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coding logic and syntax in R

4 messages · Pravin, Brian Ripley, Eric Lecoutre +1 more

Original
Brian Ripley
#
On Wed, 24 Dec 2003, Pravin wrote:

            

      
      
      
    

        
BTW, that is quite a small dataset these days: not even 10 million is `huge'.

            

      
      
      
    

        
Let me see if I understand you.  This was really

111110111
111011111
111111111
111111...

and you want

111110000
111000000
111111111
111111...

So let's treat it as a matrix (extending to 4 complete sets):

x <- as.numeric(strsplit("111110111111011111111111111111111011", NULL)[[1]])
xx <- matrix(x, ncol=9, byrow=TRUE)

Then a simple loop

for(i in 2:9) xx[,i] <- xx[,i] & xx[,i-1]

give me the second matrix, which I can read out as a vector as

as.vector(t(xx))
[1] 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0

or in what I understand as your format

paste(t(xx), collapse="")
[1] "111110000111000000111111111111111000"

Doing this with 81000 random 0/1's took a fraction of a second.

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Eric Lecoutre
    

    

      
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In R, always begin to try to obtain result on a little unit.
Begin to make a function that will make replacements for ONE vector (of size 9)

FillWith=function(vec,SearchForOne=0,ReplaceNextValues=0)
{
  pp=which(vec==SearchForOne)
if (length(pp)>0) vec[pp:length(vec)]=ReplaceNextValues
return(vec)
}

Verify it works:

 > FillWith(c(1,1,0,1,1))
[1] 1 1 0 0 0


Then try to apply it with your data, using one of the ?apply functions.
Here, tapply seems to be adequate.

 > data=c(rep(1,9),rep(1,4),0,rep(1,4))
 > data
  [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
 > data=cbind(data,groups=((1:length(data)-1)%/%9))
 > data
       data groups
  [1,]    1      0
  [2,]    1      0
  [3,]    1      0
  [4,]    1      0
  [5,]    1      0
  [6,]    1      0
  [7,]    1      0
  [8,]    1      0
  [9,]    1      0
[10,]    1      1
[11,]    1      1
[12,]    1      1
[13,]    1      1
[14,]    0      1
[15,]    1      1
[16,]    1      1
[17,]    1      1
[18,]    1      1

 > tapply(data[,1],data[,2],FUN=FillWith)
$"0"
[1] 1 1 1 1 1 1 1 1 1

$"1"
[1] 1 1 1 1 0 0 0 0 0

And then come back to a vector with unlist().

Eric

        
At 08:27 24/12/2003, Pravin wrote:

            

      
      
      
    

        
--------------------------------------------------
L'erreur est certes humaine, mais un vrai d?sastre
n?cessite un ou deux ordinateurs. Citation anonyme
--------------------------------------------------
Eric Lecoutre
Informaticien/Statisticien
Institut de Statistique / UCL

TEL (+32)(0)10473050       lecoutre at stat.ucl.ac.be
URL http://www.stat.ucl.ac.be/ISpersonnel/lecoutre

  
  


  


      

    

    

      
      

        


  

        

      Renaud Lancelot
    

    

      
      #
    

  


  

      
Pravin a ?crit :

            

      
      
      
    

        

      
      
    

        
Pravin a ?crit :

 > Hello,
 >
 >
 >
 > I am a beginner in R programming and recently heard about this 
mailing list.
 > Currently, I am trapped into a simple problem for which I just can't 
find a
 > solution. I have a huge dataset (~81,000 observations) that has been
 > analyzed and the final result is in the form of 0 and 1(one column).
 >
 >
 >
 > I need to write a code to process this column in a little complicated 
way.
 >
 > These 81,000 observations are actually 9,000 sets (81,000/9).
 >
 > So, in each set whenever zero appears, rest all observations become zero.
 >
 >
 >
 > For example;
 >
 > If the column has:
 >
 > 111110111111011111111111111111111....
 >
 > The output should look like:
 >
 > 111110000111000000111111111111111...
 >
 >
 >
 > I hope this makes sense.
 >
 >
 >
 > Thank you in anticipation,
 >
 >
 >
 > Pravin
 >
 >
 >
 > Pravin Jadhav
 >
 >
 >
 >
 > 	[[alternative HTML version deleted]]
 >
 > ______________________________________________
 > R-help at stat.math.ethz.ch mailing list
 > https://www.stat.math.ethz.ch/mailman/listinfo/r-help
 >
 >
Here is an example:

set.seed(101)
v <- sample(c(0, 1), size = 36, replace = TRUE, prob = c(.05, .95))
L <- length(v) / 9
idx <- rep(seq(L), each = 9)

fn <- function(x){
         ok <- FALSE
         for(i in seq(length(x))){
           if(x[i] == 0) ok <- TRUE
           x[i] <- if(ok) 0 else 1
           }
         x
   }

cbind(idx, v, recod = unlist(tapply(v, idx, fn)))
    idx v recod
11   1 1     1
12   1 1     1
13   1 1     1
14   1 1     1
15   1 1     1
16   1 1     1
17   1 1     1
18   1 1     1
19   1 1     1
21   2 1     1
22   2 1     1
23   2 1     1
24   2 1     1
25   2 1     1
26   2 1     1
27   2 1     1
28   2 1     1
29   2 1     1
31   3 1     1
32   3 1     1
33   3 1     1
34   3 0     0
35   3 1     0
36   3 1     0
37   3 1     0
38   3 1     0
39   3 1     0
41   4 1     1
42   4 1     1
43   4 1     1
44   4 1     1
45   4 1     1
46   4 1     1
47   4 1     1
48   4 1     1
49   4 1     1
 >

Merry Christmas !

Renaud


-- 
Dr Renaud Lancelot
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Ambassade de France - SCAC
BP 834 Antannarivo 101
Madagascar

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