Ruby Array - Delete first 10 digits

Question

I have an array in Ruby and I would like to delete the first 10 digits in the array.

array = [1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q", 30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]

It would ideally return

['a', 'b', 'c', 'd', 'a', 'z', 'e', 'q', 0, 'a', 4, t, 7, m, 5 , 1, 2, q, s, 1, 13, 46, 31]

By removing the first 10 digits (1,3,2,4,5,1,7,2,1,3).

Note that 21(2 and 1) and 30(3 and 0) both have 2 digits

Here's what I've tried

digits = array.join().scan(/\d/).first(10).map{|s|s.to_i}
=> [1,3,2,4,5,1,7,2,1,3]
elements = array - digits

This is what I got

["a", "b", "c", "d", "a", "z", "e", 21, "q", 30, "a", "t", "m", "q", "s", "l", 13, 46, 31]

Now it looks like it took the difference instead of subtracting.

I have no idea where to go from here. and now I'm lost. Any help is appreciated.

if the last digit is a part of a bigger number (lets, say it is 2 from 3245), what do you want to have in the array? Just 45 or you want to remove the whole number? — Ivaylo Petrov
– Ivaylo Petrov, Commented Jul 17, 2014 at 16:41
I don't want to remove the whole number. Only the first 10 digits, not numbers. Some numbers like 47 have 2 digits. For example 21 and 30 showed up, but they were supposed to be deleted. — George Lucas
– George Lucas, Commented Jul 17, 2014 at 16:43
The answer I got suggests that digits are supposed to be removed, not numbers. For example from [1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q", 3245, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13] I think that George Lucas wants to get ["a", "b", "c", "d", "a", "z", "e", "q", 245, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13]. This is my understanding... And the accepted answer produces ["a", "b", "c", "d", "a", "z", "e", "q", "a", "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]. Is that the expected behavior? — Ivaylo Petrov
– Ivaylo Petrov, Commented Jul 17, 2014 at 19:01
@Ivaylo - That answer is correct. Digits are supposed to be removed but not numbers. For example say we had a two digit number (50), if the 10th digit is 5 and the 11th digit is 0, than the digit 5 would be removed but the digit 0 would remain. — George Lucas
– George Lucas, Commented Jul 17, 2014 at 19:11

TomDunning · Accepted Answer · 2014-07-18 10:29:55Z

3

To delete 10 numbers:

10.times.each {array.delete_at(array.index(array.select{|i| i.is_a?(Integer)}.first))}
array

To delete 10 digits:

array = [1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q", 30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
i = 10
while (i > 0) do
    x = array.select{|item| item.is_a?(Integer)}.first
    if x.to_s.length > i
      y = array.index(x)
      array[y] = x.to_s[0, (i-1)].to_i
    else
      array.delete_at(array.index(x))
    end
    i -= x.to_s.length
end
array

edited Jul 18, 2014 at 10:29

answered Jul 17, 2014 at 16:44

TomDunning

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6 Comments

George Lucas Over a year ago

That method returns 10.

fbonetti Over a year ago

10.times { array.delete_if { |x| x.is_a?(Integer) } }

Cary Swoveland Over a year ago

TypeError: no implicit conversion from nil to integer when there are fewer than 10 digits in the array.

steenslag Over a year ago

This runs i.is_a?(Integer) 99 times before erroring.

steenslag Over a year ago

@fbonetti The first run deletes all integers.

|

Stefan · Accepted Answer · 2014-07-17 19:08:47Z

3

Unfortunately not a one-liner:

count = 10
array.each_with_object([]) { |e, a|
  if e.is_a?(Integer) && count > 0
    str = e.to_s                     # convert integer to string
    del = str.slice!(0, count)       # delete up to 'count' characters
    count -= del.length              # subtract number of actually deleted characters
    a << str.to_i unless str.empty?  # append remaining characters as integer if any
  else
    a << e
  end
}
#=> ["a", "b", "c", "d", "a", "z", "e", "q", 0, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]

answered Jul 17, 2014 at 19:08

Stefan

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1 Comment

George Lucas Over a year ago

There might be easier ways to solve this, but nonetheless this answered the question. Therefore it is a good solution. Good answer, the output was as expected.

Cary Swoveland · Accepted Answer · 2014-07-17 21:18:13Z

I would be inclined to do it like this.

Code

def doit(array, max_nbr_to_delete)
  cnt = 0
  array.map do |e|
    if (e.is_a? Integer) && cnt < max_nbr_to_delete
      cnt += e.to_s.size
      if cnt <= max_nbr_to_delete
        nil
      else
        e.to_s[cnt-max_nbr_to_delete..-1].to_i
      end
    else
      e
    end
  end.compact
end

Examples

array = [ 1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q",
         30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]

doit(array, 10)
  #=> ["a", "b", "c", "d", "a", "z", "e", "q", 0, "a", 4,
  #    "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
doit(array, 100)
  #=> ["a", "b", "c", "d", "a", "z", "e", "q", "a", "t", "m", "q", "s", "l"]

Explanation

Each element e of the array that is not an integer is mapped to e.

For each non-negative integer n having d digits, suppose cnt is the number of digits that map has already been removed from the string. There are three possibilities:

if cnt >= max_nbr_to_delete, no more digits are to be removed, so e (itself) is returned
if cnt + d <= max_nbr_to_delete all d digits of e are to be removed, which is done by mapping e to nil and subsequently removing nil elements
if cnt < max_nbr_to_delete and cnt + d > max_nbr_to_delete, e.to_s[cnt+d-max_nbr_to_delete..-1].to_i is returned (i.e. the first cnt+d-max_nbr_to_delete digits of e are removed).

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