So what Im trying to accomplish is write a (shorter) condition that makes sure each element is different from the other array. This is confusing but I hope this example clears it up.
array = [1, 2, 3]
new_array = array.shuffle
until array[0] != new_array[0] &&
array[1] != new_array[1] &&
array[2] != new_array[2]
new_array = array.shuffle
end
So what Im doing is making sure that every single element/index pair does not match in the other array.
# [1, 2, 3] => [3, 1, 2] yayyyy
# [1, 2, 3] => [3, 2, 1] not what I want because the 2 didnt move
Is there a better way to do what I want to do? Ive looked up the .any? and .none? but I cant seem to figure out how to implement them. Thanks!
I would do this:
array.zip(new_array).all? { |left, right| left != right }
Here are two approaches that do not involve repeated sampling until a valid sample is obtained:
Sample from the population of valid permutations
Construct the population from which you are sampling:
array = [1, 2, 3, 4]
population = array.permutation(array.size).reject do |a|
a.zip(array).any? { |e,f| e==f }
end
#=> [[2, 1, 4, 3], [2, 3, 4, 1], [2, 4, 1, 3], [3, 1, 4, 2], [3, 4, 1, 2],
# [3, 4, 2, 1], [4, 1, 2, 3], [4, 3, 1, 2], [4, 3, 2, 1]]
Then just choose one at random:
10.times { p population.sample }
# [4, 3, 1, 2]
# [3, 4, 1, 2]
# [3, 4, 1, 2]
# [4, 3, 1, 2]
# [2, 1, 4, 3]
# [2, 1, 4, 3]
# [4, 1, 2, 3]
# [2, 1, 4, 3]
# [4, 3, 1, 2]
# [3, 4, 1, 2]
Sequentially sample for each position in the array
def sample_no_match(array)
a = array.each_index.to_a.shuffle
last_ndx = a[-1]
a.dup.map do |i|
if a.size == 2 && a[-1] == last_ndx
select = a[-1]
else
select = (a-[i]).sample
end
a.delete(select)
array[select]
end
end
10.times.each { p sample_no_match(array) }
# [2, 4, 3, 1]
# [4, 3, 1, 2]
# [2, 1, 3, 4]
# [1, 3, 4, 2]
# [1, 3, 2, 4]
# [1, 3, 2, 4]
# [1, 4, 3, 2]
# [3, 4, 2, 1]
# [1, 3, 4, 2]
# [1, 3, 4, 2]
I have been unable to prove or disprove that the second method produces a random sample. We can, however, determine relative frequencies of outcomes:
n = 500_000
h = n.times.with_object(Hash.new(0)) { |_,h| h[sample_no_match(array)] += 1 }
h.keys.each { |k| h[k] = (h[k]/(n.to_f)).round(4) }
h #=> {[1, 2, 3, 4]=>0.0418, [2, 1, 3, 4]=>0.0414, [1, 4, 2, 3]=>0.0418,
# [3, 4, 2, 1]=>0.0417, [4, 3, 2, 1]=>0.0415, [3, 1, 4, 2]=>0.0419,
# [2, 3, 1, 4]=>0.0420, [4, 2, 3, 1]=>0.0417, [3, 2, 1, 4]=>0.0413,
# [4, 2, 1, 3]=>0.0417, [2, 1, 4, 3]=>0.0419, [1, 3, 2, 4]=>0.0415,
# [1, 2, 4, 3]=>0.0418, [1, 3, 4, 2]=>0.0417, [2, 4, 1, 3]=>0.0414,
# [3, 4, 1, 2]=>0.0412, [1, 4, 3, 2]=>0.0423, [4, 1, 3, 2]=>0.0411,
# [3, 2, 4, 1]=>0.0411, [2, 4, 3, 1]=>0.0418, [3, 1, 2, 4]=>0.0419,
# [4, 3, 1, 2]=>0.0412, [4, 1, 2, 3]=>0.0421, [2, 3, 4, 1]=>0.0421}
avg = (h.values.reduce(:+)/h.size.to_f).round(4)
#=> 0.0417
mn, mx = h.values.minmax
#=> [0.0411, 0.0423]
([avg-mn,mx-avg].max/avg).round(6)
#=> 0.014388
which means that the maximum deviation from the average was only 1.4% percent of the average.
