Given a sentence, I want to count all the duplicated words:
It is an exercice from Exercism.io Word count
For example for the input "olly olly in come free"
plain
olly: 2
in: 1
come: 1
free: 1
I have this test for exemple:
def test_with_quotations
phrase = Phrase.new("Joe can't tell between 'large' and large.")
counts = {"joe"=>1, "can't"=>1, "tell"=>1, "between"=>1, "large"=>2, "and"=>1}
assert_equal counts, phrase.word_count
end
this is my method
def word_count
phrase = #phrase.downcase.split(/\W+/)
counts = phrase.group_by{|word| word}.map {|k,v| [k, v.count]}
Hash[*counts.flatten]
end
For the test above I have this failure when I run it in the terminal:
2) Failure:
PhraseTest#test_with_apostrophes [word_count_test.rb:69]:
--- expected
+++ actual
## -1 +1 ##
-{"first"=>1, "don't"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
+{"first"=>1, "don"=>2, "t"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
My problem is to remove all chars except 'apostrophe...
the regex in the method almost works...
phrase = #phrase.downcase.split(/\W+/)
but it remove the apostrophes...
I don't want to keep the single quote around a word, 'Hello' => Hello
but Don't be cruel => Don't be cruel
Maybe something like:
string.scan(/\b[\w']+\b/i).each_with_object(Hash.new(0)){|a,(k,v)| k[a]+=1}
The regex employs word boundaries (\b).
The scan outputs an array of the found words and for each word in the array they are added to the hash, which has a default value of zero for each item which is then incremented.
Turns out my solution whilst finding all items and ignoring case will still leave the items in the case they were found in originally.
This would now be a decision for Nelly to either accept as is or to perform a downcase on the original string or the array item as it is added to the hash.
I'll leave that decision up to you :)
Given:
irb(main):015:0> phrase
=> "First: don't laugh. Then: don't cry."
Try:
irb(main):011:0> Hash[phrase.downcase.scan(/[a-z']+/)
.group_by{|word| word.downcase}
.map{|word, words|[word, words.size]}
]
=> {"first"=>1, "don't"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
With your update, if you want to remove single quotes, do that first:
irb(main):038:0> p2
=> "Joe can't tell between 'large' and large."
irb(main):039:0> p2.gsub(/(?<!\w)'|'(?!\w)/,'')
=> "Joe can't tell between large and large."
Then use the same method.
But you say -- gsub(/(?<!\w)'|'(?!\w)/,'') will remove the apostrophe in 'Twas the night before. Which I reply you will eventually need to build a parser that can determine the distinction between an apostrophe and a single quote if /(?<!\w)'|'(?!\w)/ is not sufficient.
You can also use word boundaries:
irb(main):041:0> Hash[p2.downcase.scan(/\b[a-z']+\b/)
.group_by{|word| word.downcase}
.map{|word, words|[word, words.size]}
]
=> {"joe"=>1, "can't"=>1, "tell"=>1, "between"=>1, "large"=>2, "and"=>1}
But that does not solve 'Tis the night either.
Another way:
str = "First: don't 'laugh'. Then: 'don't cry'."
reg = /
[a-z] #single letter
[a-z']+ #one or more letters or apostrophe
[a-z] #single letter
'? #optional single apostrophe
/ix #case-insensitive and free-spacing regex
str.scan(reg).group_by(&:itself).transform_values(&:count)
#=> {"First"=>1, "don't"=>2, "laugh"=>1, "Then"=>1, "cry'"=>1}
Related
I want to convert all the words(alphabetic) in the string to their abbreviations like i18n does. In other words I want to change "extraordinary" into "e11y" because there are 11 characters between the first and the last letter in "extraordinary". It works with a single word in the string. But how can I do the same for a multi-word string? And of course if a word is <= 4 there is no point to make an abbreviation from it.
class Abbreviator
def self.abbreviate(x)
x.gsub(/\w+/, "#{x[0]}#{(x.length-2)}#{x[-1]}")
end
end
Test.assert_equals( Abbreviator.abbreviate("banana"), "b4a", Abbreviator.abbreviate("banana") )
Test.assert_equals( Abbreviator.abbreviate("double-barrel"), "d4e-b4l", Abbreviator.abbreviate("double-barrel") )
Test.assert_equals( Abbreviator.abbreviate("You, and I, should speak."), "You, and I, s4d s3k.", Abbreviator.abbreviate("You, and I, should speak.") )
Your mistake is that your second parameter is a substitution string operating on x (the original entire string) as a whole.
