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 + " " }
Related
String:
string = "this;is;a;string;yes"
I can split the string and append each element to an array like this
arr = []
string.split(";").each do |x|
arr << x
end
Is there an easy way to take the first third and fourth values other than something like this.
x = 0
string.split(";").each do |x|
if x == 0 or x == 2 or x == 3 then arr << x end
x += 1
end
Sure. Use Array#values_at:
string = "this;is;a;string;yes"
string.split(";").values_at(0, 2, 3)
# => ["this", "a", "string"]
See it on repl.it: https://repl.it/#jrunning/FussyRecursiveSpools
I am trying to write a function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false. Only lower case letters (a-z) will be used. No punctuation or digits will be included. For example:
str1 = 'rkqodlw'; str2 = 'world' should return true.
str1 = 'cedewaraaossoqqyt'; str2 = 'codewars' should return true.
str1 = 'katas'; str2 = 'steak' should return false.
This is my code:
def scramble(s1, s2)
#sorts strings into arrays
first = s1.split("").sort
second = s2.split("").sort
correctLetters = 0
for i in 0...first.length
#check for occurrences of first letter
occurrencesFirst = first.count(s1[i])
for j in 0...second.length
#scan through second string
occurrencesSecond = second.count(s2[j])
#if letter to be tested is correct and occurrences of first less than occurrences of second
#meaning word cannot be formed
if (s2[j] == s1[i]) && occurrencesFirst < occurrencesSecond
return false
elsif s2[j] == s1[i]
correctLetters += 1
elsif first.count(s1[s2[j]]) == 0
return false
end
end
end
if correctLetters == 0
return false
end
return true
end
I need help optimising this code. Please give me suggestions.
Here is one efficient and Ruby-like way of doing that.
Code
def scramble(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
Examples
scramble('abecacdeba', 'abceae')
#=> true
scramble('abecacdeba', 'abweae')
#=> false
Explanation
The three steps are as follows.
str1 = 'abecacdeba'
str2 = 'abceae'
h1 = char_counts(str1)
#=> {"a"=>3, "b"=>2, "e"=>2, "c"=>2, "d"=>1}
h2 = char_counts(str2)
#=> {"a"=>2, "b"=>1, "c"=>1, "e"=>2}
h2.all? { |ch, nbr| nbr <= h1[ch] }
#=> true
The last statement is equivalent to
2 <= 3 && 1 <= 2 && 1 <= 2 && 2 <=2
The method char_counts constructs what is sometimes called a "counting hash". To understand how char_counts works, see Hash::new, especially the explanation of the effect of providing a default value as an argument of new. In brief, if a hash is defined h = Hash.new(0), then if h does not have a key k, h[k] returns the default value, here 0 (and the hash is not changed).
Suppose, for different data,
h1 = { "a"=>2 }
h2 = { "a"=>1, "b"=>2 }
Then we would find that 1 <= 2 #=> true but 2 <= 0 #=> false, so the method would return false. The second comparison is 2 <= h1["b"]. As h1 does not have a key "b", h1["b"] returns the default value, 0.
The method char_counts is effectively a short way of writing the method expressed as follows.
def char_counts(str)
h = {}
str.each_char do |ch|
h[ch] = 0 unless h.key?(ch) # instead of Hash.new(0)
h[ch] = h[c] + 1 # instead of h[c][ += 1
end
h # no need for this if use `each_with_object`
end
See Enumerable#each_with_object, String#each_char (preferable to String.chars, as the latter produces an unneeded temporary array whereas the former returns an enumerator) and Hash#key? (or Hash#has_key?, Hash#include? or Hash#member?).
An Alternative
def scramble(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
I have found the method Array#difference to be so useful I proposed it be added to the Ruby Core (here). The response has been, er, underwhelming.
One way:
def scramble(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
Another way:
def scramble(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
Yet another:
def scramble(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
Since this appears to be from codewars, I submitted my first two there. Both got accepted and the first one was a bit faster. Then I was shown solutions of others and saw someone using ('a'..'z') and it's fast, so I include that here.
The codewars "performance tests" aren't shown explicitly but they're all up to about 45000 letters long. So I benchmarked these solutions as well as Cary's (yours was too slow to be included) on shuffles of the alphabet repeated to be about that long (and doing it 100 times):
user system total real
Stefan 1 0.812000 0.000000 0.812000 ( 0.811765)
Stefan 2 2.141000 0.000000 2.141000 ( 2.127585)
Other 0.125000 0.000000 0.125000 ( 0.122248)
Cary 1 2.562000 0.000000 2.562000 ( 2.575366)
Cary 2 3.094000 0.000000 3.094000 ( 3.106834)
Moral of the story? String#count is fast here. Like, ridiculously fast. Almost unbelievably fast (I actually had to run extra tests to believe it). It counts through about 1.9 billion letters per second (100 times 26 letters times 2 strings of ~45000 letters, all in 0.12 seconds). Note that the difference to my own first solution is just that I do s2.chars.uniq, and that increases the time from 0.12 seconds to 0.81 seconds. Meaning this double pass through one string takes about six times as long as the 52 passes for counting. The counting is about 150 times faster. I did expect it to be very fast, because it presumably just searches a byte in an array of bytes using C code (edit: looks like it does), but this speed still surprised me.
