Re: [R] expand.grid game
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start, calculation, end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
__
R-help@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.
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
__
R-help@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.
Re: [R] Signif. codes
No, it does not mean that the numbers have zero chance of being wrong. The extent to which the estimate can be wrong (which is a very bad and imprecise expression) is indicated by the standard error. The p-value close to zero implies that the intercept of the underlying population from which your sample was drawn is significantly different from zero with a probability that approaches certainty (One minus p2e-16). Remember that your data is assumed to be a random sample drawn from an underlying, larger population. Thus, the sample can never PERFECTLY represent the underlying population (only the underlying population itself can). However, the regression model gives you an estimate for what the data-generating process in the underlying population was (i.e., it gives you probability distributions for the true coefficients of the population, assuming that the assumptions for OLS regression are met). So, given the observed mean in your sample (i.e., your data), the probability that the true mean of the intercept in the underlying population is zero approaches zero. Another way to look at this is that it would be extremely unlikely (next to impossible) to draw a random sample from the population that has a zero intercept. By the OLS assumptions, the probability density for the true intercept in the population will be distributed normally around the estimate for the intercept, with the mean equal to the estimated intercept and standard deviation equal to the standard error of the intercept. Daniel - cuncta stricte discussurus - -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of John Paul Telthorst Sent: Monday, December 21, 2009 1:13 AM To: r-help@r-project.org Subject: [R] Signif. codes My question is about the Signif. codes and the p-value, specifically, the output when I run summary(nameofregression.lm) So you get this little key: Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 And on a regression I ran, next to the intercept data, I get '***' Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 7.956520.59993 13.262 2e-16 *** day.f2 -0.043480.84843 -0.0510.959 day.f3 -0.130430.84843 -0.1540.878 day.f4 -0.217390.84843 -0.2560.798 day.f5 0.021740.84843 0.0260.980 day.f6 -0.152170.84843 -0.1790.858 day.f7 0.149860.84390 0.1780.859 Does this mean that these numbers have a 0% chance of being wrong? Is there a way to change this to the .05 level of significance? Thanks, John [[alternative HTML version deleted]] __ R-help@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.
[R] [R-pkgs] new version of randtoolbox
Dear useRs, We are happy to announce a new version of the randtoolbox* package (version 1.09) on CRAN. The package is dedicated to Random Number Generation and in a less exhaustive way to RNG tests. The package is source-controlled at r-forge as part of the Rmetrics project http://r-forge.r-project.org/projects/rmetrics/ . We invite you to take a look at the NEWS section to know the latest development http://cran.r-project.org/web/packages/randtoolbox/NEWS . Christophe Dutang and Petr Savicky * for a license issue, the package has been splitted into 2 packages: randtoolbox and rngWELL. -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@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.
Re: [R] expand.grid game
Wow!
system.time({
all = blockparts(rep(9,8),17)
print( dim(all[,all[1,]!=0])[2] ) # remove leading 0s
})
## 229713
user system elapsed
0.160 0.068 0.228
In some ways I think this is close to Hadley's suggestion, though I
didn't know how to implement it.
Thanks a lot to everybody who participated, I have learned interesting
things from a seemingly innocuous question.
Best regards,
baptiste
2009/12/21 Robin Hankin rk...@cam.ac.uk:
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start, calculation,
end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
__
R-help@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.
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
__
R-help@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.
Re: [R] expand.grid game
Hello again everybody.
I fired off my reply before reading the correspondence about
the leading zeros.
You can also assume that there is at least one block
at the leading position, [so that position can take 0,1,2,...,8 additional
blocks] and distribute the remaining 16 blocks amongst all 8
places:
system.time(dim(blockparts(c(8,rep(9,7)),16)))
user system elapsed
0.056 0.016 0.069
this is faster! :-)
best wishes
rksh
baptiste auguie wrote:
Wow!
system.time({
all = blockparts(rep(9,8),17)
print( dim(all[,all[1,]!=0])[2] ) # remove leading 0s
})
## 229713
user system elapsed
0.160 0.068 0.228
In some ways I think this is close to Hadley's suggestion, though I
didn't know how to implement it.
Thanks a lot to everybody who participated, I have learned interesting
things from a seemingly innocuous question.
Best regards,
baptiste
2009/12/21 Robin Hankin rk...@cam.ac.uk:
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start, calculation,
end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
__
R-help@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.
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
__
R-help@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.
Re: [R] expand.grid game
I wonder whether this answers Baptiste's question as asked.
1: An 8-digit number can have some digits equal to 0;
see Baptiste's comment maxi - 9 # digits from 0 to 9
2: According to the man-page fror blockparts in partitions,
all sets of a=(a1,...,an) satisfying Sum[ai] = n subject
to 0 ai = yi are given in lexicographical order.
So it would seem that blockparts would not count 8-digit numbers
which have some zero digits.
One could presumably fake it by looping over the number of
non-zero digits, from 2 to 8 -- something like:
all - 0
for(i in (2:8)){
jj - blockparts(rep(9,8),17)
all - all + dim(jj)
}
Or am I missing something?!
Ted.
On 21-Dec-09 07:57:32, Robin Hankin wrote:
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up
with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic
case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start,
calculation, end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 21-Dec-09 Time: 08:45:09
-- XFMail --
__
R-help@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.
Re: [R] expand.grid game
OOPS!! See correction below!
On 21-Dec-09 08:45:13, Ted Harding wrote:
I wonder whether this answers Baptiste's question as asked.
1: An 8-digit number can have some digits equal to 0;
see Baptiste's comment maxi - 9 # digits from 0 to 9
2: According to the man-page fror blockparts in partitions,
all sets of a=(a1,...,an) satisfying Sum[ai] = n subject
to 0 ai = yi are given in lexicographical order.
So it would seem that blockparts would not count 8-digit numbers
which have some zero digits.
One could presumably fake it by looping over the number of
non-zero digits, from 2 to 8 -- something like:
all - 0
for(i in (2:8)){
jj - blockparts(rep(9,8),17)
jj - blockparts(rep(9,i),17)
all - all + dim(jj)
}
Or am I missing something?!
Ted.
On 21-Dec-09 07:57:32, Robin Hankin wrote:
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not
sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up
with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic
case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start,
calculation, end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 21-Dec-09 Time: 08:45:09
-- XFMail --
__
R-help@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.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 21-Dec-09 Time: 08:49:59
-- XFMail --
__
R-help@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.
Re: [R] expand.grid game
Hi Ted.
you've found a bug in the documentation for blockparts().
It should read 0 = ai = yi. I'll fix it before the next
major release (which will include sampling without replacement from
a multiset, Insha'Allah).
Best wishes
rksh
(Ted Harding) wrote:
I wonder whether this answers Baptiste's question as asked.
1: An 8-digit number can have some digits equal to 0;
see Baptiste's comment maxi - 9 # digits from 0 to 9
2: According to the man-page fror blockparts in partitions,
all sets of a=(a1,...,an) satisfying Sum[ai] = n subject
to 0 ai = yi are given in lexicographical order.
So it would seem that blockparts would not count 8-digit numbers
which have some zero digits.
One could presumably fake it by looping over the number of
non-zero digits, from 2 to 8 -- something like:
all - 0
for(i in (2:8)){
jj - blockparts(rep(9,8),17)
all - all + dim(jj)
}
Or am I missing something?!
Ted.
On 21-Dec-09 07:57:32, Robin Hankin wrote:
Hi
library(partitions)
jj - blockparts(rep(9,8),17)
dim(jj)
gives 318648
HTH
rksh
baptiste auguie wrote:
Dear list,
In a little numbers game, I've hit a performance snag and I'm not sure
how to code this in C.
The game is the following: how many 8-digit numbers have the sum of
their digits equal to 17?
The brute-force answer could be:
maxi - 9 # digits from 0 to 9
N - 5 # 8 is too large
test - 17 # for example's sake
sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
N-1 == test)
## 3675
Now, if I make N = 8, R stalls for some time and finally gives up
with:
Error: cannot allocate vector of size 343.3 Mb
I thought I could get around this using Reduce() to recursively apply
rowSum to intermediate results, but it doesn't seem to help,
a=list(1:maxi)
b=rep(list(0:maxi), N-1)
foo - function(a, b, fun=sum, ...){
switch(fun,
'sum' = rowSums(do.call(expand.grid, c(a, b))),
'mean' = rowMeans(do.call(expand.grid, c(a, b))),
apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic
case
}
sum(Reduce(foo, list(b), init=a) == test)
## 3675 # OK
Same problem with N=8.
Now, if N was fixed I could write a little C code to do this
calculation term-by-term, something along those lines,
test = 0;
for (i1=1, i1=9, i1++) {
for (i2=0, i2=9, i2++) {
[... other nested loops ]
test = test + (i1 + i2 + [...] == 17);
} [...]
}
but here the number of for loops, N, should remain a variable.
In despair I coded this in R as a wicked eval(parse()) construct, and
it does produce the expected result after an awfully long time.
makeNestedLoops - function(N=3){
startLoop - function(ii, start=1, end=9){
paste(for (i, ii, in seq(,start,, ,end,)) {\n, sep=)
}
first - startLoop(1)
inner.start - lapply(seq(2, N), startLoop, start=0)
calculation - paste(test - test + (, paste(i, seq(1, N),
sep=, collapse=+), ==17 )\n)
end - replicate(N, }\n)
code.to.run - do.call(paste, c(list(first), inner.start,
calculation, end))
cat(code.to.run)
invisible(code.to.run)
}
test - 0
eval(parse(text = makeNestedLoops(8)) )
## 229713
I hope I have missed a better way to do this in R. Otherwise, I
believe what I'm after is some kind of C or C++ macro expansion,
because the number of loops should not be hard coded.
Thanks for any tips you may have!
Best regards,
baptiste
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 21-Dec-09 Time: 08:45:09
-- XFMail --
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
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Re: [R] plotting polynomial regression line
Thanks very much it solved my problem. cheers, Amit On Sun, Dec 20, 2009 at 7:56 PM, David Winsemius dwinsem...@comcast.net wrote: On Dec 20, 2009, at 1:35 PM, Amit wrote: Dear All, I am trying to plot polynomial regression line to a scatterplot. I did following so far: x=c(1:9335) y=read.table(gp.txt,header=T,sep=\t) length(y$PCC) # y$PCC has values between 1 to 0 in decreasing order [1] 9335 plot(x,y$PCC,col=red) #scatterplot between x and y$PCC reg=lm(y$PCC~poly(x,6)) # calculating polynomial fit with degree 6 abline(reg,col=blue) Warning message: In abline(reg, col = blue) : only using the first two of 7regression coefficients After the above warning a line is drawn in the graph parallel to the y-axis. But I was expecting a curve line through the scatterplot. Am I doing something wrong? Please help! abline is designed to draw lines of the form y=a+bx, ... hence the name. (And as documented on its help page for regression objects). An effective method would be with lines and predict: plot(cars, main=Stopping Distance versus Speed) lines(cars$speed, predict(lm(dist~poly(speed, 6) ,data=cars) ), col=red) Best Amit ___ David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@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.