Instead of using the form of gsub where the second parameter is a substitution string, use the form of gsub where the second parameter is a block (listed, for example, third on this page). Now you are receiving each substring into your block and can operate on that substring individually.
def short_form(str)
str.gsub(/[[:alpha:]]{4,}/) { |s| "%s%d%s" % [s[0], s.size-2, s[-1]] }
end
The regex reads, "match four or more alphabetic characters".
short_form "abc" # => "abc"
short_form "a-b-c" #=> "a-b-c"
short_form "cats" #=> "c2s"
short_form "two-ponies-c" #=> "two-p4s-c"
short_form "Humpty-Dumpty, who sat on a wall, fell over"
#=> "H4y-D4y, who sat on a w2l, f2l o2r"
I would recommend something along the lines of this:
class Abbreviator
def self.abbreviate(x)
x.gsub(/\w+/) do |word|
# Skip the word unless it's long enough
next word unless word.length > 4
# Do the same I18n conversion you do before
"#{word[0]}#{(word.length-2)}#{word[-1]}"
end
end
end
The accepted answer isn't bad, but it can be made a lot simpler by not matching words that are too short in the first place:
def abbreviate(str)
str.gsub(/([[:alpha:]])([[:alpha:]]{3,})([[:alpha:]])/i) { "#{$1}#{$2.size}#{$3}" }
end
abbreviate("You, and I, should speak.")
# => "You, and I, s4d s3k."
Alternatively, we can use lookbehind and lookahead, which makes the Regexp more complex but the substitution simpler:
def abbreviate(str)
str.gsub(/(?<=[[:alpha:]])[[:alpha:]]{3,}(?=[[:alpha:]])/i, &:size)
end
I answered my own question. Forgot to initialize count = 0
I have a bunch of sentences in a paragraph.
a = "Hello there. this is the best class. but does not offer anything." as an example.
To figure out if the first letter is capitalized, my thought is to .split the string so that a_sentence = a.split(".")
I know I can "hello world".capitalize! so that if it was nil it means to me that it was already capitalized
EDIT
Now I can use array method to go through value and use '.capitalize!
And I know I can check if something is .strip.capitalize!.nil?
But I can't seem to output how many were capitalized.
EDIT
a_sentence.each do |sentence|
if (sentence.strip.capitalize!.nil?)
count += 1
puts "#{count} capitalized"
end
end
It outputs:
1 capitalized
Thanks for all your help. I'll stick with the above code I can understand within the framework I only know in Ruby. :)
Try this:
b = []
a.split(".").each do |sentence|
b << sentence.strip.capitalize
end
b = b.join(". ") + "."
# => "Hello there. This is the best class. But does not offer anything."
Your post's title is misleading because from your code, it seems that you want to get the count of capitalized letters at the beginning of a sentence.
Assuming that every sentence is finishing on a period (a full stop) followed by a space, the following should work for you:
split_str = ". "
regex = /^[A-Z]/
paragraph_text.split(split_str).count do |sentence|
regex.match(sentence)
end
And if you want to simply ensure that each starting letter is capitalized, you could try the following:
paragraph_text.split(split_str).map(&:capitalize).join(split_str) + split_str
There's no need to split the string into sentences:
str = "It was the best of times. sound familiar? Out, damn spot! oh, my."
str.scan(/(?:^|[.!?]\s)\s*\K[A-Z]/).length
#=> 2
The regex could be written with documentation by adding x after the closing /:
r = /
(?: # start a non-capture group
^|[.!?]\s # match ^ or (|) any of ([]) ., ! or ?, then one whitespace char
) # end non-capture group
\s* # match any number of whitespace chars
\K # forget the preceding match
[A-Z] # match one capital letter
/x
a = str.scan(r)
#=> ["I", "O"]
a.length
#=> 2
Instead of Array#length, you could use its alias, size, or Array#count.
You can count how many were capitalized, like this:
a = "Hello there. this is the best class. but does not offer anything."
a_sentence = a.split(".")
a_sentence.inject(0) { |sum, s| s.strip!; s.capitalize!.nil? ? sum += 1 : sum }
# => 1
a_sentence
# => ["Hello there", "This is the best class", "But does not offer anything"]
And then put it back together, like this:
"#{a_sentence.join('. ')}."
# => "Hello there. This is the best class. But does not offer anything."
EDIT
As #Humza sugested, you could use count:
a_sentence.count { |s| s.strip!; s.capitalize!.nil? }
# => 1
This is my expected result.
Input a string and get three returned string.
I have no idea how to finish it with Regex in Ruby.
this is my roughly idea.
match(/(.*?)(_)(.*?)(\d+)/)
Input and expected output
# "R224_OO2003" => R224, OO, 2003
# "R2241_OOP2003" => R2244, OOP, 2003
If the example description I gave in my comment on the question is correct, you need a very straightforward regex:
r = /(.+)_(.+)(\d{4})/
Then:
"R224_OO2003".scan(r).flatten #=> ["R224", "OO", "2003"]
"R2241_OOP2003".scan(r).flatten #=> ["R2241", "OOP", "2003"]
Assuming that your three parts consist of (R and one or more digits), then an underbar, then (one or more non-whitespace characters), before finally (a 4-digit numeric date), then your regex could be something like this:
^(R\d+)_(\S+)(\d{4})$
The ^ indicates start of string, and the $ indicates end of string. \d+ indicates one or more digits, while \S+ says one or more non-whitespace characters. The \d{4} says exactly four digits.
To recover data from the matches, you could either use the pre-defined globals that line up with your groups, or you could could use named captures.
