histogram frequency weighing

lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05))$counts
lenh  # original data

l.rle <- rle(lenh)
# determine where '0's are
Zero <- which(l.rle$values == 0)
# if last entry in rle was 0, delete from offsets since we are changing +1
if (tail(l.rle$values,1) == 0) Zero <- Zero[-length(Zero)]
l.offsets <- cumsum(l.rle$lengths)  # offsets into original vector# modify original input
lenh[l.offsets[Zero+1]] <- lenh[l.offsets[Zero  + 1]] / (l.rle$lengths[Zero]+1)
lenh  # modified data

Fellow R-helpers,

Suppose we create a histogram as follows (although it could be any vector
with zeroes in it):


R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05))
R> lenh$counts
 [1]  0  0  0  0  0  1  0  3  0  1  0  4  0  2  0  5  0  6  0 10  0  9  0  4  0
[26]  1  0  6  0  7  0  6  0  8  0  7  0  3  0  6  0  6  0  4  0  9  0  7  0  5
[51]  0  2  0  8  0  3  0  4  0  1  0  1  0  3  0  1  0  1  0  0  0  1  0  4  0
[76]  0  0  1  0  0


and we wanted to apply a weighing scheme where frequencies immediately
following (and only those) empty class intervals (0) should be adjusted by
averaging them over the number of preceding empty intervals + 1.  For
example, the first frequency that would need to be adjusted in 'lenh' is
element 6 (1), which has 5 preceding empty intervals, so its adjusted
count would be 1/6.  Similarly, the second one would be element 8 (3),
which has 1 preceding empty interval, so its adjusted count would be 3/2.
Can somebody please provide a hint to implement such a weighing scheme?

I thought about some very contrived ways to accomplish this, involving
'which' and 'diff', but I sense a function might already be available to
do this efficiently.  I couldn't find relevant info in the usual channels.
Thanks in advance for any pointers.


Cheers,

--
Seb

I thought about some very contrived ways to accomplish this, involving
'which' and 'diff', but I sense a function might already be available to
do this efficiently.

I think this should do it:

 Dear list,

I try to do the following:
I have an list of length n, with elements done by smooth.spline
(SmoothList).
Now I have a matrix with n rows and m columns with x-values(Xarray)
Now I want ot predict the y-values. Therefor I want to take the first
element of SmoothList and the first row of Xarray and predict for each
element in Xarray the y value. And then take the second element of
SmoothList and second row of Xarray, third row of SmoothList and third
row of Xarray and so on....

I tried following:

NewValues <- function(x,LIST){predict(LIST,x)}
apply(Xarray, 2, NewValues,SmoothList)

But it don't work.

Could anybody help please ?

Gunther

 Dear list,

I try to do the following:
I have an list of length n, with elements done by smooth.spline 
(SmoothList).
Now I have a matrix with n rows and m columns with x-values(Xarray) 
Now I want ot predict the y-values. Therefor I want to take the first 
element of SmoothList and the first row of Xarray and predict for each 
element in Xarray the y value. And then take the second element of 
SmoothList and second row of Xarray, third row of SmoothList and third 
row of Xarray and so on....

I tried following:

NewValues <- function(x,LIST){predict(LIST,x)} apply(Xarray, 2, 
NewValues,SmoothList)

But it don't work.

Could anybody help please ?

Gunther

Ok.
I tried this too, but it still doesn't work.
Here some more information to try out, but just an excerpt of Xarray

x <- c(0.11,0.25,0.45,0.65,0.80,0.95,1)
Y <-
matrix(c(15,83,57,111,150,168,175,37,207,142,277,375,420,437),nrow=2)

sm <- function(y,x){smooth.spline(x,y)} 
SmoothList <- apply(Y,1,sm,x)
NewValues <- function(x,LIST){predict(LIST,x)} 
Xarray <-
matrix(c(0.15,0.56,0.66,0.45,0.19,0.17,0.99,0.56,0.77,0.41,0.11,0.63,0
.42,0. 43),nrow=2)


apply(Xarray, 2, NewValues,SmoothList)
apply(Xarray, 2, NewValues,LIST=SmoothList)