Re: [R] Problems in installing MCMCglmm package
Sunny Srivastava wrote: Dear R-Helpers, I am having troubles with installing with MCMCglmm package and I get the following error with a package Matrix Warning in library(pkg, character.only = TRUE, logical.return = TRUE, lib.loc = lib.loc) : there is no package called 'Matrix' Error: package 'Matrix' could not be loaded Execution halted ERROR: lazy loading failed for package 'MCMCglmm' I tried installing the package Matrix - but it was unavailable. I am not able to understand why am I encountering this error - Any help is appreciated Start with a recent version of R. If you install R-2.10.1, Matrix should be included. If the installation fails, you need to look for the error messages why it failed. Uwe Ligges Thx, S. R version 2.8.1 (2008-12-22) x86_64-pc-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods [7] base other attached packages: [1] lattice_0.17-26 MASS_7.2-45 loaded via a namespace (and not attached): [1] grid_2.8.1 tools_2.8.1 [[alternative HTML version deleted]] __ R-help@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. __ R-help@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.
Re: [R] Reading multiple Input Files
See the FAQs, particularly
How can I save the result of each iteration in a loop into a separate
file? which applies for the other way round as well.
Uwe Ligges
Maithili Shiva wrote:
Dear R helpers,
Suppose I am dealing with no of interest rates at a time and the no of interest rates I am selecting for my analysis is random i.e. it can be 2, can be 10 or even higher. The R-code I had written (with the guidance of R helpers obviously and I am really grateful to all of you) as of now is sort of hard coding when it comes to reading interest rates as an input e.g.
## INPUT
rate_1 = read.csv('rate1_range.csv')
rate_2 = read.csv('rate2_range.csv')
rate_3 = read.csv('rate3_range.csv')
rate_4 = read.csv('rate4_range.csv')
prob_1 = read.csv('rate1_probs.csv')
prob_2 = read.csv('rate2_probs.csv')
prob_3 = read.csv('rate3_probs.csv')
prob_4 = read.csv('rate4_probs.csv')
However, since I am not sure how many interest rates I will be dealing with to
begin with, I have tried to call these inputs using a loop as follows.
In my R working directory, following files are there which are to be read as input..
rate1_range.csv
rate2_range.csv
rate3_range.csv
rate4_range.csv
rate1_probs.csv
rate2_probs.csv
rate3_probs.csv
rate4_probs.csv
My R Code
# ___
n = no_of_Interest_Rates # This 'n' will be suppllied separately can be anything.
n= 4 # As mentioned, this input will be added seperately.
for (i in 1:n)
{
rate_[i]= read.csv('rate[i]_range.csv')
prob_[i] = read.csv('rate[i]_probs.csv')
}
# End of code.
However when I excute this code (here, n = 4), I get following error.
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) :
cannot open file 'rate[i]_range.csv': No such file or directory
Obviously as usual I have written some stupid code.
Please guide
Regards
Maithili
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[R] Reading multiple Input Files
Hi
I'm not sure if I understood your problem correctly..
Perhaps you are looking for something like this?
Best regards,
Roger
Result - c()
yourPath - c:\\temp
filesInDir - dir(yourPath)
for(file in filesInDir){
fileInfo - paste(yourPath,file,sep=\\)
tmp - read.csv(fileInfo)
# assign(file,tmp)
# Result - c(Result,tmp)
}
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[R] how can generate h(u)=min{a,log(u)} for 0u1 in R ?
Hello,
How can generate a sample from h(u)=min{a,log(u)} for 0u1 in R,please?
thank you
khazaei
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Re: [R] by function ??
or if Dataset is a data.table : Dataset = data.table(Dataset) Dataset[,abs(ratio-median(ratio)),by=LEAID] LEAIDV1 [1,] 6307 0.0911905 [2,] 6307 0.0488095 [3,] 6307 0.0488095 [4,] 6307 0.1088095 [5,] 8300 0.2021538 [6,] 8300 0.000 [7,] 8300 0.060 rather than : Dataset$abs - with(Dataset, ave(ratio, LEAID, FUN=function(x)abs(x-median(x This is less code and more natural (to me anyway) e.g. it doesn't require use of function() or ave(). data.table knows that if the j expression returns a vector it should silently repeat the groups to match the length of the j result (which it is doing here). If the j expression returns a scalar you would just get 2 rows in this example. Note that the 'by' expression must evaluation to integer, or a list of integer vectors, so in this case LEAID must either be integer already or coerced to integer using by=as.integer(LEAID). To give the aggregate expression a name, just wrap with the DT function. This is also how to return multiple aggregate functions from each subset (some may return vectors, others may return vectors) by listing them inside DT() : Dataset[,DT(ratio,scaled=abs(ratio-median(ratio)),sum=sum(ratio)),by=LEAID] LEAID ratioscaled sum [1,] 6307 0.720 0.0911905 3.262381 [2,] 6307 0.7623810 0.0488095 3.262381 [3,] 6307 0.860 0.0488095 3.262381 [4,] 6307 0.920 0.1088095 3.262381 [5,] 8300 0.5678462 0.2021538 2.167846 [6,] 8300 0.770 0.000 2.167846 [7,] 8300 0.830 0.060 2.167846 William Dunlap wdun...@tibco.com wrote in message news:77eb52c6dd32ba4d87471dcd70c8d7000243c...@na-pa-vbe03.na.tibco.com... -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of L.A. Sent: Saturday, December 12, 2009 12:39 PM To: r-help@r-project.org Subject: Re: [R] by function ?? Thanks for all the help, They all worked, But I'm stuck again. I've tried searching, but I not sure how to word my search as nothing came up. Here is my new hurdle, my data has 7 abservations and my results have 2 answers: Here is my data LEAID ratio 3 6307 0.720 1 6307 0.7623810 2 6307 0.860 4 6307 0.920 5 8300 0.5678462 7 8300 0.770 6 8300 0.830 median-summaryBy(ratio ~ LEAID, data = Dataset, FUN = median) print(median) LEAID ratio.median 1 63070.8111905 2 83000.770 Now what I want is a way to compute abs(ratio- median)by LEAID for each observation to produce something like this LEAID ratio abs 3 6307 0.720 .0912 1 6307 0.7623810 .0488 2 6307 0.860 .0488 4 6307 0.920 .1088 5 8300 0.5678462 .2022 7 8300 0.770 . 6 8300 0.830 .0600 Try ave(), as in Dataset$abs - with(Dataset, ave(ratio, LEAID, FUN=function(x)abs(x-median(x Dataset LEAID ratio abs 3 6307 0.720 0.0911905 1 6307 0.7623810 0.0488095 2 6307 0.860 0.0488095 4 6307 0.920 0.1088095 5 8300 0.5678462 0.2021538 7 8300 0.770 0.000 6 8300 0.830 0.060 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com Thanks, L.A. Ista Zahn wrote: Hi, I think you want by(TestData[ , RATIO], LEAID, median) -Ista On Tue, Dec 8, 2009 at 8:36 PM, L.A. ro...@millect.com wrote: I'm just learning and this is probably very simple, but I'm stuck. I'm trying to understand the by(). This works. by(TestData, LEAID, summary) But, This doesn't. by(TestData, LEAID, median(RATIO)) ERROR: could not find function FUN HELP! Thanks, LA -- View this message in context: http://n4.nabble.com/by-function-tp955789p955789.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@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. -- View this message in context: http://n4.nabble.com/by-function-tp955789p962666.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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,
Re: [R] how can generate h(u)=min{a,log(u)} for 0u1 in R ?
Hi, see the example below. There must be a way to do this with apply or
tapply, but it always returned an error incorrect number of dimensions. At
least the code below works
n=100
a=rnorm(n)
u=runif(n)
f=function(x){min(x[,1],log(x[,2]))}
x=data.frame(a,u)
apply(x,1,f) #does not work
#this works
y=NULL
for(i in 1:n) y[i]=f(x[i,])
HTH,
Daniel
-
cuncta stricte discussurus
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of khaz...@ceremade.dauphine.fr
Sent: Monday, December 21, 2009 5:17 AM
To: r-help@r-project.org
Subject: [R] how can generate h(u)=min{a,log(u)} for 0u1 in R ?
Hello,
How can generate a sample from h(u)=min{a,log(u)} for 0u1 in R,please?
thank you
khazaei
__
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Re: [R] how can generate h(u)=min{a,log(u)} for 0u1 in R ?
Use pmin like this:
set.seed(123)
pmin(-0.1, log(runif(25)))
[1] -1.2462628 -0.2378700 -0.8940966 -0.1244104 -0.100 -3.0888020
[7] -0.6384592 -0.1138195 -0.5952313 -0.7839153 -0.100 -0.7911258
[13] -0.3892415 -0.5575096 -2.2737578 -0.100 -1.4020672 -3.1686692
[19] -1.1149834 -0.100 -0.1170516 -0.3670090 -0.4454955 -0.100
[25] -0.4220431
On Mon, Dec 21, 2009 at 5:16 AM, khaz...@ceremade.dauphine.fr wrote:
Hello,
How can generate a sample from h(u)=min{a,log(u)} for 0u1 in R,please?
thank you
khazaei
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[R] Playing with rgl: a Youtube video
I've just posted a demo made with the rgl package to Youtube, visible here: http://www.youtube.com/watch?v=prdZWQD7L5c For future reference, here are the steps I used: 1. Design a shape to be displayed, and then play with the animation functions to make it change over time. Use play3d to do it live in R, movie3d to write the individual frames of the movie to .png files. 2. Use the ffmpeg package (not an R package, a separate project at http://ffmpeg.org) to convert the .png files to an .mp4 file. The individual frames totalled about 1 GB; the compressed movie is about 45 MB. 3. Upload to Youtube. I'm not a musician, so I had to use one of their licensed background tracks, I couldn't write my own. I spent a lot of time picking one and then adjusting the timing of the video to compensate. Each render/upload cycle at full resolution took about an hour and a half. It's a lot faster to render in a smaller window with fewer frames per second, but it's still tedious. It's easier to synchronize if you actually have a copy of the music locally, but Youtube doesn't let you download their music. So the timing isn't perfect, but it's good enough for me! Duncan Murdoch __ R-help@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.
Re: [R] Which hist cell each value falls in?
cut() might work if you want a character version...
set.seed(221)
x-rnorm(50)
y-rnorm(1)
xh-hist(x)
cut(y, breaks=xh$breaks)
#or
as.character(cut(y, breaks=xh$breaks))
#or for a 'level number' version (a bit like which())
as.numeric(cut(y, breaks=xh$breaks))
Jim Lemon j...@bitwrit.com.au 18/12/2009 07:57:28
On 12/18/2009 06:35 AM, Doug Hill wrote:
Hi, all. I'm using hist() to obtain a vector of break values in an
interval.
I then want to be able to identify which cell any value from another
vector
falls in.
E.g. applying
breaks
[1] -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0
to
x
[1] -3.74519666 -0.38183630 -1.22884247 -0.20971824 -0.30533939
-0.36271207
[7] -2.27513499 -2.23688653 -1.98827155 -1.48666274 -1.26155084
-0.15234555
[13] -0.09497287 0.34488440
would give
xcells
[1] 1 8 6 8 8 8 4 4 5 6 6 8 8 9
where:
x= breaks[1] - cell 1
breaks[1] x= breaks[2] - cell 2
breaks[2] x= breaks[3] - cell 3, etc.
