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I need help to write even_odd method that accepts an array of whole number.
It should return an array of 2 arrays
The first nested array should contain only the odd numbers
The second nested array should contain only the even numbers
If there are no even or odd numbers, the respective inner array should be empty
Output should look like this : -
even_odd([3, 5, 8, 2, 4, 6])
[[3, 5], [2, 4, 6, 8]]
even_odd([3, 5])
[[3, 5], []]
even_odd([2, 4])
[[], [2, 4]]
I am new to ruby programming, I have tried below but not getting the result :-
def even_odd(numbers)
arr1, arr2 = []
idx = 0
while idx < numbers.length
if numbers[idx] % 2 == 0
puts arr1[idx]
elsif
puts arr2[idx]
end
idx += 1
end
end
puts even_odd([2, 3, 6])
Error :-
main.rb:6:in `even_odd': undefined method `[]' for nil:NilClass (NoMethodError)
from main.rb:13:in `<main>'
I would do this
def even_odd(numbers)
numbers.sort.partition(&:odd?)
end
even_odd([3, 5, 8, 2, 4, 6])
# => [[3, 5], [2, 4, 6, 8]]
even_odd([3, 5])
# => [[3, 5], []]
even_odd([2, 4])
# => [[], [2, 4]]
puts is a print statement in ruby, not an append one. It also doesn't run a function/method. You'll also want to call the index on the numbers array inside the if...else block.
This should do the trick:
def even_odd(numbers)
arr1, arr2 = [], []
idx = 0
while idx < numbers.length
if numbers[idx] % 2 == 0
arr1 << numbers[idx]
elsif
arr2 << numbers[idx]
end
idx += 1
end
return arr1, arr2
end
arrays = even_odd([2, 3, 6])
puts arrays
I'm trying to operate on certain elements of an array while referencing their index in the block. Operating on the whole array is easy
arr = [1, 2, 3, 4, 5, 6, 7, 8]
arr.each_with_index { |num, index| puts "#{num}, "#{index}" }
But what if I want to work just with elements 4, 6 to return
4, 3
6, 5
I can create a new array composed of certain elements of the original and run the block on that, but then the index changes.
How can I select the elements and their index?
Just put a condition on it:
indice = [3, 5]
arr.each_with_index do
|num, index| puts "#{num}, #{index}" if indice.include?(index)
end
This is another style:
indice = [3, 5]
arr.each_with_index do
|num, index|
next unless indice.include?(index)
puts "#{num}, #{index}"
end
I cannot tell from the question whether you are given values in the array and want to obtain their indices, or vice-versa. I therefore will suggest one method for each task. I will use this array for examples:
arr = [1, 2, 3, 4, 5, 6, 7, 8]
Values to Indices
If you are given values:
vals = [4, 6]
you can retrieve the number-index pairs like this:
vals.map { |num| [num, arr.index(num)] }
#=> [[4, 3], [6, 5]]
or print them directly:
vals.each { |num| puts "#{num}, #{arr.index(num)}" }
# 4, 3
# 6, 5
#=> [4, 6]
If an element of vals is not present in arr:
vals = [4, 99]
vals.map { |num| [num, arr.index(num)] }
#=> [[4, 3], [99, nil]]
Indices to Values
If you are given indices:
indices = [3, 5]
you can retrieve the index-value pairs like this:
indices.zip(arr.values_at(*indices))
#=> [[3, 4], [5, 6]]
and then print in whatever format you like.
If an index is out-of-range, nil will be returned:
indices.zip(arr.values_at(*[3, 99]))
#=> [[3, 4], [5, nil]]
I'm trying to find the mode of an Array. Mode = the element(s) that appear with the most frequency.
I know there are lots of tricks with #enumerable, but I'm not there yet in my learning. The exercise I'm doing assumes I can solve this problem without understanding enumerable.
I've written out my game plan, but I'm stuck on the 2nd part. I'm not sure if it's possible to compare a hash key against an array, and if found, increment the value.
def mode(array)
# Push array elements to hash. Hash should overwrite dup keys.
myhash = {}
array.each do |x|
myhash[x] = 0
end
# compare Hash keys to Array. When found, push +=1 to hash's value.
if myhash[k] == array[x]
myhash[k] += 1
end
# Sort hash by value
# Grab the highest hash value
# Return key(s) per the highest hash value
# rejoice!
end
test = [1, 2, 3, 3, 3, 4, 5, 6, 6, 6]
mode(test) # => 3, 6 (because they each appear 3 times)
You can create a hash with a default initial value:
myhash = Hash.new(0)
Then increment specific occurrences:
myhash["foo"] += 1
myhash["bar"] += 7
myhash["bar"] += 3
p myhash # {"foo"=>1, "bar"=>10}
With that understanding, if you replace your initial hash declaration and then do the incrementing in your array.each iterator, you're practically done.
myhash.sort_by{|key,value| value}[-1]
gives the last entry in the sorted set of hash values, which should be your mode. Note that there may be multiple modes, so you can iterate backwards while the value portion remains constant to determine them all.
