I'm learning Ruby since yesterday evening.
Here's a method I made that's supposed to print out multiples of any numbers up to any quantity.
def DisplayMultiples(multiplesOf, count)
i = multiplesOf
while i <= count
if i % multiplesOf == 0
puts i
end
i += 1
end
end
Any suggestions on how to improve the code to something more fitting to the Ruby style? I'm coming from a C# background so I'd like to switch things up a bit.
Edit:
Where can I find documentation for methods/classes? For example, the first answer I received use the .times method (is it a method?). I can't find the documentation for that because I don't know what type it is, since Ruby doesn't have types.
Any suggestions?
I'd do:
def display_multiples(multiplesOf, count)
count.times {|x| puts x * multiplesOf }
end
def display_multiples(number, limit)
0.step(limit, number){|n| puts n}
end
def display_multiples(multiplesOf, count)
(count/multiplesOf).times {|x| puts (x+1) * multiplesOf }
end
As for documentation, see ruby-doc.org and gotapi.com.
def display_multiples(of,nb)
p (of..of*nb).step(of).to_a
end
display_multiples(3,5) #display [3, 6, 9, 12, 15]
Related
Total beginner here, so I apologize if a) this question isn't appropriate or b) I haven't asked it properly.
I'm working on simple practice problems in Ruby and I noticed that while I arrived at a solution that works, when my solution runs in a visualizer, it gives premature returns for the array. Is this problematic? I'm also wondering if there's any reason (stylistically, conceptually, etc.) why you would want to use a while-loop vs. a for-loop with range for a problem like this or fizzbuzz.
Thank you for any help/advice!
The practice problem is:
# Write a method which collects all numbers between small_num and big_num into
an array. Ex: range(2, 5) => [2, 3, 4, 5]
My solution:
def range(small_num, big_num)
arr = []
(small_num..big_num).each do |num|
arr.push(num)
end
return arr
end
The provided solution:
def range(small_num, big_num)
collection = []
i = small_num
while i <= big_num
collection << i
i += 1
end
collection
end
Here's a simplified version of your code:
def range(small_num, big_num)
arr = [ ]
(small_num..big_num).each do |num|
arr << num
end
arr
end
Where the << or push function does technically have a return value, and that return value is the modified array. This is just how Ruby works. Every method must return something even if that something is "nothing" in the form of nil. As with everything in Ruby even nil is an object.
You're not obligated to use the return values, though if you did want to you could. Here's a version with inject:
def range(small_num, big_num)
(small_num..big_num).inject([ ]) do |arr, num|
arr << num
end
end
Where the inject method takes the return value of each block and feeds it in as the "seed" for the next round. As << returns the array this makes it very convenient to chain.
The most minimal version is, of course:
def range(small_num, big_num)
(small_num..big_num).to_a
end
Or as Sagar points out, using the splat operator:
def range(small_num, big_num)
[*small_num..big_num]
end
Where when you splat something you're in effect flattening those values into the array instead of storing them in a sub-array.
I got the task to create a lottery program which outputs 6 random numbers from 1 to 49 without duplicates. I'm not allowed to use the shuffle-Method of Arrays and as a tip I recieved that I should create an array with 49 entries and "shuffle" the numbers. I thought about this tip but it's not really helping.
This is what I got till now, I hope someone understands my code. I'm still stuck in a more Java way of writing than in ruby.
# Generating the Array
l_num = Array.new(6)
i = 0
while (i == 0)
l_num << [rand(1...49), rand(1...49), rand(1...49), rand(1...49), rand(1...49), rand(1...49)]
if (l_num.uniq.length == l_num.length)
i += 1
end
end
#Output
puts 'Todays lottery numbers are'
l_num.each { |a| print a, " " }
I picked up the uniq-Methode, because I read that it could elimante doubles that way, but I don't think it works in this case. In previous versions of my code I got a Error because I was trying to override the allready created array, I understand why Ruby is giving me an error but I have no clue how to do it in another way.
I hope someone is able to provide me some code-pieces, methods, solutions or tips for this task, thanks in advance.
