I have a string of integers that can be separated by a comma, hyphen or both. e.g.
str = '3,7-17,21'
I need to take those numbers and reduce them by 1 without losing their position in that string.
When I split, I lose commas and hyphens so their positions change.
Is there a nice way to turn my string into 2,6-16,20?
I started on the following but just ended up in a rabbit hole:
def reduce_hours(hours)
hours
.enum_for(:scan, /\d+/)
.map { Regexp.last_match(0) }
.map(&:to_i)
.map { |hour| hour - 1 }
.then { hours }
end
hours = '3,7-17,21'
It just returns the original string. When I debug, it does match the numbers individually and reduce them by 1. It just doesn't return those changes to the original string. Am I close with that snippet?
enum_for(:scan, ...), Regexp.last_match(0), then – you're overthinking it.
You can just use gsub to replace each numeric substring by its predecessor:
str = '3,7-17,21'
str.gsub(/\d+/) { |hour| hour.to_i - 1 }
#=> "2,6-16,20"
The to_i round trip is needed, because strings can't be decremented directly.
The opposite direction is even easier: (via String#succ)
"2,6-16,20".gsub(/\d+/, &:succ)
#=> "3,7-17,21"
Try this:
def process_comma(str)
str.split(',').map do |sub_str|
sub_str.include?('-') ? process_hiphen(sub_str) : (sub_str.to_i - 1).to_s
end.join(',')
end
def process_hiphen(str)
str.split('-').map do |sub_str|
(sub_str.to_i - 1).to_s
end.join('-')
end
str = '3,7-17,21'
process_comma(str) # "2,6-16,20"
You can go by each possible delimiter.
hours = '3,7-17,21'
def reduce_hours(hours)
hours
.split('-')
.map { |comma_seperated_strings|
comma_seperated_strings
.split(',')
.map(&:to_i)
.map{|x| x-1}
.map(&:to_s)
.join(',')
}
.join('-')
end
puts reduce_hours(hours)
Related
I want to find if the ending of a string overlaps with the beginning of separate string. For example if I have these two strings:
string_1 = 'People say nothing is impossible, but I'
string_2 = 'but I do nothing every day.'
How do I find that the "but I" part at the end of string_1 is the same as the beginning of string_2?
I could write a method to loop over the two strings, but I'm hoping for an answer that has a Ruby string method that I missed or a Ruby idiom.
Set MARKER to some string that never appears in your string_1 and string_2. There are ways to do that dynamically, but I assume you can come up with some fixed such string in your case. I assume:
MARKER = "###"
to be safe for you case. Change it depending on your use case. Then,
string_1 = 'People say nothing is impossible, but I'
string_2 = 'but I do nothing every day.'
(string_1 + MARKER + string_2).match?(/(.+)#{MARKER}\1/) # => true
string_1 = 'People say nothing is impossible, but I'
string_2 = 'but you do nothing every day.'
(string_1 + MARKER + string_2).match?(/(.+)#{MARKER}\1/) # => false
You can use a simple loop and test at the end:
a=string_1.split(/\b/)
idx=0
while (idx<=a.length) do
break if string_2.start_with?(a[idx..-1].join)
idx+=1
end
p a[idx..-1].join if idx<a.length
Since this starts at 0, the longest sub string overlap is found.
You can use the same logic in a .detect block on the same array:
> a[(0..a.length).detect { |idx| string_2.start_with?(a[idx..-1].join) }..-1].join
=> "but I"
Or, as pointed out in comments, you can use the strings vs the array
string_1[(0..string_1.length).detect { |idx| string_2.start_with?(string_1[idx..-1]) }..-1]
Here's a solution that works by comparing the end of string_1 to the start of string_2—using the greatest common length as a starting point—with at least one matching character. It returns the index (from the end of string_1 or the beginning of string_2) if any matching character(s) are found, which can be used to extract the matching portion.
class String
def oindex(other)
[length, other.length].min.downto(1).detect do |i|
end_with?(other[0, i])
end
end
end
string_1 = 'People say nothing is impossible, but I'
string_2 = 'but I do nothing every day.'
if (idx = string_1.oindex(string_2))
puts "Last #{idx} characters match: #{string_1[-idx..-1]}"
end
Here's an alternative that finds all the indexes of the first character of the other string in the string, and uses those indexes as starting points to check for matches:
class String
def each_index(other)
return enum_for(__callee__, other) unless block_given?
i = -1
yield i while i = index(other, i.succ)
end
def oindex(other)
each_index(other.chr).detect do |i|
other.start_with?(self[i..-1]) and break length - i
end
end
end
This should be more efficient than checking every index, especially on longer strings with shorter matches, but I haven't benchmarked it.
