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I can't seem to find a ruby-way to transform this
[0, 1, 2, 3]
into
[0, 0.5, 1, 1.5, 2, 2.5, 3]
e.g, insert new elements based on existing ones applying some rule.
EDIT: I'm asking about a general case, not necessarily a 0.5 increment. Say, when elements are not successive:
[1, 3, 12] => [1, 2, 3, 7.5, 12]
So the accepted answer is just perfect here, thanks.
arr = [0, 1, 2, 3]
def rule(x, y)
(x + y) / 2.0
end
arr.each_cons(2).flat_map{|x, y| [x, rule(x, y), y]} # => [0, 0.5, 1, 1, 1.5, 2, 2, 2.5, 3]
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I'm new to rails and I was wondering how I compare the first value of array "a" if it is greater than the first value of array "b"?
Example:
a = [1, 2, 3]
b = [3, 2, 1]
How do I check if a[0] is greater than b[0].
You can use the first method:
a = [1, 2, 3]
b = [3, 2, 1]
a.first > b.first #=> false
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Rearrange the negative elements of the array so that they are in non-descending order. Leave the remaining elements in their places.
It will show like that: [-1, 1, -3, 6, -5, -9] -> [-9, 1, -5, 6, -3, -1]
How can I do this without using additional arrays?
There are many ways to do that. Here is one.
arr = [-1, 1, -3, 6, -5, -9]
negs = arr.select { |n| n < 0 }.sort
#=> [-9, -5, -3, -1]
arr.map { |n| n >= 0 ? n : negs.shift }
#=> [-9, 1, -5, 6, -3, -1]
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The Problem
Given N input vector = [x1, x2, ..., xl] and each of them have same length L. Define F = summation of N input vector.
And the aim is to find a set of integer T = [n1, n2, ..., nN] represents delay of each input vector so that the maximum value of F is minimized.
Example
Assume 2 Input vector [0, 3, -1, 1, 0], [0, 2, -1, 0, 0]:
if T = [0, 0], then F = [0, 5, -2, 1, 0] and max(F) = 5 which is obviously not a good result.
if T = [1, 0], so that F = [0, 2, 2, -1, 1, 0], max(F) = 2. This T is we aim to find. The computing process below may help understand this problem:
T = [1, 0]
Vector1 with delay 1 [0, 3, -1, 1, 0]
Vector2 with delay 0 [0, 2, -1, 0, 0]
F = [0, 2, 2, -1, 1, 0]
The Question
Any idea how to delay vector T using dynamic programming?
Any advise would be greatly appreciated.
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If I have a hash that looks like this:
C = [[1, 1, 1, 1],
[1, 2, 1, 1],
[1, 3, 1, 7],
[1, 1, 4, 1]]
What is a fast way to sum the columns and produce the following result:
C = [4, 7, 7, 10]
Edit: The way I was doing it coming from a C background was to parse thru the result and summing manually, that's why I asked. didn't know where else to look for.
arr = [[1, 1, 1, 1],
[1, 2, 1, 1],
[1, 3, 1, 7],
[1, 1, 4, 1]]
arr.transpose.map{|e| e.inject(:+)}
# => [4, 7, 7, 10]
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I am new to ruby on rails .
I was programming ruby and want to try checking if 2 numbers of array add up to Input number in ruby.
eg,
array A[]= {3, 1, 8, 11, 5, 7}
given integer say N = 6
answer will be 1,5.
I know how to program it in java,C++ but i am stuck in ruby coding,
Can anyone please help me.Thanks in advance
You can use Array#combination:
ary = [3, 1, 8, 11, 5, 7]
n = 6
ary.combination(2).detect { |a, b| a + b == n }
#=> [1, 5]
combination(2) creates an array of all combinations of length 2, i.e. [3,1], [3,8], [3,11] etc.
detect { |a, b| a + b == n } returns the first pair with sum n
You can use find_all instead of detect to return all pairs with sum n.
a = [3, 1, 8, 11, 4, 5, 7, 2]
> a.combination(2).select {|i| i.inject(:+) == 6 }
#=> [[1, 5], [4, 2]]
a = [3, 1, 8, 11, 5, 7]
p a.combination(2).find{|i| i.inject(:+) == 6}
# >> [1, 5]