[R] Asking Favor

[Michael Bedward]

Here's a way with less code...

# assume original matrix is X
u <- unique( sort(X) )
ucdf <- cumsum( tabulate(X)[u] )
Xcdf <- matrix(ucdf[ match(X, u) ], 8)

Michael

On 16 September 2010 14:58, chuan zun liang <chuan_zl at yahoo.com.my> wrote:
> Hello Friend:
> Thanks a lot. I kindly told that Micheal already misunderstand what i
> want.But never mind.I also can share with you what I want.
> I got a set of matrix(image) as below:
> x<-t(matrix(y,8))
>> x
>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]   52   55   61   66   70   61   64   73
> [2,]   63   59   55   90  109   85   69   72
> [3,]   62   59   68  113  144  104   66   73
> [4,]   63   58   71  122  154  106   70   69
> [5,]   67   61   68  104  126   88   68   70
> [6,]   79   65   60   70   77   68   58   75
> [7,]   85   71   64   59   55   61   65   83
> [8,]   87   79   69   68   65   76   78   94
> y just the matrix change to vector.
> y<-c(52,55,61,66,70,61,64,73,63,59,55,90,109,85,69,72,62,59,68,113,144,104,66,73,63,58,71,122,154,106,70,69,67,61,68,104,126,88,68,70,79,65,60,70,77,68,58,75,85,71,64,59,55,61,65,83,87,79,69,68,65,76,78,94)
> Then, I want doing for histogram equalization for my image.What I want is I
> want to replace 52 by  value of the cumulative distribution function. For
> example,frequency 52 is 1 and frequency for 55 is 3, so cumulative frequency
> for 55 is 4.So, I want replace 52 by 1 and 55 by 4 and so on until 94
> replace 94.
> This is the script I get(just a part for histogram equalization):
> cdfv<-function(p)
> {a<-cumsum(unlist(lapply(split(p,f=p),length)))
>  b<-sort(unique(p))
>  for(i in c(1:37))
>  {x<-replace(x,x==b[i],a[i])}
> c(x)}
> matrix(cdfv(y),8)
> Thank you.
> Chuan
>
>
> ________________________________
> From: Michael Bedward <michael.bedward at gmail.com>
> To: chuan zun liang <chuan_zl at yahoo.com.my>; Rhelp <r-help at r-project.org>
> Sent: Wednesday, 15 September 2010 16:08:20
> Subject: Re: [R] Asking Favor
>
> Hello Chuan,
>
> If you just want a matrix with the numbers 1 to 64 arranged by row...
>
> m <- matrix(1:64, ncol=8, byrow=TRUE)
>
> But perhaps I don't understand your question properly ?
>
> Michael
>
>
> On 16 September 2010 12:46, chuan zun liang <chuan_zl at yahoo.com.my> wrote:
>> Dear Prof:
>>
>> My name is Chuan.I from Malaysia.I am a beginner for R.I need favor for
>> Prof.
>>
>> This is my data:
>>
>> y<-c(52,55,61,66,70,61,64,73,63,59,55,90,109,85,69,72,62,59,68,113,144,104,66,73,63,58,71,122,154,106,70,69,67,61,68,104,126,88,68,70,79,65,60,70,77,68,58,75,85,71,64,59,55,61,65,83,87,79,69,68,65,76,78,94)
>>
>>
>> and I form it into matrix as below:     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
>> [,8]
>> [1,]   52   63     62    63    67   79   85   87
>> [2,]   55   59     59    58    61   65   71   79
>> [3,]   61   55     68    71    68   60   64   69
>> [4,]   66   90   113  122  104   70   59   68
>> [5,]   70  109  144  154  126   77   55   65
>> [6,]   61   85   104  106    88   68   61   76
>> [7,]   64   69     66    70    68   58   65   78
>>
>> [8,]   73   72     73    69    70   75   83   94
>>
>> But i want replace 52 by 1, 63 by 2, 63 by 5, ........94 by 64. Is it
>> possible I
>> do it in R.Thank a lot.
>>
>> Chuan
>>
>>
>>        [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>