This suggests that the second method is a reasonable way of producing pseudo-random samples.
Initially, the first line of this method was:
a = array.each_index.to_a
By looking at the frequency distribution for outcomes, however, it was clear that that method did not produce a pseudo-random sample; hence, the need to shuffle a.
Here's one possibility:
until array.zip(new_array).reject{ |x, y| x == y }.size == array.size
new_array = array.shuffle
end
Note, though, that it will break for arrays like [1] or [1, 1, 1, 2, 3], where the number of instances of 1 exceeds half the size of the array. Recommend Array#uniq or similar, along with checking for arrays of sizes 0 or 1, depending on how trustworthy your input is!
Related
I have written the following program to find the permutation of all the elements in an array. The values are created properly but the problem occurs when I try to assign the generated sequence into a new array. The old values will get cleared and the new values are copied as per the array size
def find_perm(nums, answer, set)
if nums.empty?
p set
answer.push(set)
p answer.object_id
p answer
return true
end
for i in (0..nums.length - 1) do
new_nums = nums.clone
new_nums.delete_at(i)
set.push(nums[i])
find_perm(new_nums, answer, set)
set.pop
end
end
def permute(nums)
answer = []
set = []
element = find_perm(nums, answer, set)
return element
end
permute([1,2,3])
This are the observations that I have found out while debugging:
[1, 2, 3]
47167191669680
[[1, 2, 3]]
[1, 3, 2]
47167191669680
[[1, 3, 2], [1, 3, 2]]
[2, 1, 3]
47167191669680
[[2, 1, 3], [2, 1, 3], [2, 1, 3]]
[2, 3, 1]
47167191669680
[[2, 3, 1], [2, 3, 1], [2, 3, 1], [2, 3, 1]]
[3, 1, 2]
47167191669680
[[3, 1, 2], [3, 1, 2], [3, 1, 2], [3, 1, 2], [3, 1, 2]]
[3, 2, 1]
47167191669680
[[3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1]]
The problem was each time I was pushing the same set array into the answer array so each element in the answer array will have same reference (same object_id).
Solution: Is to clone the set array during the each push to the answer array so that each element have different reference.
The solution:
def fact(n)
return 1 if n == 1
n*fact(n-1)
end
def find_perm(nums, answer, set)
if nums.empty?
answer.push(set.clone)
end
for i in (0..nums.length - 1) do
new_nums = nums.clone
new_nums.delete_at(i)
set.push(nums[i])
find_perm(new_nums, answer, set)
set.pop
return answer if fact(nums.count) == answer.count
end
end
def permute(nums)
answer = []
set = []
element = find_perm(nums, answer, set)
return element
end
p permute([1,2,3])
I believe the approach you are taking is similar to the following.
Suppose we wish to obtain the permutations of the elements of the array
arr = [1, 2, 3, 4]
Begin with the array
[4]
This array has only a single perumutation:
perms3 = [[4]]
(3 in perms3 denotes the index of 4 in arr.) Now obtain the permuations of
[3, 4]
We see that is
perms2 = [[3, 4], [4, 3]]
We simply take each element of perms3 ([4] is the only one) and create two permuations by inserting 3 before 4 and then 3 after 4;
Now suppose the array were
[2, 3, 4]
Then
perms1 = [[2, 3, 4], [3, 2, 4], [3, 4, 2], [2, 4, 3], [4, 2, 3], [4, 3, 2]]
We create three 3-element arrays from [3, 4], one by inserting 2 before 3, one by inserting 2 between 3 and 4 and one by inserting 2 after 4. Simlarly, three 3-element arrays are generated from [4, 3] in a simlar way. This generates six arrays. (Indeed, 3! #=> 6).