There are many, many ways you could do this. Here are a few.
#1
array = [3,1,4,5,4,3]
a = array.uniq #=> [3, 1, 4, 5]
.map {|e| [e, array.count(e)]}
#=> [[3, 2], [1, 1], [4, 2], [5, 1]]
.sort_by {|_,cnt| -cnt} #=> [[3, 2], [4, 2], [1, 1], [5, 1]]
a.take_while {|_,cnt| cnt == a.first.last}
#=> [[3, 2], [4, 2]]
.map(&:first) #=> [3, 4]
#2
array.sort #=> [1, 3, 3, 4, 4, 5]
.chunk {|e| e}
#<Enumerator: #<Enumerator::Generator:0x000001021820b0>:each>
.map { |e,a| [e, a.size] } #=> [[1, 1], [3, 2], [4, 2], [5, 1]]
.sort_by { |_,cnt| -cnt } #=> [[4, 2], [3, 2], [1, 1], [5, 1]]
.chunk(&:last)
#<Enumerator: #<Enumerator::Generator:0x00000103037e70>:each>
.first #=> [2, [[4, 2], [3, 2]]]
.last #=> [[4, 2], [3, 2]]
.map(&:first) #=> [4, 3]
#3
h = array.each_with_object({}) { |e,h|
(h[e] || 0) += 1 } #=> {3=>2, 1=>1, 4=>2, 5=>1}
max_cnt = h.values.max #=> 2
h.select { |_,cnt| cnt == max_cnt }.keys
#=> [3, 4]
#4
a = array.group_by { |e| e } #=> {3=>[3, 3], 1=>[1], 4=>[4, 4], 5=>[5]}
.map {|e,ees| [e,ees.size]}
#=> [[3, 2], [1, 1], [4, 2], [5, 1]]
max = a.max_by(&:last) #=> [3, 2]
.last #=> 2
a.select {|_,cnt| cnt == max}.map(&:first)
#=> [3, 4]
In your approach, you have first initialized a hash containing keys taken from the unique values of the array, with the associated values all set to zero. For example, the array [1,2,2,3] would create the hash {1: 0, 2: 0, 3: 0}.
After this, you plan to count the instances of each of the values in the array by incrementing the value for the associated key in the hash by one for each instance. So, after finding the number 1 in the array, the hash would look like so: {1: 1, 2: 0, 3: 0}. You clearly need to do this for each value in the array, so given your approach and current level of understanding, I would suggest looping through the array again:
array.each do |x|
myhash[x] += 1
end
As you can see, we don't need to check that myhash[k] == array[x] since we have already created a key:value pair for each number in the array.
However, while this approach will work, it's not very efficient: we're having to loop through the array twice. The first time to initialize all the key:value pairs to some default (zero, in this case), and the second to count the frequencies of each number.
Since the default value for each key will be zero, we can remove the need to initialize the defaults by using a different hash constructor. myhash = {} will return nil if we access a key that doesn't exist, but myhash = Hash.new(0) will return 0 if we access a non-existent key (note that you could provide any other value or variable, if required).
By providing a default value of zero, we can get rid of the first loop entirely. When the second loop finds a key that doesn't exist, it will use the default provided and automatically initialize it.
def mode(array)
array.group_by{ |e| e }.group_by{ |k, v| v.size }.max.pop.map{ |e| e.shift }
end
Using the simple_stats gem:
test = [1, 2, 3, 3, 3, 4, 5, 6, 6, 6]
test.modes #=> [3, 6]
If it is an unsorted array, we can sort the array in descending order
array = array.sort!
Then use the sorted array to create a hash default 0 and with each element of the array as a key and number of occurrence as the value
hash = Hash.new(0)
array.each {|i| hash[i] +=1 }
Then mode will be the first element if the hash is sorted in descending order of value(number of occurrences)
mode = hash.sort_by{|key, value| -value}.first[0]
I'm generating an array of grouped indexes. The indexes are points within an array that meet my grouping requirements. For example I'm grouping indexes from a grid where things are horizontally "close" to each other. This is kind of what I'll be working with.
[[0,1,2],[3],[4,5],[5,6],[7,8],[8,9]]
I would like to merge by common indexes. So the result should look like.
