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Isolate Digits in a number

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Raviteja
Raviteja el 7 de Abr. de 2011
Take c input as
c= 0010100101100
Note: Please don't take c as char
isolate each digit into index of "nary"
i.e.,
>> naray =
0 0 1 0 1 0 0 1 0 1 1 0 0
i.e.,
nary(1)=0
nary(2)=0
nary(3)=1
nary(4)=0
nary(5)=1
nary(6)=0
nary(7)=0
nary(8)=1
nary(9)=0
nary(10)=1
nary(11)=1
nary(12)=0
nary(13)=0
  1 comentario
David Young
David Young el 7 de Abr. de 2011
Could you say why you want to do this, please? I ask because it looks a bit like it could be a homework problem. Also, is it assumed that nary is to have 13 elements? The number of leading zeros in 0010100101100 is lost, of course, so there's no information in c about how many leading zeros are to be stored in nary.

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Respuesta aceptada

Matt Fig
Matt Fig el 7 de Abr. de 2011
If c is not a char, then what is it? How could you even enter c in MATLAB if it is not a char? You will lose the leading zeros if c is not a char.
>> c = 0010100101100
c =
10100101100

Más respuestas (3)

Paulo Silva
Paulo Silva el 7 de Abr. de 2011
So if c isn't char it could be a vector:
nary='0010100101100' - '0'; %create the vector with the data provided
%now you can call each value from 1 until 13
nary(1)
nary(2)
nary(3)
nary(4)
nary(5)
nary(6)
nary(7)
nary(8)
nary(9)
nary(10)
nary(11)
nary(12)
nary(13)
  1 comentario
Raviteja
Raviteja el 7 de Abr. de 2011
Your answer working! Thanks!

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David Young
David Young el 7 de Abr. de 2011
Well, since Matt's answer has been accepted (!), I guess I might as well offer mine, for non-char input:
nary = floor(c*10.^(-13:0))-10*floor(c*10.^(-14:-1));
  2 comentarios
Walter Roberson
Walter Roberson el 7 de Abr. de 2011
Ah, but it doesn't handle negative binary numbers ;-)
If you change the floor() to fix() then for negative c, each of the 1 values will come out negative, which would arguably be a fair solution.
David Young
David Young el 11 de Abr. de 2011
Indeed!

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Walter Roberson
Walter Roberson el 7 de Abr. de 2011
nary = sprintf('%d',c) - '0';
Subject, though, to the problems the others noted about losing leading zeros.
Note: this solution is also not valid if there are more than 53 bits from the first 1 in the number.

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