Ruby Array - Delete first 10 digits - ruby

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.

Answer 1

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

Answer 2

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]

Answer 3

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).