Here are a couple of ways to do that. The first converts the two strings to arrays and then compares sequences from those arrays. The second operates on the two strings directly, comparing substrings.
#1 Convert strings to arrays and compare sequences from those arrays
Here's a simple alternative that requires the strings to be converted to arrays of words. It assumes all pairs of words are separated by one space.
def begins_with_ends?(end_str, begin_str)
end_arr = end_str.split
begin_arr = begin_str.split
!!begin_arr.each_index.find { |i| begin_arr[0,i+1] == end_arr[-1-i..-1] }
end
!!obj converts obj to false when it's "falsy" (nil or false) and to true when it's "truthy" (not "falsy"). For example, !!3 #=> true and !!nil #=> false.
end_str = 'People say nothing is impossible, but I when I'
begin_str = 'but I when I do nothing every day.'
begins_with_ends?(end_str, begin_str)
#=> true
Here the match is on the second word "I" in begin_str. Often, however, the last word of end_str only matches (at most) a single word in begin_str
#2 Compare substrings
I've implemented the following algorithm.
Set start_search to 0.
Attempt to match the last word of end_str (value of target) in begin_str, beginning at offset start_search. If no match is found return false; else let idx be the index of start_str where the last character of target appears.
Return true if the string comprised of the first idx characters of begin_str equals the string comprised by the last idx characters of end_str; else set start_search = idx + 2 and repeat step 2.
def begins_with_ends?(end_str, begin_str)
target = end_str[/[[:alnum:]]+\z/]
start_idx = 0
loop do
idx = begin_str.index(/\b#{target}\b/, start_idx)
return false if idx.nil?
idx += target.size
return true if end_str[-idx..-1] == begin_str[0, idx]
start_idx = idx + 2
end
end
begins_with_ends?(end_str, begin_str)
#=> true
This approach recognizes different numbers of spaces between the same two words in both strings (in which case there is no match).
Perhaps something like this would meet your needs?
string_1.split(' ') - string_2.split(' ')
=> ["People", "say", "is", "impossible,"]
Or this is more convoluted, but would give you the exact overlap:
string_2.
chars.
each_with_index.
map { |_, i| string_1.match(string_2[0..i]) }.
select { |s| s }.
max { |x| x.length }.
to_s
=> "but I"
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'm attempting to write a function that takes a string and returns it with all vowels removed. Below is my code.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] == "a"
i = i + 1
elsif new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
When I run the code, it returns the exact string that I entered for (str). For example, if I enter "apple", it returns "apple".
This was my original code. It had the same result.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
I need to know what I am doing wrong using this methodology. What am I doing wrong?
Finding the bug
Let's see what's wrong with your original code by executing your method's code in IRB:
$ irb
irb(main):001:0> str = "apple"
#=> "apple"
irb(main):002:0> new = str.split(" ")
#=> ["apple"]
Bingo! ["apple"] is not the expected result. What does the documentation for String#split say?
split(pattern=$;, [limit]) → anArray
Divides str into substrings based on a delimiter, returning an array of these substrings.
If pattern is a String, then its contents are used as the delimiter when splitting str. If pattern is a single space, str is split on whitespace, with leading whitespace and runs of contiguous whitespace characters ignored.
Our pattern is a single space, so split returns an array of words. This is definitely not what we want. To get the desired result, i.e. an array of characters, we could pass an empty string as the pattern:
irb(main):003:0> new = str.split("")
#=> ["a", "p", "p", "l", "e"]
"split on empty string" feels a bit hacky and indeed there's another method that does exactly what we want: String#chars
chars → an_array
Returns an array of characters in str. This is a shorthand for str.each_char.to_a.
