Ruby RegEx Commas Within Quotes [duplicate] - ruby

I would like to find a regex that will pick out all commas that fall outside quote sets.
For example:
'foo' => 'bar',
'foofoo' => 'bar,bar'
This would pick out the single comma on line 1, after 'bar',
I don't really care about single vs double quotes.
Has anyone got any thoughts? I feel like this should be possible with readaheads, but my regex fu is too weak.

This will match any string up to and including the first non-quoted ",". Is that what you are wanting?
/^([^"]|"[^"]*")*?(,)/
If you want all of them (and as a counter-example to the guy who said it wasn't possible) you could write:
/(,)(?=(?:[^"]|"[^"]*")*$)/
which will match all of them. Thus
'test, a "comma,", bob, ",sam,",here'.gsub(/(,)(?=(?:[^"]|"[^"]*")*$)/,';')
replaces all the commas not inside quotes with semicolons, and produces:
'test; a "comma,"; bob; ",sam,";here'
If you need it to work across line breaks just add the m (multiline) flag.

The below regexes would match all the comma's which are present outside the double quotes,
,(?=(?:[^"]*"[^"]*")*[^"]*$)
DEMO
OR(PCRE only)
"[^"]*"(*SKIP)(*F)|,
"[^"]*" matches all the double quoted block. That is, in this buz,"bar,foo" input, this regex would match "bar,foo" only. Now the following (*SKIP)(*F) makes the match to fail. Then it moves on to the pattern which was next to | symbol and tries to match characters from the remaining string. That is, in our output , next to pattern | will match only the comma which was just after to buz . Note that this won't match the comma which was present inside double quotes, because we already make the double quoted part to skip.
DEMO
The below regex would match all the comma's which are present inside the double quotes,
,(?!(?:[^"]*"[^"]*")*[^"]*$)
DEMO

While it's possible to hack it with a regex (and I enjoy abusing regexes as much as the next guy), you'll get in trouble sooner or later trying to handle substrings without a more advanced parser. Possible ways to get in trouble include mixed quotes, and escaped quotes.
This function will split a string on commas, but not those commas that are within a single- or double-quoted string. It can be easily extended with additional characters to use as quotes (though character pairs like « » would need a few more lines of code) and will even tell you if you forgot to close a quote in your data:
function splitNotStrings(str){
var parse=[], inString=false, escape=0, end=0
for(var i=0, c; c=str[i]; i++){ // looping over the characters in str
if(c==='\\'){ escape^=1; continue} // 1 when odd number of consecutive \
if(c===','){
if(!inString){
parse.push(str.slice(end, i))
end=i+1
}
}
else if(splitNotStrings.quotes.indexOf(c)>-1 && !escape){
if(c===inString) inString=false
else if(!inString) inString=c
}
escape=0
}
// now we finished parsing, strings should be closed
if(inString) throw SyntaxError('expected matching '+inString)
if(end<i) parse.push(str.slice(end, i))
return parse
}
splitNotStrings.quotes="'\"" // add other (symmetrical) quotes here

Try this regular expression:
(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*=>\s*(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*,
This does also allow strings like “'foo\'bar' => 'bar\\',”.

MarkusQ's answer worked great for me for about a year, until it didn't. I just got a stack overflow error on a line with about 120 commas and 3682 characters total. In Java, like this:
String[] cells = line.split("[\t,](?=(?:[^\"]|\"[^\"]*\")*$)", -1);
Here's my extremely inelegant replacement that doesn't stack overflow:
private String[] extractCellsFromLine(String line) {
List<String> cellList = new ArrayList<String>();
while (true) {
String[] firstCellAndRest;
if (line.startsWith("\"")) {
firstCellAndRest = line.split("([\t,])(?=(?:[^\"]|\"[^\"]*\")*$)", 2);
}
else {
firstCellAndRest = line.split("[\t,]", 2);
}
cellList.add(firstCellAndRest[0]);
if (firstCellAndRest.length == 1) {
break;
}
line = firstCellAndRest[1];
}
return cellList.toArray(new String[cellList.size()]);
}

