I am new to ruby and regular expressions and trying to figure out how to attack seperating the attached string of baseball players into first/last name combinations.
This is a sample string:
"JohnnyCuetoJ.J.PutzBrianMcCann"
This is the desired output:
Johnny Cueto
J.J. Putz
Brian McCann
I have figured out how to separate by capital letters which gets me close, but the outlier names like J.J. and McCann mess that pattern up. Anyone have ideas on the best way to approach this?
If you don't have to do it in one single gsub than it gets a bit easier.
string = "JohnnyCuetoJ.J.PutzBrianMcCann"
string.gsub!(/([A-Z][^A-Z]+)/, '\1 ') # separate by capital letters
string.gsub!(/(\.) ([A-Z]\.)/, '\1\2') # paste together "J. J." -> "J.J."
string.gsub!(/Mc /, 'Mc') # Remove the space in "Mc "
string.strip # Remove the extra space after "Cann "
...and of course you can put this on a single line by chaining the gsub calls, but that will basically kill the readability of the code (but on the other hand, how readable is a block of regexen anyway?)
Related
Using Ruby 1.8.7
I need to grab everything up to a certain word - and I would like to match against words in an array. Example:
match_words = ['title','author','pages']
item = "Title: Jurassic Park\n"
item += "Author: Michael Crichton\n"
if item =~ /title: (.*)#{match any word in match_words array}/i
#do something here
end
So, this would ideally return "Jurassic Park\n". I am currently matching on newlines but have found that the data I will be matching against might have newlines in strange places, like the middle of the sentence. So, I think matching to the next match_word would be a good idea.
Is this possible, or maybe can be done another way?
Try this on for size
item.scan(/(title|author|pages):\s*?(.+)/i)
What this says is find all the results that start (case-insensitive) with either title, author or pages, are then followed by a colon and option white space and then characters. Capture the label and then the characters following the whitespace. The scan method will match as many times as it can.
Just iterate over the match words and do the regex compare as you normally would.
match_words.each do |word|
if item =~ /#{word}/ # Plus case sensitivity, start/end of item, etc.
# etc.
end
end
But if you know that the things you care about are at the beginning of the lines, then split the input string on \n and just use start_with instead of bothering with the regex--that partially depends on what the real data looks like.
First, create a | separated list of keywords from match_words.
Then, use string.scan to split the string apart, giving you an array of arrays with your results. See the end of this tutorial for a reference.
Here's my best shot:
keywords = match_words.join('|')
results = item.scan(/(#{keywords}):\s*(.+?)\s*(?= (#{keywords}):)/im)
Results: [["Title", "Jurassic Park"], ["Author", "Michael Crichton"]]
Don't forget to use the /m switch to indicate that you want . to match newlines.
To explain the pattern: we look for a keyword, then use a "look ahead" (?= ) to find the next keyword without capturing it. We capture all characters in between using a "lazy" expression .+?, so that we don't capture other keywords.
I am trying to use regular expressions to match some text.
The following pattern is what I am trying to gather.
#Identifier('VariableA', 'VariableB', 'VariableX', ..., 'VariableZ')
I would like to grab a dynamic number of variables rather than a fixed set of two or three.
Is there any way to do this? I have an existing Regular Expression:
\#(\w+)\W+(\w+)\W+(\w+)\W+(\w+)
This captures the Identifier and up to three variables.
Edit: Is it just me, or are regular expressions not as powerful as I'm making them out to be?
You want to use scan for this sort of thing. The basic pattern would be this:
s.scan(/\w+/)
That would give you an array of all the contiguous sequences for word characters:
>> "#Identifier('VariableA', 'VariableB', 'VariableX', 'VariableZ')".scan(/\w+/)
=> ["Identifier", "VariableA", "VariableB", "VariableX", "VariableZ"]
You say you might have multiple instances of your pattern with arbitrary stuff surrounding them. You can deal with that with nested scans:
s.scan(/#(\w+)\(([^)]+?)\)/).map { |m| [ m.first, m.last.scan(/\w+/) ] }
That will give you an array of arrays, each inner array will have the "Identifier" part as the first element and that "Variable" parts as an array in the second element. For example:
>> s = "pancakes #Identifier('VariableA', 'VariableB', 'VariableX', 'VariableZ') pancakes #Pancakes('one','two','three') eggs"
>> s.scan(/#(\w+)\(([^)]+?)\)/).map { |m| [ m.first, m.last.scan(/\w+/) ] }
=> [["Identifier", ["VariableA", "VariableB", "VariableX", "VariableZ"]], ["Pancakes", ["one", "two", "three"]]]
If you might be facing escaped quotes inside your "Variable" bits then you'll need something more complex.
