I am currently working on a challenge where the guidelines are as follows:
You have been asked to help study the population of birds migrating across the continent. Each type of bird you are interested in will be identified by an integer value. Each time a particular kind of bird is spotted, its id number will be added to your array of sightings. You would like to be able to find out which type of bird is most common given a list of sightings. Your task is to print the type number of that bird and if two or more types of birds are equally common, choose the type with the smallest ID number.
For example, assume your bird sightings are of types arr = [1, 1, 2, 2, 3]. There are two each of types 1 and 2, and one sighting of type 3. Pick the lower of the two types seen twice: type 1.
I have written code that passes most of the tests but times out on the exceedingly large inputs, I would like your advice on how to optimize it.
My code is as follows:
def migratoryBirds(arr)
sorted = Hash[arr.map { |x| [x, arr.select { |y| y==x }.count] }]
return sorted.max_by { |k,v| v }[0]
end
Your sorted hash can be written a bit more concise as:
sorted = arr.map { |x| [x, arr.count(x)] }.to_h
For the example array [1, 1, 2, 2, 3] this is equivalent to:
[
[1, arr.count(1)], # counts all 1's in arr
[1, arr.count(1)], # counts all 1's in arr (again)
[2, arr.count(2)], # counts all 2's in arr
[2, arr.count(2)], # counts all 2's in arr (again)
[3, arr.count(3)] # counts all 3's in arr
].to_h
Not only does it count 1 and 2 twice. It also has to traverse the entire array again for each count call (or select in your code).
A better approach is to traverse the array once and use a hash for counting the occurrences:
arr = [1, 1, 2, 2, 3]
sorted = Hash.new(0)
arr.each { |x| sorted[x] += 1 }
sorted #=> {1=>2, 2=>2, 3=>1}
This can also be written in a single line via each_with_object:
sorted = arr.each_with_object(Hash.new(0)) { |x, h| h[x] += 1 }
#=> {1=>2, 2=>2, 3=>1}
Ruby 2.7 even has a dedicated method tally to count occurrences:
sorted = arr.tally
#=> {1=>2, 2=>2, 3=>1}
Related
I am trying to print all the different sums of all combinations in this array [1,2,3]. I want to first push every sum result to a new array b, then print them using b.uniq so that non of the sum results are repeated.
However, with the code I have, the 3 repeats itself, and I think it is because of the way it is pushed into the array b.
Is there a better way of doing this?
a = [1,2,3]
b = []
b.push a
b.push a.combination(2).collect {|a,b| (a+b)}
b.push a.combination(3).collect {|a,b,c| (a+b+c)}
puts b.uniq
p b #[[1, 2, 3], [3, 4, 5], [6]]
Can someone please help me with this? I am still new in ruby.
Because an Array of arbitrary length can be summed using inject(:+), we can create a more general solution by iterating over the range 1..n, where n is the length of the Array.
(1..(a.size)).flat_map do |n|
a.combination(n).map { |c| c.inject(&:+) }
end.uniq
#=> [1, 2, 3, 4, 5, 6]
By using flat_map, we can avoid getting the nested Array result, and can call uniq directly on it. Another option to ensure uniqueness would be to pass the result to a Set, for which Ruby guarantees uniqueness internally.
require "set"
sums = (1..(a.size)).flat_map do |n|
a.combination(n).map { |c| c.inject(&:+) }
end
Set.new(sums)
#=> #<Set: {1, 2, 3, 4, 5, 6}>
This will work for an any Array, as long as all elements are Fixnum.
If all you want is an array of the possible sums, flatten the array before getting the unique values.
puts b.flatten.uniq
What is happening is uniq is running over a multi-dimensional array. This causes it to look for duplicate arrays in your array. You'll need the array to be flattened first.
