How to tranlsate number to binary?

ruby-talk
1

Hi, I am a ruby newbie. I want to write a program which will scan a
input file, change every digital number into 4 bits len's binary
number.
For example, for input.txt as

903
1047

I will get a new file as

100100000011
0001000001000111

Would anyboby kindly help to tell me how to do this?

···

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Posted via http://www.ruby-forum.com/.

2

I don't know what a "4 bits len's binary number" is.

But here's how you convert decimal to binary in Ruby:

irb(main):004:0> 903.to_s(2)
=> "1110000111"
irb(main):005:0> 1+2+4+128+256+512
=> 903

Hope that helps,
-Harold

···

On 3/26/07, Ak 756 <macro.peng@gmail.com> wrote:

Hi, I am a ruby newbie. I want to write a program which will scan a
input file, change every digital number into 4 bits len's binary
number.
For example, for input.txt as

903
1047

I will get a new file as

100100000011
0001000001000111

Would anyboby kindly help to tell me how to do this?

3

For example, for input.txt as

903
1047

I will get a new file as

100100000011
0001000001000111

  >> 0x903.to_s(2)
  => "100100000011"
  >> 0x1047.to_s(2)
  => "1000001000111"

And so it was:

  puts IO.read('input.txt').map { |x| x.to_i(16).to_s(2) }.join

_why

···

On Mon, Mar 26, 2007 at 10:26:35AM +0900, Ak 756 wrote:

4

I assume that by "4 bits len binary number" he means that he wants the
binary number to be a multiple of 4 characters long. So the binary number
would have "0" prepended to it if the size wasn't divisible by 4. Here is
an overly verbose way to convert the num to the binary number string.

irb(main):001:0> def int_to_binary(num)
irb(main):002:1> binary_num = num.to_s(2)
irb(main):003:1> if ((binary_num.size % 4) > 0)
irb(main):004:2> binary_num = ('0' * (4 - (binary_num.size % 4))) +
binary_num
irb(main):005:2> end
irb(main):006:1> binary_num
irb(main):007:1> end
=> nil
irb(main):008:0> int_to_binary(903)
=> "001110000111"
irb(main):009:0> int_to_binary(15)
=> "1111"
irb(main):010:0> int_to_binary(16)
=> "00010000"
irb(main):011:0>

···

On 3/25/07, Harold Hausman <hhausman@gmail.com> wrote:

On 3/26/07, Ak 756 <macro.peng@gmail.com> wrote:
> Hi, I am a ruby newbie. I want to write a program which will scan a
> input file, change every digital number into 4 bits len's binary
> number.
> For example, for input.txt as
>
> 903
> 1047
>
> I will get a new file as
>
> 100100000011
> 0001000001000111
>
> Would anyboby kindly help to tell me how to do this?

I don't know what a "4 bits len's binary number" is.

But here's how you convert decimal to binary in Ruby:

irb(main):004:0> 903.to_s(2)
=> "1110000111"
irb(main):005:0> 1+2+4+128+256+512
=> 903

Hope that helps,
-Harold

5

=> "000110010011"
     to get leading zeros.

And so , the following
puts IO.read('input.txt').map { |x|
format("%0#{x.chomp.length*4}b",x.to_i(16)) }.join("\n")

Prasad

···

On 3/26/07, _why <why@ruby-lang.org> wrote:

On Mon, Mar 26, 2007 at 10:26:35AM +0900, Ak 756 wrote:
> For example, for input.txt as
>
> 903
> 1047
>
> I will get a new file as
>
> 100100000011
> 0001000001000111
>

  >> 0x903.to_s(2)
  => "100100000011"
  >> 0x1047.to_s(2)
  => "1000001000111"

And so it was:

  puts IO.read('input.txt').map { |x| x.to_i(16).to_s(2) }.join

_why

format("%0#{"193".length*4}b", 0x193)

6

Mike Moore wrote:

irb(main):008:0> int_to_binary(903)
=> "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

···

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Posted via http://www.ruby-forum.com/\.

7

903.to_s.split(//).map{|n| sprintf("%04d",n.to_i.to_s(2))}.join

Alex Gutteridge

Bioinformatics Center
Kyoto University

···

On 26 Mar 2007, at 12:32, Ak 756 wrote:

Mike Moore wrote:

irb(main):008:0> int_to_binary(903)
=> "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

--
Posted via http://www.ruby-forum.com/\.

