[Rd] Option "installWithVers" seems to impact new.packages() badly?
In Rdevel, SVN version built this morning around 10am central european time, it looks like install.packages(new.packages(),installWithVers=TRUE) seem to ignore the version information -- that is, it reinstalls current versions of packages. This did not happen before I used "installWithVers=TRUE" option, that is I could use update.packages() and install.packages(new.packages()) to keep a current, platform-complete, installed base of CRAN. best, -tony [EMAIL PROTECTED] Muttenz, Switzerland. "Commit early,commit often, and commit in a repository from which we can easily roll-back your mistakes" (AJR, 4Jan05). __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
[Rd] bug in gsub with perl=TRUE (PR#8164)
Full_Name: Richard Mott
Version: 2.0.1
OS: Linux toad 2.6.9 #4 SMP Mon Feb 21 16:20:16 GMT 2005 x86_64 AMD Opteron(tm)
Processor 848 AuthenticAMD GNU/Linux
Submission from: (NULL) (129.67.46.247)
gsub with perl=TRUE does not work properly. It pads/truncates the resulting
string to
the length of the input string:
my.formula <- "log10(Biochem.ALP)^2+1 ~ Family + GENDER"
> gsub("^.+~", "transformed.y ~", my.formula )
[1] "transformed.y ~ Family + GENDER"
> gsub("^.+~", "transformed.y ~", my.formula, perl=TRUE )
[1] "transformed.y ~ Family + GENDER\0\006\0\0\r\377\0\0\0" # padded
my.formula <- "Biochem.ALP ~ Family + GENDER"
> gsub("^.+~", "transformed.y ~", my.formula, perl=TRUE )
[1] "transformed.y ~ Family + GEND" # truncated
> gsub("^.+~", "transformed.y ~", my.formula )
[1] "transformed.y ~ Family + GENDER"
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Re: [Rd] bug in gsub with perl=TRUE (PR#8164)
[EMAIL PROTECTED] wrote:
> Full_Name: Richard Mott
> Version: 2.0.1
This version is completely outdated.
Please try with a *recent* version of R when reporting bugs, in this
case R-2.2.0 beta (or in worst case R-2.1.1, the current release).
The bug reported below has been fixed some months ago ...
Uwe Ligges
> OS: Linux toad 2.6.9 #4 SMP Mon Feb 21 16:20:16 GMT 2005 x86_64 AMD
> Opteron(tm) Processor 848 AuthenticAMD GNU/Linux
> Submission from: (NULL) (129.67.46.247)
>
>
> gsub with perl=TRUE does not work properly. It pads/truncates the resulting
> string to
> the length of the input string:
>
> my.formula <- "log10(Biochem.ALP)^2+1 ~ Family + GENDER"
>
>
>>gsub("^.+~", "transformed.y ~", my.formula )
>
> [1] "transformed.y ~ Family + GENDER"
>
>
>>gsub("^.+~", "transformed.y ~", my.formula, perl=TRUE )
>
> [1] "transformed.y ~ Family + GENDER\0\006\0\0\r\377\0\0\0" # padded
>
> my.formula <- "Biochem.ALP ~ Family + GENDER"
>
>>gsub("^.+~", "transformed.y ~", my.formula, perl=TRUE )
>
> [1] "transformed.y ~ Family + GEND" # truncated
>
>>gsub("^.+~", "transformed.y ~", my.formula )
>
> [1] "transformed.y ~ Family + GENDER"
>
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[Rd] Subscripting fails if name of element is "" (PR#8161)
Dear all,
I resend this mail because it was blocked: I submitted a bug from the r-bug
webpage and hypatia seems to block mail that is send from a different IP
than that usually associated with the email. Looks like it is currently
impossible to correctly submit bugs from the website. However, here is the
original bug report:
(PR#8161)
Dear all,
The following shows cases where accessing elements via their name fails (if
the
name is a string of length zero).