I have a pair of arrays,
array_1 = [1,2,3,4,5]
array_2 = [10,9,8,7,6]
and I'm trying to subtract the nth element of one array from the (n-1)-th element of the second array, starting with the n-th element, yielding an array of:
[9-1, 8-2, 7-3, 6-4] = [8, 6, 4, 2]
I wrote it in a procedural fashion:
array_1.pop
array_2.shift
[array_2,array_1].transpose.map { |a,b| a-b }
but I do not wish to alter the arrays. Is there a method or another way to go about this?
Another way:
enum1 = array_1.to_enum
enum2 = array_2.to_enum
enum2.next
arr = []
loop do
arr << enum2.next - enum1.next
end
arr
#=> [8, 6, 4, 2]
Use the non-destructive drop for the receiver, and zip, which will stop when the receiver runs out of an element even if the argument has more.
array_2.drop(1).zip(array_1).map{|a, b| a - b}
I think you may be overthinking it a bit; as long as both arrays are the same length, you can just iterate over the indices you care about, and reference the other array by index - offset.
array_1 = [1,2,3,4,5]
array_2 = [10,9,8,7,6]
n = 1
(n...array_1.length).map {|i| array_2[i] - array_1[i - 1] }
You can set n to whatever number you like and compute from that point onwards, so even if the arrays were tremendously large, you don't have to generate any intermediate arrays, and you don't have to perform any unnecessary work.
def peel array
output = []
while ! array.empty? do
output << array.shift
mutate! array
end
output.flatten
end
I have not included the mutate! method, because I am only interested in removing the output variable. The mutate! call is important because we cannot iterate over the array using each because array is changing.
EDIT: I am getting an array as output, which is what I want. The method works correctly, but I think there is a way to collect the array.shift values without using a temp variable.
EDIT #2: OK, here is the mutate! method and test case:
def mutate! array
array.reverse!
end
a = (1..5).to_a
peel( a ).should == [ 1, 5, 2, 4, 3 ]
It doesn't matter if peel modifies array. I guess it should be called peel!. Yes, mutate! must be called after each element is removed.
All this reversing makes me dizzy.
def peel(array)
indices = array.size.times.map do |i|
i = -i if i.odd?
i = i/2
end
array.values_at(*indices) # indices will be [0, -1, 1, -2, 2] in the example
end
a = (1..5).to_a
p peel(a) #=>[1, 5, 2, 4, 3]
Another approach:
def peel(array)
mid = array.size/2
array[0..mid]
.zip(array[mid..-1].reverse)
.flatten(1)
.take(array.size)
end
Usage:
peel [1,2,3,4,5,6]
#=> [1, 6, 2, 5, 3, 4]
peel [1,2,3,4,5]
#=> [1, 5, 2, 4, 3]
Here's a way using parallel assignment:
def peel array
n = array.size
n.times {|i| (n-2-2*i).times {|j| array[n-1-j], array[n-2-j] = array[n-2-j], array[n-1-j]}}
array
end
peel [1,2,3,4,5] # => [1,5,2,4,3]
peel [1,2,3,4,5,6] # => [1,6,2,5,3,4]
What I'm doing here is a series of pairwise exchanges. By way of example, for [1,2,3,4,5,6], the first 6-2=4 steps (6 being the size of the array) alter the array as follows:
[1,2,3,4,6,5]
[1,2,3,6,4,5]
[1,2,6,3,4,5]
[1,6,2,3,4,5]
The 1, 6 and the 2 are in now the right positions. We repeat these steps, but this time only 6-4=2 times, to move the 5 and 3 into the correct positions:
[1,6,2,3,5,4]
[1,6,2,5,3,4]
The 4 is pushed to the end, it's correct position, so we are finished.
