Take for instance, I have a string like this:
options = "Cake or pie, ice cream, or pudding"
I want to be able to split the string via or, ,, and , or.
The thing is, is that I have been able to do it, but only by parsing , and , or first, and then splitting each array item at or, flattening the resultant array afterwards as such:
options = options.split(/(?:\s?or\s)*([^,]+)(?:,\s*)*/).reject(&:empty?);
options.each_index {|index| options[index] = options[index].sub("?","").split(" or "); }
The resultant array is as such: ["Cake", "pie", "ice cream", "pudding"]
Is there a more efficient (or easier) way to split my string on those three delimiters?
What about the following:
options.gsub(/ or /i, ",").split(",").map(&:strip).reject(&:empty?)
replaces all delimiters but the ,
splits it at ,
trims each characters, since stuff like ice cream with a leading space might be left
removes all blank strings
First of all, your method could be simplified a bit with Array#flatten:
>> options.split(',').map{|x|x.split 'or'}.flatten.map(&:strip).reject(&:empty?)
=> ["Cake", "pie", "ice cream", "pudding"]
I would prefer using a single regex:
>> options.split /\s*, or\s+|\s*,\s*|\s+or\s+/
=> ["Cake", "pie", "ice cream", "pudding"]
You can use | in a regex to give alternatives, and putting , or first guarantees that it won’t produce an empty item. Capturing the whitespace with the regex is probably best for efficiency, since you don’t have to scan the array again.
As Zabba points out, you may still want to reject empty items, prompting this solution:
>> options.split(/,|\sor\s/).map(&:strip).reject(&:empty?)
=> ["Cake", "pie", "ice cream", "pudding"]
As "or" and "," does the same thing, the best approach is to tell the regex that multiple cases should be treated the same as a single case:
options = "Cake or pie, ice cream, or pudding"
regex = /(?:\s*(?:,|or)\s*)+/
options.split(regex)
Related
I've printed the code, wit ruby
string = "hahahah"
pring string.gsub("a","b")
How do I add more letter replacements into gsub?
string.gsub("a","b")("h","l") and string.gsub("a","b";"h","l")
didnt work...
*update I have tried this too but without any success .
letters = {
"a" => "l"
"b" => "n"
...
"z" => "f"
}
string = "hahahah"
print string.gsub(\/w\,letters)
You're overcomplicating. As with most method calls in Ruby, you can simply chain #gsub calls together, one after the other:
str = 'adfh'
print str.gsub("a","b").gsub("h","l") #=> 'bdfl'
What you're doing here is applying the second #gsub to the result of the first one.
Of course, that gets a bit long-winded if you do too many of them. So, when you find yourself stringing too many together, you'll want to look for a regex solution. Rubular is a great place to tinker with them.
The way to use your hash trick with #gsub and a regex expression is to provide a hash for all possible matches. This has the same result as the two #gsub calls:
print str.gsub(/[ah]/, {'a'=>'b', 'h'=>'l'}) #=> 'bdfl'
The regex matches either a or h (/[ah]/), and the hash is saying what to substitute for each of them.
All that said, str.tr('ah', 'bl') is the simplest way to solve your problem as specified, as some commenters have mentioned, so long as you are working with single letters. If you need to work with two or more characters per substitution, you'll need to use #gsub.
I am looking to build a ruby regex to match multiple occurrences of a pattern and return them in an array. The pattern is simply: [[.+]]. That is, two left brackets, one or more characters, followed by two right brackets.
This is what I have done:
str = "Some random text[[lead:first_name]] and more stuff [[client:last_name]]"
str.match(/\[\[(.+)\]\]/).captures
The regex above doesn't work because it returns this:
["lead:first_name]] and another [[client:last_name"]
When what I wanted was this:
["lead:first_name", "client:last_name"]
I thought if I used a noncapturing group that for sure it should solve the issue:
str.match(/(?:\[\[(.+)\]\])+/).captures
But the noncapturing group returns the same exact wrong output. Any idea on how I can resolve my issue?
The problem with your regex is that the .+ part is "greedy", meaning that if the regex matches both a smaller and larger part of the string, it will capture the larger part (more about greedy regexes).
In Ruby (and most regex syntaxes), you can qualify your + quantifier with a ? to make it non-greedy. So your regex would become /(?:\[\[(.+?)\]\])+/.
However, you'll notice this still doesn't work for what you want to do. The Ruby capture groups just don't work inside a repeating group. For your problem, you'll need to use scan:
"[[a]][[ab]][[abc]]".scan(/\[\[(.+?)\]\]/).flatten
=> ["a", "ab", "abc"]
Try this:
=> str.match(/\[\[(.*)\]\].*\[\[(.*)\]\]/).captures
=> ["lead:first_name", "client:last_name"]
With many occurrences:
=> str
=> "Some [[lead:first_name]] random text[[lead:first_name]] and more [[lead:first_name]] stuff [[client:last_name]]"
=> str.scan(/\[(\w+:\w+)\]/)
=> [["lead:first_name"], ["lead:first_name"], ["lead:first_name"], ["client:last_name"]]
I want to remove words from a string which are there in some set. One way is iterate over this set and remove the particular word using str.gsub("subString", ""). Does this kind of function already exits ?