-----Urspr?ngliche Nachricht-----
Von: Petr Pikal [mailto:petr.pikal at precheza.cz] 
Gesendet: Montag, 18. September 2006 08:43
An: Gunther H?ning; r-help at stat.math.ethz.ch
Betreff: Re: [R] Question on apply()

Hi

not much information about what can be wrong. As nobody knows your
Xarray and SmoothList it is hard to guess. You even omitted to show
what "does not work" So here are few guesses.

predict usually expects comparable data
apply(Xarray, 2, NewValues,LIST=SmoothList)


HTH
Petr




On 18 Sep 2006 at 8:05, Gunther H?ning wrote:

From:           	Gunther H?ning <gunther.hoening at ukmainz.de>
To:             	<r-help at stat.math.ethz.ch>
Date sent:      	Mon, 18 Sep 2006 08:05:28 +0200
Subject:        	[R] Question on apply()

 Dear list,

I try to do the following:
I have an list of length n, with elements done by smooth.spline
(SmoothList). Now I have a matrix with n rows and m columns with
x-values(Xarray) Now I want ot predict the y-values. Therefor I want
to take the first element of SmoothList and the first row of Xarray
and predict for each element in Xarray the y value. And then take
the second element of SmoothList and second row of Xarray, third row
of SmoothList and third row of Xarray and so on....

I tried following:

NewValues <- function(x,LIST){predict(LIST,x)} apply(Xarray, 2,
NewValues,SmoothList)

But it don't work.

Could anybody help please ?

Gunther

Petr Pikal
petr.pikal at precheza.cz

Ok.
I tried this too, but it still doesn't work.
Here some more information to try out, but just an excerpt of Xarray

x <- c(0.11,0.25,0.45,0.65,0.80,0.95,1)
Y <-
matrix(c(15,83,57,111,150,168,175,37,207,142,277,375,420,437),nrow=2)

sm <- function(y,x){smooth.spline(x,y)} SmoothList <- apply(Y,1,sm,x) 
NewValues <- function(x,LIST){predict(LIST,x)} Xarray <- 
matrix(c(0.15,0.56,0.66,0.45,0.19,0.17,0.99,0.56,0.77,0.41,0.11,0.63,0
.42,0. 43),nrow=2)


apply(Xarray, 2, NewValues,SmoothList) apply(Xarray, 2, 
NewValues,LIST=SmoothList)



-----Urspr?ngliche Nachricht-----
Von: Petr Pikal [mailto:petr.pikal at precheza.cz]
Gesendet: Montag, 18. September 2006 08:43
An: Gunther H?ning; r-help at stat.math.ethz.ch
Betreff: Re: [R] Question on apply()

Hi

not much information about what can be wrong. As nobody knows your 
Xarray and SmoothList it is hard to guess. You even omitted to show 
what "does not work" So here are few guesses.

predict usually expects comparable data apply(Xarray, 2, 
NewValues,LIST=SmoothList)


HTH
Petr




On 18 Sep 2006 at 8:05, Gunther H?ning wrote:

From:           	Gunther H?ning <gunther.hoening at ukmainz.de>
To:             	<r-help at stat.math.ethz.ch>
Date sent:      	Mon, 18 Sep 2006 08:05:28 +0200
Subject:        	[R] Question on apply()

 Dear list,

I try to do the following:
I have an list of length n, with elements done by smooth.spline 
(SmoothList). Now I have a matrix with n rows and m columns with
x-values(Xarray) Now I want ot predict the y-values. Therefor I want 
to take the first element of SmoothList and the first row of Xarray 
and predict for each element in Xarray the y value. And then take 
the second element of SmoothList and second row of Xarray, third row 
of SmoothList and third row of Xarray and so on....

I tried following:

NewValues <- function(x,LIST){predict(LIST,x)} apply(Xarray, 2,
NewValues,SmoothList)

But it don't work.