Hi Doug,
The function below does more or less what you want, except that the
bins
are numbered from zero (meaning that a value is below the range of bins
as x[1] is) to length(breaks) (meaning that a value is above the range
of bins).
whichBin-function(x,breaks,right=TRUE) {
lenx-length(x)
# any leftovers must be out of range
wb-rep(lenx,lenx)
if(right) wb[x=breaks[1]]-0
else wb[xbreaks[1]]-0
for(bin in 1:(length(breaks)-1)) {
if(right) wb[xbreaks[bin] x=breaks[bin+1]]-bin
else wb[x=breaks[bin] xbreaks[bin+1]]-bin
}
return(wb)
}
Jim
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Re: [R] as.Date question
I don't know the details of why this is happening, but in any case it is a good idea to round() your times if you want to convert them to dates. Just as you would round() a numeric value to convert it to an integer. as.Date(round(zzz1, days)) 2009/12/21 Jason Morgan jwm-r-h...@skepsi.net: Hello, On 2009.12.20 18:06:17, MAL wrote: All! This piece of code: zzz1 - as.POSIXct(1999-03-18, tz=CET) zzz2 - as.POSIXlt(1999-03-18, tz=CET) zzz1 == zzz2 as.Date(zzz1) as.Date(zzz2) yields TRUE for zzz1==zzz2, but the two dates returned by as.Date are different: as.Date(zzz1) [1] 1999-03-17 as.Date(zzz2) [1] 1999-03-18 I'm using R 2.10.0. Would be glad for any clarifications. Thanks! I don't know why as.Date() is giving different results, but if look at the value of the variables, they are equal: zzz1 - as.POSIXct(1999-03-18, tz=CET) zzz2 - as.POSIXlt(1999-03-18, tz=CET) zzz1 == zzz2 [1] TRUE as.Date(zzz1) [1] 1999-03-17 as.Date(zzz2) [1] 1999-03-18 zzz1 [1] 1999-03-18 CET zzz2 [1] 1999-03-18 CET Maybe someone here can explain the behavior of as.Date(). Cheers, ~Jason -- Jason W. Morgan Graduate Student Department of Political Science *The Ohio State University* 154 North Oval Mall Columbus, Ohio 43210 __ R-help@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. -- Felix Andrews / 安福立 Postdoctoral Fellow Integrated Catchment Assessment and Management (iCAM) Centre Fenner School of Environment and Society [Bldg 48a] The Australian National University Canberra ACT 0200 Australia M: +61 410 400 963 T: + 61 2 6125 4670 E: felix.andr...@anu.edu.au CRICOS Provider No. 00120C -- http://www.neurofractal.org/felix/ __ R-help@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.
Re: [R] Object of type 'closure' not subsettable
Thanks Barry for the clarification. With regards to the following; d2[[i]] - file[[i]] - mean renamed to d2[[i]] - f[[i]] - m The object f contains the following so clearly f is subsettable [[1]] V1 1 10 2 10 3 10 [[2]] V1 1 11 2 11 3 11 [[3]] V1 1 12 2 12 3 12 My plan is to subtract m from each subset of f and to store the output as each individual subsets. So, output1 - f[[1]] - m output2 - f[[2]] - m output3 - f[[3]] - m If I run the following to achieve the desired result, there is an error as written in the subject heading. d2[[i]] - f[[i]] - m If I run the following, the error is gone but I'm not getting the output for each individual file I require d2 - f[[i]] - m The issue is, how do I make d2 subsettable? Thanks. Muhammad Muhammad Rahiz | Doctoral Student in Regional Climate Modeling Climate Research Laboratory, School of Geography the Environment Oxford University Centre for the Environment South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194 Mobile: +44 (0)7854-625974 Email: muhammad.ra...@ouce.ox.ac.uk Barry Rowlingson wrote: On Sun, Dec 20, 2009 at 7:40 PM, Muhammad Rahiz muhammad.ra...@ouce.ox.ac.uk wrote: Hi all, How can I overcome the error object of type 'closure' not subsettable I ran the following script seq - paste(seq(1914, 1916, by=1), *.y, sep=.) # make sequence c - 3 # total number of files d2 - file # creates dummy file No it doesn't. It copies the object called 'file' into an object called 'd2'. What's the object called 'file'? If you've not created one already, its the 'file' function that R uses to read stuff from files. So when you do: d2[[i]] - file[[i]] - mean you are trying to subset from d2 (and from 'file'). If I do this: file[[2]] I get your error message: Error in file[[2]] : object of type 'closure' is not subsettable So clearly you aren't doing what you think you're doing. Some hints: 1. Read a good introduction to R. You are making a number of fundamental mistakes here. 2. Run each line separately and check what value you get back by printing it. 3. Don't give your objects the same name as R functions (you're using 'seq', 'file', and 'mean'). Although this may work, it will confuse people later... Barry __ R-help@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.
Re: [R] Object of type 'closure' not subsettable
On Mon, Dec 21, 2009 at 1:43 PM, Muhammad Rahiz muhammad.ra...@ouce.ox.ac.uk wrote: Thanks Barry for the clarification. With regards to the following; d2[[i]] - file[[i]] - mean renamed to d2[[i]] - f[[i]] - m The object f contains the following so clearly f is subsettable [[1]] V1 1 10 2 10 3 10 [[2]] V1 1 11 2 11 3 11 [[3]] V1 1 12 2 12 3 12 My plan is to subtract m from each subset of f and to store the output as each individual subsets. So, output1 - f[[1]] - m output2 - f[[2]] - m output3 - f[[3]] - m If I run the following to achieve the desired result, there is an error as written in the subject heading. d2[[i]] - f[[i]] - m If I run the following, the error is gone but I'm not getting the output for each individual file I require d2 - f[[i]] - m The issue is, how do I make d2 subsettable? Well you haven't shown us this time how you initialised d2. Create it as an empty list: d2 - list() and then you can do: d2[[1]]=c(1,2,3) d2[[2]]=c(4,5,6) d2 [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 As I said, read one of the basic R documents linked on the documentation section of the R web site, and play with lists and vectors for a while. Barry __ R-help@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.
Re: [R] Object of type 'closure' not subsettable
Hi You were instucted to 1. Read a good introduction to R. You are making a number of fundamental mistakes here. 2. Run each line separately and check what value you get back by printing it. 3. Don't give your objects the same name as R functions (you're using 'seq', 'file', and 'mean'). Although this may work, it will confuse people later... which you clearly did not do. r-help-boun...@r-project.org napsal dne 21.12.2009 14:43:44: Thanks Barry for the clarification. With regards to the following; d2[[i]] - file[[i]] - mean renamed to d2[[i]] - f[[i]] - m What is d2? What does str(d2) tell about the object. for everybody except you the error message is d2[[i]] - f[[i]] - m Error: object 'f' not found So what do you expect? Barry clearly explained that d2 is probably a function based on what you explained that you did. d2-file d2[[1]] Error in d2[[1]] : object of type 'closure' is not subsettable Besides of reading a tiny part of R intro docs especially about objects and how to manipulate them you could consult also apply family of functions (lapply, sapply, ...) and also ave which probably do what you want without twisting R to C programming. ave(some.data, some.levels, FUN=function(x) x-mean(x)) shall do what you want Regards Petr The object f contains the following so clearly f is subsettable [[1]] V1 1 10 2 10 3 10 [[2]] V1 1 11 2 11 3 11 [[3]] V1 1 12 2 12 3 12 My plan is to subtract m from each subset of f and to store the output as each individual subsets. So, output1 - f[[1]] - m output2 - f[[2]] - m output3 - f[[3]] - m If I run the following to achieve the desired result, there is an error as written in the subject heading. d2[[i]] - f[[i]] - m If I run the following, the error is gone but I'm not getting the output for each individual file I require d2 - f[[i]] - m The issue is, how do I make d2 subsettable? Thanks. Muhammad Muhammad Rahiz | Doctoral Student in Regional Climate Modeling Climate Research Laboratory, School of Geography the Environment Oxford University Centre for the Environment South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194Mobile: +44 (0)7854-625974 Email: muhammad.ra...@ouce.ox.ac.uk Barry Rowlingson wrote: On Sun, Dec 20, 2009 at 7:40 PM, Muhammad Rahiz muhammad.ra...@ouce.ox.ac.uk wrote: Hi all, How can I overcome the error object of type 'closure' not subsettable I ran the following script seq - paste(seq(1914, 1916, by=1), *.y, sep=.) # make sequence c - 3 # total number of files d2 - file # creates dummy file No it doesn't. It copies the object called 'file' into an object called 'd2'. What's the object called 'file'? If you've not created one already, its the 'file' function that R uses to read stuff from files. So when you do: d2[[i]] - file[[i]] - mean you are trying to subset from d2 (and from 'file'). If I do this: file[[2]] I get your error message: Error in file[[2]] : object of type 'closure' is not subsettable So clearly you aren't doing what you think you're doing. Some hints: 1. Read a good introduction to R. You are making a number of fundamental mistakes here. 2. Run each line separately and check what value you get back by printing it. 3. Don't give your objects the same name as R functions (you're using 'seq', 'file', and 'mean'). Although this may work, it will confuse people later... Barry __ R-help@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. __ R-help@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.
Re: [R] Playing with rgl: a Youtube video
On Mon, Dec 21, 2009 at 12:42 PM, Duncan Murdoch murd...@stats.uwo.ca wrote: I've just posted a demo made with the rgl package to Youtube, visible here: http://www.youtube.com/watch?v=prdZWQD7L5c For future reference, here are the steps I used: 1. Design a shape to be displayed, and then play with the animation functions to make it change over time. Use play3d to do it live in R, movie3d to write the individual frames of the movie to .png files. 2. Use the ffmpeg package (not an R package, a separate project at http://ffmpeg.org) to convert the .png files to an .mp4 file. The individual frames totalled about 1 GB; the compressed movie is about 45 MB. 3. Upload to Youtube. I'm not a musician, so I had to use one of their licensed background tracks, I couldn't write my own. I spent a lot of time picking one and then adjusting the timing of the video to compensate. Each render/upload cycle at full resolution took about an hour and a half. It's a lot faster to render in a smaller window with fewer frames per second, but it's still tedious. It's easier to synchronize if you actually have a copy of the music locally, but Youtube doesn't let you download their music. So the timing isn't perfect, but it's good enough for me! Very nice video, Duncan! Paul __ R-help@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.
[R] Question About Repeat Random Sampling from a Data Frame
Good Morning:
I've read many, many posts on the r-help system and I feel compelled to quickly
admit that I am relatively new to R, I do have several reference books around
me, but I cannot count myself among the fortunate who seem to strong
programming intuition.
I have a data set consisting of 1637 observations of five variables: tensile
strength, yield strength, elongation, hardness and a character indicator with
three levels: (Y)es, (N)o, and (F)ail.
My objective is to randomly sample various subsets from this data set and then
evaluate these subsets using simple parameters among them tests for normality,
shape and skewness. The data set is ordered by the character variable prior to
sampling, and the samples are weighted to mirror representation in an overall,
physical process.
I am sampling the data set using this code:
sample - dataset[sample(1:1637, 500,
prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace =
TRUE),]
What I would like to do is iterate this process to create many (say 500 or
more) sampled sets of n=500 and then evaluate each set for the parameters of
interest. I would actually be evaluating each variable within each subset for
my characteristic of interest. I am familiar with sampling and saving single
columns of data to do this sort of thing, but I am not sure how to accomplish
this with a multiple-variable data set.