Some notes on the expression:
# # A literal "#".
( # Open a group
\w+ # One more more ("+") word characters ("\w").
) # Close the group.
\( # A literal "(", parentheses are used for group so we escape it.
( # Open a group.
[ # Open a character class.
^) # The "^" at the beginning of a [] means "not", the ")" isn't escaped because it doesn't have any special meaning inside a character class.
] # Close a character class.
+? # One more of the preceding pattern but don't be greedy.
) # Close the group.
\) # A literal ")".
You don't really need [^)]+? here, just [^)]+ would do but I use the non-greedy forms by habit because that's usually what I mean. The grouping is used to separate the #Identifier and Variable parts so that we can easily get the desired nested array output.
But alex thinks that you meant you wanted to capture the same thing four times. If you want to capture the same pattern, but different things, then you may want to consider two things:
Iteration. In perl, you can say
while ($variable =~ /regex/g) {
the 'g' stands for 'global', and means that each time the regex is called, it matches the /next/ instance.
The other option is recursion. Write your regex like this:
/(what you want)(.*)/
Then, you have backreference 1 containing the first thing, which you can push to an array, and backreference 2 which you'll then recurse over until it no longer matches.
You may use simply (\w+).
Given the input string
#Identifier('VariableA', 'VariableB', 'VariableX', 'VariableZ')
The results would be:
Identifier
VariableA
VariableB
VariableX
VariableZ
This would work for an arbitrary number of variables.
For future reference, it's easy and fun to play around with regexp ideas on Rubular.
So you are asking if there is a way to capture both the identifier and an arbitrary number of variables. I am afraid that you can only do this with regex engines that support captures. Note here that captures and capturing groups are not the one and the same thing. You want to remember all the "variables". This can't be done with simple capturing groups.
I am unaware whether Ruby supports this or not, but I am sure that .NET and the new PERL 6 support it.
In your case you could use two regexes. One to capture the identifier e.g. ^\s*#(\w+)
and another one to capture all variables e.g. result = subject.scan(/'[^']+'/)
I have the following
address.gsub(/^\d*/, "").gsub(/\d*-?\d*$/, "").gsub(/\# ?\d*/,"")
Can this be done in one gsub? I would like to pass a list of patterns rather then just one pattern - they are all being replaced by the same thing.
You could combine them with an alternation operator (|):
address = '6 66-666 #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/, "")
# " 66-666 "
address = 'pancakes 6 66-666 # pancakes #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/,"")
# "pancakes 6 66-666 pancakes "
You might want to add little more whitespace cleanup. And you might want to switch to one of:
/\A\d*|\d*-?\d*\z|\# ?\d*/
/\A\d*|\d*-?\d*\Z|\# ?\d*/
depending on what your data really looks like and how you need to handle newlines.
Combining the regexes is a good idea--and relatively simple--but I'd like to recommend some additional changes. To wit:
address.gsub(/^\d+|\d+(?:-\d+)?$|\# *\d+/, "")
Of your original regexes, ^\d* and \d*-?\d*$ will always match, because they don't have to consume any characters. So you're guaranteed to perform two replacements on every line, even if that's just replacing empty strings with empty strings. Of my regexes, ^\d+ doesn't bother to match unless there's at least one digit at the beginning of the line, and \d+(?:-\d+)?$ matches what looks like an integer-or-range expression at the end of the line.
Your third regex, \# ?\d*, will match any # character, and if the # is followed by a space and some digits, it'll take those as well. Judging by your other regexes and my experience with other questions, I suspect you meant to match a # only if it's followed by one or more digits, with optional spaces intervening. That's what my third regex does.
If any of my guesses are wrong, please describe what you were trying to do, and I'll do my best to come up with the right regex. But I really don't think those first two regexes, at least, are what you want.
EDIT (in answer to the comment): When working with regexes, you should always be aware of the distinction between a regex the matches nothing and a regex that doesn't match. You say you're applying the regexes to street addresses. If an address doesn't happen to start with a house number, ^\d* will match nothing--that is, it will report a successful match, said match consisting of the empty string preceding the first character in the address.
That doesn't matter to you, you're just replacing it with another empty string anyway. But why bother doing the replacement at all? If you change the regex to ^\d+, it will report a failed match and no replacement will be performed. The result is the same either way, but the "matches noting" scenario (^\d*) results in a lot of extra work that the "doesn't match" scenario avoids. In a high-throughput situation, that could be a life-saver.
The other two regexes bring additional complications: \d*-?\d*$ could match a hyphen at the end of the string (e.g. "123-", or even "-"); and \# ?\d* could match a hash symbol anywhere in string, not just as part of an apartment/office number. You know your data, so you probably know neither of those problems will ever arise; I'm just making sure you're aware of them. My regex \d+(?:-\d+)?$ deals with the trailing-hyphen issue, and \# *\d+ at least makes sure there are digits after the hash symbol.
