Let's say that I have an array like this:
[1,2,3,4,5,6,7]
how can I multiply every other number of this array except the first by 2
so my new array looks like this
[1,4,3,8,5,12,7]
You can use map and with_index:
[1,2,3,4,5,6,7].map.with_index{|v,i| i % 2 == 0 ? v : v * 2 }
# => [1, 4, 3, 8, 5, 12, 7]
[1,2,3,4,5,6,7].each_slice(2).flat_map{|k, l| [k, *(l * 2 if l)]}
# => [1, 4, 3, 8, 5, 12, 7]
Related
I need to add together two arrays of numbers and print the result as a total number.
arr1 = [1, 2, 3]
arr2 = [4, 5, 6]
arr1 + arr2
Just gives me = 1 2 3 4 5 6, but I want 21.
you can use the sum method
arr1 = [1, 2, 3]
arr2 = [4, 5, 6]
(arr1 + arr2).sum
This is another way
arr1 = [1, 2, 3]
arr2 = [4, 5, 6]
p (arr1 + arr2).reduce(:+)
Output
21
New to Ruby (coming from Python) and try to experiment this exercise:
(mixed the array items by taking first, last in rotating fashion)
Expected Output to be - [1, 7, 2, 6, 3, 5, 4]. But I did not expect 'nil' at the end... The orig. array can contain even or odd size of numbers.
Can someone shed the light of this unexpected? Thanks in advance.
[Updates - re-write the example from Ruby Cookbook p.162 Array ]
nums = (1..7).to_a # [1, 2, 3, 4, 5, 6, 7]
mixed = []
#middle = nums.length / 2
#index = 0
until nums.empty?
mixed << nums.shift(). #get 1st element out
mixed << nums.pop() #get last element out
#index += 1
end
print mixed # Got [1, 7, 2, 6, 3, 5, 4, nil]
What's happening is that the total num of elements in the array is odd so the last value is put into mixed on 'shift' and then there is no element left in the array. This will solve your issue:
nums = (1..7).to_a # [1, 2, 3, 4, 5, 6, 7]
mixed = []
#middle = nums.length / 2
#index = 0
until nums.empty?
mixed << nums.shift()
mixed << nums.pop() unless nums.empty?
#index += 1
end
print mixed # Got [1, 7, 2, 6, 3, 5, 4]
Another way is: If the num of elements is odd then run the loop till n-1 and then get the last element out using shift/pop (doesn't matter if you use shift or pop at the end, you will get the same element.)
The Cookbook method can be made non-destructive (avoid modifying nums) as follows.
def doit(nums)
nums.size.times.map { |i| i.even? ? nums[i/2] : nums[-i/2] }
end
doit [1, 2, 3, 4, 5, 6, 7]
#=> [1, 7, 2, 6, 3, 5, 4]
doit [1, 2, 3, 5, 6, 7]
#=> [1, 7, 2, 6, 3, 5]
Here is another (non-destructive) way to do that.
def doit(nums)
n = nums.size/2
nums.first(n).zip(nums.last(n).reverse).flatten.tap do |a|
a << nums[n] if nums.size.odd?
end
end
doit [1, 2, 3, 4, 5, 6, 7]
#=> [1, 7, 2, 6, 3, 5, 4]
doit [1, 2, 3, 5, 6, 7]
#=> [1, 7, 2, 6, 3, 5]
Input:
a = [4, 5, 5, 5, 6, 6, 4, 1, 4, 4, 3, 6, 6, 3, 6, 1, 4, 5, 5, 5]
How to list out no of pairs in an array.
Output:
9
Description
#no 1(1 pair)
#no 3(1 pair)
#no 4(2 pairs)
#no 5(3 pairs)
#no 6(2 pairs)
#so total 9 pairs
Here is another option:
a.group_by(&:itself).transform_values{ |v| v.size / 2 }.values.sum
#=> 9
How it works.
First group the elements by value:
a.group_by(&:itself) #=> {4=>[4, 4, 4, 4, 4], 5=>[5, 5, 5, 5, 5, 5], 6=>[6, 6, 6, 6, 6], 1=>[1, 1], 3=>[3, 3]}
Then transforming the keys to the pair count:
a.group_by(&:itself).transform_values{ |v| v.size / 2 } #=> {4=>2, 5=>3, 6=>2, 1=>1, 3=>1}
So, get the values of the hash:
a.group_by(&:itself).transform_values{ |v| v.size / 2 }.values #=> [2, 3, 2, 1, 1]
Finally, sum the values, which is the first line of code posted above.
arr = [4, 5, 5, 5, 6, 6, 4, 1, 4, 4, 3, 6, 6, 3, 6, 1, 4, 5, 5, 5]
hash = Hash.new(0)
arr.each { |e| hash[e] += 1 }
hash.values.reduce(0) { |s, n| s += n / 2 } // => 9
Since from what I can gather you are basically removing integers the moment they got paired once so technically it's just an integer division by two.
[1] How to count identical string elements in a Ruby array
[2] Reduce Hash Values
I have done like this, It works
b = []
a.uniq.each { |i| b.push(a.count(i)/2)}
b.sum
Simple question, but somehow I can't think of a solution. How can I delete a single smallest element in an array of random integers?
a = [7, 5, 3, 2, 1, 4]
b = [2, 2, 1, 1, 2]
This is what I come up with:
def remove_it(num)
num.delete(num.sort[0])
end
Code works with a, but not b. It deletes both 1's in b. I only need to delete one 1.
How can I delete one smallest number in an array and keep the order?
Easy-peasy. Use .delete_at + .index:
def remove_it(num)
num.delete_at(num.index(num.min))
num
end
a = [7, 5, 3, 2, 1, 4]
b = [2, 2, 1, 1, 2]
remove_it(a) # => [7, 5, 3, 2, 4]
remove_it(b) # => [2, 2, 1, 2]
Example:
a = [1, 3, 4, 5]
b = [2, 3, 1, 5, 6]
How do I get the last value 5 in array a or last value 6 in array b without using a[3] and b[4]?
Use -1 index (negative indices count backward from the end of the array):
a[-1] # => 5
b[-1] # => 6
or Array#last method:
a.last # => 5
b.last # => 6
One other way, using the splat operator:
*a, last = [1, 3, 4, 5]
a => [1, 3, 4]
last => 5