To use the match globals just use $1, $2, and $3. In general, you can figure out the number to use by counting the left parentheses of the specific group.
To use the named captures, include ? right after the left paren of a particular group. For example:
x = "R2241_OOP2003"
match_data = /^(?<first>R\d+)_(?<second>\S+)(?<third>\d{4})$/.match(x)
puts match_data['first'], match_data['second'], match_data['third']
yields
R2241
OOP
2003
as expected.
As long as your pattern covers all possibilities, then you just need to use the match object to return the 3 strings:
my_match = "R224_OO2003".match(/(.*?)(_)(.*?)(\d+)/)
#=> #<MatchData "R224_OO2003" 1:"R224" 2:"_" 3:"OO" 4:"2003">
puts my_match[0] #=> "R224_OO2003"
puts my_match[1] #=> "R224"
puts my_match[2] #=> "_"
puts my_match[3] #=> "00"
puts my_match[4] #=> "2003"
A MatchData object contains an array of each match group starting at index [1]. As you can see, index [0] returns the entire string. If you don't want the capture the "_" you can leave it's parentheses out.
Also, I'm not sure you are getting what you want with the part:
(.*?)
this basically says one or more of any single character followed by zero or one of any single character.
I tried to write a function which will be able to randomly change letters in word except first and last one.
def fun(string)
z=0
s=string.size
tab=string
a=(1...s-1).to_a.sample s-1
for i in 1...(s-1)
puts tab[i].replace(string[a[z]])
z=z+1
end
puts tab
end
fun("sample")
My output is:
p
l
a
m
sample
Anybody know how to make it my tab be correct?
it seems to change in for block, because in output was 'plamp' so it's random as I wanted but if I want to print the whole word (splampe) it doesn't working. :(
What about:
def fun(string)
first, *middle, last = string.chars
[first, middle.shuffle, last].join
end
fun("sample") #=> "smalpe"
s = 'sample'
[s[0], s[1..-2].chars.shuffle, s[-1]].join
# => "slpmae"
Here is my solution:
def fun(string)
first = string[0]
last = string[-1]
middle = string[1..-2]
puts "#{first}#{middle.split('').shuffle.join}#{last}"
end
fun('sample')
there are some problems with your function. First, when you say tab=string, tab is now a reference to string, so, when you change characters on tab you change the string characters too. I think that for clarity is better to keep the index of sample (1....n)to reference the position in the original array.
I suggest the usage of tab as a new array.
def fun(string)
if string.length <= 2
return
z=1
s=string.size
tab = []
tab[0] = string[0]
a=(1...s-1).to_a.sample(s-1)
(1...s-1).to_a.each do |i|
tab[z] = string[a[i - 1]]
z=z+1
end
tab.push string[string.size-1]
tab.join('')
end
fun("sample")
=> "spalme"
Another way, using String#gsub with a block:
def inner_jumble(str)
str.sub(/(?<=\w)\w{2,}(?=\w)/) { |s| s.chars.shuffle.join }
end
inner_jumble("pneumonoultramicroscopicsilicovolcanoconiosis") # *
#=> "poovcanaiimsllinoonroinuicclprsciscuoooomtces"
inner_jumble("what ho, fellow coders?")
#=> "waht ho, folelw coedrs?"
(?<=\w) is a ("zero-width") positive look-behind that requires the match to immediately follow a word character.
(?=\w) is a ("zero-width") positive look-ahead that requires the match to be followed immediately by a word character.
You could use \w\w+ in place of \w{2,} for matching two or more consecutive word characters.
If you only want it to apply to individual words, you can use gsub or sub.
*A lung disease caused by inhaling very fine ash and sand dust, supposedly the longest word in some English dictionaries.
I have a string composed by words divided by'#'. For instance 'this#is#an#example' and I need to extract the last word or the last two words according to the second to last word.
If the second to last is 'myword' I need the last two words otherwise just the last one.
'this#is#an#example' => 'example'
'this#is#an#example#using#myword#also' => 'myword#also'
Is there a better way than splitting and checking the second to last? perhaps using regular expression?
Thanks.
You can use the end-of-line anchor $ and make the myword# prefix optional:
str = 'this#is#an#example'
str[/(?:#)((myword#)?[^#]+)$/, 1]
#=> "example"
str = 'this#is#an#example#using#myword#also'
str[/(?:#)((myword#)?[^#]+)$/, 1]
#=> "myword#also"
However, I don't think using a regular expression is "better" in this case. I would use something like Santosh's (deleted) answer: split the line by # and use an if clause.
def foo(str)
*, a, b = str.split('#')
if a == 'myword'
"#{a}##{b}"
else
b
end
end
str = 'this#is#an#example#using#myword#also'
array = str.split('#')
array[-2] == 'myword' ? array[-2..-1].join('#') : array[-1]
With regex:
'this#is#an#example'[/(myword\#)*\w+$/]
# => "example"
'this#is#an#example#using#myword#also'[/(myword\#)*\w+$/]
# => "myword#also"