Ruby getting the longest word of a sentence - ruby

I'm trying to create method named longest_word that takes a sentence as an argument and The function will return the longest word of the sentence.
My code is:
def longest_word(str)
words = str.split(' ')
longest_str = []
return longest_str.max
end

The shortest way is to use Enumerable's max_by:
def longest(string)
string.split(" ").max_by(&:length)
end

Using regexp will allow you to take into consideration punctuation marks.
s = "lorem ipsum, loremmm ipsummm? loremm ipsumm...."
first longest word:
s.split(/[^\w]+/).max_by(&:length)
# => "loremmm"
# or using scan
s.scan(/\b\w+\b/).max_by(&:length)
# => "loremmm"
Also you may be interested in getting all longest words:
s.scan(/\b\w+\b/).group_by(&:length).sort.last.last
# => ["loremmm", "ipsummm"]

It depends on how you want to split the string. If you are happy with using a single space, than this works:
def longest(source)
arr = source.split(" ")
arr.sort! { |a, b| b.length <=> a.length }
arr[0]
end
Otherwise, use a regular expression to catch whitespace and puntuaction.

def longest_word(sentence)
longest_word = ""
words = sentence.split(" ")
words.each do |word|
longest_word = word unless word.length < longest_word.length
end
longest_word
end
That's a simple way to approach it. You could also strip the punctuation using a gsub method.

Funcional Style Version
str.split(' ').reduce { |r, w| w.length > r.length ? w : r }
Another solution using max
str.split(' ').max { |a, b| a.length <=> b.length }

sort_by! and reverse!
def longest_word(sentence)
longw = sentence.split(" ")
longw.sort_by!(&:length).reverse!
p longw[0]
end
longest_word("once upon a time long ago a very longword")

If you truly want to do it in the Ruby way it would be:
def longest(sentence)
sentence.split(' ').sort! { |a, b| b.length <=> a.length }[0]
end

This is to strip the word from the extra chars
sen.gsub(/[^0-9a-z ]/i, '').split(" ").max_by(&:length)

Find Longest word in a string
sentence = "Hi, my name is Mesut. There is longestword here!"
def longest_word(string)
long = ""
string.split(" ").each do |sent|
if sent.length >= long.length
long = sent
end
end
return long
end
p longest_word(sentence)

Related

How to I get the right output when there is more than vowel in each word? My code only works with one vowel in each word

