I am trying to multiply any number of unknown arguments together to make a total.
def multiply(*num)
num.each { |i| puts num * num}
end
multiply(2,3,4)
multiply(2,3,4,5,6,7)
Another attempt:
def multiply(num)
num.to_i
i = 0
while i < num.length
total = num * num
return total
end
end
multiply(2,3,4)
multiply(2,3,4,5,6,7)
I keep running into errors:
(eval):73: undefined local variable or method `num_' for main:Object (NameError)
from (eval):81
Some saying that Array needs to be Integer.
I've tried doing something I thought was suppose to be very simple program to write.
def multiply(*num) captures all arguments given to the method in an array. If you run multiply(1, 2, 3, 4) then num will equal [1, 2, 3, 4]. So in both your attempts you're trying to multiply a whole array with itself (num * num), which is not going to work.
Others have suggested solutions to how to write the multiply method correctly, but let me give you a correct version of your first attempt and explain it properly:
def multiply(*numbers)
result = 1
numbers.each { |n| result = result * n }
result
end
As in your attempt I capture all arguments in an array, I call mine numbers. Then I declare a variable that will hold the result of all the multiplications. Giving it the value of 1 is convenient because it will not affect the multiplications, but I could also have shifted off the first value off of the numbers array, and there are other solutions.
Then I iterate over all the numbers with each, and for each number I multiply it with the current result and store this as the new result (I could have written this shorter with result *= n).
Finally I return the result (the last value of a method is what will be returned by the method).
There are shorter ways of doing the same thing, others have suggested using numbers.reduce { |n, result| n * result }, or even numbers.reduce(:*), but even though they are shorter, they are pretty cryptic and I assume they don't really help you get things working.
What you want is the #inject method
def multiple(*nums)
nums.inject(:*)
end
Inject will combine all of the elements in nums, using the operation specified ( * in this case ).
You can try this:
num.reduce(1) {|x, y| x*y }
Related
I was doing this question on codewars: "Given an array, find the int that appears an odd number of times. There will always be only one integer that appears an odd number of times."
Code:
def find_it(seq)
int = []
for a in seq do
count = 0
for b in seq do
if a == b
count += 1
end
end
if count % 2.0 != 0
int << b
end
end
puts int.uniq[0].to_i
end
It was tested against a couple inputs, but the answers were wrong for these two arrays:
find_it([1,1,2,-2,5,2,4,4,-1,-2,5]) - returns 5 instead of -1
find_it([1,1,1,1,1,1,10,1,1,1,1]) - returns 1 instead of 10
What went wrong with my code?
if count % 2.0 != 0
int << b
end
The problem you have here is that your pushing b instead of a into the integer array, so what's happening is that instead of the value that you counted being pushed in, your pushing in the last value of b which is the last value element in the array regardless as long as the condition that the counter is an odd number, although b and counter have nothing to do with each other. so to fix it you replace b with a so that it pushes in the value you are testing comparing with the other elements in the second loop
fix:
if count % 2.0 != 0
int << a
end
a similar yet simpler code that does a similar job except in a shorter and more understandable way is:
def find_it(seq)
numberWithOddCount = 0
seq.each do |currentElement|
counter = 0
seq.each { |elementToCompare| counter += 1 if currentElement == elementToCompare}
numberWithOddCount = currentElement if counter % 2 != 0
end
numberWithOddCount
end
Just added a few tid-bits that you could also utilize to shorten and simplify code.
Happy Coding!
Note:
You could utilize built in ruby methods in creative ways to make the code do what you want in very few lines (or even one line) such as what #iGian did in the questions comments, but if your still new to ruby then its best to utilize those methods one by one when learning them otherwise you'll be confused. But if your willing to take the time now to learn them, I suggest you take his code and separate each method execution into its own line and output what each method had done to know what's doing what. and practice using each separately.
#aimen_alt is right about your mistake
but let's decompose your problem.
First, you need to calculate the appearances of each number.
Second, you need to find the one with the odd count of the appearances.
Accordingly to the problem, there is only one such number, so you can return it right away.