[[0,1,2],[3],[4,5,6],[7,8,9]]
It feels like it should be an inject :+ on pairs if any inner items match. But I don't see the Ruby way to do this.
x.sort.inject([]) do |y, new|
(((y.last || []) & new).length > 0) ? y[0..-2].push(y.last | new) : y.push(new)
end.map(&:sort)
Knowing Ruby, there's probably a more concise way to do this, but this should give you what you want:
foo = [[0,1,2],[3],[4,5],[5,6],[7,8],[8,9]]
foo.inject([]) {|result,element|
if (result and existing = result.find_index{|a| !(element & [*a]).empty?})
tmp = result[existing]
result.delete_at(existing)
result << (tmp | element).sort
else
result << element
end
}.sort
Output:
=> [[0, 1, 2], [3], [4, 5, 6], [7, 8, 9]]
Logic:
For each element in the original array, check the newly-built array-so-far (result) for any entry which contains any of the same numbers as the next element using array intersection -- !(element & [*a]).empty? ...
if found, remove said entry from the result, union it with the new element from the original array -- (tmp | element) -- then add it back to the result
if not found, simply concatenate the element from the original array to the result
Someone might find a more compact method, but this works...
array = [[0,1,2],[3],[4,5],[5,6],[7,8],[8,9]]
(0...array.length).each do |a|
(a+1...array.length).each do |b|
unless array[a].to_a & array[b].to_a == []
array[a].push(array[b]).flatten!.uniq!.sort!
array.delete_at(b)
b -= 1
end
end
end
p array
=> [[0, 1, 2], [3], [4, 5, 6], [7, 8, 9]]
The solutions so far seem overly-complicated to me. I suggest this (assuming each element of arr is non-empty and contains only integers):
arr = [[0, 1, 2], [3],
[4, 5], [5, 6],
[7, 8, 9], [9, 10], [10, 11],
[12, 13], [13], [13, 14]]
arr.each_with_object([]) do |a,b|
if b.any? && b.last.last == a.first
b[-1] += a[1..-1]
else
b << a
end
end
#=> [[0, 1, 2], [3], [4, 5, 6], [7, 8, 9, 10, 11], [12, 13, 14]]
You could alternatively do it by stepping through arr with an enumerator:
enum = arr.each
b = [enum.next]
loop do
a = enum.next
if b.last.last == a.first
b[-1] += a[1..-1]
else
b << a
end
end
b
I have two arrays:
#a = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
#b = [a, b, c]
I need to replace n-th column in a with b like:
swap_column(0)
#=> [a, 2, 3]
[b, 5, 6]
[c, 8, 9]
(This is for using Cramer's rule for solving equations system, if anybody wonders.)
The code I've come up with:
def swap_column(n)
#a.map.with_index { |row, j| row[n] = #b[j] }
end
How do I get rid of assignment here so that map returns the modified matrix while leaving #a intact?
What you wanted is dup. Also, you had the return value of the map.with_index block wrong.
def swap_column(i)
#a.map.with_index{|row, j| row = row.dup; row[i] = #b[j]; row}
end
or
def swap_column(i)
#a.map.with_index{|row, j| row.dup.tap{|row| row[i] = #b[j]}}
end
The answer by sawa is good and the main point is you need to dup your inner arrays for this to work properly. The only reason for this additional post is to point out that often when you are using with_index so that you can directly 1:1 index into another array you can simplify the code by using zip.
def swap_column(n)
#a.zip(#b).map {|r,e| r.dup.tap{|r| r[n] = e}}
end
What zip does is combine your two arrays into a new array where each element is an array made of the two corresponding elements of the initial arrays. In this case it would be an array of an array and an element you want to later use for replacement. We then map over those results and automatically destructure each element into the two pieces. We then dup the array piece and tap it to replace the nth element.
You can use transpose to do the following:
class M
attr :a, :b
def initialize
#a = [[1,2,3],
[4,5,6],
[7,8,9]
]
#b = [:a, :b, :c]
end
def swap_column(n)
t = #a.transpose
t[0] = #b
t.transpose
end
end
m = M.new
=> #<M:0x007ffdc2952e38 #a=[[1, 2, 3], [4, 5, 6], [7, 8, 9]], #b=[:a, :b, :c]>
m.swap_column(0)
=> [[:a, 2, 3], [:b, 5, 6], [:c, 8, 9]]
m # m is unchanged
=> #<M:0x007ffdc2952e38 #a=[[1, 2, 3], [4, 5, 6], [7, 8, 9]], #b=[:a, :b, :c]>