For example, I am currently iterating this using a clunky process:
mysamples-list()
for (i in 1:10){
mysamples[[i]] - dataset[
sample(1:1637,100,prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace
= TRUE), ]
}
But this leaves me with the additional task of defining each mysample[i]
iteration and converting it to a form on which I can apply a standard
statistical test like mean() or skewness() to the variable columns within each
subset. I have attempted to iteratively convert these lists using this code:
mat-matrix(nrow=100,ncol=5)
for (i in 1:length(mysamples))
{mat[i]-do.call('rbind',mysamples[i])}
but running the code generates the error message: number of items to replace is
not a multiple of replacement length. I have tried unsuccessfully, by reading
many, many helpful r-help emails on this error, to understand my probably
obvious mistake.
Based on the small amount that I think I know about R it seems to me that
sampling the data frame and containing the samples in a list is likely a pretty
inefficient way to do this task. Any help that any of you could provide to
assist me in iteratively sampling the data frame, and storing the samples in a
form on which I can apply other statistical tests would be greatly appreciated.
Thank you very much for taking the time to consider my questions.
Adam
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question About Repeat Random Sampling from a Data Frame
On Mon, Dec 21, 2009 at 4:12 PM, Adam Carr adamlc...@yahoo.com wrote:
Good Morning:
I've read many, many posts on the r-help system and I feel compelled to
quickly admit that I am relatively new to R, I do have several reference
books around me, but I cannot count myself among the fortunate who seem to
strong programming intuition.
I have a data set consisting of 1637 observations of five variables: tensile
strength, yield strength, elongation, hardness and a character indicator with
three levels: (Y)es, (N)o, and (F)ail.
My objective is to randomly sample various subsets from this data set and
then evaluate these subsets using simple parameters among them tests for
normality, shape and skewness. The data set is ordered by the character
variable prior to sampling, and the samples are weighted to mirror
representation in an overall, physical process.
I am sampling the data set using this code:
sample - dataset[sample(1:1637, 500,
prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace
= TRUE),]
What I would like to do is iterate this process to create many (say 500 or
more) sampled sets of n=500 and then evaluate each set for the parameters of
interest. I would actually be evaluating each variable within each subset for
my characteristic of interest. I am familiar with sampling and saving single
columns of data to do this sort of thing, but I am not sure how to accomplish
this with a multiple-variable data set.
For example, I am currently iterating this using a clunky process:
mysamples-list()
for (i in 1:10){
mysamples[[i]] - dataset[
sample(1:1637,100,prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace
= TRUE), ]
}
But this leaves me with the additional task of defining each mysample[i]
iteration and converting it to a form on which I can apply a standard
statistical test like mean() or skewness() to the variable columns within
each subset. I have attempted to iteratively convert these lists using this
code:
mat-matrix(nrow=100,ncol=5)
for (i in 1:length(mysamples))
{mat[i]-do.call('rbind',mysamples[i])}
but running the code generates the error message: number of items to replace
is not a multiple of replacement length. I have tried unsuccessfully, by
reading many, many helpful r-help emails on this error, to understand my
probably obvious mistake.
Based on the small amount that I think I know about R it seems to me that
sampling the data frame and containing the samples in a list is likely a
pretty inefficient way to do this task. Any help that any of you could
provide to assist me in iteratively sampling the data frame, and storing the
samples in a form on which I can apply other statistical tests would be
greatly appreciated.
Thank you very much for taking the time to consider my questions.
Adam
[[alternative HTML version deleted]]
That's pretty much how I tend to do those things. what you seem to be
missing is the ?apply family:
mysamples.means-lapply(mysamples,function(x)mean(x[,1]))
Hope that gets you on your way. If you want more help, I'd suggest
including an example data set in your follow-up messages.
/Gustaf
--
Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE
skype:gustaf_rydevik
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[R] physical distributions
Hi there, I was looking for a Lorentz-distribution in R but I didn't find any physical distribution. Is there s.th which I can use for fitting. thanks Markus __ R-help@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.
[R] Help，Suggest me some methods to identify t raining set and test set!!!
I want to split my whole dateset to training set and test set, building model in training set, and validate model using test set. Now, How can I split my dataset to them reasonally. Please give me a hand, It is better to give me some R code. and I see some ways like using SOM to project whole independents to 2-dimensions and find some to be training set and others are test set. like below. I also want to do this. and my date is in xls accessory. Please help me. and excel file is 218*47 matrix, 47 are indepents. I want to project it to 2D and label the corresponding sample label like that picture below. thank you! http://n4.nabble.com/file/n976245/SOM%2Btraining%2Bset%2Band%2Btest%2Bset.jpg SOM+training+set+and+test+set.jpg http://n4.nabble.com/file/n976245/matlab218x47.xls matlab218x47.xls -- View this message in context: http://n4.nabble.com/Help-Suggest-me-some-methods-to-identify-training-set-and-test-set-tp976245p976245.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
Re: [R] Question About Repeat Random Sampling from a Data Frame
On Dec 21, 2009, at 10:12 AM, Adam Carr wrote:
Good Morning:
I've read many, many posts on the r-help system and I feel compelled
to quickly admit that I am relatively new to R, I do have several
reference books around me, but I cannot count myself among the
fortunate who seem to strong programming intuition.
I have a data set consisting of 1637 observations of five variables:
tensile strength, yield strength, elongation, hardness and a
character indicator with three levels: (Y)es, (N)o, and (F)ail.
My objective is to randomly sample various subsets from this data
set and then evaluate these subsets using simple parameters among
them tests for normality, shape and skewness. The data set is
ordered by the character variable prior to sampling, and the samples
are weighted to mirror representation in an overall, physical process.
I am sampling the data set using this code:
sample - dataset[sample(1:1637, 500,
prob
=
c
(rep
(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace =
TRUE),]
What I would like to do is iterate this process to create many (say
500 or more) sampled sets of n=500 and then evaluate each set for
the parameters of interest. I would actually be evaluating each
variable within each subset for my characteristic of interest. I am
familiar with sampling and saving single columns of data to do this
sort of thing, but I am not sure how to accomplish this with a
multiple-variable data set.
For example, I am currently iterating this using a clunky process:
mysamples-list()
for (i in 1:10){
mysamples[[i]] -
dataset
[ sample
(1
:
1637,100
,prob
=
c
(rep
(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace =
TRUE), ]
}
Using lists to store intermediate results is not considered clunky in
R. (You might want to provide statistical justification for the
otherwise puzzling sampling strategy.)
But this leaves me with the additional task of defining each
mysample[i] iteration and converting it to a form on which I can
apply a standard statistical test like mean() or skewness() to the
variable columns within each subset. I have attempted to iteratively
convert these lists using this code:
mat-matrix(nrow=100,ncol=5)
for (i in 1:length(mysamples))
{mat[i]-do.call('rbind',mysamples[i])}
It would help if you explained what you are attempting here in
ordinary English. There are 10 elements in mysamples, each of which is
a 100 x 5 dataframe, and mat is just one 100 x 5 matrix, which you
seem to be referencing incorrectly, given the fact that it has two,
rather than one, dimension. Furthermore, those dataframes may not be
of a uniform class, since you said you had character variable. Do you
really want these all in a character type matrix, which would be what
is likely to happen given R's requirement that matrix element be of
only one class? What you say above suggests not.
but running the code generates the error message: number of items to
replace is not a multiple of replacement length.
Because of the way you are referencing the matrix, probably. If you
wanted a 10 x 100 x 5 array, then create an array. In R, as far as I
can tell anyway, matrices are necessarily of 2 dimensions. Tables and
arrays can be of higher dimension.
I have tried unsuccessfully, by reading many, many helpful r-help
emails on this error, to understand my probably obvious mistake.
Sorting out such problems is best done with smaller test objects. I
was surprised to see that you thought it was necessary to convert
dataframes to matrices in order to calculate descriptive statistics.
Nothing could be farther from the truth. Furthermore, it for some
other more valid reason you wanted a list of matrices, there is a
perfectly good function that will convert a dataframe to a matrix,
data.matrix(), remembering of course that if there is a single
character variable in the dataframe, that the entire matrix will be of
type character.
Based on the small amount that I think I know about R it seems to me
that sampling the data frame and containing the samples in a list is
likely a pretty inefficient way to do this task. Any help that any
of you could provide to assist me in iteratively sampling the data
frame, and storing the samples in a form on which I can apply other
statistical tests would be greatly appreciated.
Thank you very much for taking the time to consider my questions.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Help，Suggest me some methods to identify tr aining set and test set!!!
Not ellegant.. but... MyDF-data.frame(cbind(x=runif(10), y=rnorm(10))) TrainingSize=5 TrainingSize_list-sample(1:nrow(MyDF))[1:TrainingSize] TrainingSize_list MyDF.training-MyDF[(1:nrow(MyDF) %in% TrainingSize_list),] MyDF.training MyDF.test-MyDF[ ! (1:nrow(MyDF) %in% TrainingSize_list),] MyDF.test bests milton On Mon, Dec 21, 2009 at 9:09 AM, bbslover dlu...@yeah.net wrote: I want to split my whole dateset to training set and test set, building model in training set, and validate model using test set. Now, How can I split my dataset to them reasonally. Please give me a hand, It is better to give me some R code. and I see some ways like using SOM to project whole independents to 2-dimensions and find some to be training set and others are test set. like below. I also want to do this. and my date is in xls accessory. Please help me. and excel file is 218*47 matrix, 47 are indepents. I want to project it to 2D and label the corresponding sample label like that picture below. thank you! http://n4.nabble.com/file/n976245/SOM%2Btraining%2Bset%2Band%2Btest%2Bset.jpg SOM+training+set+and+test+set.jpg http://n4.nabble.com/file/n976245/matlab218x47.xls matlab218x47.xls -- View this message in context: http://n4.nabble.com/Help-Suggest-me-some-methods-to-identify-training-set-and-test-set-tp976245p976245.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Help，Suggest me some methods to identify tr aining set and test set!!!
Hi, On Mon, Dec 21, 2009 at 9:09 AM, bbslover dlu...@yeah.net wrote: I want to split my whole dateset to training set and test set, building model in training set, and validate model using test set. Now, How can I split my dataset to them reasonally. Please give me a hand, It is better to give me some R code. and I see some ways like using SOM to project whole independents to 2-dimensions and find some to be training set and others are test set. like below. I also want to do this. and my date is in xls accessory. Please help me. and excel file is 218*47 matrix, 47 are indepents. I want to project it to 2D and label the corresponding sample label like that picture below. I noticed Max already pointed you to the caret package. Load the library and look at the help for the createFolds function, eg: library(caret) ?createFolds -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@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.
[R] Maximum spacing method
Hi! I'm trying to compare maximum likelihood estimator with maximum spacing estimator (in Generalized extreme value distribution). My problem is the following: i know how to determinate maximum likelihood estimator on R, but i dont find a way to do the maximum spacing estimator...:-( Any help would be great:-) -- View this message in context: http://n4.nabble.com/Maximum-spacing-method-tp976335p976335.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] Suppressing certain debugging output
keywords: debug, do.call, debugging in, suppress, arguments, list,
screen, console, output
Hi,
I would like to control the output to the console while debugging
functions that are called by do.call( FCN, ARGUMENTS ) when the
arguments are quite large. Is there a simple way to suppress (or
compress) this output?