You can go your way and do it in O(N^2) complexity by scanning your sequence for each item in the sequence (so N items in the sequence multiply by the size of the sequence N = N*N). You can do it linearly* by constructing a Hash and than you'll be able to get the key with odd value:
def find_it(seq)
numbers = {}
seq.each do |item|
numbers[item] = numbers[item].to_i + 1
end
numbers.select{ |k,v| v.odd? }.first.first
end
to be more idiomatic you can use group_by to group the numbers themselves:
seq = [1, 2, 6, 1, 2]
seq.group_by{ |item| item }
#=> {1=>[1, 1], 2=>[2, 2], 6=>[6]}
You can see that each value is an Array, and you just need to get one with the odd amount of items:
seq = [1, 2, 6, 1, 2]
seq.group_by{ |item| item }.select{ |k, v| v.size.odd? }
#=> {6=>[6]}
And the last thing you would like to do is to get the value of the key:
seq.group_by{ |item| item }.select{ |k, v| v.size.odd? }.keys.first
So, the final solution would be
def find_it(seq)
seq.group_by{ |item| item }
.select{ |k, v| v.size.odd? }
.keys
.first
end
as #pascalbetz mentioned:
def find_it(seq)
seq.group_by{ |item| item }
.find{ |k, v| v.size.odd? }
.first
end
def find_it(seq)
seq.group_by{|x| x}.select{|k, v| (v.count % 2.0 !=0)}.first[0]
end
The above code will take a sequence in an array. Here we are grouping by elements:
For example:
[1,1,2,-2,5,2,4,4,-1,-2,5].group_by{|x| x}
# => {1=>[1, 1], 2=>[2, 2], -2=>[-2, -2], 5=>[5, 5], 4=>[4, 4], -1=>[-1]}
after getting the above results, we are finding the whose elements count not odd with the select condition.
ex:
[1,1,2,-2,5,2,4,4,-1,-2,5].group_by{|x| x}.select{|k, v| (v.count % 2.0 !=0)}
we will get the results as {-1=>[-1]}
we are taking the key as result element.
What about this one
def find_it(seq)
seq.reduce(:^)
end
^ -> this symbol is bitwise XOR.
reduce function is taking each value and doing whatever work assigned inside. In this case, it's taking each element and doing an XOR operation. the first element is doing XOR with zero and the next element doing XOR with the previous result and so on.
In this way, we found the odd element.
How XOR operation work
0 ^ 2 = 2
4 ^ 4 = 0
If you want to know more about XOR. kindly refer to this.
Thank you for all the detailed answers, I'm going over everyone's answers now. I'm new to Ruby, and I'm still in the process of learning the methods/rules of using them/Big O notation, so I much appreciated everyone's input. Codewar listed some top ranked solutions. This seems to be the fastest so far:
def find_it(seq)
seq.detect { |n| seq.count(n).odd? }
end
I am using this inject method to make a running total of values into an array. I am trying to figure out why I am getting an error.
def running_totals(myarray)
results = []
myarray.inject([]) do |sum,n|
results << sum + n
end
results
end
p running_totals([1,2,3,4,5])
I am getting the error
in `+': no implicit conversion of Fixnum into Array (TypeError)
When breaking this method down, isn't this the same as adding two integers and adding that into an array? I'm a bit confused here. Thanks for the help.
In the first iteration sum will be an array (as you specified an array as the default when calling inject([])) and you try to add a number to it. in the results << sum + n statement
Instead, set the initial value to 0, then add, then add the result to the array, then make sure you let sum get passed into the next iteration of inject.
def running_totals(myarray)
results = []
myarray.inject(0) do |sum,n| # First iteration sum will be 0.
sum += n # Add value to sum.
results << sum # Push to the result array.
sum # Make sure sum is passed to next iteration.
end
results
end
p running_totals([1,2,3,4,5]) #=> [1, 3, 6, 10, 15]
The result of results << sum + n is an array results and it's this that's replacing the sum value and so the next iteration you're trying to add a fixnum n into an array sum ... plus it doesn't help that you're initializing the value of sum to be an array.
Make sure that the last executed statement in your inject block is what you want the accumulated value to be.
def running_totals(myarray)
results = []
results << myarray.inject do |sum, n|
results << sum
sum + n
end
results
end
p running_totals([1,2,3,4,5])
=> [1, 3, 6, 10, 15]
Note that I moved the result of the inject into results array as well, so that the final value is also included, otherwise you'd only have the four values and would be missing the final (15) value.
The return value of the inject block is passed as the first argument the next time the block is called, so those have to match. In your code, you're passing an array as an intital value, and then returning an array; so far, so good. But inside the code block you treat that array parameter (sum) as a number, which won't work. Try this:
def running_totals(myarray)
myarray.inject([]) do |results,n|
results << n + (results.last || 0)
end
end
The [] passed as an argument to inject becomes the first value of results; the first array element (1 in your example) becomes the first value of n. Since results is empty, results.last is nil and the result of (results.last || 0) is 0, which we add to n to get 1, which we push onto results and then return that newly-modified array value from the block.
