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I am trying to conceptualize the iteration of two loops
numbers_array = [1,2,3,4,5,6,7,8,9,10]
add_to_array = [1,2,3,4]
While the numbers_array iterates, add_to_array iterates simultaneously adding both elements together at the same time. The caveat is once add_to_array reaches the end, it starts over adding its element to the next index in numbers_array. So at numbers_array[4] we would be adding add_to_array[0] then adding numbers_array[5] to add_to_array[1] and so on. This process would repeat until we reach the end of the numbers_array.
Any input would be greatly appreciated!
You are looking for Enumerable#zip and Enumerable#cycle:
numbers_array = [1,2,3,4,5,6,7,8,9,10]
#⇒ [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
add_to_array = [1,2,3,4]
#⇒ [1, 2, 3, 4]
numbers_array.zip(add_to_array.cycle)
#⇒ [[1, 1], [2, 2], [3, 3], [4, 4], [5, 1],
# [6, 2], [7, 3], [8, 4], [9, 1], [10, 2]]
Now do whatever you want with the array returned. E.g. to reduce the zipped result summing elements, map ro Enumerable#sum:
numbers_array.zip(add_to_array.cycle).map(&:sum)
#⇒ [2, 4, 6, 8, 6, 8, 10, 12, 10, 12]
It works by using the % operator to cycle through the indexes.
numbers_array = [1,2,3,4,5,6,7,8,9,10]
add_to_array = [1,2,3,4]
numbers_array.map.with_index do |n, i|
n + add_to_array[i % add_to_array.length]
end
A cool method that's similar, if you didn't want to start over at the next array, would be .zip
https://apidock.com/ruby/Array/zip
add_to_array.zip(*numbers_array.each_slice(add_to_array.size)).
map { |a| a.sum { |e| e.to_i } }
#=> [16, 20, 13, 16]
e.to_i is needed to convert nil values to zeros. See NilClass#to_i.
Another option:
numbers_array.map { |e| e + add_to_array.rotate!.last }
# => [2, 4, 6, 8, 6, 8, 10, 12, 10, 12]
Drawback: add_to_array is mutated by rotate!
I have a find_num method that returns the index of a specified number in an ordered array, e.g.
find_num(6, [1, 4, 6, 9, 13]) #=> 2
however my spec also requires that if the number is not available it finds the position of the next highest number so ...
find_num(8, [1, 4, 6, 9, 13]) #=> 3
as 9 is the next available number.
Having trouble implementing this... I have thought of iterating through the whole array but I am told to take into account the array could be large...
You can pass a block to index and it ...
[...] returns the index of the first object for which the block returns true. Returns nil if no match is found.
Examples:
[1, 4, 6, 9, 13].index { |n| n >= 6 } #=> 2
[1, 4, 6, 9, 13].index { |n| n >= 8 } #=> 3
[1, 4, 6, 9, 13].index { |n| n >= 15 } #=> nil
Because this requires the array to be ordered, you can also use bsearch_index which performs a binary search.
you can also find a index of any element in array like this.
2.1.8 :040 > [1, 4, 6, 9, 13].index(6)
=> 2
2.1.8 :041 > [1, 4, 6, 9, 13].index(15)
=> nil
def find_num(n,a)
a.each_with_index.to_a.sort_by(&:first).find { |nbr,_| nbr >= n }.last
end
find_num(6, [1, 4, 6, 9, 13])
#=> 2
find_num(8, [1, 4, 6, 9, 13]) #=> 3
#=> 3
The steps for
n = 8
a = [1, 4, 6, 9, 13]
are as follows.
b = a.each_with_index.to_a
#=> [[1, 0], [4, 1], [6, 2], [9, 3], [13, 4]]
c = b.sort_by(&:first)
#=> [[1, 0], [4, 1], [6, 2], [9, 3], [13, 4]]
d = c.find { |nbr,_| nbr >= n }
#=> [9, 3]
d.last
#=> 3
I am looking for a way to perform a certain operation (for instance delete_if) on an array and return both the deleted elements, and the remaining elements.
For example
a = [1,2,3,4,5,6,7,8,9,10]
a.delete_if {|x| x.even? } #=> [[1, 3, 5, 7, 9]]
But what I am looking for is something like
a = [1,2,3,4,5,6,7,8,9,10]
a.some_operation #=> [[1,3,5,7,9],[2,4,6,8,10]]
How would I go about doing this?