Let's give it a try:
irb(main):004:0> new = str.chars
#=> ["a", "p", "p", "l", "e"]
Perfect, just as advertised.
Another bug
With the new method in place, your code still doesn't return the expected result (I'm going to omit the IRB prompt from now on):
vowel("apple") #=> "elpp"
This is because
result = new[i] + result
prepends the character to the result string. To append it, we have to write
result = result + new[i]
Or even better, use the append method String#<<:
result << new[i]
Let's try it:
def vowel(str)
result = ""
new = str.chars
i = 0
while i < new.length
if new[i] != "a"
result << new[i]
end
i = i + 1
end
return result
end
vowel("apple") #=> "pple"
That looks good, "a" has been removed ("e" is still there, because you only check for "a").
Now for some refactoring.
Removing the explicit loop counter
Instead of a while loop with an explicit loop counter, it's more idiomatic to use something like Integer#times:
new.length.times do |i|
# ...
end
or Range#each:
(0...new.length).each do |i|
# ...
end
or Array#each_index:
new.each_index do |i|
# ...
end
Let's apply the latter:
def vowel(str)
result = ""
new = str.chars
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
return result
end
Much better. We don't have to worry about initializing the loop counter (i = 0) or incrementing it (i = i + 1) any more.
Avoiding character indices
Instead of iterating over the character indices via each_index:
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
we can iterate over the characters themselves using Array#each:
new.each do |char|
if char != "a"
result << char
end
end
Removing the character array
We don't even have to create the new character array. Remember the documentation for chars?
This is a shorthand for str.each_char.to_a.
String#each_char passes each character to the given block:
def vowel(str)
result = ""
str.each_char do |char|
if char != "a"
result << char
end
end
return result
end
The return keyword is optional. We could just write result instead of return result, because a method's return value is the last expression that was evaluated.
Removing the explicit string
Ruby even allows you to pass an object into the loop using Enumerator#with_object, thus eliminating the explicit result string:
def vowel(str)
str.each_char.with_object("") do |char, result|
if char != "a"
result << char
end
end
end
with_object passes "" into the block as result and returns it (after the characters have been appended within the block). It is also the last expression in the method, i.e. its return value.
You could also use if as a modifier, i.e.:
result << char if char != "a"
Alternatives
There are many different ways to remove characters from a string.
Another approach is to filter out the vowel characters using Enumerable#reject (it returns a new array containing the remaining characters) and then join the characters (see Nathan's answer for a version to remove all vowels):
def vowel(str)
str.each_char.reject { |char| char == "a" }.join
end
For basic operations like string manipulation however, Ruby usually already provides a method. Check out the other answers for built-in alternatives:
str.delete('aeiouAEIOU') as shown in Gagan Gami's answer
str.tr('aeiouAEIOU', '') as shown in Cary Swoveland's answer
str.gsub(/[aeiou]/i, '') as shown in Avinash Raj's answer
Naming things
Cary Swoveland pointed out that vowel is not the best name for your method. Choose the names for your methods, variables and classes carefully. It's desirable to have a short and succinct method name, but it should also communicate its intent.
vowel(str) obviously has something to do with vowels, but it's not clear what it is. Does it return a vowel or all vowels from str? Does it check whether str is a vowel or contains a vowel?
remove_vowels or delete_vowels would probably be a better choice.
Same for variables: new is an array of characters. Why not call it characters (or chars if space is an issue)?
Bottom line: read the fine manual and get to know your tools. Most of the time, an IRB session is all you need to debug your code.
I should use regex.
str.gsub(/[aeiou]/i, "")
> string= "This Is my sAmple tExt to removE vowels"
#=> "This Is my sAmple tExt to removE vowels"
> string.delete 'aeiouAEIOU'
#=> "Ths s my smpl txt t rmv vwls"
You can create a method like this:
def remove_vowel(str)
result = str.delete 'aeiouAEIOU'
return result
end
remove_vowel("Hello World, This is my sample text")
# output : "Hll Wrld, Ths s my smpl txt"
Live Demo
Assuming you're trying to learn about the basics of programming, rather than finding the quickest one-liner to do this (which would be to use a regular expression as Avinash has said), you have a number of problems with your code you need to change.
new = str.split(" ")
This line is likely the culprit, because it splits the string based on spaces. So your input string would have to be "a p p l e" to have the effect you're looking for.
new = str.split("")
You should also remove the duplicate i = i+1 once you've changed that.
