I'm trying to learn ruby and having a hard time figuring out what each individual part of this code is doing. Specifically, how does the global subbing determine whether two sequential numbers are both one of these values [13579] and how does it add a dash (-) in between them?
def DashInsert(num)
num_str = num.to_s
num_str.gsub(/([13579])(?=[13579])/, '\1-')
end
num_str.gsub(/([13579])(?=[13579])/, '\1-')
() called capturing group, which captures the characters matched by the pattern present inside the capturing group. So the pattern present inside the capturing group is [13579] which matches a single digit from the given set of digits. That corresponding digit was captured and stored inside index 1.
(?=[13579]) Positive lookahead which asserts that the match must be followed by the character or string matched by the pattern inside the lookahead. Replacement will occur only if this condition is satisfied.
\1 refers the characters which are present inside the group index 1.
Example:
> "13".gsub(/([13579])(?=[13579])/, '\1-')
=> "1-3"
You may start with some random tests:
def DashInsert(num)
num_str = num.to_s
num_str.gsub(/([13579])(?=[13579])/, '\1-')
end
10.times{
x = rand(10000)
puts "%6i: %6s" % [x,DashInsert(x)]
}
Example:
9633: 963-3
7774: 7-7-74
6826: 6826
7386: 7-386
2145: 2145
7806: 7806
9499: 949-9
4117: 41-1-7
4920: 4920
14: 14
And now to check the regex.
([13579]) take any odd number and remember it (it can be used later with \1
(?=[13579]) Check if the next number is also odd, but don't take it (it still remains in the string)
'\1-' Output the first odd num and ab a - to it.
In other word:
Puts a - between each two odds numbers.
Related
I have this code running inside a buffer (used to unescape a JS string in Ruby):
elsif hex_substring =~ /^\\u[0-9a-fA-F]{1,4}/
hex_substring.scan(/^((\\u[\da-fA-F]{4}){1,})/) do |match|
hex_byte = match[0]
buffer << JSON.load(%Q("#{hex_byte}"))
hex_index += hex_byte.length
end
...
I have a concern that the scan() is matching a bit too much:
hex_substring.scan(/^((\\u[\da-fA-F]{4}){1,})/)
# => [["\\ud83c\\udfec", "\\udfec"]]
I am using only "\\ud83c\\udfec", not "\\udfec".
Is there a way in Ruby or in regex to grab only the first part?
You should use a single grouping construct here, the one to match 1 or more occurrences of four hex chars, and omit the inner capturing group that resulted in an extra item in the resulting array:
.scan(/^(?:\\u[\da-fA-F]{4})+/)
Note that + is a simpler and shorter way to write {1,} (one or more occurrences).
Details
^ - start of string
(?: - start of a non-capturing group (what it matches won't be added to the final scan result):
\\u - a \u substring
[\da-fA-F]{4} - four hex chars
)+ - 1 or more occurrences (of the group pattern sequence).
I'm facing a very strange behaviour with ruby String#scan method return. I have this code below and I can't find out why "scan" doesn't return 2 elements.
str = "10011011001"
regexp = "0110"
p str.scan(/(#{regexp})/)
==> [["0110"]]
String "str" clearly contains 2 occurrences of pattern "0110".
I want to fetch all the occurences of my regexp in str of course.
The reason is that after finding the first result, the regex engine continues its walk at the position after this first result. So the zero at the end of the first result can't be reuse for an other result.
The way to get overlapping results is to put your pattern in a lookahead and in a capture group (a lookahead is only a zero-width assertion (a test) and doesn't consume any characters). In this way the regex engine advance always one character at a time and can test all positions in the string even something is captured in the group:
(?=(yourpattern))
Then your result is in the capture group 1
With your example:
p str.scan(/(?=(0110))/)
[["0110"], ["0110"]]
str = "10011011001"
match = "0110"
str.chars.each_cons(match.size).map(&:join).select { |cons| cons == match }
Should do it.
The following code from a book is supposed to transfer "FOO92OBAR" to "FOO92_O_BAR":
gsub(/([a-z\d])([A-Z])/, '\1_\2')
Can anyone explain how this works?
([a-z\d]) looks for a lowercase letter (a-z) or a number (\d means a digit). The () around the whole thing assign the result to regex subgroup 1.