Using Enumerable#partition:
a = [1,2,3,4,5,6,7,8,9,10]
a.partition &:even?
# => [[2, 4, 6, 8, 10], [1, 3, 5, 7, 9]]
The first element of the Enumerable#partition return value contains the elements that are evaluated to true in the block. So you need to use odd? to get what you want.
a.partition &:odd?
# => [[1, 3, 5, 7, 9], [2, 4, 6, 8, 10]]
You might be looking for something like this:
a = [1,2,3,4,5,6,7,8,9,10]
a.group_by { |x| x.even? }.values
#=> [[1, 3, 5, 7, 9], [2, 4, 6, 8, 10]]
I know the idiomatic way to do a for loop in Ruby is to use an Enumerator like .each, but I'm running into a problem: I'd like to iterate over a subset of an Array and modify those elements. Calling .map! with a subset like ary[0..2] or .slice(0..2) doesn't seem to do it; presumably because that slicing operator is creating a new Array?
Desired behavior with for instead of iterator:
iter_ind = [2,3,4]
my_ary = [1,3,5,7,9,11]
for j in iter_ind
my_ary[j] = my_ary[j] + 1
# some other stuff like an exchange operation maybe
end
=> [1, 3, 6, 8, 10, 11]
Things that don't work:
irb(main):032:0> ar[2..4].map! {|el| el = el+1}
=> [6, 8, 10]
irb(main):033:0> ar
=> [1, 3, 5, 7, 9, 11]
irb(main):034:0> ar.slice(2..4).map! {|el| el = el+1}
=> [6, 8, 10]
irb(main):035:0> ar
=> [1, 3, 5, 7, 9, 11]
irb(main):036:0> ar[2..4].collect! {|el| el = el+1}
=> [6, 8, 10]
irb(main):037:0> ar
=> [1, 3, 5, 7, 9, 11]
Try this.
In example below I implemented something that could be named map_with_index. each_with_index if no block given returns iterator. I use it to map our array.
ary = [1, 3, 5, 7, 9, 11]
ary.each_with_index.map { |elem, index| index.between?(2, 4) ? elem += 1 : elem }
# => [1, 3, 6, 8, 10, 11]
You may also try the following:
?> ary = [1, 3, 5, 7, 9, 11]
=> [1, 3, 5, 7, 9, 11]
?> ary.map!.with_index {|item, index| index.between?(2, 4) ? item += 1 : item}
=> [1, 3, 6, 8, 10, 11]
?> ary
=> [1, 3, 6, 8, 10, 11]
You could use Array#each_index if you don't mind referencing the array twice:
ary = [1, 3, 5, 7, 9, 11]
ary.each_index { |i| ary[i] += 1 if i.between? 2, 4 }
#=> [1, 3, 6, 8, 10, 11]
Or if you don't want to iterate the whole array, this would work, too:
ary = [1, 3, 5, 7, 9, 11]
ary[2..4] = ary[2..4].map { |el| el + 1 }
ary
#=> [1, 3, 6, 8, 10, 11]
I have two arrays:
array1 = [3, 4, 4, 5, 6, 7, 8, 8]
array2 = [4, 5, 8, 8]
I want to remove those elements of array1, which are found in array2, but only in one instance. The resulting array, array3, must be like this:
array3 = [3, 4, 6, 7]
I tried:
array3 = array1 - array2
but the result was unsatisfactory:
array3 -> [3, 6, 7]
This may not be the most efficient way of doing what you want, but it works:
array1 = [3, 4, 4, 5, 6, 7, 8, 8]
array2 = [4, 5, 8, 8]
array2.each do |item|
index = array1.index item
array1.delete_at index if index
end
A non-imperative just to show other ways of doing things. Using Facets (just for convenience to get the histogram), I'd write this. O(n):
require 'facets'
array3 = array1.reduce([array2.frequency, []]) do |(h, output), x|
if h[x] && h[x] > 0
[h.update(x => h[x] - 1), output]
else
[h, output << x]
end
end[1]
#=> [3, 4, 6, 7]
To make the snippet purely functional you would use Hash#merge/Array#+ instead of Hash#update/Array:<<, but due to the nature of these data structures it would be terribly inefficient.