Aba is a German children’s game where secret messages are exchanged. In Aba,
after every vowel we add “b” and add that same vowel.
Write a method aba_translate that takes in a sentence string and returns a new
sentence representing its Aba translation. Capitalized words of the original sentence
should be properly capitalized in the new sentence.
aba_translate(“Cats and dogs”) #=> “Cabats aband dobogs”
aba_translate(“Everyone can code”) #=> “Ebeveryobonebe caban cobodebe”
aba_translate(“Africa is Africa in German”) #=> “Abafribicaba ibis Abafribicaba ibin
Gebermaban”
My code:
def aba_translate(sentence)
translation = []
words = sentence.split(" ")
vowels = "aeiou"
vowel = ""
before = ""
after = ""
full = ""
words.each do |word|
word.each_char.with_index do |char, idx|
if vowels.include?(char)
vowel = char
before = word[0...idx]
after = word[idx+1..-1]
full = before + vowel + "b" + vowel + after
translation << full
end
end
end
return translation.join(" ")
end
puts aba_translate("Cats and dogs")
puts aba_translate("Everyone can code")
puts aba_translate("Africa is Africa in German")
Your code generates a whole new word every time it sees a vowel. Instead you need to build each word character by character and make changes when it sees a vowel.
def aba_translate(sentence)
translation = []
words = sentence.split(" ")
vowels = "aeiouAEIOU"
words.each do |word|
full = ""
word.each_char.with_index do |char, idx|
full += char
if vowels.include?(char)
full = full + "b" + char.downcase
end
end
translation << full
end
return translation.join(" ")
end
Every time you find a vowel, you take the entire string before the vowel and the entire string after the vowel and add it to the result.
So, for a word like "code", that means you first produce the output c + obo + de and then the output cod + ebe.
However, what you actually need to do is simply keep the part you have processed instead of duplicating it.
You can do this by either changing your logic to keep track of up to which index you have already processed the word, or alternatively by processing it character-by-character instead of chunk-by-chunk.
However for problems like this, Regex are usually a much better solution:
VOWELS = 'aeiou'
def aba_translate(sentence)
sentence.gsub(Regexp.union(*VOWELS.chars), '\0b\0')
end
or just making VOWELS a Regexp in the first place:
VOWELS = /[aeiou]/.freeze
def aba_translate(sentence)
sentence.gsub(VOWELS, '\0b\0')
end
I think the nested loops complicates it since you can solve the problem with just one loop. Here you just need to initialize a string and a string of vowels to check each character. While you iterate through each character you shovel it into the empty string, and then you check if it is a vowel you shovel b + that vowel's lowercase version to account for the uppercase instances. Then you finally return the new string.
def aba_translate(string)
new_string = ""
vowels = "AEIOUaeiou"
string.each_char do |char|
new_string << char
if vowels.include?(char)
new_string << "b" + char.downcase
end
end
return new_string
end
Here's my beginner-friendly solution
def aba_translate(sent)
vowels = "AEIOUaeiou"
aba_sent = ""
sent.each_char do |char|
if vowels.include?(char)
aba_sent += char + "b" + char.downcase
else
aba_sent += char
end
end
return aba_sent
end

Trying to remove punctuation without using regex

I am trying to remove punctuation from an array of words without using regular expression. In below eg,
str = ["He,llo!"]
I want:
result # => ["Hello"]
I tried:
alpha_num="abcdefghijklmnopqrstuvwxyz0123456789"
result= str.map do |punc|
punc.chars {|ch|alpha_num.include?(ch)}
end
p result
But it returns ["He,llo!"] without any change. Can't figure out where the problem is.
include? block returns true/false, try use select function to filter illegal characters.
result = str.map {|txt| txt.chars.select {|c| alpha_num.include?(c.downcase)}}
.map {|chars| chars.join('')}
p result
str=["He,llo!"]
alpha_num="abcdefghijklmnopqrstuvwxyz0123456789"
Program
v=[]<<str.map do |x|
x.chars.map do |c|
alpha_num.chars.map.include?(c.downcase) ? c : nil
end
end.flatten.compact.join
p v
Output
["Hello"]
exclusions = ((32..126).map(&:chr) - [*'a'..'z', *'A'..'Z', *'0'..'9']).join
#=> " !\"\#$%&'()*+,-./:;<=>?#[\\]^_`{|}~"
arr = ['He,llo!', 'What Ho!']
arr.map { |word| word.delete(exclusions) }
#=> ["Hello", "WhatHo"]
If you could use a regular expression and truly only wanted to remove punctuation, you could write the following.
arr.map { |word| word.gsub(/[[:punct:]]/, '') }
#=> ["Hello", "WhatHo"]
See String#delete. Note that arr is not modified.

String compressor (Ruby)