Best regards
Jens Oehlschlägel
> p <- 1:3
> names(p) <- c("a","", as.character(NA))
> p
a
123
>
> for (i in names(p))
+ print(p[[i]])
[1] 1
[1] 2
[1] 3
>
> # error 1: vector subsripting with "" fails in second element
> for (i in names(p))
+ print(p[i])
a
1
NA
3
>
> # error 2: print method for list shows no name for second element
> p <- as.list(p)
>
>
> for (i in names(p))
+ print(p[[i]])
[1] 1
[1] 2
[1] 3
>
> # error 3: list subsripting with "" fails in second element
> for (i in names(p))
+ print(p[i])
$a
[1] 1
$"NA"
NULL
$"NA"
[1] 3
>
> version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor1.1
year 2005
month06
day 20
language R
# -- replication code --
p <- 1:3
names(p) <- c("a","", as.character(NA))
p
for (i in names(p))
print(p[[i]])
# error 1: vector subsripting with "" fails in second element
for (i in names(p))
print(p[i])
# error 2: print method for list shows no name for second element
p <- as.list(p)
for (i in names(p))
print(p[[i]])
# error 3: list subsripting with "" fails in second element
for (i in names(p))
print(p[i])
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Re: [Rd] Summary of translation status
On 9/28/2005 11:50 PM, Fernando Henrique Ferraz P. da Rosa wrote: > Dear R-devel & Translation Teams, > > In order to monitor the progress of the translation for the > pt_BR team I wrote a script to summarize the status of the translations. > It wasn't difficult to extend it to the other languages so I decided to > set up a page with the summaries of the translation for all languages > for which currently exist a translation. > > http://www.ime.usp.br/~feferraz/en/rtransstat.html > > If any of you find it useful I can keep it updated on a regular basis > (daily or weekly). > > > Thank you, > > > (PS: I'm resending this message because it didn't get through the > filter the first time. Sorry for the inconvenience for those that > are receiving it more than one time). Hi Fernando. That's a nice page. I'd add an explicit statement about which branch the statistics apply to. You say "Statistics based on SVN: 35706", presumably on the trunk, but soon interest will shift to the R-2-2-patches branch. (If this is automated and you have the disk space for both, perhaps both trunk and the current patch branch could be listed, but I expect the statistics will be very similar.) Duncan Murdoch __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Subscripting fails if name of element is "" (PR#8161)
On Fri, 30 Sep 2005, "Jens Oehlschlägel" wrote:
Dear all,
The following shows cases where accessing elements via their name fails (if
the
name is a string of length zero).
This looks deliberate (there is a function NonNullStringMatch that does
the matching). I assume this is because there is no other way to
indicate that an element has no name.
If so, it is a documentation bug -- help(names) and FAQ 7.14 should
specify this behaviour. Too late for 2.2.0, unfortunately.
-thomas
Best regards
Jens Oehlschlägel
p <- 1:3
names(p) <- c("a","", as.character(NA))
p
a
123
for (i in names(p))
+ print(p[[i]])
[1] 1
[1] 2
[1] 3
# error 1: vector subsripting with "" fails in second element
for (i in names(p))
+ print(p[i])
a
1
NA
3
# error 2: print method for list shows no name for second element
p <- as.list(p)
for (i in names(p))
+ print(p[[i]])
[1] 1
[1] 2
[1] 3
# error 3: list subsripting with "" fails in second element
for (i in names(p))
+ print(p[i])
$a
[1] 1
$"NA"
NULL
$"NA"
[1] 3
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor1.1
year 2005
month06
day 20
language R
# -- replication code --
p <- 1:3
names(p) <- c("a","", as.character(NA))
p
for (i in names(p))
print(p[[i]])
# error 1: vector subsripting with "" fails in second element
for (i in names(p))
print(p[i])
# error 2: print method for list shows no name for second element
p <- as.list(p)
for (i in names(p))
print(p[[i]])
# error 3: list subsripting with "" fails in second element
for (i in names(p))
print(p[i])
--
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Thomas Lumley Assoc. Professor, Biostatistics
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[Rd] Compiling R on OpenSolaris
Hi: I am trying to R (v2.1.1) to compile on OpenSolaris (build 22) using gcc version 3.4.3. But I am getting this error message: gmake[5]: Entering directory `/usr/local/R-2.1.1/src/library/tools/src' ../../../../library/tools/libs/tools.so is unchanged gmake[5]: Leaving directory `/usr/local/R-2.1.1/src/library/tools/src' gmake[4]: Leaving directory `/usr/local/R-2.1.1/src/library/tools/src' Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/usr/local/R-2.1.1 /library/tools/libs/tools.so': ld.so.1: R: fatal: relocation error: R_AMD64_PC32: file /usr/local/R-2.1.1/library/tools/libs/tools.so: symbol main: value 0x28001413f04 does not fit Execution halted gmake[3]: *** [all] Error 1 gmake[3]: Leaving directory `/usr/local/R-2.1.1/src/library/tools' I am on an Athlon64 box so I put in -m64 -mtune=athlon64 for the compiler option. Any help how I can fix this problem much appreciated. thanks ---Vincent [[alternative HTML version deleted]] __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
[Rd] by() processing on a dataframe
I want to calculate a statistic on a number of subgroups of a dataframe,
then put the results into a dataframe. (What SAS PROC MEANS does, I
think, though it's been years since I used it.)