Thanks,
- Stu
For example:
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,10)))
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,100)))
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,1000)))
The first line prints the following to the console ...
debugging in: function(x) { length(x) }
(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
debug: {
length(x)
}
Browse[1]
The second prints ...
debugging in: function(x) { length(x) }
(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
debug: {
length(x)
}
Browse[1]
Etc.
Etc.
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and provide commented, minimal, self-contained, reproducible code.
[R] an IDE for R... JGR perhaps?
Hi, I've just been browsing for IDEs for R, see e.g. http://www.linuxlinks.com/article/20080413133953606/R.html It's probably highly subjective, but which free IDE is the best one around? For the Python language, I've always been using IDLE. I like the syntax highlighting and the syntax suggestions the user gets while typing a function name. Ideally, the IDE should run on any platform. Thanks! Albert-Jan Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ [[alternative HTML version deleted]] __ R-help@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.
[R] Closing files after using write
When I run the following code in a loop I get an error after about 125 times
saying too many open files...
if(inherits(atmpt, try-error)){
output - paste(yahooSymbol,\n,sep=)
write(output, file = data, append = TRUE, sep = )
} else {
I've tried various things to close the file after write opens it, and looked
through the RBase docs... but I can't find anything that succesfully closes
the file.
This is obviously trivial... any ideas
Nick
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] an IDE for R... JGR perhaps?
Hi Albert-Jan, This question gets asked a lot, in both r-help and r-devel, so you should search the mailing list archives. The answers usually favor Emacs (GNU or X) with ESS. --Gray On Mon, Dec 21, 2009 at 11:18 AM, Albert-Jan Roskam fo...@yahoo.com wrote: Hi, I've just been browsing for IDEs for R, see e.g. http://www.linuxlinks.com/article/20080413133953606/R.html It's probably highly subjective, but which free IDE is the best one around? For the Python language, I've always been using IDLE. I like the syntax highlighting and the syntax suggestions the user gets while typing a function name. Ideally, the IDE should run on any platform. Thanks! Albert-Jan Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ [[alternative HTML version deleted]] __ R-help@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. -- Gray Calhoun Assistant Professor of Economics Iowa State University __ R-help@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.
Re: [R] Suppressing certain debugging output
On Mon, 21 Dec 2009, Stuart Andrews wrote:
keywords: debug, do.call, debugging in, suppress, arguments, list, screen,
console, output
Hi,
I would like to control the output to the console while debugging functions
that are called by do.call( FCN, ARGUMENTS ) when the arguments are quite
large. Is there a simple way to suppress (or compress) this output?
Yes. From ?options
‘deparse.max.lines’: controls the number of lines used when
deparsing in ‘traceback’, ‘browser’, and upon entry to a
function whose debugging flag is set. Initially unset, and
only used if set to a positive integer.
Thanks,
- Stu
For example:
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,10)))
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,100)))
ttt = function(x) { length(x) }; debug(ttt); do.call(ttt,
list(x=rep(0,1000)))
The first line prints the following to the console ...
debugging in: function(x) { length(x) }
(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
debug: {
length(x)
}
Browse[1]
The second prints ...
debugging in: function(x) { length(x) }
(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
debug: {
length(x)
}
Browse[1]
Etc.
Etc.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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--
Brian D. Ripley, rip...@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, UKFax: +44 1865 272595__
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Re: [R] Closing files after using write
That would work... except I've got a DBConnection I don't want to close and
open constantly
On Mon, Dec 21, 2009 at 12:28 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try:
closeAllConnections()
On Mon, Dec 21, 2009 at 3:22 PM, Nick Torenvliet
nick.torenvl...@gmail.com wrote:
When I run the following code in a loop I get an error after about 125
times
saying too many open files...
if(inherits(atmpt, try-error)){
output - paste(yahooSymbol,\n,sep=)
write(output, file = data, append = TRUE, sep = )
} else {
I've tried various things to close the file after write opens it, and
looked
through the RBase docs... but I can't find anything that succesfully
closes
the file.
This is obviously trivial... any ideas
Nick
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Closing files after using write
Try:
closeAllConnections()
On Mon, Dec 21, 2009 at 3:22 PM, Nick Torenvliet
nick.torenvl...@gmail.com wrote:
When I run the following code in a loop I get an error after about 125 times
saying too many open files...
if(inherits(atmpt, try-error)){
output - paste(yahooSymbol,\n,sep=)
write(output, file = data, append = TRUE, sep = )
} else {
I've tried various things to close the file after write opens it, and looked
through the RBase docs... but I can't find anything that succesfully closes
the file.
This is obviously trivial... any ideas
Nick
[[alternative HTML version deleted]]
__
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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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help，Suggest me some methods to identify tr aining set and test set!!!
I noticed Max already pointed you to the caret package. Load the library and look at the help for the createFolds function, eg: library(caret) ?createFolds I think that the createDataPartition function in caret might work better for you. There are a number of other packages with similar functions. Max __ R-help@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.
Re: [R] Hello all, How can I get corss-validation MSE of SVM in e1071?
The caret package is so good, I am learning it, but one problem is that nearZeroVar function can be used to identify near zero–variance variables and it only identify, how can I remove those variables that were identify, because I have many zero- or near zero- ones, it is not realistic to removel it by hand, can this function can identify and removel those ones automatically? newx - x[, -nearZeroVar(x), drop = FALSE] should do it __ R-help@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.
Re: [R] Closing files after using write
Hi Nick,
I hope you're being a little disingenuous about how hard you looked
Typing ?close will pull up the right help page. I think you're
having trouble because you don't store the file object in a variable,
so there's no convenient way to refer to it when you need to close it
(or when you want to write to it...). Do three different steps: 1)
open file. 2) write to file. 3) close file; and you should be fine.
You also don't need to open the file 125 times: open it before the loop.
--Gray
On Mon, Dec 21, 2009 at 11:22 AM, Nick Torenvliet
nick.torenvl...@gmail.com wrote:
When I run the following code in a loop I get an error after about 125 times
saying too many open files...
if(inherits(atmpt, try-error)){
output - paste(yahooSymbol,\n,sep=)
write(output, file = data, append = TRUE, sep = )
} else {
I've tried various things to close the file after write opens it, and looked
through the RBase docs... but I can't find anything that succesfully closes
the file.
This is obviously trivial... any ideas
Nick
[[alternative HTML version deleted]]
__
R-help@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.
--
Gray Calhoun
Assistant Professor of Economics
Iowa State University
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] DROP and KEEP statements in R
My preferred solution would be to use the select option of the subset function. data ( iris ) names ( iris ) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species # KEEP-like way iris_subset_by_keep - subset ( iris, select = c( Sepal.Length, Species ) ) # DROP-like way iris_subset_by_drop - subset ( iris, select = c ( -Sepal.Width, -Petal.Length, -Petal.Width) ) identical ( iris_subset_by_keep , iris_subset_by_drop ) [1] TRUE names ( iris_subset_by_keep ) [1] Sepal.Length Species p.s. data NeweRa ( drop = SAS ); set Much_Better; eRic - Original message - From: Wensui Liu liuwen...@gmail.com To: sarjin...@yahoo.com Cc: r-help@r-project.org Date: Sat, 19 Dec 2009 20:19:51 -0500 Subject: Re: [R] DROP and KEEP statements in R drop example. data(iris) summary(iris) Sepal.LengthSepal.Width Petal.LengthPetal.Width Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 Median :5.800 Median :3.000 Median :4.350 Median :1.300 Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800 Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500 Species setosa:50 versicolor:50 virginica :50 iris$Species - NULL summary(iris) Sepal.LengthSepal.Width Petal.LengthPetal.Width Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 Median :5.800 Median :3.000 Median :4.350 Median :1.300 Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800 Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500 On Sat, Dec 19, 2009 at 3:21 PM, sarjin...@yahoo.com wrote: What is equivalent to DROP or KEEP statements of SAS in R? -- This message was sent on behalf of sarjin...@yahoo.com at openSubscriber.com http://www.opensubscriber.com/messages/r-help@r-project.org/topic.html __ R-help@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. -- == WenSui Liu Blog : statcompute.spaces.live.com Tough Times Never Last. But Tough People Do. - Robert Schuller == [[alternative HTML version deleted]] __ R-help@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. __ R-help@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.
Re: [R] Question About Repeat Random Sampling from a Data Frame
Good Afternoon Dr. Winsemius: You ask some very good questions and make excellent points; my responses are below. I've tried to extract your questions and provide answers just to reduce the clutter. 1. You might want to provide statistical justification for the otherwise puzzling sampling strategy. I assume you mean my overall process of random sampling from a large data set. The data set is comprised of observations collected over four years. Although the basis for sampling would make a good four-frame Dilbert cartoon if it could be condensed enough, my answer begins with the unfortunate truth that there is a great divide between the technical and marketing groups at the business where I am employed. Many powerful marketing executives, some with technical backgrounds, feel that there is something fundamentally wrong with the manufacturing process because the data generated over the long term is not approximately normally distributed. My task was to examine this set of data, trying to keep the representation of Y, N and F approximately equal in the sample when compared to the large set, to determine if any subset exhibits the holy grail-like normal distribution characteristics. I don't feel that this is statistical justification, but it is the reason why I am doing this. 2. It would help if you explained what you are attempting here in ordinary English. There are 10 elements in mysamples, each of which is a 100 x 5 dataframe, and mat is just one 100 x 5 matrix, which you seem to be referencing incorrectly, given the fact that it has two, rather than one, dimension. Furthermore, those dataframes may not be of a uniform class, since you said you had character variable. Do you really want these all in a character type matrix, which would be what is likely to happen given R's requirement that matrix element be of only one class? What you say above suggests not. It seems from your response that I incorrectly assumed that a list is not the same as a data frame. I started down this path after reading the questions and answers to a similar problem where the r-help responder suggested a two step process and said that the list must be converted to another form in order to be available for analysis. And you are absolutely correct that I do not want each sample in a character type matrix. In plain English, I hope, I am simply trying to iterate the process of removing random samples from the large data set, and then saving these samples in a format that is available for simple analysis. For example, if I remove five hundred mysample sets, each of which is composed of a 100 x 5 sample of the large data set I am interested in determining the skewness, kurtosis, mean and standard deviation of each of the four numeric variables in each of the five hundred mysample sets. 3. Sorting out such problems is best done with smaller test objects. I was surprised to see...type character. I agree. I began to do this with a small test data set but it was late last evening and I realized that I should ask for help before proceeding on what I thought might be incorrect assumptions. I clearly misunderstood that a list needed to be converted to a data frame in order to be available for analysis. Thank you for taking the time to respond. The discussion and suggestions are very helpful. Adam From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org Sent: Mon, December 21, 2009 11:23:43 AM Subject: Re: [R] Question About Repeat Random Sampling from a Data Frame On Dec 21, 2009, at 10:12 AM, Adam Carr wrote: Good Morning: I've read many, many posts on the r-help system and I feel compelled to quickly admit that I am relatively new to R, I do have several reference books around me, but I cannot count myself among the fortunate who seem to strong programming intuition. I have a data set consisting of 1637 observations of five variables: tensile strength, yield strength, elongation, hardness and a character indicator with three levels: (Y)es, (N)o, and (F)ail. My objective is to randomly sample various subsets from this data set and then evaluate these subsets using simple parameters among them tests for normality, shape and skewness. The data set is ordered by the character variable prior to sampling, and the samples are weighted to mirror representation in an overall, physical process. I am sampling the data set using this code: sample - dataset[sample(1:1637, 500, prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace = TRUE),] What I would like to do is iterate this process to create many (say 500 or more) sampled sets of n=500 and then evaluate each set for the parameters of interest. I would actually be evaluating each variable within each subset for my characteristic of interest. I am familiar with sampling and saving single columns of data to do this sort of
Re: [R] Closing files after using write
Hey all, thanks for your input... assigning the connection and then using
unlink works just fine.