Code:
require 'benchmark'
def scramble_stefan1(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_stefan2(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
def scramble_other(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_cary1(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
def scramble_cary2(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
Benchmark.bmbm do |x|
n = 100
s1 = (('a'..'z').to_a * (45000 / 26)).shuffle.join
s2 = s1.chars.shuffle.join
x.report('Stefan 1') { n.times { scramble_stefan1(s1, s2) } }
x.report('Stefan 2') { n.times { scramble_stefan2(s1, s2) } }
x.report('Other') { n.times { scramble_other(s1, s2) } }
x.report('Cary 1') { n.times { scramble_cary1(s1, s2) } }
x.report('Cary 2') { n.times { scramble_cary2(s1, s2) } }
end
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"
Trying to convert a simple math string "1 2 3 * + 4 5 - /" to an array of integers and/or symbols like [1, 2, 3, :*, :+, 4, 5, :-, :/].
Is there a more elegant (and extendable) solution than this?
def tokenize s
arr = s.split(/ /)
symbols = %w{ + - / * }
arr.collect! do |c|
if symbols.include?(c)
c.to_sym
else
c.to_i
end
end
end
def tokenize(str)
str.split.map! { |t| t[/\d/] ? t.to_i : t.to_sym }
end
Instead of checking if the token is in a set of operations, you can just check if it contains a numeral digit (for your use case). So it's either an integer, or an operation.
Also note that Array#collect! and Array#map! are identical, and String#split by default splits on white space.
How about...
h = Hash[*%w{+ - / *}.map { |x| [x, x.to_sym] }.flatten]
And then,
arr.map { |c| h[c] || c.to_i }
So together:
def tokenize s
h = Hash[*%w{+ - / *}.map { |x| [x, x.to_sym] }.flatten]
s.split.map { |c| h[c] || c.to_i }
end
def tokenize s
s.scan(/(\d+)|(\S+)/)
.map{|num, sym| num && num.to_i || sym.to_sym}
end
I'm trying to learn Ruby, and am going through some of the Project Euler problems. I solved problem number two as such:
def fib(n)
return n if n < 2
vals = [0, 1]
n.times do
vals.push(vals[-1]+vals[-2])
end
return vals.last
end
i = 1
s = 0
while((v = fib(i)) < 4_000_000)
s+=v if v%2==0
i+=1
end
puts s
While that works, it seems not very ruby-ish—I couldn't come up with any good purely Ruby answer like I could with the first one ( puts (0..999).inject{ |sum, n| n%3==0||n%5==0 ? sum : sum+n }).
For a nice solution, why don't you create a Fibonacci number generator, like Prime or the Triangular example I gave here.
From this, you can use the nice Enumerable methods to handle the problem. You might want to wonder if there is any pattern to the even Fibonacci numbers too.
Edit your question to post your solution...
Note: there are more efficient ways than enumerating them, but they require more math, won't be as clear as this and would only shine if the 4 million was much higher.
As demas' has posted a solution, here's a cleaned up version:
class Fibo
class << self
include Enumerable
def each
return to_enum unless block_given?
a = 0; b = 1
loop do
a, b = b, a + b
yield a
end
end
end
end
puts Fibo.take_while { |i| i < 4000000 }.
select(&:even?).
inject(:+)
My version based on Marc-André Lafortune's answer:
class Some
#a = 1
#b = 2
class << self
include Enumerable
def each
1.upto(Float::INFINITY) do |i|
#a, #b = #b, #a + #b
yield #b
end
end
end
end
puts Some.take_while { |i| i < 4000000 }.select { |n| n%2 ==0 }
.inject(0) { |sum, item| sum + item } + 2
def fib
first, second, sum = 1,2,0
while second < 4000000
sum += second if second.even?
first, second = second, first + second
end
puts sum
end
You don't need return vals.last. You can just do vals.last, because Ruby will return the last expression (I think that's the correct term) by default.
fibs = [0,1]
begin
fibs.push(fibs[-1]+fibs[-2])
end while not fibs[-1]+fibs[-2]>4000000
puts fibs.inject{ |sum, n| n%2==0 ? sum+n : sum }
Here's what I got. I really don't see a need to wrap this in a class. You could in a larger program surely, but in a single small script I find that to just create additional instructions for the interpreter. You could select even, instead of rejecting odd but its pretty much the same thing.
fib = Enumerator.new do |y|
a = b = 1
loop do
y << a
a, b = b, a + b
end
end
puts fib.take_while{|i| i < 4000000}
.reject{|x| x.odd?}
.inject(:+)
That's my approach. I know it can be less lines of code, but maybe you can take something from it.
class Fib
def first
#p0 = 0
#p1 = 1
1
end
def next
r =
if #p1 == 1
2
else
#p0 + #p1
end
#p0 = #p1
#p1 = r
r
end
end
c = Fib.new
f = c.first
r = 0
while (f=c.next) < 4_000_000
r += f if f%2==0
end
puts r
I am new to Ruby, but here is the answer I came up with.
x=1
y=2
array = [1,2]
dar = []
begin
z = x + y
if z % 2 == 0
a = z
dar << a
end
x = y
y = z
array << z
end while z < 4000000
dar.inject {:+}
puts "#{dar.sum}"
def fib_nums(num)
array = [1, 2]
sum = 0
until array[-2] > num
array.push(array[-1] + array[-2])
end
array.each{|x| sum += x if x.even?}
sum
end