Lastly we generating the 4! #=> 24 permuations of [1, 2, 3, 4] by inserting 1 in four locations of each element of perm1, the first four derived from perms1[0] #=> [2, 3, 4]:
[[1, 2, 3, 4], [2, 1, 3, 4], [2, 3, 1, 4], [2, 3, 4, 1]]
We can do this in code as follows.
def my_permutations(arr)
perms = [[arr.last]]
(arr.size-2).downto(0) do |i|
x = arr[i]
perms = perms.flat_map do |perm|
(0..(perm.size)).map { |i| perm.dup.insert(i, x) }
end
end
perms
end
my_permutations(arr)
#=> [[1, 2, 3, 4], [2, 1, 3, 4], [2, 3, 1, 4], [2, 3, 4, 1], [1, 3, 2, 4],
# [3, 1, 2, 4], [3, 2, 1, 4], [3, 2, 4, 1], [1, 3, 4, 2], [3, 1, 4, 2],
# [3, 4, 1, 2], [3, 4, 2, 1], [1, 2, 4, 3], [2, 1, 4, 3], [2, 4, 1, 3],
# [2, 4, 3, 1], [1, 4, 2, 3], [4, 1, 2, 3], [4, 2, 1, 3], [4, 2, 3, 1],
# [1, 4, 3, 2], [4, 1, 3, 2], [4, 3, 1, 2], [4, 3, 2, 1]]
See Enumerable#flat_map and Array#insert. Note that we need to make a copy of the array perm before invoking insert.
We could of course have gone "forward" in arr (starting with [[1]]), rather than "backward", though the elements of the array of permutations would be ordered differently.
I'm new to Ruby and would like to know if there is a better way to solve the following problem.
I have an array that looks like this:
[6, 1, 3, 6, 2, 4, 1, 3, 2, 3]
I'd like to turn it into this:
[ [1,1], [2,2], [3,3,3], [4], [], [6,6] ]
This is my current solution (again, I'm new to Ruby):
def split_array_into_arrays(array)
max_num = array.max
arrays = Array.new(max_num) { Array.new }
array.each do |num|
arrays[num-1] << num
end
arrays
end
arrays = split_array_into_arrays([6, 1, 3, 6, 2, 4, 1, 3, 2, 3])
puts arrays.inspect
Produces:
[[1, 1], [2, 2], [3, 3, 3], [4], [], [6, 6]]
Note: I realize I am not handling possible errors.
How might an experienced Ruby developer implement this?
ar = [6, 1, 3, 6, 2, 4, 1, 3, 2, 3]
(1..ar.max).map{|n| [n]*ar.count(n)}
# => [[1, 1], [2, 2], [3, 3, 3], [4], [], [6, 6]]
Normal uniq:
[1, 2, 3, 1, 1, 4].uniq => [1, 2, 3, 4]
I want to replace the duplicate with a replacement at where it was.
Is there a method or way to achieve something like this?
[1, 2, 3, 1, 1, 4].uniq_with_replacement(-1) => [1, 2, 3, -1, -1, 4]
Thanks in advance!
Here's a one-liner:
a.fill{ |i| a.index(a[i]) == i ? a[i] : -1 }
Something like this?:
class Array
def uniq_with_replacement(v)
map.with_object([]){|value, obj| obj << (obj.include?(value) ? v : value) }
end
end
Now:
[1, 2, 3, 1, 1, 4].uniq_with_replacement(-1)
# => [1, 2, 3, -1, -1, 4]
[1, 2, 3, 1, 1, 2, 4].uniq_with_replacement(-1)
# => [1, 2, 3, -1, -1, -1, 4]
1 more:
arr = [1, 2, 3, 1, 1, 4]
value = -1
a = arr.each_with_index.to_a
#=> [[1, 0], [2, 1], [3, 2], [1, 3], [1, 4], [4, 5]]
b = (a - a.uniq(&:first)).map(&:last)
#=> [3, 4]
arr.map.with_index { |e,i| b.include?(i) ? value : e }
#=> [1, 2, 3, -1, -1, 4]
I'd like to get [[2, 1, 3], [1, 3, 2]] from [1, 2, 3] in Ruby.
