问题
Is there a method in ruby to turn fixnum like 74239
into an array like [7,4,2,3,9]
?
回答1:
You don't need to take a round trip through string-land for this sort of thing:
def digits(n)
Math.log10(n).floor.downto(0).map { |i| (n / 10**i) % 10 }
end
ary = digits(74239)
# [7, 4, 2, 3, 9]
This does assume that n
is positive of course, slipping an n = n.abs
into the mix can take care of that if needed. If you need to cover non-positive values, then:
def digits(n)
return [0] if(n == 0)
if(n < 0)
neg = true
n = n.abs
end
a = Math.log10(n).floor.downto(0).map { |i| (n / 10**i) % 10 }
a[0] *= -1 if(neg)
a
end
回答2:
Maybe not the most elegant solution:
74239.to_s.split('').map(&:to_i)
Output:
[7, 4, 2, 3, 9]
回答3:
The divmod method can be used to extract the digits one at a time
def digits n
n= n.abs
[].tap do |result|
while n > 0
n,digit = n.divmod 10
result.unshift digit
end
end
end
A quick benchmark showed this to be faster than using log to find the number of digits ahead of time, which was itself faster than string based methods.
bmbm(5) do |x|
x.report('string') {10000.times {digits_s(rand(1000000000))}}
x.report('divmod') {10000.times {digits_divmod(rand(1000000000))}}
x.report('log') {10000.times {digits(rand(1000000000))}}
end
#=>
user system total real
string 0.120000 0.000000 0.120000 ( 0.126119)
divmod 0.030000 0.000000 0.030000 ( 0.023148)
log 0.040000 0.000000 0.040000 ( 0.045285)
回答4:
You can convert to string and use the chars method:
74239.to_s.chars.map(&:to_i)
Output:
[7, 4, 2, 3, 9]
Its a bit more elegant than splitting.
回答5:
You can also use Array.new
instead of map
:
n = 74239
s = Math.log10(n).to_i + 1 # Gets the size of n
Array.new(s) { d = n % 10; n = n / 10; d }.reverse
回答6:
In Ruby 2.4, integers will have a digits method.
来源:https://stackoverflow.com/questions/12908008/turning-long-fixed-number-to-array-ruby