Here is my code in ruby for a word compression.
For any given word (e.g. abbbcca) the compressed word/output should be in the format as "letter+repetition" (for above example, output: a1b3c2a1).
Here I'm so close to the completion but my result isn't in the expected format. It's counting the whole letters in string.chars.each thus resulting output as a2b3c2a2.
Any help?
def string_compressor(string)
new_string = []
puts string.squeeze
string.squeeze.chars.each { |s|
count = 0
string.chars.each { |w|
if [s] == [w]
count += 1
end
}
new_string << "#{s}#{count}"
puts "#{new_string}"
}
if new_string.length > string.length
return string
elsif new_string.length < string.length
return new_string
else "Equal"
end
end
string_compressor("abbbcca")
'abbbcca'.chars.chunk{|c| c}.map{|c, a| [c, a.size]}.flatten.join
Adapted from a similar question.
Similar:
'abbbcca'.chars.chunk{|c| c}.map{|c, a| "#{c}#{a.size}"}.join
See chunk documentation
You can use a regular expression for that.
'abbbcca'.gsub(/(.)\1*/) { |m| "%s%d" % [m[0], m.size] }
#=> "a1b3c2a1"
The regular expression reads, "match any character, capturing it in group 1. Then match the contents of capture group 1 zero or more times".
As you said, your code counts every letter in the string, not just the one grouped next to one another.
Here's a modified version :
def display_count(count)
if count == 1
""
else
count.to_s
end
end
def string_compressor(string)
new_string = ''
last_char = nil
count = 0
string.chars.each do |char|
if char == last_char
count += 1
else
new_string << "#{last_char}#{display_count(count)}" if last_char
last_char = char
count = 1
end
end
new_string << "#{last_char}#{display_count(count)}" if last_char
new_string
end
p string_compressor('abbbcca') #=> "ab3c2a"
p string_compressor('aaaabbb') #=> "a4b3"
p string_compressor('aabb') #=> "a2b2"
p string_compressor('abc') #=> "abc"
Note that with display_count removing 1s from the string, new_string can never be longer than string. It also probably isn't a good idea to return Equal as a supposedly compressed string.
To decompress the string :
def string_decompressor(string)
string.gsub(/([a-z])(\d+)/i){$1*$2.to_i}
end
p string_decompressor("a5b11") #=> "aaaaabbbbbbbbbbb"
p string_decompressor("ab3c2a") #=> "abbbcca"

Reverse words of a string in Ruby?

I am trying to reverse the words of a string in Ruby, without using the reverse method. I want to implement the known algorithm of:
Reverse the whole string
Reverse each word in the reversed string.
Here is what I have come up with:
class String
def custom_reverse(start, limit)
i_start = start
i_end = limit - 1
while (i_start <= i_end)
tmp = self[i_start]
self[i_start] = self[i_end]
self[i_end] = tmp
i_start += 1
i_end -= 1
end
return self
end
def custom_reverse_words
self.custom_reverse(0, self.size)
i_start = 0
i_end = 0
while (i_end <= self.length)
if (i_end == self.length || self[i_end] == ' ')
self.custom_reverse(i_start, i_end)
i_start += 1
end
i_end += 1
end
end
end
test_str = "hello there how are you"
p test_str.custom_reverse_words
But the results are "yahthello ow ou er ereh"
What am I missing?
The gist of any reverse operation is to iterate over elements in the reverse order of what you'd normally do. That is, where you'd usually use the set (0..N-1) you'd instead go through (N-1..0) or more specifically N-1-i where i is 0..N-1:
class String
def reverse_words
split(/\s+/).map{|w|wl=w.length-1;(0..wl).map{|i|w[wl-i]}.join}.join(' ')
end
end
puts "this is reverse test".reverse_words.inspect
# => "siht si esrever tset"
The same principle can be applied to the words in a given string.
Interview questions of this sort are of highly dubious value. Being "clever" in production code is usually a Very Bad Idea.
Here's one way to reverse an array without using the built-in reverse:
class Array
def reverse
tmp_ary = self.dup
ret_ary = []
self.size.times do
ret_ary << tmp_ary.pop
end
ret_ary
end
end
%w[a b c].reverse # => ["c", "b", "a"]
tmp_ary.pop is the secret. pop removes elements from the end of the array.
The cleanest solution I could think of is:
class Array
def my_reverse
sort_by.with_index {|_, i| -i}
end
end
class String
def words
split(/\W+/)
end
def revert_words
words.my_reverse.join(' ')
end
def revert_each_word
words.map {|w| w.chars.my_reverse.join}.join(' ')
end
end
Once you define a simple and efficient array reverser:
def reverse_array(a)
(a.length / 2).times {|i| a[i],a[-(i+1)] = a[-(i+1)],a[i]}
a
end
You can reverse a sentence pretty straightforwardly:
def reverse_sentence(s)
reverse_array(s.split('')).join.split(" ").map{|w| reverse_array(w.split('')).join}.join(" ")
end
reverse_sentence "Howdy pardner" # => "pardner Howdy"
Here's another way:
class String
def reverse_words
split.inject([]){|str, word| str.unshift word}.join(' ')
end
def reverse_chars
each_char.inject([]){|str, char| str.unshift char}.join('')
end
end
Revised
Carey raises a good point, reverse_chars can be simplified, since string is already an Enumerable:
class String
def reverse_chars
each_char.inject(""){|str, char| str.insert(0, char) }
end
end