As others have already identified the problems with the OP's code, I will merely suggest an alternative; namely, you could use String#tr:
"Now is the time for all good people...".tr('aeiouAEIOU', '')
#=> "Nw s th tm fr ll gd ppl..."
If regex is not allowed, you can do it this way:
def remove_vowels(string)
string.split("").delete_if { |letter| %w[a e i o u].include? letter }.join
end
Say that we want to count the number of words in a document. I know we can do the following:
text.each_line(){ |line| totalWords = totalWords + line.split.size }
Say, that I just want to add some exceptions, such that, I don't want to count the following as words:
(1) numbers
(2) standalone letters
(3) email addresses
How can we do that?
Thanks.
You can wrap this up pretty neatly:
text.each_line do |line|
total_words += line.split.reject do |word|
word.match(/\A(\d+|\w|\S*\#\S+\.\S+)\z/)
end.length
end
Roughly speaking that defines an approximate email address.
Remember Ruby strongly encourages the use of variables with names like total_words and not totalWords.
assuming you can represent all the exceptions in a single regular expression regex_variable, you could do:
text.each_line(){ |line| totalWords = totalWords + line.split.count {|wrd| wrd !~ regex_variable }
your regular expression could look something like:
regex_variable = /\d.|^[a-z]{1}$|\A([^#\s]+)#((?:[-a-z0-9]+\.)+[a-z]{2,})\Z/i
I don't claim to be a regex expert, so you may want to double check that, particularly the email validation part
In addition to the other answers, a little gem hunting came up with this:
WordsCounted Gem
Get the following data from any string or readable file:
Word count
Unique word count
Word density
Character count
Average characters per word
A hash map of words and the number of times they occur
A hash map of words and their lengths
The longest word(s) and its length
The most occurring word(s) and its number of occurrences.
Count invividual strings for occurrences.
A flexible way to exclude words (or anything) from the count. You can pass a string, a regexp, an array, or a lambda.
Customisable criteria. Pass your own regexp rules to split strings if you prefer. The default regexp has two features:
Filters special characters but respects hyphens and apostrophes.
Plays nicely with diacritics (UTF and unicode characters): "São Paulo" is treated as ["São", "Paulo"] and not ["S", "", "o", "Paulo"].
Opens and reads files. Pass in a file path or a url instead of a string.
Have you ever started answering a question and found yourself wandering, exploring interesting, but tangential issues, or concepts you didn't fully understand? That's what happened to me here. Perhaps some of the ideas might prove useful in other settings, if not for the problem at hand.
For readability, we might define some helpers in the class String, but to avoid contamination, I'll use Refinements.
Code
module StringHelpers
refine String do
def count_words
remove_punctuation.split.count { |w|
!(w.is_number? || w.size == 1 || w.is_email_address?) }
end
def remove_punctuation
gsub(/[.!?,;:)](?:\s|$)|(?:^|\s)\(|\-|\n/,' ')
end
def is_number?
self =~ /\A-?\d+(?:\.\d+)?\z/
end
def is_email_address?
include?('#') # for testing only
end
end
end
module CountWords
using StringHelpers
def self.count_words_in_file(fname)
IO.foreach(fname).reduce(0) { |t,l| t+l.count_words }
end
end
Note that using must be in a module (possibly a class). It does not work in main, presumably because that would make the methods available in the class self.class #=> Object, which would defeat the purpose of Refinements. (Readers: please correct me if I'm wrong about the reason using must be in a module.)