Example string :
"Hotel Silver Stone Resorts"
Strings in set:
["Hotel" , "Resorts"]
Output should be:
" Silver Stone "
You can build a union of several patterns with Regexp::union:
words = ["Hotel" , "Resorts"]
re = Regexp.union(words)
#=> /Hotel|Resorts/
"Hotel Silver Stone Resorts".gsub(re, "")
#=> " Silver Stone "
Note that you might have to escape your words.
You can subtract one array from another in ruby. Result is that all elements from the first array are removed from the second.
Split the string on whitespace, remove all extra words in one swift move, rejoin the sentence.
s = "Hotel Silver Stone Resorts"
junk_words = ['Hotel', 'Resorts']
def strip_junk(original, junk)
(original.split - junk).join(' ')
end
strip_junk(s, junk_words) # => "Silver Stone"
It certainly looks better (to my eye). Not sure about performance characteristics (too lazy to benchmark it)
I am not sure what you wanted but as I understood
sentence = 'Hotel Silver Stone Resorts'
remove_words = ["Hotel" , "Resorts"] # you can add words to this array which you wanted to remove
sentence.split.delete_if{|x| remove_words.include?(x)}.join(' ')
=> "Silver Stone"
OR
if you have an array of strings, it's easier:
sentence = 'Hotel Silver Stone Resorts'
remove_words = ["Hotel" , "Resorts"]
(sentence.split - remove_words).join(' ')
=> "Silver Stone"
You could try something different , but I don't know if it will be faster or not (depends on the length of your strings and set)
require 'set'
str = "Hotel Silver Stone Resorts"
setStr = Set.new(str.split)
setToRemove = Set.new( ["Hotel", "Resorts"])
modifiedStr = (setStr.subtract setToRemove).to_a.join " "
Output
"Silver Stone"
It uses the Set class which is faster for retrieving single element (built on Hash).
But again, the underlying transformation with to_a may not improve speed if your strings / set are very big.
It also remove implicitly the duplicates in your string and your set (when your create the sets)
Lets say I have the following string and I want the below output without requiring csv.
this, "what I need", to, do, "i, want, this", to, work
this
what i need
to
do
i, want, this
to
work
This problem is a classic case of the technique explained in this question to "regex-match a pattern, excluding..."
We can solve it with a beautifully-simple regex:
"([^"]+)"|[^, ]+
The left side of the alternation | matches complete "quotes" and captures the contents to Group1. The right side matches characters that are neither commas nor spaces, and we know they are the right ones because they were not matched by the expression on the left.
Option 2: Allowing Multiple Words
In your input, all tokens are single words, but if you also want the regex to work for my cat scratches, "what I need", your dog barks, use this:
"([^"]+)"|[^, ]+(?:[ ]*[^, ]+)*
The only difference is the addition of (?:[ ]*[^, ]+)* which optionally adds spaces + characters, zero or more times.
This program shows how to use the regex (see the results at the bottom of the online demo):
subject = 'this, "what I need", to, do, "i, want, this", to, work'
regex = /"([^"]+)"|[^, ]+/
# put Group 1 captures in an array
mymatches = []
subject.scan(regex) {|m|
$1.nil? ? mymatches << $& : mymatches << $1
}
mymatches.each { |x| puts x }
Output
this
what I need
to
do
i, want, this
to
work
Reference
How to match (or replace) a pattern except in situations s1, s2, s3...
Article about matching a pattern unless...
I am having a string as below:
str1='"{\"#Network\":{\"command\":\"Connect\",\"data\":
{\"Id\":\"xx:xx:xx:xx:xx:xx\",\"Name\":\"somename\",\"Pwd\":\"123456789\"}}}\0"'
I wanted to extract the somename string from the above string. Values of xx:xx:xx:xx:xx:xx, somename and 123456789 can change but the syntax will remain same as above.
I saw similar posts on this site but don't know how to use regex in the above case.
Any ideas how to extract the above string.
Parse the string to JSON and get the values that way.
require 'json'
str = "{\"#Network\":{\"command\":\"Connect\",\"data\":{\"Id\":\"xx:xx:xx:xx:xx:xx\",\"Name\":\"somename\",\"Pwd\":\"123456789\"}}}\0"
json = JSON.parse(str.strip)
name = json["#Network"]["data"]["Name"]
pwd = json["#Network"]["data"]["Pwd"]
Since you don't know regex, let's leave them out for now and try manual parsing which is a bit easier to understand.
Your original input, without the outer apostrophes and name of variable is:
"{\"#Network\":{\"command\":\"Connect\",\"data\":{\"Id\":\"xx:xx:xx:xx:xx:xx\",\"Name\":\"somename\",\"Pwd\":\"123456789\"}}}\0"
You say that you need to get the 'somename' value and that the 'grammar will not change'. Cool!.
