Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]
Any help would be greatly appreciated.
On Mon, May 12, 2008 at 6:36 PM, Nadim K. [email protected] wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]
irb(main):001:0> x=1234
=> 1234
irb(main):002:0> x = x.to_s.split(‘’)
=> [“1”, “2”, “3”, “4”]
Mikel
1 Like
From: [email protected] [mailto:[email protected]]
Let’s say I have: x=1234
How can I convert that to the follow array: x=[1, 2, 3, 4]
irb is friendly, just play w it…
“1234”.split(//)
#=> [“1”, “2”, “3”, “4”]
your next task is to convert those into integers, then you’re good to
go…
kind regards -botp
On 12.05.2008 10:36, Nadim K. wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]Any help would be greatly appreciated.
x.scan /\d/
robert
Integer to Array
x=1234
y=x.to_s.scan(/d)
or
y=x.to_s.split(//)
or
y=x.to_s.split(’’)
Array to integer
z=y.to_s.to_i
Thanks and Regards,
Lokesh Agrawal
Software Engineer
“Dream is not what you see in sleep, is the the thing which does not let
you
sleep”
Hi –
On Mon, 12 May 2008, Nadim K. wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]
require ‘scanf’
“1234”.scanf("%1d" * 4)
=> [1, 2, 3, 4]
David
Hi self –
On Mon, 12 May 2008, David A. Black wrote:
“1234”.scanf("%1d" * 4)
=> [1, 2, 3, 4]
“%1d” * x.size would be better (no need to hard-code the 4).
David
Nadim K. wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]
A solution that doesn’t use strings:
result_array = []
while x > 0
result_array.unshift x % 10
x /= 10
end
result_array
This will “destroy” x, though.
HTH,
Sebastian
Hi –
On Mon, 12 May 2008, Mikel L. wrote:
On Mon, May 12, 2008 at 6:36 PM, Nadim K. [email protected] wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]irb(main):001:0> x=1234
=> 1234
irb(main):002:0> x = x.to_s.split(‘’)
=> [“1”, “2”, “3”, “4”]
That’s an array of strings rather than integers, though. See my other
reply, using scanf.
David
2008/5/12 Robert K. [email protected]:
x.scan /\d/
Ah, not really. What I meant was:
x.to_s.scan(/\d/).map {|i| i.to_i}
Cheers
robert
On May 13, 2008, at 7:26 AM, Robert K. wrote:
while x > 0
def int_split(x)robert
–
use.inject do |as, often| as.you_can - without end
def int_split(x)
return [0] if x.zero?
r = []
while x > 0
x, b = x.divmod 10
r.unshift b
end
r
end
You don’t need y since Fixnums are immediate. In fact, x is only a
label, doing y=x just adds a new label to the object referred to by
x. (Go ahead, try it with a Bignum.)
It still doesn’t work for negative numbers, but since that behavior
hasn’t been defined, it’s left as an exercise for the OP.
-Rob
2008/5/12 Sebastian H. [email protected]:
result_array.unshift x % 10
x /= 10
end
result_arrayThis will “destroy” x, though.
Well, that’s easily fixed: just work with another variable. You can
also combine division and mod:
def int_split(x)
r = []
y = x
while y > 0
y, b = y.divmod 10
r.unshift b
end
r
end
Kind regards
robert
2008/5/13 Rob B. [email protected]:
x=1234
end
y = xend
r
endYou don’t need y since Fixnums are immediate.
The reasoning is wrong but comes to the right conclusion: if you want
to retain the original value of x then it does not matter whether
values are mutable or not. It is sufficient to assign to x to loose
the original value.
In my code I don’t need y because x is a method parameter. My remark
was a reaction to Sebastian’s comment and piece of code which modified
x.
It still doesn’t work for negative numbers, but since that behavior hasn’t
been defined, it’s left as an exercise for the OP.
Exactly. 
Kind regards
robert
Hi –
On Tue, 13 May 2008, Robert K. wrote:
Let’s say I have:
x /= 10
r = []
robert
r.unshift bIn my code I don’t need y because x is a method parameter. My remark
was a reaction to Sebastian’s comment and piece of code which modified
x.
I still like scanf 
David
From: David A. Black [mailto:[email protected]]
I still like scanf
indeed.
i also like the scan+map since i don’t need the formatting (and the
require 
001:0> “1234”.scan(/\d/).map(&:to_i)
=> [1, 2, 3, 4]
kind regards -botp
Mikel L. wrote:
On Mon, May 12, 2008 at 6:36 PM, Nadim K. [email protected] wrote:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]
Most of the other ways are better, but none-the-less:
x=1234
array = []
results = x.to_s.split(‘’)
results.each{|x| array << x.to_i}
array #=> [1,2,3,4]
Regards,
- Mac
Hi,
Rob B. wrote:
def int_split(x)
return [0] if x.zero?
r = []
while x > 0
x, b = x.divmod 10
r.unshift b
end
r
end
Here is a little short version:
def int_split(x)
r=[x];r[0,1]=*r[0].divmod(10)while r[0]>9;r
end
Regards,
Park H.
From: Peña, Botp [mailto:[email protected]]
scan+map since i don’t need the formatting
(and the require
001:0> “1234”.scan(/\d/).map(&:to_i)
=> [1, 2, 3, 4]
arggh, 1.9 is getting better,
002:0> “1234”.each_char.map(&:to_i)
=> [1, 2, 3, 4]
very readable, imho
kind regards -botp
Nadim K. wrote in post #673293:
Let’s say I have:
x=1234
How can I convert that to the follow array:
x=[1, 2, 3, 4]Any help would be greatly appreciated.
This may help…
irb(main):049:0> x = 1234.to_s.split(//)
=> [“1”, “2”, “3”, “4”]
irb(main):050:0> y = x.map{|y| y.to_i}
=> [1, 2, 3, 4]
I am using Ruby 2.2
A solution that doesn’t use strings:
result_array = []
while x > 0
result_array.unshift x % 10
x /= 10
end
result_arrayThis will “destroy” x, though.
HTH,
Sebastian
Same thing only different
Positive numbers only
def splitnum(n)
z = Math.log10(n).floor
Array.new(z+1){|i| (n/10**(z-i))%10}
end
#OR
def splitnum2(n)
z = Math.log10(n).floor
(0…z).map{|i| (n/10**(z-i))%10}
end
Harry
Online Karnaugh map solver with circuit