I have the following string:
str = "peep toe t-shirt blue"
I need to split the items using space:
str.split(" ")
And I get
[pee, toe, t-shirt, blue]
The questions is that peep toe must not be splitted, because it's a name that makes sense together.
Is there a way to solve this?
I think it's good to use split and then fix up the data later:
str = "peep toe t-shirt blue"
words = str.split(" ")
phrases = []
i = 0
while i < words.size
if words[i] == 'peep' && words[i + 1] == 'toe'
phrases << 'peep toe'
i += 2
else
phrases << words[i]
i += 1
end
end
p phrases
Related
I'm trying to work through a level 5 kata by using while loops. Essentially the problem is to turn each letter rotors[n] number of times and then move on to the next rotors number until you get an output word.
flap_display(["CAT"],[1,13,27])
should output ["DOG"]
Here's what I have so far
def flap_display(lines, rotors)
stuff = "ABCDEFGHIJKLMNOPQRSTUVWXYZ?!##&()|<>.:=-+*/0123456789"
i = 0
j = 0
new_word = lines
while i < rotors.length
while j < new_word[0].length
new_word[0][j] = stuff[stuff.index(new_word[0][j]) + rotors[i]]
j += 1
end
i += 1
j = 0
end
new_word
end
This technically traverses the stuff string and assigns the right letters. However it fails two important things: it does not skip each letter when it rotates to the correct position (C should stop rotating when it hits D, A when it hits O etc) and it does not account for reaching the end of the stuff list and eventually returns a nil value for stuff[stuff.index(new_word[0][j]) + rotors[i]]. How can I fix these two problems using basic loops and enumerables or maybe a hash?
A fuller statement of the problem is given here. This is one Ruby-like way it could be done.
FLAPS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ ?!##&()|<>.:=-+*/0123456789"
NBR_FLAPS = FLAPS.size
def flap_display(str, rot)
rot_cum = rot.each_with_object([]) { |n,a| a << a.last.to_i + n }
str.gsub(/./) { |c| FLAPS[(c.ord + rot_cum.shift - 65) % NBR_FLAPS] }
end
flap_display("CAT", [1,13,27])
#=> "DOG"
flap_display("DOG", [-1,-13,-27])
#=> "CAT"
flap_display("CAT", [5,37,24])
#=> "H*&"
'A'.ord #=> 65 and rot_cum contains the cumulative values of rot:
arr = [1, 13, 27]
rot_cum = arr.each_with_object([]) { |n,a| a << a.last.to_i + n }
#=> [1, 14, 41]
I've written a.last.to_i rather than a.last to deal with the case where a is empty, so a.last #=> nil, meaning a.last.to_i => nil.to_i => 0. See NilClass#to_i. Those opposed to such trickery could write:
rot_cum = arr.drop(1).each_with_object([arr.first]) { |n,a| a << a.last + n }
def encrypt(string)
alphabet = ("a".."b").to_a
result = ""
idx = 0
while idx < string.length
character = string[idx]
if character == " "
result += " "
else
n = alphabet.index(character)
n_plus = (n + 1) % alphabet.length
result += alphabet[n_plus]
end
idx += 1
end
return result
end
puts encrypt("abc")
puts encrypt("xyz")
I'm trying to get "abc" to print out "bcd" and "xyz" to print "yza". I want to advance the letter forward by 1. Can someone point me to the right direction?
All I had to do was change your alphabet array to go from a to z, not a to b, and it works fine.
def encrypt(string)
alphabet = ("a".."z").to_a
result = ""
idx = 0
while idx < string.length
character = string[idx]
if character == " "
result += " "
else
n = alphabet.index(character)
n_plus = (n + 1) % alphabet.length
result += alphabet[n_plus]
end
idx += 1
end
return result
end
puts encrypt("abc")
puts encrypt("xyz")
Another way to solve the issue, that I think is simpler, personally, is to use String#tr:
ALPHA = ('a'..'z').to_a.join #=> "abcdefghijklmnopqrstuvwxyz"
BMQIB = ('a'..'z').to_a.rotate(1).join #=> "bcdefghijklmnopqrstuvwxyza"
def encrypt(str)
str.tr(ALPHA,BMQIB)
end
def decrypt(str)
str.tr(BMQIB,ALPHA)
end
encrypt('pizza') #=> "qjaab"
decrypt('qjaab') #=> "pizza"
Alternatively if you don't want to take up that memory storing the alphabet you could use character codings and then just use arithmetic operations on them to shift the letters:
def encrypt(string)
result = ""
idx = 0
while idx < string.length
result += (string[idx].ord == 32 ? (string[idx].chr) : (string[idx].ord+1).chr)
idx += 1
end
result
end
Other strange thing about ruby is that you do not need to explicitly return something at the end of the method body. It just returns the last thing by default. This is considered good style amongst ruby folks.