Thanks
On Mon, Dec 21, 2009 at 12:33 PM, Gray Calhoun gray.calh...@gmail.comwrote:
Hi Nick,
I hope you're being a little disingenuous about how hard you looked
Typing ?close will pull up the right help page. I think you're
having trouble because you don't store the file object in a variable,
so there's no convenient way to refer to it when you need to close it
(or when you want to write to it...). Do three different steps: 1)
open file. 2) write to file. 3) close file; and you should be fine.
You also don't need to open the file 125 times: open it before the loop.
--Gray
On Mon, Dec 21, 2009 at 11:22 AM, Nick Torenvliet
nick.torenvl...@gmail.com wrote:
When I run the following code in a loop I get an error after about 125
times
saying too many open files...
if(inherits(atmpt, try-error)){
output - paste(yahooSymbol,\n,sep=)
write(output, file = data, append = TRUE, sep = )
} else {
I've tried various things to close the file after write opens it, and
looked
through the RBase docs... but I can't find anything that succesfully
closes
the file.
This is obviously trivial... any ideas
Nick
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Gray Calhoun
Assistant Professor of Economics
Iowa State University
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] by function ??
Well, I'm back again. Thanks for all the help. Besides working, it's helping me begin to understand how these functions work. I still have trouble reading or following the process of a function, which brings my next question: Dataset: LEAID ratio 3 6307 0.720 1 6307 0.7623810 2 6307 0.860 4 6307 0.920 5 8300 0.5678462 7 8300 0.770 6 8300 0.830 Here I want to create confidence intervals around the median. My hang-up now is how to do pbinom() by LEAID. Here's my script. Dataset$obs - with(Dataset, ave(ratio, LEAID, FUN=function(x)length(x))) obs -by(Dataset[ , ratio], LEAID, length) obsn -by(Dataset[ , obs], LEAID, FUN=function(x)1:max(x)) q - obsn n - obs pbinom -pbinom(q, n, .5) pbinom - with(Dataset, by(obs, LEAID, FUN=function(x)pbinom(q, n, .5))) x - pbinom x.nolag=x[2:n] x.lag1=x[1:(obs-1)] my.new=cbind(x.nolag,x.lag1) x.fill= cbind(0,0) my.dat.fr=rbind(x.fill, my.new) my.dat.fr pr=1 - (2 * my.dat.fr) cil=ifelse(pr = .95, Dataset[,ratio], 0) cil max=sort(cil) LowerCI=round(max(max),digits=2) LowerCI cn=ifelse(pr = .95, Dataset[,ObsNumber], 0) cmax=sort(cn) c=max(cmax) prb=ifelse(pr = .95, pr, 999) con=sort(prb) Confidence=min(con) o= n -( c - 1) ch=ifelse (q = o , Dataset[,Ratio], 0) UpperCI=round(max(ch),digits=2) UpperCI I hope your not growing tired of my questions, I try reading all I can find, but I often can't seem understand it until I can actually see it in process. Thanks, L.A. -- View this message in context: http://n4.nabble.com/by-function-tp955789p976414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] mgcv with discrete covariates, interaction term
I'm trying to build a model with an overall smooth function for all the data, plus an additional smooth function for *some* of the data. If ind is my indicator variable (0 for some x, 1 for others), I imagined I could write it like this: gam.obj - gam(y ~ s(x) + ind*s(x)) but this does not work. Does mgcv have a way of doing this? If not directly, is there a reasonable workaround? __ R-help@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.
[R] Object of type 'closure' not subsettable - OK
Thanks Petr, Barry David for your useful comments. Given d2[[i]] - f[[i]] - m The problem lies in how I define the object, d2[[i]], which I managed to resolve by d2 - list() -- Muhammad Rahiz | Doctoral Student in Regional Climate Modeling Climate Research Laboratory, School of Geography the Environment Oxford University Centre for the Environment South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194 Mobile: +44 (0)7854-625974 Email: muhammad.ra...@ouce.ox.ac.uk __ R-help@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.
Re: [R] physical distributions
Is the Lorentz distribution another name for the Cauchy distribution
with density
f(x) = \frac{1}{\pi} \frac{1}{1+x^2}
possibly with location and scale parameters?
albyn
On Mon, Dec 21, 2009 at 05:04:05PM +0100, Markus Häge wrote:
Hi there,
I was looking for a Lorentz-distribution in R but I didn't find any
physical distribution. Is there s.th which I can use for fitting.
thanks
Markus
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[R] qq plot for 3-parameter lognormal distribution
Hi: Could you please guide me how to do a qq plot for a 3-parameter lognormal curve. Here is an example from VGAM package to fit lognormal curve. I need to do a qq plot to check the fit. Thanks. lambda = 4 y = lambda + rlnorm(n - 1000, mean=1.5, sd=exp(-0.8)) fit = vglm(y ~ 1, lognormal3, trace=TRUE) Coef(fit) summary(fit) reez __ [[elided Yahoo spam]] __ R-help@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.
Re: [R] Reading multiple Input Files
The easiest way to do this is to use an lapply:
#in working directory
filelist - list.files()
ranges - lapply(filelist[grep(.*range\\.csv, filelist)], read.csv)
probs - lapply(filelist[grep(.*probs\\.csv, filelist)], read.csv)
Hope this helps,
Greg
On 12/20/09 10:24 PM, Maithili Shiva wrote:
Dear R helpers,
Suppose I am dealing with no of interest rates at a time and the no of
interest rates I am selecting for my analysis is random i.e. it can be 2, can
be 10 or even higher. The R-code I had written (with the guidance of R
helpers obviously and I am really grateful to all of you) as of now is sort
of hard coding when it comes to reading interest rates as an input e.g.
## INPUT
rate_1 = read.csv('rate1_range.csv')
rate_2 = read.csv('rate2_range.csv')
rate_3 = read.csv('rate3_range.csv')
rate_4 = read.csv('rate4_range.csv')
prob_1 = read.csv('rate1_probs.csv')
prob_2 = read.csv('rate2_probs.csv')
prob_3 = read.csv('rate3_probs.csv')
prob_4 = read.csv('rate4_probs.csv')
However, since I am not sure how many interest rates I will be dealing with
to begin with, I have tried to call these inputs using a loop as follows.
In my R working directory, following files are there which are to be read as
input..
rate1_range.csv
rate2_range.csv
rate3_range.csv
rate4_range.csv
rate1_probs.csv
rate2_probs.csv
rate3_probs.csv
rate4_probs.csv
My R Code
# ___
n = no_of_Interest_Rates # This 'n' will be suppllied separately can
be anything.
n= 4 # As mentioned, this input will be added seperately.
for (i in 1:n)
{
rate_[i]= read.csv('rate[i]_range.csv')
prob_[i] = read.csv('rate[i]_probs.csv')
}
# End of code.
However when I excute this code (here, n = 4), I get following error.
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) :
cannot open file 'rate[i]_range.csv': No such file or directory
Obviously as usual I have written some stupid code.
Please guide
Regards
Maithili
The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.
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Re: [R] How to put text outside an xyplot?
Hi
Marius Hofert wrote:
Dear R-users,
I have a plot created with the code below. I tried to put some text
to the right of the color key. I worked with grid.text and also
viewport(), but couldn't achieve this. I know this should be simple,
but I just couldn't figure out how to do it. How does this work?
The output in the panel is clipped to the panel. You could get around
that by pushing a viewport that turns the clipping off ...
pushViewport(viewport(clip=off))
grid.text(text to the right of the color key,
1.2,0.5,rot=90)
popViewport()
... (though that way you lose access to the axis coordinates in the
panel) or, because lattice generates a version of the panel viewport
with clipping already turned off, draw the plot first then revisit the
panel viewport with the clipping off ...
xyplot(data[,2]~data[,1],aspect=1,
panel=function(...){
panel.xyplot(...,type=n)
lpoints(data[,1],data[,2],pch=1,col=colorvec)
},
xlab=x,ylab=y,
legend=list(right=list(fun=draw.colorkey,args=list(key=list(col=mygray(10),at=1:10
)
trellis.focus(panel, 1, 1, clip.off=TRUE)
grid.text(text to the right of the color key,
1.2, 0.5, rot=90)
trellis.unfocus()
... either way, there is still a possibility you won't see anything if
that x=1.2 for the text is off the right-hand side of the page, so you
might want to use a different coordinate system, or just draw the text
relative to the right-hand side of the page rather than relative to the
panel ...
xyplot(data[,2]~data[,1],aspect=1,
panel=function(...){
panel.xyplot(...,type=n)
lpoints(data[,1],data[,2],pch=1,col=colorvec)
},
xlab=x,ylab=y,
legend=list(right=list(fun=draw.colorkey,args=list(key=list(col=mygray(10),at=1:10
)
grid.text(text to the right of the color key,
x=unit(1, npc) - unit(1, mm),
just=bottom, rot=90)
Hope that helps.