This is possible using by(), but it seems cumbersome and fragile. Is
there a more straightforward way than this?
Here's a simple example showing my current strategy:
> dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
c(2,2,2,2)), value = rnorm(8))
> dataset
gp1 gp2 value
1 1 1 0.9493232
2 1 1 -0.0474712
3 1 2 -0.6808021
4 1 2 1.9894999
5 2 3 2.0154786
6 2 3 0.4333056
7 2 4 -0.4746228
8 2 4 0.6017522
>
> handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
+ gp2 = subset$gp2[1], statistic = mean(subset$value))
>
> bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
>
> result <- do.call('rbind', bylist)
> result
gp1 gp2 statistic
11 1 0.45092598
11 1 2 0.65434890
12 2 3 1.22439210
13 2 4 0.06356469
tapply() is inappropriate because I don't have all possible combinations
of gp1 and gp2 values, only some of them:
> tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
1 23 4
1 0.450926 0.6543489 NA NA
2 NANA 1.224392 0.06356469
In the real case, I only have a very sparse subset of all the
combinations, and tapply() and by() both die for lack of memory.
Any suggestions on how to do what I want, without using SAS?
Duncan Murdoch
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Re: [Rd] by() processing on a dataframe
I'm not entirely sure what you want, but maybe this does the trick?
data.frame.by <- function(data, variables, fun, ...) {
if (length(variables) == 0 ) {
df <- data.frame(results = 0)
df$results <- list(fun(data$value, ...))
return(df)
}
sorted <- sort.df(data, variables)[,c(variables), drop=FALSE]
duplicates <- duplicated(sorted[,variables, drop=FALSE])
index <- cumsum(!duplicates)
results <- by(data, index, fun, ...)
cols <- sorted[!duplicates,variables, drop=FALSE]
cols$results <- array(results)
cols
}
sort.df <- function(data, vars) {
data[do.call("order", data[,vars, drop=FALSE]), ,drop=FALSE]
}
dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
c(2,2,2,2)), value = rnorm(8))
data.frame.by(dataset, c("gp1", "gp2"), function(data) mean(data$value))
data.frame.by(dataset, "gp1", function(data) tapply(data$value, data$gp2, mean))
data.frame.by(dataset, "gp1", function(data) lm(gp2 ~ value, data)) #
doesn't print, but everything is there ok
(note that the results column will be a list if necessary - this may
be a serious abuse of data frames, but I'm not sure and no one replied
when I queried the list)
Hadley
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Re: [Rd] by() processing on a dataframe
Duncan Murdoch <[EMAIL PROTECTED]> writes:
> I want to calculate a statistic on a number of subgroups of a dataframe,
> then put the results into a dataframe. (What SAS PROC MEANS does, I
> think, though it's been years since I used it.)
>
> This is possible using by(), but it seems cumbersome and fragile. Is
> there a more straightforward way than this?
>
> Here's a simple example showing my current strategy:
>
> > dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
> c(2,2,2,2)), value = rnorm(8))
> > dataset
>gp1 gp2 value
> 1 1 1 0.9493232
> 2 1 1 -0.0474712
> 3 1 2 -0.6808021
> 4 1 2 1.9894999
> 5 2 3 2.0154786
> 6 2 3 0.4333056
> 7 2 4 -0.4746228
> 8 2 4 0.6017522
> >
> > handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
> + gp2 = subset$gp2[1], statistic = mean(subset$value))
> >
> > bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
> >
> > result <- do.call('rbind', bylist)
> > result
> gp1 gp2 statistic
> 11 1 0.45092598
> 11 1 2 0.65434890
> 12 2 3 1.22439210
> 13 2 4 0.06356469
>
> tapply() is inappropriate because I don't have all possible combinations
> of gp1 and gp2 values, only some of them:
>
> > tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
> 1 23 4
> 1 0.450926 0.6543489 NA NA
> 2 NANA 1.224392 0.06356469
>
>
>
> In the real case, I only have a very sparse subset of all the
> combinations, and tapply() and by() both die for lack of memory.
