I have some sequences in a string denoted by "#number" (/#\d/)
I want to remove any redundant sequences, where #2 is followed by #2,
I only want to remove them if another identical #number sequence is found directly after somewhere in the text, so for #2lorem#2ipsum the 2nd #2 is removed, but for #2lorem#1ipsum#2dolor nothing is removed because #1 is between the two #2 sequences.
"#2randomtext#2randomtext#2randomtext#1bla#2bla2#2bla2"
becomes:
"#2randomtextrandomtextrandomtext#1bla#2bla2bla2
"#2randomtext#2randomtext#2randomtext#1bla#2bla2#2bla2".gsub /(?<=(#\d))([^#]*)\1/,'\2'
=> "#2randomtextrandomtextrandomtext#1bla#2bla2bla2"
You can split it into tokens:
my_string = "#2randomtext#2randomtext#2randomtext#1bla#2bla2#2bla2"
tokens = my_string.scan /(#\d+)?((?:(?!#\d+).)*)/
#=> [["#2", "randomtext"], ["#2", "randomtext"], ["#2", "randomtext"], ["#1", "bla"], ["#2", "bla2"], ["#2", "bla2"]]
Then chunk, map and join:
tokens.chunk{|x| x[0].to_s}.map{|n, v| [n, v.map(&:last)]}.join
#=> "#2randomtextrandomtextrandomtext#1bla#2bla2bla2"
my_string = "#2randomtext#2randomtext#2randomtext#1bla#2bla2#2bla2"
prev_sequence = String.new
penultimate_index = my_string.length - 2
for i in 0..penultimate_index
if my_string[i] == '#'
new_sequence = "##{my_string[i+1]}"
if new_sequence == prev_sequence
my_string.slice!( i, 2 )
else
prev_sequence = new_sequence
end
end
end
puts my_string
easy... split your string into an array, and then compare the number that comes right after that. If it's the same, remove it/them. The complicated (through not that much), is that you can't remove entries from an array while looping through them... so what you need to do is make a recursive function... here's the pseudo:
-= Global values =-
Decalre StringArray and set it to OriginalString.SplitOn("#")
-= Method RemoveLeadingDuplicates =-
Declare Counter
Declare RemoveIndex
loop for each string in StringArray
if previous lead == current lead
Set RemoveIndex
break from loop
else
previous lead = current lead
Increase Counter By 1
end loop
if RemoveIndex is not null
Remove the item at specified index from the array
Call RemoveLeadingDuplicates
Return
Related
Where as str[] will replace a character, str.insert will insert a character at a position. But it requires two lines of code:
str = "COSO17123456"
str.insert 4, "-"
str.insert 7, "-"
=> "COSO-17-123456"
I was thinking how to do this in one line of code. I came up with the following solution:
str = "COSO17123456"
str.each_char.with_index.reduce("") { |acc,(c,i)| acc += c + ( (i == 3 || i == 5) ? "-" : "" ) }
=> "COSO-17-123456
Is there a built-in Ruby helper for this task? If not, should I stick with the insert option rather than combining several iterators?
Use each to iterate over an array of indices:
str = "COSO17123456"
[4, 7].each { |i| str.insert i, '-' }
str #=> "COSO-17-123456"
You can uses slices and .join:
> [str[0..3], str[4..5],str[6..-1]].join("-")
=> "COSO-17-123456"
Note that the index after the first one (between 3 and 4) will be different since you are not inserting earlier insertion first. ie, more natural (to me anyway...)
You will insert at the absolute index of the original string -- not the moving relative index as insertions are made.
If you want to insert at specific absolute index values, you can also use ..each_with_index and control the behavior character by character:
str2 = ""
tgts=[3,5]
str.split("").each_with_index { |c,idx| str2+=c; str2+='-' if tgts.include? idx }
Both of the above create a new string.
String#insert returns the string itself.
This means you can chain the method calls, which can be a prettier and more efficient if you only have to do it a couple of times like in your example:
str = "COSO17123456".insert(4, "-").insert(7, "-")
puts str
COSO-17-123456
Your reduce version can be therefore more concisely written as:
[4,7].reduce(str) { |str, idx| str.insert(idx, '-') }
I'll bring one more variation to the table, String#unpack:
new_str = str.unpack("A4A2A*").join('-')
# or with String#%
new_str = "%s-%s-%s" % str.unpack("A4A2A*")
I have the following function which accepts text and a word count and if the number of words in the text exceeded the word-count it gets truncated with an ellipsis.
#Truncate the passed text. Used for headlines and such
def snippet(thought, wordcount)
thought.split[0..(wordcount-1)].join(" ") + (thought.split.size > wordcount ? "..." : "")
end
However what this function doesn't take into account is extremely long words, for instance...
"Helloooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
world!"
I was wondering if there's a better way to approach what I'm trying to do so it takes both word count and text size into consideration in an efficient way.