([A-Z]) then looks for an uppercase letter, assigning the result to group 2. So the whole thing looks for a lowercase-or-digit followed by an uppercase letter. The second part, '\1_\2', means "regex group 1 followed by regex group 2"
gsub replaces every time it sees a lowercase-or-digit followed by an uppercase letter with (the first thing)_(the second thing).
So actually FOO92OBAR will be FOO92_OBAR.
For FOO92OBAR to become FOO92_O_BAR, the replace part should be '\1_\2_' (since only the O is the second part.. BAR is not matched, so not replaced at all).
It works using regular expressions.
The two parameters of gsub are the match expression and the replacement. Because the match /([a-z\d])([A-Z])/ contains groups (identified by (...)), then you can reference a match in the replacement using \ID where the ID is the number of the group, starting from 1.
That said, the code gsub(/([a-z\d])([A-Z])/, '\1_\2')
# take any combination of
([a-z\d])([A-Z])
# which means any combinations of a (1) lower-case char or (2) digit
([a-z\d])
# followed by an (1) upper case letter
([A-Z])
# if any, replace it with
\1_\2
# that represents the first group
\1
# followed by _
# followed by the second group
\2
Please note that your example will generate FOO92_OBAR, not FOO92_O_BAR
2.1.5 :001 > string = "FOO92OBAR"
=> "FOO92OBAR"
2.1.5 :002 > string.gsub(/([a-z\d])([A-Z])/, '\1_\2')
=> "FOO92_OBAR"
The explanation is because there is only one case of a "lower-case char or digit" (and that is a digit) followed by an upper case char.
2.1.5 :003 > string.scan(/([a-z\d])([A-Z])/)
=> [["2", "O"]]
Regular expressions are case sensitive by default.
I need a very simple string validator that would show where is first symbol not corresponding to the desired format. I want to use regex but in this case I have to find the place where the string stops corresponding to the expression and I can't find a method that would do that.
(It's got to be a fairly simple method... maybe there isn't one?)
For example if I have regex:
/^Q+E+R+$/
with string:
"QQQQEEE2ER"
The desired result should be 7
An idea: what you can do is to tokenize your pattern and write it with optional nested capturing groups:
^(Q+(E+(R+($)?)?)?)?
Then you only need to count the number of capture groups you obtain to know where the regex engine stops in the pattern and you can determine the offset of the match end in the string with the whole match length.
As #zx81 notices it in his comment, if one of the elements can match the next element (example Q can match the element E), things become different.
Let's say that Q is \w (and can match E and R). For the string QQQEEERRR the precedent pattern will give only one capturing group (the greedy \w+ matches all) when ^(\w+)(E+)(R+)$ will give three groups: QQQEE, E, RRR
To obtain the same result you need to add an alternation:
^((?:\w+(?=E)|\w+)(E+(R+($)?)?)?)?
In the alternation, the case where E exists must be tested first, and only if this branch fails (with the lookahead), then the other branch where E doesn't exist is used.
Thus the full pattern can be rewritten like this to deal with this specific case:
^((?:Q+(?=E)|Q+)((?:E+(?=R)|E+)((?:R+(?=$)|R+)($)?)?)?)?
Perhaps could you take a look to the gem amatch too.
This is an interesting task that can be accomplished with a neat regex trick:
^(?:(?=(Q+)))?(?:(?=(Q+E+)))?(?:(?=(Q+E+R+)))?(?:(?=(Q+E+R+$)))?
We have four optional lookaheads checking various parts of the pattern and capturing the partial matches to Groups 1, 2, 3 and 4 incrementally.
Group 1 contains Q+ if it can be matched, in your example QQQQ.
Group 2 contains Q+E+ if it can be matched, in your example EEE.
Group 3 contains Q+E+R+ if it can be matched, in your example nil.
Group 3 contains Q+E+R+$ if it can be matched, in your example nil.
In your code, check which is the last Group that is set by testing !$1.nil?, !$2.nil? and so on.
The last one set gives you the length that is matchable, so in your example $2.length gives you the 7 you wanted.
Incidentally, the fact that Group 2 is the last one set also tells you that we fail on R+.
For your example, you could do the following.
Code
Change your regex from:
/^Q+E+R+$/
to
R = /^(Q*)(E*)(R*)/
and then apply the following method to the string:
def nbr_matched_chars(str)
str.scan(R).flatten.reduce(0) {|t,e| return t if e.nil?; t+e.size }
end
str matches the original regex if and only if nbr_matched_chars(str) == str.size.
