thr3ads.net - R help - [R] coding logic and syntax in R [Dec 2003]

If this information is useful, please help other people find it:
Share via:

Pravin

2003-Dec-24 07:27 UTC

[R] coding logic and syntax in R

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]]

Prof Brian Ripley

2003-Dec-24 08:09 UTC

head link

[R] coding logic and syntax in R

On Wed, 24 Dec 2003, Pravin wrote:
> 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
BTW, that is quite a small dataset these days: not even 10 million is
`huge'.
> 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...
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.


-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

Eric Lecoutre

2003-Dec-24 08:09 UTC

head link

[R] coding logic and syntax in R

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:>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


--------------------------------------------------
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

2003-Dec-24 09:00 UTC

head link

[R] coding logic and syntax in R

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
> 
> 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
v?t?rinaire ?pid?miologiste
Ambassade de France - SCAC
BP 834 Antannarivo 101
Madagascar

t?l. +261 (0)32 04 824 55 (cell)
      +261 (0)20 22 494 37 (home)




-- 
Dr Renaud Lancelot
v?t?rinaire ?pid?miologiste
Ambassade de France - SCAC
BP 834 Antannarivo 101
Madagascar

t?l. +261 (0)32 04 824 55 (cell)
      +261 (0)20 22 494 37 (home)

Adaikalavan RAMASAMY

2003-Dec-26 06:01 UTC

head link

[R] coding logic and syntax in R

In addition to the previous replies, try this

x <- as.numeric(strsplit("111110111111011111111111111", NULL)[[1]])
g <- rep(1:3, each=9)                   # set numbering
rbind(x, g)                             # to check

y <- unlist( tapply(x, g, cummin) )> y11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 38 39
 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

tapply() applies a given function, in this case cummin(), to the sets defined by
'g'.
cummin() returns the cummulative minimum
Here, the names of vector y is a combination of set number and observation in
set number.

--
Adaikalavan Ramasamy 

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at
stat.math.ethz.ch] On Behalf Of Renaud Lancelot
Sent: Wednesday, December 24, 2003 5:00 PM
To: Pravin
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] coding logic and syntax in R


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
> 
> 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
v?t?rinaire ?pid?miologiste
Ambassade de France - SCAC
BP 834 Antannarivo 101
Madagascar

t?l. +261 (0)32 04 824 55 (cell)
      +261 (0)20 22 494 37 (home)




-- 
Dr Renaud Lancelot
v?t?rinaire ?pid?miologiste
Ambassade de France - SCAC
BP 834 Antannarivo 101
Madagascar

t?l. +261 (0)32 04 824 55 (cell)
      +261 (0)20 22 494 37 (home)

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Apparently Analagous Threads

Search for more seemingly similar threads

R help - Dec 2003 - coding logic and syntax in R

[R] coding logic and syntax in R

[R] coding logic and syntax in R

[R] coding logic and syntax in R

[R] coding logic and syntax in R

[R] coding logic and syntax in R

Apparently Analagous Threads