Is this a Rails project?
Why not use the following helper:
truncate("Once upon a time in a world far far away", :length => 17)
If not, just reuse the code.
This is probably a two step process:
Truncate the string to a max length (no need for regex for this)
Using regex, find a max words quantity from the truncated string.
Edit:
Another approach is to split the string into words, loop through the array adding up
the lengths. When you find the overrun, join 0 .. index just before the overrun.
Hint: regex ^(\s*.+?\b){5} will match first 5 "words"
The logic for checking both word and char limits becomes too convoluted to clearly express as one expression. I would suggest something like this:
def snippet str, max_words, max_chars, omission='...'
max_chars = 1+omision.size if max_chars <= omission.size # need at least one char plus ellipses
words = str.split
omit = words.size > max_words || str.length > max_chars ? omission : ''
snip = words[0...max_words].join ' '
snip = snip[0...(max_chars-3)] if snip.length > max_chars
snip + omit
end
As other have pointed out Rails String#truncate offers almost the functionality you want (truncate to fit in length at a natural boundary), but it doesn't let you independently state max char length and word count.
First 20 characters:
>> "hello world this is the world".gsub(/.+/) { |m| m[0..20] + (m.size > 20 ? '...' : '') }
=> "hello world this is t..."
First 5 words:
>> "hello world this is the world".gsub(/.+/) { |m| m.split[0..5].join(' ') + (m.split.size > 5 ? '...' : '') }
=> "hello world this is the world..."
I am currently working on a Ruby Problem quiz but I'm not sure if my solution is right. After running the check, it shows that the compilation was successful but i'm just worried it is not the right answer.
The problem:
A string S consisting only of characters '(' and ')' is called properly nested if:
S is empty,
S has the form "(U)" where
U is a properly nested string,
S has
the form "VW" where V and W are
properly nested strings.
For example, "(()(())())" is properly nested and "())" isn't.
Write a function
def nesting(s)
that given a string S returns 1 if S
is properly nested and 0 otherwise.
Assume that the length of S does not
exceed 1,000,000. Assume that S
consists only of characters '(' and
')'.
For example, given S = "(()(())())"
the function should return 1 and given
S = "())" the function should return
0, as explained above.
Solution:
def nesting ( s )
# write your code here
if s == '(()(())())' && s.length <= 1000000
return 1
elsif s == ' ' && s.length <= 1000000
return 1
elsif
s == '())'
return 0
end
end
Here are descriptions of two algorithms that should accomplish the goal. I'll leave it as an exercise to the reader to turn them into code (unless you explicitly ask for a code solution):
Start with a variable set to 0 and loop through each character in the string: when you see a '(', add one to the variable; when you see a ')', subtract one from the variable. If the variable ever goes negative, you have seen too many ')' and can return 0 immediately. If you finish looping through the characters and the variable is not exactly 0, then you had too many '(' and should return 0.
Remove every occurrence of '()' in the string (replace with ''). Keep doing this until you find that nothing has been replaced (check the return value of gsub!). If the string is empty, the parentheses were matched. If the string is not empty, it was mismatched.
You're not supposed to just enumerate the given examples. You're supposed to solve the problem generally. You're also not supposed to check that the length is below 1000000, you're allowed to assume that.
The most straight forward solution to this problem is to iterate through the string and keep track of how many parentheses are open right now. If you ever see a closing parenthesis when no parentheses are currently open, the string is not well-balanced. If any parentheses are still open when you reach the end, the string is not well-balanced. Otherwise it is.
Alternatively you could also turn the specification directly into a regex pattern using the recursive regex feature of ruby 1.9 if you were so inclined.
My algorithm would use stacks for this purpose. Stacks are meant for solving such problems
Algorithm
Define a hash which holds the list of balanced brackets for
instance {"(" => ")", "{" => "}", and so on...}
Declare a stack (in our case, array) i.e. brackets = []
Loop through the string using each_char and compare each character with keys of the hash and push it to the brackets
Within the same loop compare it with the values of the hash and pop the character from brackets
In the end, if the brackets stack is empty, the brackets are balanced.
def brackets_balanced?(string)
return false if string.length < 2
brackets_hash = {"(" => ")", "{" => "}", "[" => "]"}
brackets = []
string.each_char do |x|
brackets.push(x) if brackets_hash.keys.include?(x)
brackets.pop if brackets_hash.values.include?(x)
end
return brackets.empty?
end
You can solve this problem theoretically. By using a grammar like this:
S ← LSR | LR
L ← (
R ← )
The grammar should be easily solvable by recursive algorithm.
That would be the most elegant solution. Otherwise as already mentioned here count the open parentheses.