>
> Any suggestions on how to do what I want, without using SAS?
Have you tried aggregate()?
Alternatively, you migth split on interaction(, drop=TRUE)
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [Rd] by() processing on a dataframe
Check out summaryBy in the doBy package at:
http://genetics.agrsci.dk/~sorenh/misc
e.g.
summaryBy(value ~ gp1 + gp2, data = dataset)
On 9/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> I want to calculate a statistic on a number of subgroups of a dataframe,
> then put the results into a dataframe. (What SAS PROC MEANS does, I
> think, though it's been years since I used it.)
>
> This is possible using by(), but it seems cumbersome and fragile. Is
> there a more straightforward way than this?
>
> Here's a simple example showing my current strategy:
>
> > dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
> c(2,2,2,2)), value = rnorm(8))
> > dataset
> gp1 gp2 value
> 1 1 1 0.9493232
> 2 1 1 -0.0474712
> 3 1 2 -0.6808021
> 4 1 2 1.9894999
> 5 2 3 2.0154786
> 6 2 3 0.4333056
> 7 2 4 -0.4746228
> 8 2 4 0.6017522
> >
> > handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
> + gp2 = subset$gp2[1], statistic = mean(subset$value))
> >
> > bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
> >
> > result <- do.call('rbind', bylist)
> > result
>gp1 gp2 statistic
> 11 1 0.45092598
> 11 1 2 0.65434890
> 12 2 3 1.22439210
> 13 2 4 0.06356469
>
> tapply() is inappropriate because I don't have all possible combinations
> of gp1 and gp2 values, only some of them:
>
> > tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
> 1 23 4
> 1 0.450926 0.6543489 NA NA
> 2 NANA 1.224392 0.06356469
>
>
>
> In the real case, I only have a very sparse subset of all the
> combinations, and tapply() and by() both die for lack of memory.
>
> Any suggestions on how to do what I want, without using SAS?
>
> Duncan Murdoch
>
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Re: [Rd] by() processing on a dataframe
On Fri, 2005-09-30 at 13:22 -0400, Duncan Murdoch wrote:
> I want to calculate a statistic on a number of subgroups of a dataframe,
> then put the results into a dataframe. (What SAS PROC MEANS does, I
> think, though it's been years since I used it.)
>
> This is possible using by(), but it seems cumbersome and fragile. Is
> there a more straightforward way than this?
>
> Here's a simple example showing my current strategy:
>
> > dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
> c(2,2,2,2)), value = rnorm(8))
> > dataset
>gp1 gp2 value
> 1 1 1 0.9493232
> 2 1 1 -0.0474712
> 3 1 2 -0.6808021
> 4 1 2 1.9894999
> 5 2 3 2.0154786
> 6 2 3 0.4333056
> 7 2 4 -0.4746228
> 8 2 4 0.6017522
> >
> > handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
> + gp2 = subset$gp2[1], statistic = mean(subset$value))
> >
> > bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
> >
> > result <- do.call('rbind', bylist)
> > result
> gp1 gp2 statistic
> 11 1 0.45092598
> 11 1 2 0.65434890
> 12 2 3 1.22439210
> 13 2 4 0.06356469
>
> tapply() is inappropriate because I don't have all possible combinations
> of gp1 and gp2 values, only some of them:
>
> > tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
> 1 23 4
> 1 0.450926 0.6543489 NA NA
> 2 NANA 1.224392 0.06356469
>
>
>
> In the real case, I only have a very sparse subset of all the
> combinations, and tapply() and by() both die for lack of memory.
>
> Any suggestions on how to do what I want, without using SAS?
>
> Duncan Murdoch
Duncan,
Does this do what you want?