This is the strategy I would take:
lottery_numbers = []
begin
# add 1 because otherwise it gives you numbers from 0-48
number = rand(49)+1
lottery_numbers.push(number) unless lottery_numbers.include?(number)
end while lottery_numbers.size < 6
puts "lottery numbers:"
puts lottery_numbers.join(" ")
Rubyists tend to initialize arrays with [] as opposed to the verbose Array.new
Charles, I think the easiest would be the following:
Array.new(49) { |x| x + 1 }.sample(6)
However this is what I believe you was prohibited to do? If yes, try a more "manual" solution:
Array.new(49) { |x| x + 1 }.sort { rand() - 0.5 }.take(6)
Otherwise, try implementing one completely "manual" solution. For example:
require 'set'
result = Set.new
loop do
result << 1 + rand(49)
break if result.size == 6
end
puts result.to_a.inspect
Here's one way to do it you can't use sample:
a = *1..49
#=> [1, 2, 3,..., 49]
6.times.map { a.delete_at(rand(a.size))+1 }
# => [44, 41, 15, 19, 46, 17]
min_lottery_number = 1
max_lottery_number = 49
total_size = 6
(min_lottery_number..max_lottery_number).to_a.sample(total_size)
this is the question
Shuffle. Now that you’ve finished your
new sorting algorithm, how about the
opposite? Write a shuffle method that
takes an array and returns a totally
shuffled version. As always, you’ll
want to test it, but testing this one
is trickier: How can you test to make
sure you are getting a perfect
shuffle? What would you even say a
perfect shuffle would be? Now test for
it.
This is my code answer:
def shuffle arr
x = arr.length
while x != 0
new_arr = []
rand_arr = (rand(x))
x--
new_arr.push rand_arr
arr.pop rand_arr
end
new_arr
end
puts (shuffle ([1,2,3]))
What are my mistakes? Why doesn't this code work?
Here's a far more Rubyish version:
class Array
def shuffle!
size.downto(1) { |n| push delete_at(rand(n)) }
self
end
end
puts [1,2,3].shuffle!
Here's a more concise way of writing it:
def shuffle(arr)
new_arr = []
while (arr.any?) do
new_arr << arr.delete_at(rand(arr.length))
end
new_arr
end
And some tests:
5.times do
puts shuffle((1..5).to_a).join(',')
end
>> 4,2,1,3,5
>> 3,2,1,4,5
>> 4,2,5,1,3
>> 5,2,1,4,3
>> 4,3,1,5,2
Beside minor other errors you seems not to understand what pop and push are doing (taking or adding some items from the end of the array).
You are probably trying to write something like below.
def shuffle arr
x = arr.length
new_arr = []
while x != 0
randpos = rand(x)
x = x-1
item = arr[randpos]
new_arr.push item
arr[randpos] = arr[x]
arr.pop
end
new_arr
end
puts (shuffle ([1,2,3]))
You're getting your indexes mixed up with your values. When you do new_arr.push rand_arr, you're putting whatever random index you came up with as a value on the end of new_arr. What you meant to do is new_arr.push arr[rand_arr], where arr[rand_arr] is the value at the index rand_arr in arr.
Ruby 1.8.7 and 1.9.2 have a built-in Array#shuffle method.
A variant of Mark Thomas's answer. His algorithm can be quite slow with a large array, due to delete operation performance.
class Array
def shuffle!
size.downto(1) do |n|
index=rand(n)
# swap elements at index and the end
self[index], self[size-1] = self[size-1],self[index]
end
self
end
end
puts [1,2,3].shuffle!
This algorithm is O(size), while Mark's algorithm is O(size^2). On my computer, Mark's answer takes 400 seconds to shuffle an array of 1,000,000 elements on my machine, versus 0.5 seconds with my method.
How do I do this type of for loop in Ruby?
for(int i=0; i<array.length; i++) {
}
array.each do |element|
element.do_stuff
end
or
for element in array do
element.do_stuff
end
If you need index, you can use this:
array.each_with_index do |element,index|
element.do_stuff(index)
end
limit = array.length;
for counter in 0..limit
--- make some actions ---
end
the other way to do that is the following
3.times do |n|
puts n;
end
thats will print 0, 1, 2, so could be used like array iterator also
Think that variant better fit to the author's needs
I keep hitting this as a top link for google "ruby for loop", so I wanted to add a solution for loops where the step wasn't simply '1'. For these cases, you can use the 'step' method that exists on Numerics and Date objects. I think this is a close approximation for a 'for' loop.