Could anybody help please ?

Gunther

Petr Pikal
petr.pikal at precheza.cz

Hi,

I tried both ideas, but it isn't that what I'm looking for.
I want to avoid for loop, because the matrix is of big size(1200*1200
entries)

With a loop I would do:

for ( i in seq(along = SmoothList))
{
 Xarry[i,] <- predict(SmoothList[[i]],Xarry[i,])$y
} 

Actually I want to do more than just to predict a value, but it isn't
important for the initial question...

Gunther

-----Urspr?ngliche Nachricht-----
Von: Petr Pikal [mailto:petr.pikal at precheza.cz] 
Gesendet: Montag, 18. September 2006 11:44
An: Gunther H?ning
Cc: r-help at stat.math.ethz.ch
Betreff: Re: AW: [R] Question on apply() with more information...

Hi

If I am correct apply do not choose from SmoothList as you expected.
Instead probably

lapply(SmoothList, predict,Xarray)
or
mapply(predict,SmoothList, Xarray)

can give you probably what you want.

HTH
Petr


On 18 Sep 2006 at 9:26, Gunther H?ning wrote:

From:           	Gunther H?ning <gunther.hoening at ukmainz.de>
To:             	"'Petr Pikal'" <petr.pikal at precheza.cz>,
 <r-help at stat.math.ethz.ch>
Subject:        	AW: [R] Question on apply() with more information...
Date sent:      	Mon, 18 Sep 2006 09:26:01 +0200

Ok.
I tried this too, but it still doesn't work.
Here some more information to try out, but just an excerpt of Xarray

x <- c(0.11,0.25,0.45,0.65,0.80,0.95,1)
Y <-
matrix(c(15,83,57,111,150,168,175,37,207,142,277,375,420,437),nrow=2
)

sm <- function(y,x){smooth.spline(x,y)} SmoothList <-
apply(Y,1,sm,x) NewValues <- function(x,LIST){predict(LIST,x)}
Xarray <-
matrix(c(0.15,0.56,0.66,0.45,0.19,0.17,0.99,0.56,0.77,0.41,0.11,0.63
,0 .42,0. 43),nrow=2)


apply(Xarray, 2, NewValues,SmoothList) apply(Xarray, 2, 
NewValues,LIST=SmoothList)



-----Urspr?ngliche Nachricht-----
Von: Petr Pikal [mailto:petr.pikal at precheza.cz]
Gesendet: Montag, 18. September 2006 08:43
An: Gunther H?ning; r-help at stat.math.ethz.ch
Betreff: Re: [R] Question on apply()

Hi

not much information about what can be wrong. As nobody knows your
Xarray and SmoothList it is hard to guess. You even omitted to show
what "does not work" So here are few guesses.

predict usually expects comparable data apply(Xarray, 2, 
NewValues,LIST=SmoothList)


HTH
Petr




On 18 Sep 2006 at 8:05, Gunther H?ning wrote:

From:           	Gunther H?ning <gunther.hoening at ukmainz.de>
To:             	<r-help at stat.math.ethz.ch>
Date sent:      	Mon, 18 Sep 2006 08:05:28 +0200
Subject:        	[R] Question on apply()

 Dear list,

I try to do the following:
I have an list of length n, with elements done by smooth.spline
(SmoothList). Now I have a matrix with n rows and m columns with
x-values(Xarray) Now I want ot predict the y-values. Therefor I
want to take the first element of SmoothList and the first row of
Xarray and predict for each element in Xarray the y value. And
then take the second element of SmoothList and second row of
Xarray, third row of SmoothList and third row of Xarray and so
on....

I tried following:

NewValues <- function(x,LIST){predict(LIST,x)} apply(Xarray, 2,
NewValues,SmoothList)

But it don't work.

Could anybody help please ?

Gunther

Petr Pikal
petr.pikal at precheza.cz

Petr Pikal
petr.pikal at precheza.cz