> set.seed(1)
> df <- data.frame(gp1 = rep(1:2, c(4,4)),
gp2 = rep(1:4, c(2,2,2,2)),
value = rnorm(8))
> df
gp1 gp2 value
1 1 1 -0.6264538
2 1 1 0.1836433
3 1 2 -0.8356286
4 1 2 1.5952808
5 2 3 0.3295078
6 2 3 -0.8204684
7 2 4 0.4874291
8 2 4 0.7383247
> means <- aggregate(df$value, list(gp1 = df$gp1, gp2 = df$gp2), mean)
> means
gp1 gp2 x
1 1 1 -0.2214052
2 1 2 0.3798261
3 2 3 -0.2454803
4 2 4 0.6128769
> merge(df, means, by = c("gp1", "gp2"))
gp1 gp2 value x
1 1 1 -0.6264538 -0.2214052
2 1 1 0.1836433 -0.2214052
3 1 2 -0.8356286 0.3798261
4 1 2 1.5952808 0.3798261
5 2 3 0.3295078 -0.2454803
6 2 3 -0.8204684 -0.2454803
7 2 4 0.4874291 0.6128769
8 2 4 0.7383247 0.6128769
HTH,
Marc Schwartz
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Re: [Rd] by() processing on a dataframe
On 9/30/2005 1:41 PM, Peter Dalgaard wrote:
> Duncan Murdoch <[EMAIL PROTECTED]> writes:
>
>> I want to calculate a statistic on a number of subgroups of a dataframe,
>> then put the results into a dataframe. (What SAS PROC MEANS does, I
>> think, though it's been years since I used it.)
>>
>> This is possible using by(), but it seems cumbersome and fragile. Is
>> there a more straightforward way than this?
>>
>> Here's a simple example showing my current strategy:
>>
>> > dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
>> c(2,2,2,2)), value = rnorm(8))
>> > dataset
>>gp1 gp2 value
>> 1 1 1 0.9493232
>> 2 1 1 -0.0474712
>> 3 1 2 -0.6808021
>> 4 1 2 1.9894999
>> 5 2 3 2.0154786
>> 6 2 3 0.4333056
>> 7 2 4 -0.4746228
>> 8 2 4 0.6017522
>> >
>> > handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
>> + gp2 = subset$gp2[1], statistic = mean(subset$value))
>> >
>> > bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
>> >
>> > result <- do.call('rbind', bylist)
>> > result
>> gp1 gp2 statistic
>> 11 1 0.45092598
>> 11 1 2 0.65434890
>> 12 2 3 1.22439210
>> 13 2 4 0.06356469
>>
>> tapply() is inappropriate because I don't have all possible combinations
>> of gp1 and gp2 values, only some of them:
>>
>> > tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
>> 1 23 4
>> 1 0.450926 0.6543489 NA NA
>> 2 NANA 1.224392 0.06356469
>>
>>
>>
>> In the real case, I only have a very sparse subset of all the
>> combinations, and tapply() and by() both die for lack of memory.
>>
>> Any suggestions on how to do what I want, without using SAS?
>
> Have you tried aggregate()?
aggregate() has a few problems:
- it applies the function to every column in the dataframe. In my
case it only makes sense to apply it to some of them. (This may not be
a killer, but it certainly makes things inefficient and tricky.)
- I'd like to look at the whole subset to figure out the function (but
I can probably work around this)
- It uses too much memory. E.g. try
> df <- data.frame(x=rnorm(1000), y=rnorm(1000), z=rnorm(1000),
w=rnorm(1000))
> aggregate(df, list(df$x,df$y,df$z), mean)
Error: cannot allocate vector of size 3906250 Kb
In addition: Warning messages:
1: Reached total allocation of 1007Mb: see help(memory.size)
2: Reached total allocation of 1007Mb: see help(memory.size)
This should have returned the same dataframe (there are 1000 subsets),
but it tried to construct a billion of them.
On 9/30/2005 1:48 PM, Don MacQueen wrote:
> Look at the summarize() function in the Hmisc package.
It seems to want a matrix, not a data.frame. The real situation has
mixed types (character, factors, numeric) so it can't be a matrix.
> (and I this is an r-help question, not an r-devel question, I would
think)
Yes, that's where I should have posted. Sorry. However, this is
starting to look like a development problem...
Peter again:
> Alternatively, you migth split on interaction(, drop=TRUE)
Looking at the code, it appears that will construct the full product
interaction, then subset to the non-empty cases... Yes, it does that.
Looks like I'll have to write my own.
Duncan
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Re: [Rd] by() processing on a dataframe
On 9/30/2005 1:41 PM, hadley wickham wrote:
> I'm not entirely sure what you want, but maybe this does the trick?
>
> data.frame.by <- function(data, variables, fun, ...) {
> if (length(variables) == 0 ) {
> df <- data.frame(results = 0)
> df$results <- list(fun(data$value, ...))