#SocialCensus, The example you gave in the comment to MarkusQ, where you throw in ' alongside the ", doesn't work with the example MarkusQ gave right above that if we change sam to sam's: (test, a "comma,", bob, ",sam's,",here) has no match against (,)(?=(?:[^"']|["|'][^"']")$). In fact, the problem itself, "I don't really care about single vs double quotes", is ambiguous. You have to be clear what you mean by quoting either with " or with '. For example, is nesting allowed or not? If so, to how many levels? If only 1 nested level, what happens to a comma outside the inner nested quotation but inside the outer nesting quotation? You should also consider that single quotes happen by themselves as apostrophes (ie, like the counter-example I gave earlier with sam's). Finally, the regex you made doesn't really treat single quotes on par with double quotes since it assumes the last type of quotation mark is necessarily a double quote -- and replacing that last double quote with ['|"] also has a problem if the text doesn't come with correct quoting (or if apostrophes are used), though, I suppose we probably could assume all quotes are correctly delineated.
MarkusQ's regexp answers the question: find all commas that have an even number of double quotes after it (ie, are outside double quotes) and disregard all commas that have an odd number of double quotes after it (ie, are inside double quotes). This is generally the same solution as what you probably want, but let's look at a few anomalies. First, if someone leaves off a quotation mark at the end, then this regexp finds all the wrong commas rather than finding the desired ones or failing to match any. Of course, if a double quote is missing, all bets are off since it might not be clear if the missing one belongs at the end or instead belongs at the beginning; however, there is a case that is legitimate and where the regex could conceivably fail (this is the second "anomaly"). If you adjust the regexp to go across text lines, then you should be aware that quoting multiple consecutive paragraphs requires that you place a single double quote at the beginning of each paragraph and leave out the quote at the end of each paragraph except for at the end of the very last paragraph. This means that over the space of those paragraphs, the regex will fail in some places and succeed in others.
Examples and brief discussions of paragraph quoting and of nested quoting can be found here http://en.wikipedia.org/wiki/Quotation_mark .

Related

Removing all whitespace from a string in Ruby

How can I remove all newlines and spaces from a string in Ruby?
For example, if we have a string:
"123\n12312313\n\n123 1231 1231 1"
It should become this:
"12312312313123123112311"
That is, all whitespaces should be removed.
You can use something like:
var_name.gsub!(/\s+/, '')
Or, if you want to return the changed string, instead of modifying the variable,
var_name.gsub(/\s+/, '')
This will also let you chain it with other methods (i.e. something_else = var_name.gsub(...).to_i to strip the whitespace then convert it to an integer). gsub! will edit it in place, so you'd have to write var_name.gsub!(...); something_else = var_name.to_i. Strictly speaking, as long as there is at least one change made,gsub! will return the new version (i.e. the same thing gsub would return), but on the chance that you're getting a string with no whitespace, it'll return nil and things will break. Because of that, I'd prefer gsub if you're chaining methods.
gsub works by replacing any matches of the first argument with the contents second argument. In this case, it matches any sequence of consecutive whitespace characters (or just a single one) with the regex /\s+/, then replaces those with an empty string. There's also a block form if you want to do some processing on the matched part, rather than just replacing directly; see String#gsub for more information about that.
The Ruby docs for the class Regexp are a good starting point to learn more about regular expressions -- I've found that they're useful in a wide variety of situations where a couple of milliseconds here or there don't count and you don't need to match things that can be nested arbitrarily deeply.
As Gene suggested in his comment, you could also use tr:
var_name.tr(" \t\r\n", '')
It works in a similar way, but instead of replacing a regex, it replaces every instance of the nth character of the first argument in the string it's called on with the nth character of the second parameter, or if there isn't, with nothing. See String#tr for more information.
You could also use String#delete:
str = "123\n12312313\n\n123 1231 1231 1"
str.delete "\s\n"
#=> "12312312313123123112311"
You could use String#delete! to modify str in place, but note delete! returns nil if no change is made
Alternatively you could scan the string for digits /\d+/ and join the result:
string = "123\n\n12312313\n\n123 1231 1231 1\n"
string.scan(/\d+/).join
#=> "12312312313123123112311"
Please note that this would also remove alphabetical characters, dashes, symbols, basically everything that is not a digit.