start = Date.new(2013,06,30)
stop = Date.new(2011,06,30)
# step back in time over two years, one week at a time
start.step(stop, -7).each do |d|
puts d
end
The equivalence would be
for i in (0...array.size)
end
or
(0...array.size).each do |i|
end
or
i = 0
while i < array.size do
array[i]
i = i + 1 # where you may freely set i to any value
end
array.each_index do |i|
...
end
It's not very Rubyish, but it's the best way to do the for loop from question in Ruby
To iterate a loop a fixed number of times, try:
n.times do
#Something to be done n times
end
If you don't need to access your array, (just a simple for loop) you can use upto or each :
Upto:
2.upto(4) {|i| puts i}
2
3
4
Each:
(2..4).each {|i| puts i}
2
3
4
What? From 2010 and nobody mentioned Ruby has a fine for /in loop (it's just nobody uses it):
ar = [1,2,3,4,5,6]
for item in ar
puts item
end
['foo', 'bar', 'baz'].each_with_index {|j, i| puts "#{i} #{j}"}
Ruby's enumeration loop syntax is different:
collection.each do |item|
...
end
This reads as "a call to the 'each' method of the array object instance 'collection' that takes block with 'blockargument' as argument". The block syntax in Ruby is 'do ... end' or '{ ... }' for single line statements.
The block argument '|item|' is optional but if provided, the first argument automatically represents the looped enumerated item.
What's the slickest, most Ruby-like way to do this?
[1, 3, 10, 5].diff
should produce
[2, 7, -5]
that is, an array of first order differences. I've come up with a solution which I'll add below, but it requires ruby 1.9 and isn't all that slick. what else is possible?
I like this functional style:
module Enumerable
def diff
each_cons(2).map {|pair| pair.reverse.reduce :-}
end
end
EDIT: I just realized that the reverse is totally unnecessary. If this were a functional language, I would have used pattern matching, but Ruby doesn't support pattern matching. It does, however, support destructuring bind, which is a good enough approximation for pattern matching in this case.
each_cons(2).map {|first, second| second - first}
No smiley, though.
I like how this sounds if you just read it out loud from left to right: "For each pair, apply the difference between the first and second elements of the pair." In fact, I normally don't like the name collect and prefer map instead, but in this case that reads even better:
each_cons(2).collect {|first, second| second - first}
"For each pair, collect the difference between its elements." Sounds almost like a definition of first order difference.
Yet another way..Seems the shortest so far:)
module Enumerable
def diff
self[1..-1].zip(self).map {|x| x[0]-x[1]}
end
end
The concept comes from functional programming, of course:
module Enumerable
def diff
self.inject([0]) { |r,x| r[-1] += x; r << -x } [1..-2]
end
end
[1,3,10,5].diff
Note that you don't need any separate intermediate variables here
Here's the fastest way I could find (faster than all the others suggested here as of this moment, in both 1.8 and 1.9):
module Enumerable
def diff
last=nil
map do |x|
r = last ? x - last : nil
last = x
r
end.compact
end
end
With this close runner-up:
module Enumerable
def diff
r = []
1.upto(size-1) {|i| r << self[i]-self[i-1]}
r
end
end
Of the others here, testr's self-described "feeble" attempt is the next fastest, but it's still slower than either of these.
And if speed is no object, here's my aesthetic favorite:
module Enumerable
def diff!
[-shift+first] + diff! rescue []
end
def diff
dup.diff!
end
end
But this is (for reasons I don't entirely understand) an order of magnitude slower than any other suggestion here!
Minor variation on Jörg W Mittag's:
module Enumerable
def diff
each_cons(2).map{|a,b| b-a}
end
end
# Attempt, requires ruby 1.9.
module Enumerable
def diff
each_cons(2).with_object([]){|x,array| array << x[1] - x[0]}
end
end
Example:
[1,3,10,5].diff
=> [2, 7, -5]
Another way to do it.
module Enumerable
def diff
result = []
each_with_index{ |x, i|
return result if (i == (self.length-1))
result << self[i+1] - x
}
end
end
My feeble attempt...
module Enumerable
def diff
na = []
self.each_index { |x| r << self[x]-self[x-1] if x > 0 }
na
end
end
p [1,3,10,5].diff #returned [2, 7, -5]