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I'm learning Rust and I wanted to write program that would take a (randomly generated) bowling sheet and generate the score. I now know that I will have to use Rusts Vecs instead of arrays but I got stuck at accessing the value from the nested arrays so I would like to find the solution before I re-write it.
What I wanted to do is access the individual values and run some logic on them, but I got stuck at the "accessing values" part, this is what I came up with (but it doesn't work):
let sheet: [[u32; 2]; 10] = [[1, 3], [0, 6], [9, 0], [0, 5], [5, 3], [4, 2], [1, 4], [2, 3], [3, 0], [4, 4]];
for frame in 0..sheet.len() {
for score in 0..sheet[frame].len() {
println!("{}", sheet[frame[score]]);
}
}
You should clarify what exactly do you mean by 'accessing individual values', but from your code i'd assume that you just want to iterate over every score. Here's how you do it with for loops:
let sheet = [[1, 3], [0, 6], [9, 0], [0, 5], [5, 3], [4, 2], [1, 4], [2, 3], [3, 0], [4, 4]];
for frame in sheet {
// On the first iteration frame will be == [1, 3], then [0, 6], etc
for score in frame {
// on first iteration score will be == 1, then 3, then 0, etc
println!("{}", score);
}
}
I'm hoping to get an array into a specific format and do so in a one-liner if possible. Using each_slice(2).to_a is helpful in splitting the array, however I would like to keep second coordinate of each array as the first coordinate of the next array.
Initial Array
# format is [x1, y1, x2, y2, x3, y3, x4, y4]...
full_line_coords = [[1, 1, 2, 2, 3, 3, 4, 4], [5, 5, 6, 6, 7, 7, 8, 8]]
Desired Output
# desired format is [[[x1, y1], [x2, y2]], [[x2, y2], [x3, y3]], [[x3, y3], [x4, y4]]]...
desired = [[[1, 1], [2, 2]], [[2, 2], [3, 3]], [[3, 3], [4, 4]]]
Success without one-liner
# without one line:
desired = []
temp_array = []
full_line_coords.each do |x|
temp_array << x.each_slice(2).to_a
end
temp_array.each do |x|
i = 0
until i == x.length - 1
desired << [x[i], x[i+1]]
i += 1
end
end
p desired
# => [[[1, 1], [2, 2]], [[2, 2], [3, 3]], [[3, 3], [4, 4]], [[5, 5], [6, 6]], [[6, 6], [7, 7]], [[7, 7], [8, 8]]]
Unsure how to make this as one-line, found it simple enough to do the split, but not keeping the end/start coordinates in each array (as below).
One-liner attempt
attempt = full_line_coords.each { |x| p x.each_slice(2).to_a.each_slice(2).to_a } # p to show this is where i'd like the array to be in 'desired' format if possible.
# => [[[1, 1], [2, 2]], [[3, 3], [4, 4]]]
# [[[5, 5], [6, 6]], [[7, 7], [8, 8]]]
Background/Reasoning
The only reason I wish to keep it is a one-liner is because I want to return the "parent" object itself, not just the resulting attributes.
"Parent" objects being a list of links: #<Link:0x803a2e8>, with many attributes, including "segments".
links.each do |l|
puts l.segments
end
# Gives an array of XY coordinates, including all vertices. e.g. [1, 1, 2, 2, 3, 3, 4, 4]
I am then looking to use the "desired" output in some other defined methods but return the link object itself at the end #<Link:0x803a2e8>, not just products from the link's attributes.
Many thanks.
This is the first option I found:
full_line_coords.flat_map { |e| e.each_slice(2).each_cons(2).to_a }
Find the methods in Enumerable class and Array class
Input
full_line_coords = [[1, 1, 2, 2, 3, 3, 4, 4], [5, 5, 6, 6, 7, 7, 8, 8]]
Code
p full_line_coords.map { |var| var.slice_when { |x, y| x != y }
.each_cons(2)
.to_enum
.map(&:itself) }
Output
[[[[1, 1], [2, 2]], [[2, 2], [3, 3]], [[3, 3], [4, 4]]], [[[5, 5], [6, 6]], [[6, 6], [7, 7]], [[7, 7], [8, 8]]]]
I'd like to prevent producing pairs with the same items when producing a random set of pairs in a Ruby array.