How do I split this certain kind of string into an array in ruby [duplicate]

I would like to find a regex that will pick out all commas that fall outside quote sets.
For example:
'foo' => 'bar',
'foofoo' => 'bar,bar'
This would pick out the single comma on line 1, after 'bar',
I don't really care about single vs double quotes.
Has anyone got any thoughts? I feel like this should be possible with readaheads, but my regex fu is too weak.
This will match any string up to and including the first non-quoted ",". Is that what you are wanting?
/^([^"]|"[^"]*")*?(,)/
If you want all of them (and as a counter-example to the guy who said it wasn't possible) you could write:
/(,)(?=(?:[^"]|"[^"]*")*$)/
which will match all of them. Thus
'test, a "comma,", bob, ",sam,",here'.gsub(/(,)(?=(?:[^"]|"[^"]*")*$)/,';')
replaces all the commas not inside quotes with semicolons, and produces:
'test; a "comma,"; bob; ",sam,";here'
If you need it to work across line breaks just add the m (multiline) flag.
The below regexes would match all the comma's which are present outside the double quotes,
,(?=(?:[^"]*"[^"]*")*[^"]*$)
DEMO
OR(PCRE only)
"[^"]*"(*SKIP)(*F)|,
"[^"]*" matches all the double quoted block. That is, in this buz,"bar,foo" input, this regex would match "bar,foo" only. Now the following (*SKIP)(*F) makes the match to fail. Then it moves on to the pattern which was next to | symbol and tries to match characters from the remaining string. That is, in our output , next to pattern | will match only the comma which was just after to buz . Note that this won't match the comma which was present inside double quotes, because we already make the double quoted part to skip.
DEMO
The below regex would match all the comma's which are present inside the double quotes,
,(?!(?:[^"]*"[^"]*")*[^"]*$)
DEMO
While it's possible to hack it with a regex (and I enjoy abusing regexes as much as the next guy), you'll get in trouble sooner or later trying to handle substrings without a more advanced parser. Possible ways to get in trouble include mixed quotes, and escaped quotes.
This function will split a string on commas, but not those commas that are within a single- or double-quoted string. It can be easily extended with additional characters to use as quotes (though character pairs like « » would need a few more lines of code) and will even tell you if you forgot to close a quote in your data:
function splitNotStrings(str){
var parse=[], inString=false, escape=0, end=0
for(var i=0, c; c=str[i]; i++){ // looping over the characters in str
if(c==='\\'){ escape^=1; continue} // 1 when odd number of consecutive \
if(c===','){
if(!inString){
parse.push(str.slice(end, i))
end=i+1
}
}
else if(splitNotStrings.quotes.indexOf(c)>-1 && !escape){
if(c===inString) inString=false
else if(!inString) inString=c
}
escape=0
}
// now we finished parsing, strings should be closed
if(inString) throw SyntaxError('expected matching '+inString)
if(end<i) parse.push(str.slice(end, i))
return parse
}
splitNotStrings.quotes="'\"" // add other (symmetrical) quotes here
Try this regular expression:
(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*=>\s*(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*,
This does also allow strings like “'foo\'bar' => 'bar\\',”.
MarkusQ's answer worked great for me for about a year, until it didn't. I just got a stack overflow error on a line with about 120 commas and 3682 characters total. In Java, like this:
String[] cells = line.split("[\t,](?=(?:[^\"]|\"[^\"]*\")*$)", -1);
Here's my extremely inelegant replacement that doesn't stack overflow:
private String[] extractCellsFromLine(String line) {
List<String> cellList = new ArrayList<String>();
while (true) {
String[] firstCellAndRest;
if (line.startsWith("\"")) {
firstCellAndRest = line.split("([\t,])(?=(?:[^\"]|\"[^\"]*\")*$)", 2);
}
else {
firstCellAndRest = line.split("[\t,]", 2);
}
cellList.add(firstCellAndRest[0]);
if (firstCellAndRest.length == 1) {
break;
}
line = firstCellAndRest[1];
}
return cellList.toArray(new String[cellList.size()]);
}
#SocialCensus, The example you gave in the comment to MarkusQ, where you throw in ' alongside the ", doesn't work with the example MarkusQ gave right above that if we change sam to sam's: (test, a "comma,", bob, ",sam's,",here) has no match against (,)(?=(?:[^"']|["|'][^"']")$). In fact, the problem itself, "I don't really care about single vs double quotes", is ambiguous. You have to be clear what you mean by quoting either with " or with '. For example, is nesting allowed or not? If so, to how many levels? If only 1 nested level, what happens to a comma outside the inner nested quotation but inside the outer nesting quotation? You should also consider that single quotes happen by themselves as apostrophes (ie, like the counter-example I gave earlier with sam's). Finally, the regex you made doesn't really treat single quotes on par with double quotes since it assumes the last type of quotation mark is necessarily a double quote -- and replacing that last double quote with ['|"] also has a problem if the text doesn't come with correct quoting (or if apostrophes are used), though, I suppose we probably could assume all quotes are correctly delineated.
MarkusQ's regexp answers the question: find all commas that have an even number of double quotes after it (ie, are outside double quotes) and disregard all commas that have an odd number of double quotes after it (ie, are inside double quotes). This is generally the same solution as what you probably want, but let's look at a few anomalies. First, if someone leaves off a quotation mark at the end, then this regexp finds all the wrong commas rather than finding the desired ones or failing to match any. Of course, if a double quote is missing, all bets are off since it might not be clear if the missing one belongs at the end or instead belongs at the beginning; however, there is a case that is legitimate and where the regex could conceivably fail (this is the second "anomaly"). If you adjust the regexp to go across text lines, then you should be aware that quoting multiple consecutive paragraphs requires that you place a single double quote at the beginning of each paragraph and leave out the quote at the end of each paragraph except for at the end of the very last paragraph. This means that over the space of those paragraphs, the regex will fail in some places and succeed in others.
Examples and brief discussions of paragraph quoting and of nested quoting can be found here http://en.wikipedia.org/wiki/Quotation_mark .