Here's a neat way to do it using inject:
class String
def valid_parentheses?
valid = true
self.gsub(/[^\(\)]/, '').split('').inject(0) do |counter, parenthesis|
counter += (parenthesis == '(' ? 1 : -1)
valid = false if counter < 0
counter
end.zero? && valid
end
end
> "(a+b)".valid_parentheses? # => true
> "(a+b)(".valid_parentheses? # => false
> "(a+b))".valid_parentheses? # => false
> "(a+b))(".valid_parentheses? # => false
You're right to be worried; I think you've got the very wrong end of the stick, and you're solving the problem too literally (the info that the string doesn't exceed 1,000,000 characters is just to stop people worrying about how slow their code would run if the length was 100times that, and the examples are just that - examples - not the definitive list of strings you can expect to receive)
I'm not going to do your homework for you (by writing the code), but will give you a pointer to a solution that occurs to me:
The string is correctly nested if every left bracket has a right-bracket to the right of it, or a correctly nested set of brackets between them. So how about a recursive function, or a loop, that removes the string matches "()". When you run out of matches, what are you left with? Nothing? That was a properly nested string then. Something else (like ')' or ')(', etc) would mean it was not correctly nested in the first place.
Define method:
def check_nesting str
pattern = /\(\)/
while str =~ pattern do
str = str.gsub pattern, ''
end
str.length == 0
end
And test it:
>ruby nest.rb (()(())())
true
>ruby nest.rb (()
false
>ruby nest.rb ((((()))))
true
>ruby nest.rb (()
false
>ruby nest.rb (()(((())))())
true
>ruby nest.rb (()(((())))()
false
Your solution only returns the correct answer for the strings "(()(())())" and "())". You surely need a solution that works for any string!
As a start, how about counting the number of occurrences of ( and ), and seeing if they are equal?
I'm trying to read through a file, find a certain pattern and then grabbing a set number of lines of text after the line that contains that pattern. Not really sure how to approach this.
If you want the n number of lines after the line matching pattern in the file filename:
lines = File.open(filename) do |file|
line = file.readline until line =~ /pattern/ || file.eof;
file.eof ? nil : (1..n).map { file.eof ? nil : file.readline }.compact
end
This should handle all cases, like the pattern not present in the file (returns nil) or there being less than n lines after the matching lines (the resulting array containing the last lines of the file.)
First parse the file into lines. Open, read, split on the line break
lines = File.open(file_name).read.split("\n")
Then get index
index = line.index{|x| x.match(/regex_pattern/)}
Where regex_pattern is the pattern that you are looking for. Use the index as a starting point and then the second argument is the number of lines (in this case 5)
lines[index, 5]
It will return an array of 'lines'
You could combine it a bit more to reduce the number of lines. but I was attempting to keep it readable.
If you're not tied to Ruby, grep -A 12 trivet will show the 12 lines after any line with trivet in it. Any regex will work in place of "trivet"
matched = false;
num = 0;
res = "";
new File(filename).each_line { |line|
if (matched) {
res += line+"\n";
num++;
if (num == num_lines_desired) {
break;
}
} elsif (line.match(/regex/)) {
matched = true;
}
}
This has the advantage of not needing to read the whole file in the event of a match.
When done, res will hold the desired lines.
in rails (only difference is how I generate the file object)
file = File.open(File.join(Rails.root, 'lib', 'file.json'))
#convert file into an array of strings, with \n as the separator
line_ary = file.readlines
line_count = line_ary.count
i = 0
#or however far up the document you want to be...you can get very fancy with this or just do it manually
hsh = {}
line_count.times do |l|
child_id = JSON.parse(line_ary[i])
i += 1
parent_ary = JSON.parse(line_ary[i])
i += 1
hsh[child_id] = parent_ary
end
haha I've said too much that should definitely get you started
I have a text file from which I want to create a Hash for faster access. My text file is of format (space delimited)
author title date popularity
I want to create a hash in which author is the key and the remaining is the value as an array.
created_hash["briggs"] = ["Manup", "Jun,2007", 10]
Thanks in advance.
require 'date'
created_hash = File.foreach('test.txt', mode: 'rt', encoding: 'UTF-8').
reduce({}) {|hsh, l|
name, title, date, pop = l.split
hsh.tap {|hsh| hsh[name] = [title, Date.parse(date), pop.to_i] }
}
I threw some type conversion code in there, just for fun. If you don't want that, the loop body becomes even simpler:
k, *v = l.split
hsh.tap {|hsh| hsh[k] = v }
You can also use readlines instead of foreach. Note that IO#readlines reads the entire file into an array first. So, you need enough memory to hold both the entire array and the entire hash. (Of course, the array will be eligible for garbage collection as soon as the loop finishes.)
Just loop through each line of the file, use the first space-delimited item as the hash key and the rest as the hash value. Pretty much exactly as you described.
created_hash = {}
file_contents.each_line do |line|
data = line.split(' ')
created_hash[data[0]] = data.drop 1
end