For example:
[1,1,2,2,3,4].shuffle.each_slice(2).to_a
might produce:
[[1, 1], [3, 4], [2, 2]]
I'd like to be able to ensure that it produces a result such as:
[[4, 1], [1, 2], [3, 2]]
Thanks in advance for the help!
arr = [1,1,2,2,3,4]
loop do
sliced = arr.shuffle.each_slice(2).to_a
break sliced if sliced.none? { |a| a.reduce(:==) }
end
Here are three ways to produce the desired result (not including the approach of sampling repeatedly until a valid sample is found). The following array will be used for illustration.
arr = [1,4,1,2,3,2,1]
Use Array#combination and Array#sample
If pairs sampled were permitted to have the same number twice, the sample space would be
arr.combination(2).to_a
#=> [[1, 4], [1, 1], [1, 2], [1, 3], [1, 2], [1, 1], [4, 1], [4, 2],
# [4, 3], [4, 2], [4, 1], [1, 2], [1, 3], [1, 2], [1, 1], [2, 3],
# [2, 2], [2, 1], [3, 2], [3, 1], [2, 1]]
The pairs containing the same value twice--here [1, 1] and [2, 2]--are not wanted so they are simple removed from the above array.
sample_space = arr.combination(2).reject { |x,y| x==y }
#=> [[1, 4], [1, 2], [1, 3], [1, 2], [4, 1], [4, 2], [4, 3],
# [4, 2], [4, 1], [1, 2], [1, 3], [1, 2], [2, 3], [2, 1],
# [3, 2], [3, 1], [2, 1]]
We evidently are to sample arr.size/2 elements from sample_space. Depending on whether this is to be done with or without replacement we would write
sample_space.sample(arr.size/2)
#=> [[4, 3], [1, 2], [1, 3]]
for sampling without replacement and
Array.new(arr.size/2) { sample_space.sample }
#=> [[1, 3], [4, 1], [2, 1]]
for sampling with replacement.
Sample elements of each pair sequentially, Method 1
This method, like the next, can only be used to sample with replacement.
Let's first consider sampling a single pair. We could do that by selecting the first element of the pair randomly from arr, remove all instances of that element in arr and then sample the second element from what's left of arr.
def sample_one_pair(arr)
first = arr.sample
[first, second = (arr-[first]).sample]
end
To draw a sample of arr.size/2 pairs we there execute the following.
Array.new(arr.size/2) { sample_one_pair(arr) }
#=> [[1, 2], [4, 3], [1, 2]]
Sample elements of each pair sequentially, Method 2
This method is a very fast way of sampling large numbers of pairs with replacement. Like the previous method, it cannot be used to sample without replacement.
First, compute the cdf (cumulative distribution function) for drawing an element of arr at random.
counts = arr.group_by(&:itself).transform_values { |v| v.size }
#=> {1=>3, 4=>1, 2=>2, 3=>1}
def cdf(sz, counts)
frac = 1.0/sz
counts.each_with_object([]) { |(k,v),a|
a << [k, frac * v + (a.empty? ? 0 : a.last.last)] }
end
cdf_first = cdf(arr.size, counts)
#=> [[1, 0.429], [4, 0.571], [2, 0.857], [3, 1.0]]
This means that there is a probability of 0.429 (rounded) of randomly drawing a 1, 0.571 of drawing a 1 or a 4, 0.857 of drawing a 1, 4 or 2 and 1.0 of drawing one of the four numbers. We therefore can randomly sample a number from arr by obtaining a (pseudo-) random number between zero and one (p = rand) and then determine the first element of counts_cdf, [n, q] for which p <= q:
def draw_random(cdf)
p = rand
cdf.find { |n,q| p <= q }.first
end
draw_random(counts_cdf) #=> 1
draw_random(counts_cdf) #=> 4
draw_random(counts_cdf) #=> 1
draw_random(counts_cdf) #=> 1
draw_random(counts_cdf) #=> 2
draw_random(counts_cdf) #=> 3
In simulation models, incidentally, this is the standard way of generating pseudo-random variates from discrete probability distributions.
Before drawing the second random number of the pair we need to modify cdf_first to reflect that fact that the first number cannot be drawn again. Assuming there will be many pairs to generate randomly, it is most efficient to construct a hash cdf_second whose keys are the first values drawn randomly for the pair and whose values are the corresponding cdf's.
cdf_second = counts.keys.each_with_object({}) { |n, h|
h[n] = cdf(arr.size - counts[n], counts.reject { |k,_| k==n }) }
#=> {1=>[[4, 0.25], [2, 0.75], [3, 1.0]],
# 4=>[[1, 0.5], [2, 0.833], [3, 1.0]],
# 2=>[[1, 0.6], [4, 0.8], [3, 1.0]],
# 3=>[[1, 0.5], [4, 0.667], [2, 1.0]]}
If, for example, a 2 is drawn for the first element of the pair, the probability is 0.6 of drawing a 1 for the second element, 0.8 of drawing a 1 or 4 and 1.0 of drawing a 1, 4, or 3.