Paul
Cheers,
Marius
library(lattice) library(grid) myvec=1:10
data=data.frame(x=myvec,y=myvec) mygray=function(l)
gray(seq(0.2,0.8,length=l))
colors=level.colors(data[1:10,2],at=do.breaks(c(1,10),10),col.regions=mygray)
colorvec=colors[1:10]
#trellis.device(postscript,color=F,horizontal=F,onefile=F)
xyplot(data[,2]~data[,1],aspect=1, panel=function(...){
panel.xyplot(...,type=n)
lpoints(data[,1],data[,2],pch=1,col=colorvec) grid.text(text to the
right of the color key,1.2,0.5,rot=90)#does not work. },
xlab=x,ylab=y,
legend=list(right=list(fun=draw.colorkey,args=list(key=list(col=mygray(10),at=1:10
) #dev.off()
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--
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Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
p...@stat.auckland.ac.nz
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[R] plotting
Hi: Could you please guide me how to plot weekly.t2. Below are my values and I am only able to plot weekly t1 but failing to add a second line representng weekly t2. Thanks. weekly.t1 [1] 228.5204 326.1224 387.2449 415.4082 443.6735 454.6939 451.5306 475.1020 483.2653 483.0612 488.9362 488.3673 493.4694 515.3061 509.7959 [16] 501.0204 513.5714 516.5306 509.4898 516.4286 509.7959 498.6735 492.9592 473.6735 487.7551 482.9293 weekly.t2 [1] 71.57143 95.41667 110.11905 120.65476 124.3 130.11905 132.44048 137.38095 141.60714 135.35714 131.89873 126.7 125.47619 [14] 133.57143 134.34524 137.38095 134.04762 144.40476 148.27381 147.08333 146.42857 137.5 127.14286 127.20238 130.41667 127.67857 plot(c(1:26), weekly.t1, type=l, xlab=Week since treatment start, ylab= Daily Treatment Dose, main=RIOTT doses, col=blue, axes = FALSE) points(c(1:26), weekly.t2, type=l, col=red) axis(side=1, at= c(1:26), as.character(c(1:26))) axis(side=2, at = seq(from=70, to=700, by=10)) -- View this message in context: http://n4.nabble.com/plotting-tp976461p976461.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] Inter-observer reliability, nested categorical data
Hi, I am having problems working out what to do with my data, I have a series of positions (26 for each limb!), nested within a limb (10 of them), nested within a posture (30 of them). I then have 5 observers rating themso far I have tried lmer and glm but R doesn't want to play. I'm not sure if I'm trying to do the right thing, or if it's my data it doesn't like! Any help with this problem would be much appreciated, just a point in the right direction, or another study that has had to deal with this kind of data! Thanks, Julia. -- View this message in context: http://n4.nabble.com/Inter-observer-reliability-nested-categorical-data-tp976392p976392.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
Re: [R] plotting
Hi teo, try lines() instead of points(). HTH Stephan teo schrieb: Hi: Could you please guide me how to plot weekly.t2. Below are my values and I am only able to plot weekly t1 but failing to add a second line representng weekly t2. Thanks. weekly.t1 [1] 228.5204 326.1224 387.2449 415.4082 443.6735 454.6939 451.5306 475.1020 483.2653 483.0612 488.9362 488.3673 493.4694 515.3061 509.7959 [16] 501.0204 513.5714 516.5306 509.4898 516.4286 509.7959 498.6735 492.9592 473.6735 487.7551 482.9293 weekly.t2 [1] 71.57143 95.41667 110.11905 120.65476 124.3 130.11905 132.44048 137.38095 141.60714 135.35714 131.89873 126.7 125.47619 [14] 133.57143 134.34524 137.38095 134.04762 144.40476 148.27381 147.08333 146.42857 137.5 127.14286 127.20238 130.41667 127.67857 plot(c(1:26), weekly.t1, type=l, xlab=Week since treatment start, ylab= Daily Treatment Dose, main=RIOTT doses, col=blue, axes = FALSE) points(c(1:26), weekly.t2, type=l, col=red) axis(side=1, at= c(1:26), as.character(c(1:26))) axis(side=2, at = seq(from=70, to=700, by=10)) __ R-help@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.
[R] Plot odds ratios on log scale
Hi, I have the following table of odds ratios (or), lower limits(ll) and upper limits(ul), which I would like to plot as horizontal lines beginning at the lower limit, ending at the upper limits and with a dot at the odds ratio on an x-axis on a log10 scale. The y axis would be the study sites. From what I can figure out, it looks like the plotCI function will do everything except give me an x-axis that is on a log10 scale and I can't get the logaxis function in the log10 package to work. Study or ll ul order UCSF0.7 0.550.891 MDA 0.760.710.932 UK 0.680.510.893 Mayo0.5 0.280.874 Thanks for any suggestions! Terri [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Help,Suggest me some methods to identify t raining set and test set!!!
Ot should be noted that the performance of single split into training + test does not perform satisfactorily unless N 10,000 in many cases. Frank Max Kuhn wrote: I noticed Max already pointed you to the caret package. Load the library and look at the help for the createFolds function, eg: library(caret) ?createFolds I think that the createDataPartition function in caret might work better for you. There are a number of other packages with similar functions. Max __ R-help@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. -- Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University __ R-help@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.
Re: [R] Plot odds ratios on log scale
On Dec 21, 2009, at 2:54 PM, Rice, Terri wrote: Hi, I have the following table of odds ratios (or), lower limits(ll) and upper limits(ul), which I would like to plot as horizontal lines beginning at the lower limit, ending at the upper limits and with a dot at the odds ratio on an x-axis on a log10 scale. The y axis would be the study sites. From what I can figure out, it looks like the plotCI function will do everything except give me an x-axis that is on a log10 scale and I can't get the logaxis function in the log10 package to work. Study or ll ul order UCSF0.7 0.550.891 MDA 0.760.710.932 UK 0.680.510.893 Mayo0.5 0.280.874 Thanks for any suggestions! Terri Terri, You can build your own easily, using plot() and then either arrows() or segments() to create the CI boundary lines. Just use 'log = x' in the call to plot to create the log scaled x axis. Presuming that your data above are contained in a data frame called 'DF': DF Study or ll ul order 1 UCSF 0.70 0.55 0.89 1 2 MDA 0.76 0.71 0.93 2 3UK 0.68 0.51 0.89 3 4 Mayo 0.50 0.28 0.87 4 # Get the range of values for the x axis xlim - with(DF, range(or - ll, or + ul)) xlim [1] 0.05 1.69 # Plot the points. set the range of the x axis and set to log10 scale plot(order ~ or, data = DF, xlim = xlim, pch = 19, log = x) # Add the CI's with(DF, arrows(or - ll, order, or + ul, order, code = 3, angle = 90)) See ?arrows for help on the options for formatting the lines. Alternatively, since it appears you are doing a meta-analysis of sorts, you might want to look at the metaplot() and forestplot() functions in Thomas Lumley's 'rmeta' package on CRAN. HTH, Marc Schwartz __ R-help@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.
Re: [R] StructTS standard errors
Sorry for asking again, but I did not receive any answers - I know it's a busy time... ;-) Shell I conclude that it is not possible to obtain standard errors from StructTS() or its output? Thanks, Giovanni Petris - Original Message - From: Giovanni Petris gpet...@uark.edu Date: Thursday, December 17, 2009 10:13 am Subject: [R] StructTS standard errors Hello, Does anybody know if (and how) it is possible to obtain standard errors of estimated variances from StructTS? (R 2.10.0). Thank you in advance, Giovanni __ R-help@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.htmland provide commented, minimal, self-contained, reproducible code. __ R-help@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.
Re: [R] Maximum spacing method
mxm.margarida wrote: Hi! I'm trying to compare maximum likelihood estimator with maximum spacing estimator (in Generalized extreme value distribution). My problem is the following: i know how to determinate maximum likelihood estimator on R, but i dont find a way to do the maximum spacing estimator...:-( Any help would be great:-) You can see an implementation of mps for the normal inverse Gaussian in fBasics on R-Forge. Have a look at the function .nigFit.mps in the file dist-nigFit.R David Scott -- _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 Director of Consulting, Department of Statistics __ R-help@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.
Re: [R] StructTS standard errors
Alternate hypotheses: HA1: There is no one in the readership who is familiar with the function (unlikely), or among those lacking knowledge of what package it might reside in (such as myself) chose not to look up what you might have provided on the first posting: a self-contained example to work with. HA2: Your question also suggests a certain statistical confusion. Do your really want standard errors of estimated variances? That would be somewhat unusual. Or do you want to take the square-roots of the variances to get standard errors of the estimated coefficients? On Dec 21, 2009, at 4:26 PM, Giovanni Petris wrote: Sorry for asking again, but I did not receive any answers - I know it's a busy time... ;-) Shell I conclude that it is not possible to obtain standard errors from StructTS() or its output? Thanks, Giovanni Petris - Original Message - From: Giovanni Petris gpet...@uark.edu Date: Thursday, December 17, 2009 10:13 am Subject: [R] StructTS standard errors Hello, Does anybody know if (and how) it is possible to obtain standard errors of estimated variances from StructTS? (R 2.10.0). Thank you in advance, Giovanni -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@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.
Re: [R] Plot odds ratios on log scale
On 21-Dec-09 21:19:27, Marc Schwartz wrote:
On Dec 21, 2009, at 2:54 PM, Rice, Terri wrote:
Hi,
I have the following table of odds ratios (or), lower limits(ll) and
upper limits(ul), which I would like to plot as horizontal lines
beginning at the lower limit, ending at the upper limits and with a
dot at the odds ratio on an x-axis on a log10 scale. The y axis
would be the study sites.
From what I can figure out, it looks like the plotCI function will
do everything except give me an x-axis that is on a log10 scale and
I can't get the logaxis function in the log10 package to work.
Study or ll ul order
UCSF0.7 0.550.891
MDA 0.760.710.932
UK 0.680.510.893
Mayo0.5 0.280.874
Thanks for any suggestions!
Terri
Terri,
You can build your own easily, using plot() and then either arrows()
or segments() to create the CI boundary lines. Just use 'log = x' in
the call to plot to create the log scaled x axis.
Presuming that your data above are contained in a data frame called
'DF':
DF
Study or ll ul order
1 UCSF 0.70 0.55 0.89 1
2 MDA 0.76 0.71 0.93 2
3UK 0.68 0.51 0.89 3
4 Mayo 0.50 0.28 0.87 4
# Get the range of values for the x axis
xlim - with(DF, range(or - ll, or + ul))
xlim
[1] 0.05 1.69
# Plot the points. set the range of the x axis and set to log10 scale
plot(order ~ or, data = DF, xlim = xlim, pch = 19, log = x)
# Add the CI's
with(DF, arrows(or - ll, order, or + ul, order, code = 3, angle = 90))
See ?arrows for help on the options for formatting the lines.
Alternatively, since it appears you are doing a meta-analysis of
sorts, you might want to look at the metaplot() and forestplot()
functions in Thomas Lumley's 'rmeta' package on CRAN.
HTH,
Marc Schwartz
While Marc was posting the above, I was working out an example
on just those lines! It is simply an illustration of how one
can proceed. Often, it is worth while constructing one's plot
by hand, since then you can cook things exactly as you want
them (Haute Cuisine), rather than get served with what's been
prepared (Pizza Bar, even if it is a high-quality one).
Here is ths code:
## Construct the dataframe
X-rbind(
c(0.7,0.55,0.89,1),
c(0.76,0.71,0.93,2),
c(0.68,0.51,0.89,3),
c(0.5,0.28,0.87,4)
)
colnames(X)-c(OR,LL,UL,Order)
rownames(X)-c(UCSF,MDA,UK,Mayo)
Xdf-as.data.frame(X)
## Construct the plot
plot(Xdf$OR, (5-Xdf$Order), pch=20, log=x,
xlim=c(0.2,1), ylim=c(0.5,4.5),
axes=FALSE, frame.plot=TRUE,
xlab=OR with Confidence Limits, ylab=Order)
at.x - 0.2*(1:5)
axis(1,at=at.x,labels=formatC(at.x,format=fg))
for(i in (1:4)){
y - (5-Xdf$Order[i])
lines(c(Xdf$LL[i],Xdf$UL[i]),c(y,y))
text(0.2,i,labels=as.character(y),pos=4)
text(0.21,y,labels=rownames(Xdf)[i],pos=4)
}
## Especially see ?plot and ?plot.default for details
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 21-Dec-09 Time: 22:13:40
-- XFMail --
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Re: [R] Plot odds ratios on log scale
Please look at the aedotplot, in which we plot four variables: A effect, B effect, relative risk, and CI on the relative risk. It should be easily modified to plot the odds ratios. ## install.packages(HH) ## if you don't have it yet. library(aedotplot) example(aedotplot) __ R-help@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.