> return(df)
> }
>
> sorted <- sort.df(data, variables)[,c(variables), drop=FALSE]
> duplicates <- duplicated(sorted[,variables, drop=FALSE])
> index <- cumsum(!duplicates)
>
> results <- by(data, index, fun, ...)
>
> cols <- sorted[!duplicates,variables, drop=FALSE]
> cols$results <- array(results)
> cols
> }
>
>
> sort.df <- function(data, vars) {
> data[do.call("order", data[,vars, drop=FALSE]), ,drop=FALSE]
> }
>
>
> dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
> c(2,2,2,2)), value = rnorm(8))
>
> data.frame.by(dataset, c("gp1", "gp2"), function(data) mean(data$value))
> data.frame.by(dataset, "gp1", function(data) tapply(data$value, data$gp2,
> mean))
> data.frame.by(dataset, "gp1", function(data) lm(gp2 ~ value, data)) #
> doesn't print, but everything is there ok
>
> (note that the results column will be a list if necessary - this may
> be a serious abuse of data frames, but I'm not sure and no one replied
> when I queried the list)
I think this should work. Thanks!
Duncan Murdoch
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Re: [Rd] by() processing on a dataframe
And here is one more approach using the reshape package:
library(reshape)
dataset.d <- melt(dataset, id = 1:2)
cast(dataset.d, gp1 + gp2 ~ variable, mean)
On 9/30/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Check out summaryBy in the doBy package at:
>
> http://genetics.agrsci.dk/~sorenh/misc
>
> e.g.
>
> summaryBy(value ~ gp1 + gp2, data = dataset)
>
>
>
> On 9/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> > I want to calculate a statistic on a number of subgroups of a dataframe,
> > then put the results into a dataframe. (What SAS PROC MEANS does, I
> > think, though it's been years since I used it.)
> >
> > This is possible using by(), but it seems cumbersome and fragile. Is
> > there a more straightforward way than this?
> >
> > Here's a simple example showing my current strategy:
> >
> > > dataset <- data.frame(gp1 = rep(1:2, c(4,4)), gp2 = rep(1:4,
> > c(2,2,2,2)), value = rnorm(8))
> > > dataset
> > gp1 gp2 value
> > 1 1 1 0.9493232
> > 2 1 1 -0.0474712
> > 3 1 2 -0.6808021
> > 4 1 2 1.9894999
> > 5 2 3 2.0154786
> > 6 2 3 0.4333056
> > 7 2 4 -0.4746228
> > 8 2 4 0.6017522
> > >
> > > handleonegroup <- function(subset) data.frame(gp1 = subset$gp1[1],
> > + gp2 = subset$gp2[1], statistic = mean(subset$value))
> > >
> > > bylist <- by(dataset, list(dataset$gp1, dataset$gp2), handleonegroup)
> > >
> > > result <- do.call('rbind', bylist)
> > > result
> >gp1 gp2 statistic
> > 11 1 0.45092598
> > 11 1 2 0.65434890
> > 12 2 3 1.22439210
> > 13 2 4 0.06356469
> >
> > tapply() is inappropriate because I don't have all possible combinations
> > of gp1 and gp2 values, only some of them:
> >
> > > tapply(dataset$value, list(dataset$gp1, dataset$gp2), mean)
> > 1 23 4
> > 1 0.450926 0.6543489 NA NA
> > 2 NANA 1.224392 0.06356469
> >
> >
> >
> > In the real case, I only have a very sparse subset of all the
> > combinations, and tapply() and by() both die for lack of memory.
> >
> > Any suggestions on how to do what I want, without using SAS?
> >
> > Duncan Murdoch
> >
> > __
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> >
>
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Re: [Rd] Summary of translation status
Duncan Murdoch writes: > > Hi Fernando. That's a nice page. I'd add an explicit statement about > which branch the statistics apply to. You say "Statistics based on SVN: > 35706", presumably on the trunk, but soon interest will shift to the > R-2-2-patches branch. (If this is automated and you have the disk space > for both, perhaps both trunk and the current patch branch could be > listed, but I expect the statistics will be very similar.) > > Duncan Murdoch Hum that's true. I'm using the trunk branch. The process is somewhat automated, but I'd have to keep a working copy of the patch branch in order to run the status on it. Anyways I also think the statistics will probably be the same. Everytime I submit a translation I see it appearing on trunk so I think it won't matter much. -- Fernando Henrique Ferraz P. da Rosa http://www.feferraz.net __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