Examples
nbr_matched_chars("QQQQEEE2ER") #=> 7
nbr_matched_chars("QQQQEEEERR") #=> 10 (= "QQQQEEEERR".size)
nbr_matched_chars("QQAQQEEEER") #=> 2
Explanation
To see why this [evidently :-)] works, we can look at the results of invoking String#scan, followed by Array#flatten:
"QQQQEEE2ER".scan(r).flatten #=> ["QQQQ", "EEE" , nil ]
"QQQQEEEERR".scan(r).flatten #=> ["QQQQ", "EEEE", "RR"]
"QQAQQEEEER".scan(r).flatten #=> ["QQ" , nil , nil ]
I have a string named "string" that contains six lines.
I want to remove an "Z" from the end of each line (which each has) and capitalize the first character in each line (ignoring numbers and white space; e.g., "1. apple" -> "1. Apple").
I have some idea of how to do it, but have no idea how to do it in Ruby. How do I accomplish this? A loop? What would the syntax be?
Using regular expression (See String#gsub):
s = <<EOS
1. applez
2. bananaz
3. catz
4. dogz
5. elephantz
6. fruitz
EOS
puts s.gsub(/z$/i, '').gsub(/^([^a-z]*)([a-z])/i) { $1 + $2.upcase }
# /z$/i - to match a trailing `z` at the end of lines.
# /^([^a-z]*)([a-z])/i - to match leading non-alphabets and alphabet.
# capture them as group 1 ($1), group 2 ($2)
output:
1. Apple
2. Banana
3. Cat
4. Dog
5. Elephant
6. Fruit
I would approach this by breaking your problem into smaller steps. After we've solved each of the smaller problems, you can put it all back together for a more elegant solution.
Given the initial string put forth by falsetru:
s = <<EOS
1. applez
2. bananaz
3. catz
4. dogz
5. elephantz
6. fruitz
EOS
1. Break your string into an array of substrings, separated by the newline.
substrings = s.split(/\n/)
This uses the String class' split method and a regular expression. It searches for all occurrences of newline (backslash-n) and treats this as a delimiter, splitting the string into substrings based on this delimiter. Then it throws all of these substrings into an array, which we've named substrings.
2. Iterate through your array of substrings to do some stuff (details on what stuff later)
substrings.each do |substring|
.
# Do stuff to each substring
.
end
This is one form for how you iterate across an array in Ruby. You call the Array's each method, and you give it a block of code which it will run on each element in the array. In our example, we'll use the variable name substring within our block of code so that we can do stuff to each substring.
3. Remove the z character at the end of each substring
substrings.each do |substring|
substring.gsub!(/z$/, '')
end
Now, as we iterate through the array, the first thing we want to do is remove the z character at the end of each string. You do this with the gsub! method of String, which is a search-and-replace method. The first argument for this method is the regular expression of what you're looking for. In this case, we are looking for a z followed by the end-of-string (denoted by the dollar sign). The second argument is an empty string, because we want to replace what's been found with nothing (another way of saying - we just want to remove what's been found).
4. Find the index of the first letter in each substring
substrings.each do |substring|
substring.gsub!(/z$/, '')
index = substring.index(/[a-zA-Z]/)
end
The String class also has a method called index which will return the index of the first occurrence of a string that matches the regular expression your provide. In our case, since we want to ignore numbers and symbols and spaces, we are really just looking for the first occurrence of the very first letter in your substring. To do this, we use the regular expression /[a-zA-Z]/ - this basically says, "Find me anything in the range of small A to small Z or in big A to big Z." Now, we have an index (using our example strings, the index is 3).
5. Capitalize the letter at the index we have found
substrings.each do |substring|
substring.gsub!(/z$/, '')
index = substring.index(/[a-zA-Z]/)
substring[index] = substring[index].capitalize
end
Based on the index value that we found, we want to replace the letter at that index with that same letter, but capitalized.
6. Put our substrings array back together as a single-string separated by newlines.
Now that we've done everything we need to do to each substring, our each iterator block ends, and we have what we need in the substrings array. To put the array back together as a single string, we use the join method of Array class.
result = substrings.join("\n")
With that, we now have a String called result, which should be what you're looking for.
Putting It All Together
Here is what the entire solution looks like, once we put together all of the steps:
substrings = s.split(/\n/)
substrings.each do |substring|
substring.gsub!(/z$/, '')
index = substring.index(/[a-zA-Z]/)
substring[index] = substring[index].capitalize
end
result = substrings.join("\n")