Example
Let's first informally check that the helpers are working correctly:
module CheckHelpers
using StringHelpers
s = "You can reach my dog, a 10-year-old golden, at fido#dogs.org."
p s = s.remove_punctuation
#=> "You can reach my dog a 10 year old golden at fido#dogs.org."
p words = s.split
#=> ["You", "can", "reach", "my", "dog", "a", "10",
# "year", "old", "golden", "at", "fido#dogs.org."]
p '123'.is_number? #=> 0
p '-123'.is_number? #=> 0
p '1.23'.is_number? #=> 0
p '123.'.is_number? #=> nil
p "fido#dogs.org".is_email_address? #=> true
p "fido(at)dogs.org".is_email_address? #=> false
p s.count_words #=> 9 (`'a'`, `'10'` and "fido#dogs.org" excluded)
s = "My cat, who has 4 lives remaining, is at abbie(at)felines.org."
p s = s.remove_punctuation
p s.count_words
end
All looks OK. Next, put I'll put some text in a file:
FName = "pets"
text =<<_
My cat, who has 4 lives remaining, is at abbie(at)felines.org.
You can reach my dog, a 10-year-old golden, at fido#dogs.org.
_
File.write(FName, text)
#=> 125
and confirm the file contents:
File.read(FName)
#=> "My cat, who has 4 lives remaining, is at abbie(at)felines.org.\n
# You can reach my dog, a 10-year-old golden, at fido#dogs.org.\n"
Now, count the words:
CountWords.count_words_in_file(FName)
#=> 18 (9 in ech line)
Note that there is at least one problem with the removal of punctuation. It has to do with the hyphen. Any idea what that might be?
Something like...?
def is_countable(word)
return false if word.size < 2
return false if word ~= /^[0-9]+$/
return false if is_an_email_address(word) # you need a gem for this...
return true
end
wordCount = text.split().inject(0) {|count,word| count += 1 if is_countable(word) }
Or, since I am jumping to the conclusion that you can just split your entire text into an array with split(), you might need:
wordCount = 0
text.each_line do |line|
line.split.each{|word| wordCount += 1 if is_countable(word) }
end
I want my output to search and count the frequency of the words "candy" and "gram", but also the combinations of "candy gram" and "gram candy," in a given text (whole_file.)
I am currently using the following code to display the occurrences of "candy" and "gram," but when I aggregate the combinations within the %w, only the word and frequencies of "candy" and "gram" display. Should I try a different way? thanks so much.
myArray = whole_file.split
stop_words= %w{ candy gram 'candy gram' 'gram candy' }
nonstop_words = myArray - stop_words
key_words = myArray - nonstop_words
frequency = Hash.new (0)
key_words.each { |word| frequency[word] +=1 }
key_words = frequency.sort_by {|x,y| x }
key_words.each { |word, frequency| puts word + ' ' + frequency.to_s }
It sounds like you're after n-grams. You could break the text into combinations of consecutive words in the first place, and then count the occurrences in the resulting array of word groupings. Here's an example:
whole_file = "The big fat man loves a candy gram but each gram of candy isn't necessarily gram candy"
[["candy"], ["gram"], ["candy", "gram"], ["gram", "candy"]].each do |term|
terms = whole_file.split(/\s+/).each_cons(term.length).to_a
puts "#{term.join(" ")} #{terms.count(term)}"
end
EDIT: As was pointed out in the comments below, I wasn't paying close enough attention and was splitting the file on each loop which is obviously not a good idea, especially if it's large. I also hadn't accounted for the fact that the original question may've need to sort by the count, although that wasn't explicitly asked.
whole_file = "The big fat man loves a candy gram but each gram of candy isn't necessarily gram candy"
# This is simplistic. You would need to address punctuation and other characters before
# or at this step.
split_file = whole_file.split(/\s+/)
terms_to_count = [["candy"], ["gram"], ["candy", "gram"], ["gram", "candy"]]
counts = []
terms_to_count.each do |term|
terms = split_file.each_cons(term.length).to_a
counts << [term.join(" "), terms.count(term)]
end
# Seemed like you may need to do sorting too, so here that is:
sorted = counts.sort { |a, b| b[1] <=> a[1] }
sorted.each do |count|
puts "#{count[0]} #{count[1]}"
end
Strip punctuation and convert to lower-case
The first thing you probably want to do is remove all punctuation from the string holding the contents of the file and then convert what's left to lower case, the latter so you don't have worry about counting 'Cat' and 'cat' as the same word. Those two operations can be done in either order.