First, look at what delimits that value: it has quotes, then there's a colon to the left and comma to the right. However, looking at other parts, such layout is also used near the command and near the pwd. So, colon-quote-data-quote-comma is not enough. Looking further to the sides, there's a \"Name\". It never occurs anywhere in the input data except this place. This is just great! That means, that we can quickly find the whereabouts of the data just by searching for the \"Name\" text:
inputdata = .....
estposition = inputdata.index('\"Name\"')
raise "well-known marker wa not found in the input" unless estposition
now, we know:
where the part starts
and that after the "Name" text there's always a colon, a quote, and then the-interesting-data
and that there's always a quote after the interesting-data
let's find all of them:
colonquote = inputdata.index(':\"', estposition)
datastart = colonquote+3
lastquote = inputdata.index('\"', datastart)
dataend = lastquote-1
The index returns the start position of the match, so it would return the position of : and position of \. Since we want to get the text between them, we must add/subtract a few positions to move past the :\" at begining or move back from \" at end.
Then, fetch the data from between them:
value = inputdata[datastart..dataend]
And that's it.
Now, step back and look at the input data once again. You say that grammar is always the same. The various bits are obviously separated by colons and commas. Let's try using it directly:
parts = inputdata.split(/[:,]/)
=> ["\"{\\\"#Network\\\"",
"{\\\"command\\\"",
"\\\"Connect\\\"",
"\\\"data\\\"",
"\n{\\\"Id\\\"",
"\\\"xx",
"xx",
"xx",
"xx",
"xx",
"xx\\\"",
"\\\"Name\\\"",
"\\\"somename\\\"",
"\\\"Pwd\\\"",
"\\\"123456789\\\"}}}\\0\""]
Please ignore the regex for now. Just assume it says a colon or comma. Now, in parts you will get all the, well, parts, that were detected by cutting the inputdata to pieces at every colon or comma.
If the layout never changes and is always the same, then your interesting-data will be always at place 13th:
almostvalue = parts[12]
=> "\\\"somename\\\""
Now, just strip the spurious characters. Since the grammar is constant, there's 2 chars to be cut from both sides:
value = almostvalue[2..-3]
Ok, another way. Since regex already showed up, let's try with them. We know:
data is prefixed with \"Name\" then colon and slash-quote
data consists of some text without quotes inside (well, at least I guess so)
data ends with a slash-quote
the parts in regex syntax would be, respectively:
\"Name\":\"
[^\"]*
\"
together:
inputdata =~ /\\"Name\\":\\"([^\"]*)\\"/
value = $1
Note that I surrounded the interesting part with (), hence after sucessful match that part is available in the $1 special variable.
Yet another way:
If you look at the grammar carefully, it really resembles a set of embedded hashes:
\"
{ \"#Network\" :
{ \"command\" : \"Connect\",
\"data\" :
{ \"Id\" : \"xx:xx:xx:xx:xx:xx\",
\"Name\" : \"somename\",
\"Pwd\" : \"123456789\"
}
}
}
\0\"
If we'd write something similar as Ruby hashes:
{ "#Network" =>
{ "command" => "Connect",
"data" =>
{ "Id" => "xx:xx:xx:xx:xx:xx",
"Name" => "somename",
"Pwd" => "123456789"
}
}
}
What's the difference? the colon was replaced with =>, and the slashes-before-quotes are gone. Oh, and also opening/closing \" is gone and that \0 at the end is gone too. Let's play:
tmp = inputdata[2..-4] # remove opening \" and closing \0\"
tmp.gsub!('\"', '"') # replace every \" with just "
Now, what about colons.. We cannot just replace : with =>, because it would damage the internal colons of the xx:xx:xx:xx:xx:xx part.. But, look: all the other colons have always a quote before them!
tmp.gsub!('":', '"=>') # replace every quote-colon with quote-arrow
Now our tmp is:
{"#Network"=>{"command"=>"Connect","data"=>{"Id"=>"xx:xx:xx:xx:xx:xx","Name"=>"somename","Pwd"=>"123456789"}}}
formatted a little:
{ "#Network"=>
{ "command"=>"Connect",
"data"=>
{ "Id"=>"xx:xx:xx:xx:xx:xx","Name"=>"somename","Pwd"=>"123456789" }
}
}
So, it looks just like a Ruby hash. Let's try 'destringizing' it:
packeddata = eval(tmp)
value = packeddata['#Network']['data']['Name']
Done.
Well, this has grown a bit and Jonas was obviously faster, so I'll leave the JSON part to him since he wrote it already ;) The data was so similar to Ruby hash because it was obviously formatted as JSON which is a hash-like structure too. Using the proper format-reading tools is usually the best idea, but mind that the JSON library when asked to read the data - will read all of the data and then you can ask them "what was inside at the key xx/yy/zz", just like I showed you with the read-it-as-a-Hash attempt. Sometimes when your program is very short on the deadline, you cannot afford to read-it-all. Then, scanning with regex or scanning manually for "known markers" may (not must) be much faster and thus prefereable. But, still, much less convenient. Have fun.