What is the best way to delimit a csv files thats contain commas and double quotes?

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...

String gsub - Replace characters between two elements, but leave surrounding elements

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>")

How to remove the first 4 characters from a string if it matches a pattern in Ruby

I have the following string:
"h3. My Title Goes Here"
I basically want to remove the first four characters from the string so that I just get back:
"My Title Goes Here".
The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first four characters blindly.
I checked the docs and the closest thing I could find was chomp, but that only works for the end of a string.
Right now I am doing this:
"h3. My Title Goes Here".reverse.chomp(" .3h").reverse
This gives me my desired output, but there has to be a better way. I don't want to reverse a string twice for no reason. Is there another method that will work?
To alter the original string, use sub!, e.g.:
my_strings = [ "h3. My Title Goes Here", "No h3. at the start of this line" ]
my_strings.each { |s| s.sub!(/^h3\. /, '') }
To not alter the original and only return the result, remove the exclamation point, i.e. use sub. In the general case you may have regular expressions that you can and want to match more than one instance of, in that case use gsub! and gsub—without the g only the first match is replaced (as you want here, and in any case the ^ can only match once to the start of the string).
You can use sub with a regular expression:
s = 'h3. foo'
s.sub!(/^h[0-9]+\. /, '')
puts s
Output:
foo
The regular expression should be understood as follows:
^ Match from the start of the string.
h A literal "h".
[0-9] A digit from 0-9.
+ One or more of the previous (i.e. one or more digits)
\. A literal period.
A space (yes, spaces are significant by default in regular expressions!)
You can modify the regular expression to suit your needs. See a regular expression tutorial or syntax guide, for example here.
A standard approach would be to use regular expressions:
"h3. My Title Goes Here".gsub /^h3\. /, '' #=> "My Title Goes Here"
gsub means globally substitute and it replaces a pattern by a string, in this case an empty string.
The regular expression is enclosed in / and constitutes of:
^ means beginning of the string
h3 is matched literally, so it means h3
\. - a dot normally means any character so we escape it with a backslash
is matched literally

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