The second time into the block, results is the array we just returned from the first pass, [1], and n is 2. This time results.last is 1 instead of nil, so we add 1 to 2 to get 3 and push that onto the array, returning [1,3].
The third time into the block, results is [1,3], and n is 3, so it returns [1,3,6]. And so on.
According to ri, you have to return result of the computation from inject's block.
From: enum.c (C Method):
Owner: Enumerable
Visibility: public
Signature: inject(*arg1)
Number of lines: 31
Combines all elements of enum by applying a binary
operation, specified by a block or a symbol that names a
method or operator.
If you specify a block, then for each element in enum
the block is passed an accumulator value (memo) and the element.
If you specify a symbol instead, then each element in the collection
will be passed to the named method of memo.
In either case, the result becomes the new value for memo.
At the end of the iteration, the final value of memo is the
return value for the method.
If you do not explicitly specify an initial value for memo,
then uses the first element of collection is used as the initial value
of memo.
Examples:
# Sum some numbers
(5..10).reduce(:+) #=> 45
# Same using a block and inject
(5..10).inject {|sum, n| sum + n } #=> 45
# Multiply some numbers
(5..10).reduce(1, :*) #=> 151200
# Same using a block
(5..10).inject(1) {|product, n| product * n } #=> 151200
# find the longest word
longest = %w{ cat sheep bear }.inject do |memo,word|
memo.length > word.length ? memo : word
end
longest
So your sample would work if you return computation result for each iteration, something like this:
def running_totals(myarray)
results = []
myarray.inject do |sum,n|
results << sum + n
results.last # return computation result back to Array's inject
end
results
end
Hope it helps.
def random_select(array, n)
result = []
n.times do
# I do not fully understand how this line below works or why. Thank you
result.push array[rand(array.length)]
end
result
end
You are probably confused by this part:
n.times do
result.push(array[rand(array.length)])
end
n.times says it should loop n times.
result.push says to basically "push" or "put" something in the array. For example:
a = []
a.push(1)
p a #=> [1]
In array[rand(array.length)] , rand(array.length) will produce a random number as an index for the array. Why? rand(n) produces a number from 0 to n-1. rand(5) will produce either 0,1,2,3 or 4, for example.
Arrays use 0-based indexing, so if you have an array, say a = ['x', 'y', 'z'], to access 'x' you do a[0], to access y you do a[1] and so on. If you want to access a random element from a, you do a[rand(array.length)], because a.length in this case is 3, and rand(3) will produce a number that is either 0, 1 or 2. 0 is the smallest index and 2 is the largest index of our example array.
So suppose we call this method:
random_select([6,3,1,4], 2)
Try to see this code from the inside out. When the code reaches this part:
result.push(array[rand(array.length)])
it will first execute array.length which will produce 4. It will then execute rand(array.length) or rand(4) which will get a number between 0 and 3. Then, it will execute array[rand(array.length)] or array(some_random_number_between_0_and_3) which will get you a random element from the array. Finally, result.push(all_of_that_code_inside_that_got_us_a_random_array_element) will put the random element from the array in the method (in our example, it will be either 6, 3, 1 or 4) in the results array. Then it will repeat this same process once again (remember, we told it to go 2 times through the iteration).
The code can be rewritten to be much simpler, using the block-form Array constructor:
def random_select(array, n)
Array.new(n) {array.sample}
end
This creates a new array of size n and fills it with random samples from the array.
Note that the above solution, like your sample code, selects from the entire array each time which allows duplicate selections. If you don't want any duplicate selections, it's even simpler, since it is the default behavior of Array#sample:
def random_select(array, n)
array.sample(n)
end
So I really like this syntax in Lisp:
(+ 1 1 2 3 5 8 13)
=> 33
I want to add a list of items in Ruby and would like to approximate this as best as possible.
Right now, my best solution involves an array and the collect/map method.
So:
sum = 0; [1,1,2,3,5,8,13].collect { |n| sum += n }
BUT...
I would like to add methods to this which could return nil.
sum = 0; [1, booking_fee, 8,13].collect { |n| n = 0 if n.nil?; sum += n }
And it would be really nice to do this, where all of the lines in the middle refer to methods that may return nil, but I can't exactly build an array in this manner. This is just an idea of what I want my syntax to look like.
def total
Array.new do
booking_fee
rental_charges
internationalization_charges
discounts
wild_nights
end.collect { |n| n = 0 if n.nil?; sum += n }
end
Any suggestions before I try to hack away and effectuate Greenspun's Rule? (Programming is indeed a compulsion.