I think that if you combine them together in a single gsub() regex, as an alternation,
it changes the context of the starting search position.
Example, each of these lines start at the beginning of the result of the previous
regex substitution.
s/^\d*//g
s/\d*-?\d*$//g
s/\# ?\d*//g
and this
s/^\d*|\d*-?\d*$|\# ?\d*//g
resumes search/replace where the last match left off and could potentially produce a different overall output, especially since a lot of the subexpressions search for similar
if not the same characters, distinguished only by line anchors.
I think your regex's are unique enough in this case, and of course changing the order
changes the result.
I have the following data
Animals = Dog Cat Turtle \
Mouse Parrot \
Snake
I would like the regex to construct a match of just the animals with none of the backslashes: Dog Cat Turtle Mouse Parrot Snake
I've got a regex, but need some help finishing it off.
/ANIMALS\s*=\s*([^\\\n]*)/
Since you specified a language, I need to ask you this: Why are you relying on the regex for everything? Don't make the problem harder than it has to be by forcing the regex to do everything.
Try this approach instead...
Use gsub! to get rid of the backslashes.
split the string on any whitespace.
Shift out the first two tokens ("Animals", "=").
Join the array with a single space, or whatever other delimiter.
So the only regex you might need is one for the whitespace delimiter split. I don't know Ruby well enough to say exactly how you would do that, though.
How 'bout the regex \b(?!Animals\b)\w+\b which matches all words that aren't Animals? Use the scan method to collect all such matches, e.g.
matchArray = sourceString.scan(/\b(?!Animals\b)\w+\b/)
Make sure you are matching with ignore-case, because ANIMALS will not match Animals without it.
I'm not sure how to use regular expressions in a function so that I could grab all the words in a sentence starting with a particular letter. I know that I can do:
word =~ /^#{letter}/
to check if the word starts with the letter, but how do I go from word to word. Do I need to convert the string to an array and then iterate through each word or is there a faster way using regex? I'm using ruby so that would look like:
matching_words = Array.new
sentance.split(" ").each do |word|
matching_words.push(word) if word =~ /^#{letter}/
end
Scan may be a good tool for this:
#!/usr/bin/ruby1.8
s = "I think Paris in the spring is a beautiful place"
p s.scan(/\b[it][[:alpha:]]*/i)
# => ["I", "think", "in", "the", "is"]
\b means 'word boundary."
[:alpha:] means upper or lowercase alpha (a-z).
You can use \b. It matches word boundaries--the invisible spot just before and after a word. (You can't see them, but oh they're there!) Here's the regex:
/\b(a\w*)\b/
The \w matches a word character, like letters and digits and stuff like that.
You can see me testing it here: http://rubular.com/regexes/13347
Similar to Anon.'s answer:
/\b(a\w*)/g
and then see all the results with (usually) $n, where n is the n-th hit. Many libraries will return /g results as arrays on the $n-th set of parenthesis, so in this case $1 would return an array of all the matching words. You'll want to double-check with whatever library you're using to figure out how it returns matches like this, there's a lot of variation on global search returns, sadly.
As to the \w vs [a-zA-Z], you can sometimes get faster execution by using the built-in definitions of things like that, as it can easily have an optimized path for the preset character classes.
The /g at the end makes it a "global" search, so it'll find more than one. It's still restricted by line in some languages / libraries, though, so if you wish to check an entire file you'll sometimes need /gm, to make it multi-line
If you want to remove results, like your title (but not question) suggests, try:
/\ba\w*//g
which does a search-and-replace in most languages (/<search>/<replacement>/). Sometimes you need a "s" at the front. Depends on the language / library. In Ruby's case, use:
string.gsub(/(\b)a\w*(\b)/, "\\1\\2")
to retain the non-word characters, and optionally put any replacement text between \1 and \2. gsub for global, sub for the first result.
/\ba[a-z]*\b/i
will match any word starting with 'a'.
The \b indicates a word boundary - we want to only match starting from the beginning of a word, after all.
Then there's the character we want our word to start with.
Then we have as many as possible letter characters, followed by another word boundary.
To match all words starting with t, use:
\bt\w+
That will match test but not footest; \b means "word boundary".
Personally i think that regex is overkill for this application, simply running a select is more than capable of solving this particular problem.
"this is a test".split(' ').select{ |word| word[0,1] == 't' }
result => ["this", "test"]
or if you are determined to use regex then go with grep
"this is a test".split(' ').grep(/^t/)
result => ["this", "test"]
Hope this helps.