What is the clearest and most efficient way to add all the elements in 2D array by their position in Ruby. Example:
2darray = [[1,2,3],[1,2,3]]
result = [2,4,6]
I have the following code
def sum_elements_by_position(array)
total_elements = array.length
result = []
for i in 0...array.first.length
n = 0
array.each { |subarray| n += subarray[i] }
result << n
end
result
end
Assumptions: All primary elements are of the same length
For bonus points it would be great to see a solution that works primary elements of an arbitrary length
You can zip the first row with the rest of them and then do the sum:
def sum_elements_by_position(array)
array[0].zip(*array[1..-1]).map do |col|
col.inject(:+)
end
end
Here's a solution addressing when the rows aren't the same length.
def sum_cols arr
arr.reduce( [] ) do |res,row|
row.each_with_index { |e,i| res[i] ||= 0; res[i] += e }
res
end
end
irb> sum_cols [ [0,1,2], [3,4], [5,6,7,8] ]
=> [8, 11, 9, 8]
#oldergod suggested using zip based on the longest row, but finding the longest row and rejecting nils has a cost. I benchmarked the following against the above method using the example array above and found the reduce+each_with_index method more than 30% faster:
def sum_cols_using_zip arr
max_len = arr.map(&:size).max
([0] * max_len).zip(*arr).map do |col|
col.compact.inject(:+)
end
end
I'd do this:
a.transpose.map {|x| x.reduce(:+)}
Clean, simple, flexible. The .transpose turns this
[[1,2,3,4],[2,3,4,5],[3,4,5,6]]
into this
[[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6]]
and then .map applies .reduce to each subarray. And .reduce, in turn, aggregates the subvalues by adding them. Or, more precisely, by applying the + method to them.
I highly recommend reading the doc for these functions until you fully understand this example, as it's a pretty good succinct demonstration of how to think in a Rubyish way!
I'm having a trouble understanding Enumerators in Ruby.
Please correct me If I'm wrong, o.enum_for(:arg) method is supposed to convert object to Enumerator and every iteration over object o should call arg method?
What confuses me is how this line of code works
[4, 1, 2, 0].enum_for(:count).each_with_index do |elem, index|
elem == index
end
It should count how many elements are equal to their position in the array, and it works. However, I don't understand what's actually going on. Is each_with_index calling count method on every iteration? If someone could explain, it would be great.
From Programming Ruby:
count enum.count {| obj | block } → int
Returns the count of objects in enum that equal obj or for which the
block returns a true value.
So the enumerator visits each element and adds 1 to the count if the block returns true, which in this case means if the element is equal to the array index.
I haven't seen this pattern used much (if at all)—I would be more inclined to:
[4, 1, 2, 0].each_with_index.select { |elem, index| elem == index }.count
EDIT
Lets take a look at the example from your comment:
[4, 1, 2, 0].enum_for(:each_slice, 2).map do |a, b|
a + b
end
each_slice(2) takes the array 2 elements at a time and returns an array for each slice:
[4, 1, 2, 0].each_slice(2).map # => [[4,1], [2,0]]
calling map on the result lets us operate on each sub-array, passing it into a block:
[4, 1, 2, 0].enum_for(:each_slice, 2).map do |a,b|
puts "#{a.inspect} #{b.inspect}"
end
results in
4 1
2 0
a and b get their values by virtue of the block arguments being "splatted":
a, b = *[4, 1]
a # => 4
b # => 1
You could also take the array slice as the argument instead:
[4, 1, 2, 0].enum_for(:each_slice, 2).map {|a| puts "#{a.inspect}"}
[4, 1]
[2, 0]
Which lets you do this:
[4, 1, 2, 0].enum_for(:each_slice, 2).map {|a| a.inject(:+) } #=> [5,2]
Or if you have ActiveSupport (i.e. a Rails app),
[4, 1, 2, 0].enum_for(:each_slice, 2).map {|a| a.sum }
Which seems a lot clearer to me than the original example.
array.count can normally take a block, but on its own, it returns a fixnum, so it can't be chained to .with_index the way some other iterators can (try array.map.with_index {|x,i ... }, etc).
.enum_for(:count) converts it into a enumerator, which allows that chaining to take place. It iterates once over the members of array, and keeps a tally of how many of them equal their indexes. So count is really only being called once, but only after converting the array into something more flexible.