We can then sample one pair as follows.
def sample_one_pair(cdf_first, cdf_second)
first = draw_random(cdf_first)
[first, draw_random(cdf_second[first])]
end
As before, to sample arr.size/2 values with replacement, we execute
Array.new(arr.size/2) { sample_one_pair }
#=> [[2, 1], [3, 2], [1, 2]]
With replacement, you may get results like:
unique_pairs([1, 1, 2, 2, 3, 4]) # => [[4, 1], [1, 2], [1, 3]]
Note that 1 gets chosen three times, even though it's only in the original array twice. This is because the 1 is "replaced" each time it's chosen. In other words, it's put back into the collection to potentially be chosen again.
Here's a version of Cary's excellent sample_one_pair solution without replacement:
def unique_pairs(arr)
dup = arr.dup
Array.new(dup.size / 2) do
dup.shuffle!
first = dup.pop
second_index = dup.rindex { |e| e != first }
raise StopIteration unless second_index
second = dup.delete_at(second_index)
[first, second]
end
rescue StopIteration
retry
end
unique_pairs([1, 1, 2, 2, 3, 4]) # => [[4, 3], [1, 2], [2, 1]]
This works by creating a copy of the original array and deleting elements out of it as they're chosen (so they can't be chosen again). The rescue/retry is in there in case it becomes impossible to produce the correct number of pairs. For example, if [1, 3] is chosen first, and [1, 4] is chosen second, it becomes impossible to make three unique pairs because [2, 2] is all that's left; the sample space is exhausted.
This should be slower than Cary's solution (with replacement) but faster (on average) than the posted solutions (without replacement) that require looping and retrying. Welp, chalk up another point for "always benchmark!" I was wrong about all most of my assumptions. Here are the results on my machine with an array of 16 numbers ([1, 1, 2, 2, 3, 4, 5, 5, 5, 6, 7, 7, 8, 9, 9, 10]):
cary_with_replacement
93.737k (± 2.9%) i/s - 470.690k in 5.025734s
mwp_without_replacement
187.739k (± 3.3%) i/s - 943.415k in 5.030774s
mudasobwa_without_replacement
129.490k (± 9.4%) i/s - 653.150k in 5.096761s
EDIT: I've updated the above solution to address Stefan's numerous concerns. In hindsight, the errors are obvious and embarrassing! On the plus side, the revised solution is now faster than mudasobwa's solution, and I've confirmed that the two solutions have the same biases.
You can check if there any mathes and shuffle again:
a = [1,1,2,2,3,4]
# first time shuffle
sliced = a.shuffle.each_slice(2).to_a
# checking if there are matches and shuffle if there are
while sliced.combination(2).any? { |a, b| a.sort == b.sort } do
sliced = a.shuffle.each_slice(2).to_a
end
It is unlikely, be aware about possibility of infinity loop
a = [1, 2, 3]
b = [4, 5, 6]
How would I combine the two arrays in a 2D array?:
[[1, 4], [2, 5], [3, 6]]
Try Array#zip
a.zip(b)
=> [[1,4],[2,5],[3,6]]
While zip is obviously the most straightforward answer, this also works:
[a, b].transpose
=> [[1, 4], [2, 5], [3, 6]]
I have a list of elements (e.g. numbers) and I want to retrieve a list of all possible pairs. How can I do that using Ruby?
Example:
l1 = [1, 2, 3, 4, 5]
Result:
l2 #=> [[1,2], [1,3], [1,4], [1,5], [2,3], [2,4], [2,5], [3,4], [3,5], [4,5]]
In Ruby 1.8.6, you can use Facets:
require 'facets/array/combination'
i1 = [1,2,3,4,5]
i2 = []
i1.combination(2).to_a # => [[1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 5]]
In 1.8.7 and later, combination is built-in:
i1 = [1,2,3,4,5]
i2 = i1.combination(2).to_a
Or, if you really want a non-library answer:
i1 = [1,2,3,4,5]
i2 = (0...(i1.size-1)).inject([]) {|pairs,x| pairs += ((x+1)...i1.size).map {|y| [i1[x],i1[y]]}}