I have an array of words. I would like to get the letter count to print next to each word. I get the words to print with a total of all letters to print at the end. Any guidance would be great.
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If you are looking to print length of each word, below should work.
a = #[array of words]
a.each {|e| p "#{e} has #{e.length} letters"}
words = [] #array words
words.each {|w| puts "#{w} : #{w.length}"}
other way you can use Gem "words_counted" https://github.com/abitdodgy/words_counted for natural language processor
Related
To convert array to string I used Array#join and got space between the
beginning of the string, first quote mark, and the first word. I do not understand why this is happening.
I resolved with String#strip but I would like to understand
def order(words)
arr_new = []
arr = words.split(" ")
nums = ["1","2","3","4","5","6","7","8","9"]
arr.each do |word|
nums.each do |num|
if word.include? num
arr_new[num.to_i] = word
end
end
end
arr_new.join(" ").strip
end
order("is2 Thi1s T4est 3a")
Without .strip the output is:
" Thi1s is2 3a T4est"
After .strip:
"Thi1s is2 3a T4est"
The reason you're seeing the extra space is because arrays in ruby are 0 indexed, so you have an nil array element because your first insert is a index 1
x = []
x[1] = "test"
This creates an array as such:
[
nil,
"test"
]
If you created an empty array named x and assigned x[10] = "test" you'd have 10 nil values, and the word "test" in your array.
So, your array, before joining, is actually:
[nil, "Thi1s", "is2", "3a", "T4est"]
You have a couple options:
Change your strings to start with zero
Change your assignment to adjust the offset (subtract one)
Use compact before you join (this will remove nils)
Use strip as you noted
I'd suggest compact because it would address a few edge cases (such as "gaps" in your numbers.
More info in the array docs
#Jay's explanation is indeed correct.
I'll simply suggest a cleaner version of your code that doesn't have the same problem.
This assumes that the 1-9 order isn't dynamic. Aka wouldn't work if you wanted to sort by random characters for example.
def order(words)
words.split.sort_by { |word| word[/\d/].to_i }.join ' '
end
I'm trying to find out if any of the words in the array keywords:
keywords = Array['hello', 'hi', 'home']
appear in split:
split = work.split(' ')
and for each keyword that appears in split, do something.
I have this code:
for keywords in splitDescription
score += 2
end
The code just gives me 2+ for each word in split, but I only want 2+ for each keywords appearing in split.
Use set intersection on arrays (Array#&):
(splitDescription & keywords).each do |found_keyword|
# something
end
If you know there might be some words repeating and you want to iterate over each occurrence:
splitDescription.select { |word| keywords.include?(word) }.each do |word|
# something
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.
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 an array:
array = ["abhor", "rage", "mad"]
and I want to check if a string includes any word in that array - But only that word (not a substring).
string = 'I made a cake.'
count = 0
array.each do |word|
if string.include? word
count += 1
end
end
However, the above does increment count by 1 because it's picking up the word mad from made in my string. How can I search for only mad and ensure made doesn't get counted?
The array intersection operator & is useful here.
Here's two options, depending on how you define "word":
1) If a word is any sequence of non-whitespace characters, then you can do:
array & string.split
In your example, this results in the intersection of array and words in string, which is empty.
2) If a word is any sequence of alphanumeric characters including _, then you can do:
array & string.scan(/\w+/)
For example if array = ["abhor", "rage", "mad", "cake"] then #1 above will be empty (because you have cake. with a period in the string) but will return ['cake'] for method #2.
I would do it like this:
array = ["abhor", "rage", "mad"]
string = 'I made a cake.'
string.split.count{|word| array.include?(word)}
The problem with doing a simple split is that it won't take into account punctuation. What you need is a regex, which is a little more complicated.
array.each do |word|
count += 1 if string.match(/\W#{word}\W/)
end
If you're willing to go down the regex road, \b represent a word boundary. This example, which includes all of the words in your sentence and a few which are fragments, correctly returns 4.
array = ["abhor", "rage", "mad", "I", "made", "a", "cake", "cak"]
string = 'I made a cake.'
count = 0
array.each do |word|
if string =~ /\b#{word}\b/
count += 1
end
end
Try splitting up the words first.
words = string.split
count = 0
words.each do |word|
count += 1 if array.include? word
end