaces all the characters but with space, and I want the index instead of space. Either all of the special character or these seven
\/*[]:?
I need to replace this seven mainly but it would be OK if we replace all of them.
I need a simpler way.
Thanks in advance.
You can use the global variable $` and the block form of gsub:
irb> str = "I-need_to#change$all%special^characters^"
=> "I-need_to#change$all%special^characters^"
irb> str.gsub(/[^0-9A-Za-z]/) { $`.length }
=> "I1need6to9change16all20special28characters39"
Suppose I have the following string:
mystring = "start/abc123/end"
How can you splice out the abc123 with something else, while leaving the "/start/" and "/end" elements intact?
I had the following to match for the pattern, but it replaces the entire string. I was hoping to just have it replace the abc123 with 123abc.
mystring.gsub(/start\/(.*)\/end/,"123abc") #=> "123abc"
Edit: The characters between the start & end elements can be any combination of alphanumeric characters, I changed my example to reflect this.
You can do it using this character class : [^\/] (all that is not a slash) and lookarounds
mystring.gsub(/(?<=start\/)[^\/]+(?=\/end)/,"7")
For your example, you could perhaps use:
mystring.gsub(/\/(.*?)\//,"/7/")
This will match the two slashes between the string you're replacing and putting them back in the substitution.
Alternatively, you could capture the pieces of the string you want to keep and interpolate them around your replacement, this turns out to be much more readable than lookaheads/lookbehinds:
irb(main):010:0> mystring.gsub(/(start)\/.*\/(end)/, "\\1/7/\\2")
=> "start/7/end"
\\1 and \\2 here refer to the numbered captures inside of your regular expression.
The problem is that you're replacing the entire matched string, "start/8/end", with "7". You need to include the matched characters you want to persist:
mystring.gsub(/start\/(.*)\/end/, "start/7/end")
Alternatively, just match the digits:
mystring.gsub(/\d+/, "7")
You can do this by grouping the start and end elements in the regular expression and then referring to these groups in in the substitution string:
mystring.gsub(/(?<start>start\/).*(?<end>\/end)/, "\\<start>7\\<end>")
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]
Edit: I solved this by using strip! to remove leading and trailing whitespaces as I show in this video. Then, I followed up by restoring the white space at the end of each string the array by iterating through and adding whitespace. This problem varies from the "dupe" as my intent is to keep the whitespace at the end. However, strip! will remove both the leading and trailing whitespace if that is your intent. (I would have made this an answer, but as this is incorrectly marked as a dupe, I could only edit my original question to include this.)
I have an array of words where I am trying to remove any whitespace that may exist at the beginning of the word instead of at the end. rstrip! just takes care of the end of a string. I want whitespaces removed from the beginning of a string.
example_array = ['peanut', ' butter', 'sammiches']
desired_output = ['peanut', 'butter', 'sammiches']
As you can see, not all elements in the array have the whitespace problem, so I can't just delete the first character as I would if all elements started with a single whitespace char.
Full code:
words = params[:word].gsub("\n", ",").delete("\r").split(",")
words.delete_if {|x| x == ""}
words.each do |e|
e.lstrip!
end
Sample text that a user may enter on the form:
Corn on the cob,
Fibonacci,
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String#lstrip (or String#lstrip!) is what you're after.
desired_output = example_array.map(&:lstrip)
More comments about your code:
delete_if {|x| x == ""} can be replaced with delete_if(&:empty?)
Except you want reject! because delete_if will only return a different array, rather than modify the existing one.
words.each {|e| e.lstrip!} can be replaced with words.each(&:lstrip!)
delete("\r") should be redundant if you're reading a windows-style text document on a Windows machine, or a Unix-style document on a Unix machine
split(",") can be replaced with split(", ") or split(/, */) (or /, ?/ if there should be at most one space)
So now it looks like:
words = params[:word].gsub("\n", ",").split(/, ?/)
words.reject!(&:empty?)
words.each(&:lstrip!)
I'd be able to give more advice if you had the sample text available.
Edit: Ok, here goes:
temp_array = text.split("\n").map do |line|
fields = line.split(/, */)
non_empty_fields = fields.reject(&:empty?)
end
temp_array.flatten(1)
The methods used are String#split, Enumerable#map, Enumerable#reject and Array#flatten.
Ruby also has libraries for parsing comma seperated files, but I think they're a little different between 1.8 and 1.9.
> ' string '.lstrip.chop
=> "string"
Strips both white spaces...