Ruby dynamically updating value in nested array

Question

In Ruby, with a 1D Array, I can dynamically select elements by passing an Integer key in brackets, as follows:

example =  [0,1,2,[3,4]]
i = 2

example[i]
==> 2

What I would like to achieve, is to dynamically update an element in a multi-dimensional Array by passing an Array of Integers, each representing the index to select in each array. Example of what I'd like to achieve:

example =  [0,1,2,[3,4]]
path = [3, 1] (corresponds to the 4)

example[*path or some other syntax] = 9
example 
==> [0,1,2,[3,9]]

What I've tried is storing the result with each path iteration:

temp = example
path.each {|index|
  temp = temp[index]
}
temp
==> 4

This successfully identifies the element I'd like to update. However, it appears to have stored a copy, rather than to reference the original location, as:

temp = 9
example
==> [0,1,2,[3,4]]

How can I update the base array example without hardcoding path in individual brackets?

Clarification after a comment: I don't know the path length in advance, which is why hardcoding isn't viable.

No, as I want to be able to accommodate multiple path lengths. Hard-coding with brackets limits me to a 2-step path. — Jake Morris
– Jake Morris, Commented Dec 20, 2021 at 23:49
Write a method. Passing in the array object, then indices. You will have to make your method be able to accept any number of indices. — Sagar Pandya
– Sagar Pandya, Commented Dec 20, 2021 at 23:52

mu is too short · Accepted Answer · 2021-12-21 07:32:05Z

3

Your iteration is almost there, you just need to stop one step before you run through path so that you can have the array you need to modify rather than the element.

So split the path into the pieces you want:

*p, target = path
# [3], 1

Then use #inject to find the array:

ary = p.inject(example) { |i, a| a[i] }
# [3, 4]

and then do your assignment:

ary[target] = 9

Of course you'll need to add some logic to deal with the unexpected such as path leading you to a non-array element or path not matching the structure of example (consider path = [11, 6, 23] in your example).

You could also use #dig instead of #inject:

ary = example.dig(*p)
ary[target] = 9

# or
example.dig(*p)[target] = 9

That would take care of some of the problematic paths and you'd be left with deciding what to do if ary.nil?.

edited Dec 21, 2021 at 7:32

answered Dec 21, 2021 at 0:28

mu is too short

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Fravadona · Accepted Answer · 2021-12-21 00:24:34Z

2

Define a function, something along the lines of:

def deep_set! val, arr, *path
  path[0...-1].each {|i| arr = arr[i]}
  arr[path[-1]] = val
end

arr =  [0,1,2,[3,4]]
deep_set! 9, arr, 3 ,1

arr #=> [0, 1, 2, [3, 9]]

answered Dec 21, 2021 at 0:24

Fravadona

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Comments

Cary Swoveland · Accepted Answer · 2021-12-21 07:29:47Z

I assume path contains at least one element.

You could use Array#dig.

def dig_and_replace(arr, path, replacement)
  *first, last = path
  if first.empty?
    arr[last] = replacement
  else
    arr.dig(*first)[last] = replacement
  end
  arr
end

arr = [0,1,2,[3,4]]
dig_and_replace(arr,[3,1],99)
  #=> [0,1,2,[3,99]]
arr
  #=> [0,1,2,[3,99]]

dig_and_replace([0,1,2,[3,             4              ]],[3,1],
  {cat:'meow', dog:'woof'})
  #=>           [0,1,2,[3,{:cat=>"meow", :dog=>"woof"}]]
dig_and_replace([0,[1,[2,3,[4,5,[6, 7],8],9],10,11]],[1,1,2,2,1],99)
  #=>           [0,[1,[2,3,[4,5,[6,99],8],9],10,11]]
dig_and_replace([0,[1,[2,3,[4,5,[6,7],8],9],10,11]],[1,1,2],99)
  #=>           [0,[1,[2,3,     99,      9],10,11]]

Note that if

path = [1,2,3,4]
*first, last = path

then

first #=> [1,2,3]
last  #=> 4

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