8

Maybe this one is goon for you?

903.to_s.split(//).map{|x| tmp=x.to_i.to_s(2); "0"*(4-tmp.length)
+tmp}.join

···

On Mar 26, 1:32 pm, Ak 756 <macro.p...@gmail.com> wrote:

Mike Moore wrote:
> irb(main):008:0> int_to_binary(903)
> => "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

--
Posted viahttp://www.ruby-forum.com/.

9

I don't know about efficent, but you could start with something like
this and see if it's performant enough:

irb(main):002:0> 903.to_s.split("").each {|digit| p digit.to_i.to_s(2)}
"1001"
"0"
"11"

Meh, actually, I bet someone else can do better than that, but I'm at
work right now. (:

-Harold

···

On 3/26/07, Ak 756 <macro.peng@gmail.com> wrote:

Mike Moore wrote:

> irb(main):008:0> int_to_binary(903)
> => "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

10

Ak 756 wrote:

Mike Moore wrote:

irb(main):008:0> int_to_binary(903)
=> "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where '1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

"903".split("").map{|s| "%04b" % s.to_i}.join
=> "100100000011"

···

--
       vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407

11

irb(main):001:0> "903".split(//).map { |c| "%04b" % c }
=> ["1001", "0000", "0011"]
irb(main):002:0> "903".split(//).map { |c| "%04b" % c }.join
=> "100100000011"
irb(main):003:0> "903".split(//).map { |c| "%04b" % c }.join(' ')
=> "1001 0000 0011"
irb(main):004:0>

Something like that I would work,another approach would be to use pack
with an H to then process the resulting string in bits.

pth

···

On 3/25/07, Ak 756 <macro.peng@gmail.com> wrote:

Mike Moore wrote:

> irb(main):008:0> int_to_binary(903)
> => "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

--
Posted via http://www.ruby-forum.com/\.

12

Just for the record, that is commonly called BCD (for Binary Coded Decimal).

···

On 3/25/07, Ak 756 <macro.peng@gmail.com> wrote:

Mike Moore wrote:

> irb(main):008:0> int_to_binary(903)
> => "001110000111"

Hi ,thanks for advise.
I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

--
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/

13

No need to call to_i on the digit strings:

irb(main):002:0> 903.to_s.split('').map{|n| "%04b" % n }.to_s
=> "100100000011"

···

On Mar 25, 10:06 pm, Joel VanderWerf <v...@path.berkeley.edu> wrote:

Ak 756 wrote:
> I want my function to give output 100100000011 for input 903. Where
> '1001','0000','0011' is the binary value for 9,0,3
> Is there efficient way to do this?

"903".split("").map{|s| "%04b" % s.to_i}.join
=> "100100000011"

14

Gavin Kistner wrote:

···

On Mar 25, 10:06 pm, Joel VanderWerf <v...@path.berkeley.edu> wrote:

Ak 756 wrote:
> I want my function to give output 100100000011 for input 903. Where
> '1001','0000','0011' is the binary value for 9,0,3
> Is there efficient way to do this?

"903".split("").map{|s| "%04b" % s.to_i}.join
=> "100100000011"

No need to call to_i on the digit strings:

irb(main):002:0> 903.to_s.split('').map{|n| "%04b" % n }.to_s
=> "100100000011"

Woo.., that's really what's I want.
Thanks you guys-:slight_smile:

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Posted via http://www.ruby-forum.com/\.

15

Phrogz wrote:

Ak 756 wrote:

I want my function to give output 100100000011 for input 903. Where
'1001','0000','0011' is the binary value for 9,0,3
Is there efficient way to do this?

"903".split("").map{|s| "%04b" % s.to_i}.join
=> "100100000011"

No need to call to_i on the digit strings:

irb(main):002:0> 903.to_s.split('').map{|n| "%04b" % n }.to_s
=> "100100000011"

Had no idea that would work, but now it seems obvious.... Apparently #Integer is being called by #% when the format char is 'b', because it works correctly with different bases:

"%04b" % "16"
=> "10000"
"%04b" % "0x10"
=> "10000"
"%04b" % "0b10000"
=> "10000"

···

On Mar 25, 10:06 pm, Joel VanderWerf <v...@path.berkeley.edu> wrote:

--
       vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407