I have a string of digits:
s = "12345678910"
As you can see it is the numbers 1 through 10 listed in increasing order. I want to convert it to an array of those numbers:
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
How can I do it?
3 Answers
How about this:
a = ["123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899"]
b = a.first.each_char.map {|n| n.to_i }
if b.size > 8
c = b[0..8]
c += b[9..b.size].each_slice(2).map(&:join).map(&:to_i)
end
# It would yield as follows:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
For later numbers beyond 99, modify existing predicate accordingly.
1 Comment
Cary Swoveland
How would you "modify existing predicate" for an arbitrary string? That is, your code should not make assumptions about the number of digits in the last integer extracted from the string.
Assuming a monotonic sequence, here's my run at it.
input = a.first.chars
output = []
previous_int = 0
until input.empty?
temp = []
temp << input.shift until temp.join.to_i > previous_int
previous_int = temp.join.to_i
output << previous_int
end
puts output.to_s
#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
5 Comments
Phrogz
Note: the desired output asks for ints in the output, not strings.
sawa
Monotone sequence is not narrow enough of a condition to disambiguate the question.
[1, 2, 34, 567, 8910] would be another solution. It is primarily the OP's fault, but you are also wrong in making it sound like you disambiguated the question.
seph
@sawa - Along with the accompanying code it does. We're not doing rigorous proofs here, we're answering questions as best we can. Nice counterexample by the way.
Cary Swoveland
Nice one. You might consider making the sequence monotonically non-decreasing rather than increasing. You could meet @sawa's objection by stating the assumption that if
n is a natural number extracted from the string, the next number m extracted will have as few digits as possible (provided m >= n).seph
@Cary - Thank you. I'll let you take it from here if it gets reopened.
Assumptions
- the first (natural) number extracted from the string is the first character of the string converted to an integer;
- if the number
nis extracted from the string, the next number extracted,m, satisfiesn <= m(i.e., the sequence is monotonically non-decreasing); - if
nis extracted from the string, the next number extracted will have as few digits as possible (i.e., at most one greater than the number of digits inn); and - there is no need to check the validity of the string (e.g., "54632" is invalid).
Code
def split_it(str)
return [] if str.empty?
a = [str[0]]
offset = 1
while offset < str.size
sz = a.last.size
sz +=1 if str[offset,sz] < a.last
a << str[offset, sz]
offset += sz
end
a.map(&:to_i)
end
Examples
split_it("12345678910")
#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
split_it("12343636412252891407189118901")
#=> [1, 2, 3, 4, 36, 36, 41, 225, 289, 1407, 1891, 18901]
1,12or12345678910? It would be pretty simple if you've wanted only integers in 0..9 range. Otherwise it is pretty...impossible.