Changing upper-case letters to lower-case is easy:
text = whole_file.downcase
To remove the punctuation it is probably easier to decide what to keep rather than what to discard. If we only want to keep lower-case letters, you can do this:
text = whole_file.downcase.gsub(/[^a-z]/, '')
That is, substitute an empty string for all characters other than (^) lowercase letters.1
Determine frequency of individual words
If you want to count the number of times text contains the word 'candy', you can use the method String#scan on the string text and then determine the size of the array that is returned:
text.scan(/\bcandy\b/).size
scan returns an array with every occurrence of the string 'candy'; .size returns the size of that array. Here \b ensures 'candy gram' has a word "boundary" at each end, which could be whitespace or the beginning or end of a line or the file. That's to prevent `candycane' from being counted.
A second way is to convert the string text to an array of words, as you have done2:
myArray = text.split
If you don't mind, I'd like to call this:
words = text.split
as I find that more expressive.3
The most direct way to determine the number of times 'candy' appears is to use the method Enumberable#count, like this:
words.count('candy')
You can also use the array difference method, Array#-, as you noted:
words.size - (words - ['candy']).size
If you wish to know the number of times either 'candy' or 'gram' appears, you could of course do the above for each and sum the two counts. Some other ways are:
words.size - (myArray - ['candy', 'gram']).size
words.count { |word| word == 'candy' || word = 'gram' }
words.count { |word| ['candy', 'gram'].include?(word) }
Determine the frequency of all words that appear in the text
Your use of a hash with a default value of zero was a good choice:
def frequency_of_all_words(words)
frequency = Hash.new(0)
words.each { |word| frequency[word] +=1 }
frequency
end
I wrote this as a method to emphasize that words.each... does not return frequency. Often you would see this written more compactly using the method Enumerable#each_with_object, which returns the hash ("object"):
def frequency_of_all_words(words)
words.each_with_object(Hash.new(0)) { |word, h| h[word] +=1 }
end
Once you have the hash frequency you can sort it as you did:
frequency.sort_by {|word, freq| freq }
or
frequency.sort_by(&:last)
which you could write:
frequency.sort_by {|_, freq| freq }
since you aren't using the first block variable. If you wanted the most frequent words first:
frequency.sort_by(&:last).reverse
or
frequency.sort_by {|_, freq| -freq }
All of these will give you an array. If you want to convert it back to a hash (with the largest values first, say):
Hash[frequency.sort_by(&:last).reverse]
or in Ruby 2.0+,
frequency.sort_by(&:last).reverse.to_h
Count the number of times a substring appears
Now let's count the number of times the string 'candy gram' appears. You might think we could use String#scan on the string holding the entire file, as we did earlier4:
text.scan(/\bcandy gram\b/).size
The first problem is that this won't catch 'candy\ngram'; i.e., when the words are separated by a newline character. We could fix that by changing the regex to /\bcandy\sgram\b/. A second problem is that 'candy gram' might have been 'candy. Gram' in the file, in which case you might not want to count it.
A better way is to use the method Enumerable#each_cons on the array words. The easiest way to show you how that works is by example:
words = %w{ check for candy gram here candy gram again }
#=> ["check", "for", "candy", "gram", "here", "candy", "gram", "again"]
enum = words.each_cons(2)
#=> #<Enumerator: ["check", "for", "candy", "gram", "here", "candy",
# "gram", "again"]:each_cons(2)>
enum.to_a
#=> [["check", "for"], ["for", "candy"], ["candy", "gram"],
# ["gram", "here"], ["here", "candy"], ["candy", "gram"],
# ["gram", "again"]]
each_cons(2) returns an enumerator; I've converted it to an array to display its contents.
So we can write
words.each_cons(2).map { |word_pair| word_pair.join(' ') }
#=> ["check for", "for candy", "candy gram", "gram here",
# "here candy", "candy gram", "gram again"]
and lastly:
words.each_cons(2).map { |word_pair|
word_pair.join(' ') }.count { |s| s == 'candy gram' }
#=> 2
1 If you also wanted to keep dashes, for hyphenated words, change the regex to /[^-a-z]/ or /[^a-z-]/.
2 Note from String#split that .split is the same as both .split(' ') and .split(/\s+/)).
3 Also, Ruby's naming convention is to use lower-case letters and underscores ("snake-case") for variables and methods, such as my_array.