CODEWITHSUNDEEP
rbinarynumbers
Wecsley Prates
Wecsley Prates

Reputation: 51

Convert a Binary Sample in Number

I have a binary sample like this :

Z = c(0,0,0,1,0,1,1,1,0,1,0,0,1,0,1,0,1,1,1,0,1,0,1,0)

I would like to convert all the sequences of the length 4 in number, i.e :

I need to read my original binary sample and convert all the possible sequence of length 4 into numbers.

Example: The sequence 0000 will be 1, the sequence 0001 will be 2, the sequence 0010 will be 3, ..., the sequence 1111 will be 16.

The expected output should be a new sample formed by the numbers 1,2,3,...16 having the same length as the original sample : 

Z = c(0,0,0,1,0,1,1,1,0,1,0,0,1,0,1,0,1,1,1,0,1,0,1,0)

Z1 = c(2,3,6,12,8,15,14,11,5,10,3,11,6,12,8,15,14,11,6,11)

How can I do that in R ?

Upvotes: 3

Views: 135

Answers (4)

Patric
Patric

Reputation: 2131

Because you transfer 4 bit binaries to decimal number, the formula will be:

dec = x1 * 2^3 + x2 * 2^2 + x3 * 2^1 + x4 * 2^0

And this can be implemented by R with sweep multiplication

dec <- sum( X * c(8,4,2,1) )

OR vector multiplication (as @Greg Snow shown).

dec <- X %*% c(8,4,2,1)

Finally, using this computational pattern to every 4 elements of array by sapply, then the whole code as below .

Z <- c(0,0,0,1,0,1,1,1,0,1,0,0,1,0,1,0,1,1,1,0,1,0,1,0)
Z.len  <- length(Z)

# stand for 2^3, 2^2, 2^1, 2^0
Z.base <- c(8,4,2,1)

res1 <- sapply(1:(Z.len-3), FUN=function(x) sum(Z[x:(x+3)] * Z.base)+1 )
res2 <- sapply(1:(Z.len-3), FUN=function(x) (Z[x:(x+3)] %*% Z.base)+1 )
all.equal(res1, res2)
#[1] TRUE 
res1
# [1]  2  3  6 12  8 15 14 11  5 10  3  6 11  6 12  8 15 14 11  6 11

Upvotes: 0

Greg Snow
Greg Snow

Reputation: 49640

Here is another approach:

Z <- c(0,0,0,1,0,1,1,1,0,1,0,0,1,0,1,0,1,1,1,0,1,0,1,0)
Z.tmp <- embed(Z,4)
Z1 <- as.vector(Z.tmp %*% c(1,2,4,8) + 1)

Upvotes: 3

akrun
akrun

Reputation: 887251

You could also use 

library(zoo)
library(compositions)
unbinary(rollapply(z,4, FUN= paste, collapse=''))+1L
#[1]  2  3  6 12  8 15 14 11  5 10  3  6 11  6 12  8 15 14 11  6 11

Upvotes: 2

etienne
etienne

Reputation: 3678

Try :

z<-c(0,0,0,1,0,1,1,1,0,1,0,0,1,0,1,0,1,1,1,0,1,0,1,0)
y<-as.character(z)
z1<-sapply(1:(length(y)-3),function(x){strtoi(paste(y[x:(x+3)],collapse=''),2)+1})

[1]  2  3  6 12  8 15 14 11  5 10  3  6 11  6 12  8 15 14 11  6 11

The code works like this :

  • convert the vector z as a character vector (y)
  • for the length of y minus 3 (because each number requires 4 digits), apply the strtoi function

The strtoi function convert the number by specifying the base of the input number (here, 2 because it's binary). We add 1 because in binary 0000 equals 0 and not 1.

Note: the conversion to character is optional, you can directly do 

sapply(1:(length(z)-3),function(x){strtoi(paste(z[x:(x+3)],collapse=''),2)+1})

it will also be faster to use vapply :

vapply(1:(length(z)-3),function(x){strtoi(paste(z[x:(x+3)],collapse=''),2)+1},FUN.VALUE=1)

Unit: microseconds
   expr     min      lq     mean   median      uq     max neval cld
 vapply 206.866 209.111 214.3936 210.0735 211.356 338.362   100  a 
 sapply 230.278 231.882 234.0249 232.8440 234.128 273.897   100   b

Upvotes: 3

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