Your question has been answered, so here are a couple of more Ruby-like ways of doing that.
Use String#gsub with a hash
CODE_MAP = ('a'..'z').each_with_object({}) { |c,h| h[c] = c < 'z' ? c.next : 'a' }
#=> {"a"=>"b", "b"=>"c",..., "y"=>"z", "z"=>"a"}
DECODE_MAP = CODE_MAP.invert
#=> {"b"=>"a", "c"=>"b",..., "z"=>"y", "a"=>"z"}
def encrypt(word)
word.gsub(/./, CODE_MAP)
end
def decrypt(word)
word.gsub(/./, DECODE_MAP)
end
encrypt('pizza')
#=> "qjaab"
decrypt('qjaab')
#=> "pizza"
Use String#gsub with Array#rotate
LETTERS = ('a'..'z').to_a
#=> ["a", "b", ..., "z"]
def encrypt(word)
word.gsub(/./) { |c| LETTERS.rotate[LETTERS.index(c)] }
end
def decrypt(word)
word.gsub(/./) { |c| LETTERS.rotate(-1)[LETTERS.index(c)] }
end
encrypt('pizza')
#=> "qjaab"
decrypt('qjaab')
#=> "pizza"
I have a school assignment where i have to find the longest run of adjacent equal characters in a given string with Ruby. My program works fine without the last loop, but once i added it gave me the error:
(repl):47: syntax error, unexpected keyword_end
(repl):53: syntax error, unexpected end-of-input, expecting keyword_end
puts longestRun
^
Here is my Code
puts 'What is your string?'
givenString = gets.chomp
def maxBlock(str)
maxRun = 0
currentRun = 1
characterCounter = 1
if str.length == 0
maxRun = 0
#If no input, longest run is zero
elsif str.length == 1
maxRun = 1
#If string is one character, longest run is 1
elsif str.length == 2 and str[characterCounter] != str[characterCounter + 1]
maxRun = 1
#if string is two chars and they do not equal, longest run is 1
elsif str.length == 3 and str[0] != str[1] and str[1] != str[2]
maxRun = 1
#if string is three chars and they do not equal, longest run is 1
else
str.each_char do|st|
#Go through each char, compare it to the next, find longest run
if st == str[characterCounter]
currentRun++
if currentRun > maxRun
maxRun = currentRun
end
else
currentRun = 1
end
characterCounter++
end
end
end
longestRun = maxBlock(givenString)
puts longestRun
EDIT: I am a highschool student, and only have a base knowledge of programming.
EDIT: I just made a few stupid mistakes. I appreciate everyone's help. Here is my working program without the use of anything too complicated.
puts 'What is your string?'
givenString = gets.chomp
def maxBlock(str)
maxRun = 0
currentRun = 1
characterCounter = 0
if str.length == 0
maxRun = 0
#If no input, longest run is zero
elsif str.length == 1
maxRun = 1
#If string is one character, longest run is 1
elsif str.length == 2 and str[characterCounter] != str[characterCounter + 1]
maxRun = 1
#if string is two chars and they do not equal, longest run is 1
elsif str.length == 3 and str[0] != str[1] and str[1] != str[2]
maxRun = 1
#if string is three chars and they do not equal, longest run is 1
else
characterCounter += 1
str.each_char do|st|
#Go through each char, compare it to the next, find longest run
if st == str[characterCounter]
currentRun += 1
if currentRun > maxRun
maxRun = currentRun
end
else
currentRun = 1
end
characterCounter += 1
end
end
return maxRun
end
longestRun = maxBlock(givenString)
puts longestRun
String Scans and Sorting
There are algorithms for this, but Ruby offers some nice shortcuts. For example:
def longest_string str
str.scan(/((\p{Alnum})\2+)/).collect { |grp1, grp2| grp1 }.sort_by(&:size).last
end
longest_string 'foo baaar quuuux'
#=> "uuuu"
This basically just captures all runs of repeated characters, sorts the captured substrings by length, and then returns the last element of the length-sorted array.