How to get words frequency in efficient way with ruby?

Sample input:
"I was 09809 home -- Yes! yes! You was"
and output:
{ 'yes' => 2, 'was' => 2, 'i' => 1, 'home' => 1, 'you' => 1 }
My code that does not work:
def get_words_f(myStr)
myStr=myStr.downcase.scan(/\w/).to_s;
h = Hash.new(0)
myStr.split.each do |w|
h[w] += 1
end
return h.to_a;
end
print get_words_f('I was 09809 home -- Yes! yes! You was');
This works but I am kinda new to Ruby too. There might be a better solution.
def count_words(string)
words = string.split(' ')
frequency = Hash.new(0)
words.each { |word| frequency[word.downcase] += 1 }
return frequency
end
Instead of .split(' '), you could also do .scan(/\w+/); however, .scan(/\w+/) would separate aren and t in "aren't", while .split(' ') won't.
Output of your example code:
print count_words('I was 09809 home -- Yes! yes! You was');
#{"i"=>1, "was"=>2, "09809"=>1, "home"=>1, "yes"=>2, "you"=>1}
def count_words(string)
string.scan(/\w+/).reduce(Hash.new(0)){|res,w| res[w.downcase]+=1;res}
end
Second variant:
def count_words(string)
string.scan(/\w+/).each_with_object(Hash.new(0)){|w,h| h[w.downcase]+=1}
end
def count_words(string)
Hash[
string.scan(/[a-zA-Z]+/)
.group_by{|word| word.downcase}
.map{|word, words|[word, words.size]}
]
end
puts count_words 'I was 09809 home -- Yes! yes! You was'
This code will ask you for input and then find the word frequency for you:
puts "enter some text man"
text = gets.chomp
words = text.split(" ")
frequencies = Hash.new(0)
words.each { |word| frequencies[word.downcase] += 1 }
frequencies = frequencies.sort_by {|a, b| b}
frequencies.reverse!
frequencies.each do |word, frequency|
puts word + " " + frequency.to_s
end
This works, and ignores the numbers:
def get_words(my_str)
my_str = my_str.scan(/\w+/)
h = Hash.new(0)
my_str.each do |s|
s = s.downcase
if s !~ /^[0-9]*\.?[0-9]+$/
h[s] += 1
end
end
return h
end
print get_words('I was there 1000 !')
puts '\n'
You can look at my code that splits the text into words. The basic code would look as follows:
sentence = "Ala ma kota za 5zł i 10$."
splitter = SRX::Polish::WordSplitter.new(sentence)
histogram = Hash.new(0)
splitter.each do |word,type|
histogram[word.downcase] += 1 if type == :word
end
p histogram
You should be careful if you wish to work with languages other than English, since in Ruby 1.9 the downcase won't work as you expected for letters such as 'Ł'.
class String
def frequency
self.scan(/[a-zA-Z]+/).each.with_object(Hash.new(0)) do |word, hash|
hash[word.downcase] += 1
end
end
end
puts "I was 09809 home -- Yes! yes! You was".frequency

Resources