For [1, 2, 3, 4], I'd like to get [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
Rule: Within two numbers, if left one is smaller then it swap the position.
I have the following codes so far but it returns [[2, 3, 1], [2, 3, 1]]
What am I doing wrong here? I appreciate any inputs.
In amidakuji.rb
class Amidakuji
def initialize(column, rung)
#column = column
#rung = rung
#myarr = []
#per_arr = []
#build_arr = []
end
def build_initial
#arr = (1..#column).to_a
end
def swap_element
i = 0
arr = build_initial
while i < #column - 1 do
#build_arr << swap(arr, i)
i += 1
end
#build_arr
end
def swap(arr, a)
if arr[a] < arr[a + 1]
arr[a], arr[a + 1] = arr[a + 1], arr[a]
end
arr
end
end
In amidakuji_spec.rb
it 'should create an array with swapped elements' do
expect(#kuji1.swap_element).to eq ([[2, 1, 3], [1, 3, 2]])
end
Results
Failures:
expected: [[2, 1, 3], [1, 3, 2]]
got: [[2, 3, 1], [2, 3, 1]]
You can do this quite compactly by using the methods Enumerable#each_cons and Enumerable#map.
Code
def doit(arr)
(0...arr.size).each_cons(2).map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
end
Examples
doit([1,2,3]) #=> [[2, 1, 3], [1, 3, 2]]
doit([1,2,3,4]) #=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
doit([1,2,3,4,5]) #=> [[2, 1, 3, 4, 5], [1, 3, 2, 4, 5],
#=> [1, 2, 4, 3, 5], [1, 2, 3, 5, 4]]
Explanation
arr = [1,2,3,4]
b = (0...arr.size).each_cons(2)
#=> #<Enumerator: 0...4:each_cons(2)>
To view the contents of this enumerator:
b.to_a
#=> [[0, 1], [1, 2], [2, 3]]
Lastly
b.map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
#=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
In the last step, consider the first element of b that is passed to map, which assigns the following values to the block variables:
i => 0
j => 1
We then make a copy of arr, swap the elements offsets 0 and 1, making
a => [2, 1, 3, 4]
and then enter a at the end of the block, causing map to replace [0, 1] with that array.
Given what you're trying to accomplish and the output you're getting, it looks like you're reusing the same array when you want distinct arrays instead. Specifically this line:
#build_arr << swap(arr, i)
is always passing the same 'arr' to swap.
So first time, it swaps the 1 and the 2 to give you [2, 1, 3]
Second time, it swaps the 1 and the 3 give you [2, 3, 1]
You push the same array onto #build_arr twice, which is why it repeats.
I have three Ruby arrays:
[1, 2, 3, 4]
[2, 3, 4, 5]
[3, 4, 5, 6]
How can I take the average of all three numbers in position 0, then position 1, etc. and store them in a new array called 'Average'?
a = [1, 2, 3, 4]
b = [2, 3, 4, 5]
c = [3, 4, 5, 6]
a.zip(b,c)
# [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6]]
.map {|array| array.reduce(:+) / array.size }
# => [ 2,3,4,5]
Try this:
arr = ([1, 2, 3, 4] + [3, 4, 5, 6] + [2, 3, 4, 5])
arr.inject(0.0) { |sum, el| sum + el } / arr.size
The concatenation could be done in several ways, depends on how you store your arrays.
As a syntactic sugar, you could do it like this too:
arr.inject(:+).to_f / arr.size