I really don't understand your question. If you want a method that works like + does in Lisp, i.e. takes an arbitrary number of arguments and is in prefix position rather than infix, that's trivial:
def plus(*nums)
nums.inject(:+)
end
plus 1, 1, 2, 3, 5, 8, 13 # => 33
If you want to get really fancy, you could override the unary prefix + operator for Arrays:
class Array
def +#
inject(:+)
end
end
+[1, 1, 2, 3, 5, 8, 13] # => 33
Please don't do that!
I don't see how the rest of your question is related in any way to a Lisp-style addition operation.
If you want to remove nils from an Array, there's Array#compact for that.
There is already a method inject for doing what you want.
Changing nil to a number without affecting a number is easy: use to_i (or to_f if you are dealing with float).
.
[
booking_fee,
rental_charges,
internationalization_charges,
discounts,
wild_nights,
].inject(0){|sum, item| sum + item.to_i}
I am learning ruby and practicing it by solving problems from Project Euler.
This is my solution for problem 12.
# Project Euler problem: 12
# What is the value of the first triangle number to have over five hundred divisors?
require 'prime'
triangle_number = ->(num){ (num *(num + 1)) / 2 }
factor_count = ->(num) do
prime_fac = Prime.prime_division(num)
exponents = prime_fac.collect { |item| item.last + 1 }
fac_count = exponents.inject(:*)
end
n = 2
loop do
tn = triangle_number.(n)
if factor_count.(tn) >= 500
puts tn
break
end
n += 1
end
Any improvements that can be made to this piece of code?
As others have stated, Rubyists will use methods or blocks way more than lambdas.
Ruby's Enumerable is a very powerful mixin, so I feel it pays here to build an enumerable in a similar way as Prime. So:
require 'prime'
class Triangular
class << self
include Enumerable
def each
sum = 0
1.upto(Float::INFINITY) do |i|
yield sum += i
end
end
end
end
This is very versatile. Just checking it works:
Triangular.first(4) # => [1, 3, 7, 10]
Good. Now you can use it to solve your problem:
def factor_count(num)
prime_fac = Prime.prime_division(num)
exponents = prime_fac.collect { |item| item.last + 1 }
exponents.inject(1, :*)
end
Triangular.find{|t| factor_count(t) >= 500} # => 76576500
Notes:
Float::INFINITY is new to 1.9.2. Either use 1.0/0, require 'backports' or do a loop if using an earlier version.
The each could be improved by first checking that a block is passed; you'll often see things like:
def each
return to_enum __method__ unless block_given?
# ...
Rather than solve the problem in one go, looking at the individual parts of the problem might help you understand ruby a bit better.
The first part is finding out what the triangle number would be. Since this uses sequence of natural numbers, you can represent this using a range in ruby. Here's an example:
(1..10).to_a => [1,2,3,4,5,6,7,8,9,10]
An array in ruby is considered an enumerable, and ruby provides lots of ways to enumerate over data. Using this notion you can iterate over this array using the each method and pass a block that sums the numbers.
sum = 0
(1..10).each do |x|
sum += x
end
sum => 55
This can also be done using another enumerable method known as inject that will pass what is returned from the previous element to the current element. Using this, you can get the sum in one line. In this example I use 1.upto(10), which will functionally work the same as (1..10).
1.upto(10).inject(0) {|sum, x| sum + x} => 55
Stepping through this, the first time this is called, sum = 0, x = 1, so (sum + x) = 1. Then it passes this to the next element and so sum = 1, x = 2, (sum + x) = 3. Next sum = 3, x = 3, (sum + x) = 6. sum = 6, x = 4, (sum + x) = 10. Etc etc.
That's just the first step of this problem. If you want to learn the language in this way, you should approach each part of the problem and learn what is appropriate to learn for that part, rather than tackling the entire problem.
REFACTORED SOLUTION (though not efficient at all)
def factors(n)
(1..n).select{|x| n % x == 0}
end
def triangle(n)
(n * (n + 1)) / 2
end
n = 2
until factors(triangle(n)).size >= 500
puts n
n += 1
end
puts triangle(n)
It looks like you are coming from writing Ocaml, or another functional language. In Ruby, you would want to use more def to define your methods. Ruby is about staying clean. But that might also be a personal preference.
And rather than a loop do you could while (faction_count(traingle_number(n)) < 500) do but for some that might be too much for one line.