Secondary Sorting
If you want to do a secondary sort, such as first by length and then by alphabetical order, you could replace Enumerable#sort_by with the block form of Enumerable#sort. For example:
def longest_string str
str.scan(/((\p{Alnum})\2+)/).
collect { |grp1, grp2| grp1 }.
sort {|a, b| [a.size, a] <=> [b.size, b] }.
last
end
longest_string 'foo quux baar'
#=> "uu"
This is one way you could do it.
str = "111 wwwwwwwwaabbbbbbbbbbb$$$$****"
r = /
(.) # Match any character in capture group 1
\1* # Match the contents of capture group 1 zero or more times
/x # Free-spacing regex definition mode
str.gsub(r).max_by(&:size)
#=> "bbbbbbbbbbb"
I used the form of String#gsub without a second argument or block, as that returns an enumerator that generates the strings matched by the regex. I then chained that enumerator to the method Enumerable#max_by to find the longest string of consecutive characters. In other words, I used gsub merely to generate matches rather than to perform substitutions.
One could of course write str.gsub(/(.)\1*/).max_by(&:size).
Here is a simplified version that should work in all cases:
puts 'What is your string?'
given_string = gets.chomp
def max_block(str)
max_run = 0
current_run = 1
str.each_char.with_index do |st, idx|
if st == str[idx + 1]
current_run += 1
else
current_run = 1
end
max_run = current_run if current_run > max_run
end
max_run
end
longest_run = max_block(given_string)
puts longest_run
You were on the right track but Ruby can make things a lot easier for you. Notice how with_index gets rid of a lot of the complexity. Iterators, oh yeah.
I also changed your method name and variables to camel_case.
Happy coding!
I've been using the code below to find the frequency of a given word in a string using Ruby. My question is how I can adapt this to find the frequency of 2 words at a given time. As example: "baa baa baa black sheep" should return...
{"baa baa"=>2, "baa black"=>1, "black sheep"=>1}
Code:
def count_words(string)
words = string.split(' ')
frequency = Hash.new(0)
words.each { |word| frequency[word.downcase] += 1 }
return frequency
end
str = "baa baa baa black sheep"
count = Hash.new(0)
str.split.each_cons(2) do |words|
count[ words.join(' ') ] += 1
end
count
# => {"baa baa"=>2, "baa black"=>1, "black sheep"=>1}
def count_words(string)
words = string.downcase.split(' ')
frequency = Hash.new(0)
while words.size >= 2
frequency["#{words[0]} #{words[1]}"] += 1
words.shift
end
frequency
end
In this question, the asker requests a solution that would insert a space every x number of characters. The answers both involve using a regular expression. How might you achieve this without a regex?
Here's what I came up with, but it's a bit of a mouthful. Any more concise solutions?
string = "12345678123456781234567812345678"
new_string = string.each_char.map.with_index {|c,i| if (i+1) % 8 == 0; "#{c} "; else c; end}.join.strip
=> "12345678 12345678 12345678 12345678"
class String
def in_groups_of(n)
chars.each_slice(n).map(&:join).join(' ')
end
end
'12345678123456781234567812345678'.in_groups_of(8)
# => '12345678 12345678 12345678 12345678'
class Array
# This method is from
# The Poignant Guide to Ruby:
def /(n)
r = []
each_with_index do |x, i|
r << [] if i % n == 0
r.last << x
end
r
end
end
s = '1234567890'
n = 3
join_str = ' '
(s.split('') / n).map {|x| x.join('') }.join(join_str)
#=> "123 456 789 0"
This is slightly shorter but requires two lines:
new_string = ""
s.split(//).each_slice(8) { |a| new_string += a.join + " " }