Re: [R] Plot odds ratios on log scale
On Dec 21, 2009, at 4:13 PM, Ted Harding wrote: On 21-Dec-09 21:19:27, Marc Schwartz wrote: On Dec 21, 2009, at 2:54 PM, Rice, Terri wrote: Hi, I have the following table of odds ratios (or), lower limits(ll) and upper limits(ul), which I would like to plot as horizontal lines beginning at the lower limit, ending at the upper limits and with a dot at the odds ratio on an x-axis on a log10 scale. The y axis would be the study sites. From what I can figure out, it looks like the plotCI function will do everything except give me an x-axis that is on a log10 scale and I can't get the logaxis function in the log10 package to work. Study or ll ul order UCSF0.7 0.550.891 MDA 0.760.710.932 UK 0.680.510.893 Mayo0.5 0.280.874 Thanks for any suggestions! Terri Terri, You can build your own easily, using plot() and then either arrows() or segments() to create the CI boundary lines. Just use 'log = x' in the call to plot to create the log scaled x axis. Presuming that your data above are contained in a data frame called 'DF': DF Study or ll ul order 1 UCSF 0.70 0.55 0.89 1 2 MDA 0.76 0.71 0.93 2 3UK 0.68 0.51 0.89 3 4 Mayo 0.50 0.28 0.87 4 # Get the range of values for the x axis xlim - with(DF, range(or - ll, or + ul)) xlim [1] 0.05 1.69 # Plot the points. set the range of the x axis and set to log10 scale plot(order ~ or, data = DF, xlim = xlim, pch = 19, log = x) # Add the CI's with(DF, arrows(or - ll, order, or + ul, order, code = 3, angle = 90)) See ?arrows for help on the options for formatting the lines. Alternatively, since it appears you are doing a meta-analysis of sorts, you might want to look at the metaplot() and forestplot() functions in Thomas Lumley's 'rmeta' package on CRAN. HTH, Marc Schwartz While Marc was posting the above, I was working out an example on just those lines! It is simply an illustration of how one can proceed. Often, it is worth while constructing one's plot by hand, since then you can cook things exactly as you want them (Haute Cuisine), rather than get served with what's been prepared (Pizza Bar, even if it is a high-quality one). snip I nominate Ted's comments above for inclusion in the fortunes package. Now...to the kitchen...that made me hungry... :-) Regards, Marc __ R-help@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.
[R] Slow survfit -- is there a faster alternative?
Using R 2.10 on Windows: I have a filtered database of 650k event observations in a data frame with 20+ variables. I'd like to be able to quickly generate estimate and plot survival curves. However the survfit and cph() functions are extremely slow. As an example: I tried results.cox-coxph(Surv(duration, success) ~ start_time + factor1+ factor2+ variable3, data=filteredData) #(took a few seconds) plot(results.cox) #(never finished in an hour) I also tried the cph() function, with similar results. Is there some easier quick-and-dirty way of producing and plotting survival curves for large data sets? I've seen some references on this list that suggest that the underlying algorithm is O(numObs * numSuccesses) and could be sped up. Has this been done? Thanks, ___ This e-mail may contain information that is confidential, privileged or otherwise protected from disclosure. If you are not an intended recipient of this e-mail, do not duplicate or redistribute it by any means. Please delete it and any attachments and notify the sender that you have received it in error. Unless specifically indicated, this e-mail is not an offer to buy or sell or a solicitation to buy or sell any securities, investment products or other financial product or service, an official confirmation of any transaction, or an official statement of Barclays. Any views or opinions presented are solely those of the author and do not necessarily represent those of Barclays. This e-mail is subject to terms available at the following link: www.barcap.com/emaildisclaimer. By messaging with Barclays you consent to the foregoing. Barclays Capital is the investment banking division of Barclays Bank PLC, a company registered in England (number 1026167) with its registered offi! ce at 1 Churchill Place, London, E14 5HP. This email may relate to or be sent from other members of the Barclays Group. __ R-help@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.
[R] Nested For loops
Dear R experts,
Might be very simple question to ask but would be insightful. As the same story
of nested for loops. following is the code that I am using to get the
autocorrelation function of the sample data. I have tried to get rid of for
loops but since I am touching R after such a long time that I need to practice
more but I need help to revive my skills. I know that apply() or sapply() would
be beneficial. I need help...
Best Wishes,
Baloo
R code
--
acf_wal - function (infile) {
data-read.table(infile)
data_value - data[,-1]
data_value_mean - mean(data_value)
data_value_square - (data_value - data_value_mean) ^ 2
square_sum-sum(data_value_square)
entry-NROW(data_value)
deno-square_sum/entry
tab1-c()
tab2-c()
ps_value - seq(0,(floor(entry/2)),1)
for(k in 0:(floor(entry/2))){
for (i in 1:(entry-k)) {
mult-(data_value[i] - data_value_mean) * (data_value [i+k] -
data_value_mean)
tab1 - c(tab1,mult)
}
auto_avg-mean(tab1)
tab1-c()
auto_corr-auto_avg/deno
tab2-c(tab2,auto_corr)
}
table_value - cbind (ps_value, tab2)
colnames(table_value) - c(#ps, acf)
outfile-unlist(strsplit(infile, split=., fixed=TRUE))[1]
write.table(table_value,file=paste(outfile,acf.dat,sep=-),row.names=FALSE,sep=\t,quote=F)
}
--
Sample data
--
1 16.0071
2 16.7966
3 17.575
4 18.1614
5 15.982
6 16.8515
7 15.6828
8 14.9652
9 14.8623
10 14.7079
Help in this regard would be highly appreciated.
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and provide commented, minimal, self-contained, reproducible code.
[R] slow survfit -- is there a better replacement?
Using R 2.10 on Windows: I have a filtered database of 650k event observations in a data frame with 20+ variables. I'd like to be able to quickly generate estimate and plot survival curves. However the survfit and cph() functions are extremely slow. As an example: I tried results.cox-coxph(Surv(duration, success) ~ start_time + factor1+ factor2+ variable3, data=filteredData) #(took a few seconds) plot(results.cox) #(never finished in an hour) I also tried the cph() function, with similar results. Is there some easier quick-and-dirty way of producing and plotting survival curves for large data sets? I've seen some references on this list that suggest that the underlying algorithm is O(numObs * numSuccesses) and could be sped up. Has this been done? Thanks, ___ This e-mail may contain information that is confidential, privileged or otherwise protected from disclosure. If you are not an intended recipient of this e-mail, do not duplicate or redistribute it by any means. Please delete it and any attachments and notify the sender that you have received it in error. Unless specifically indicated, this e-mail is not an offer to buy or sell or a solicitation to buy or sell any securities, investment products or other financial product or service, an official confirmation of any transaction, or an official statement of Barclays. Any views or opinions presented are solely those of the author and do not necessarily represent those of Barclays. This e-mail is subject to terms available at the following link: www.barcap.com/emaildisclaimer. By messaging with Barclays you consent to the foregoing. Barclays Capital is the investment banking division of Barclays Bank PLC, a company registered in England (number 1026167) with its registered offi! ce at 1 Churchill Place, London, E14 5HP. This email may relate to or be sent from other members of the Barclays Group. __ R-help@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.
Re: [R] Help，Suggest me some methods to identify tr aining set and test set!!!
Thank you for all help. It is helpful for me. Max Kuhn wrote: I noticed Max already pointed you to the caret package. Load the library and look at the help for the createFolds function, eg: library(caret) ?createFolds I think that the createDataPartition function in caret might work better for you. There are a number of other packages with similar functions. Max __ R-help@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. -- View this message in context: http://n4.nabble.com/Help-Suggest-me-some-methods-to-identify-training-set-and-test-set-tp976245p976641.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
Re: [R] Slow survfit -- is there a faster alternative?
On Dec 21, 2009, at 8:20 PM, gregory.bron...@barclayscapital.com wrote: Using R 2.10 on Windows: I have a filtered database of 650k event observations in a data frame with 20+ variables. I'd like to be able to quickly generate estimate and plot survival curves. However the survfit and cph() functions are extremely slow. As an example: I tried results.cox-coxph(Surv(duration, success) ~ start_time + factor1+ factor2+ variable3, data=filteredData) #(took a few seconds) plot(results.cox) #(never finished in an hour) Something is wrong here. I use cph (from the Design package) on datasets numbering in the millions with crossed spline terms and the plots are virtually immediate. I also tried the cph() function, with similar results. The plot.Design function needs more than just the fit as an argument, so you are not providing enough information for good advice. When I try to plot with an object produced by coxph with no further arguments, I get an error. plot(survHb) Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ What happens when you use predict or survfit to process the fit objects: ?survfit.coxph E.g.: survHb - coxph(Surv(surv.yr, death) ~ age+nsmkr +sexMF + HbPr2 + GGT, data=hisub) plot(Hb) Error: object 'Hb' not found Error in plot(Hb) : error in evaluating the argument 'x' in selecting a method for function 'plot' #this however, succeeds --- sfit-survfit(survHb) plot(sfit) So you need to supply a process form of a fit. Is there some easier quick-and-dirty way of producing and plotting survival curves for large data sets? I've seen some references on this list that suggest that the underlying algorithm is O(numObs * numSuccesses) and could be sped up. Has this been done? Thanks, David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@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.
[R] RGtk2 - retrieve ggraphics mouse coordinates during drag-and-drop event
Hi all,
I have a gtable and a ggraphic widget.
I want to drag an element from the table onto the graphic.
When the drag object is released over the ggraphic widget, I want the
mouse coordinates inside the ggraphic to be returned.
Right now I do not know how to get the mouse ccordinates of the
ggraphic widget at any given time.
How can I access the ggraphics mouse coordinates?
Below you find my sample code (windows only due to ggraphics).
TIA,
Mark
library(gWidgets)
library(RGtk2)
options(guiToolkit=RGtk2)
w = gwindow(ggraphics example)
table - gtable(1:10, container=w)
g = ggraphics(cont=w, expand=T)
size(g) - c(500,500)
Sys.sleep(1)
plot(rnorm(20), col=red)
## convert from plt coordinates to more familiar usr
pltToUsr = function(x,y) {
plt = par(plt); usr = par(usr)
c( (usr[2]-usr[1])/(plt[2]-plt[1])*(x - plt[1]) + usr[1],
(usr[4] - usr[3])/(plt[4] - plt[3])*(y - plt[3]) + usr[3])
}
addHandlerFocus(g, handler = function(h,...) {
print(c(focus g))
})
addHandlerClicked(g, handler = function(h,...) {
x - h$x; y - h$y
xStart - x; yStart - y
pressed - TRUE
print(c(pressed at:, c(x,y)))
})
da - g...@widget@widget
callbackID - gSignalConnect(da,button-release-event, function
(w,e,...) {
allocation = w$GetAllocation()
xclick = e$GetX()
yclick = e$GetY()
x = xclick/allocation$width
y = (allocation$height - yclick)/allocation$height
xyCoordsRelease - pltToUsr(x,y)
print(c(released at:, pltToUsr(x,y)))
xMove - xyCoordsRelease[1] - xStart
yMove - xyCoordsRelease[2] - yStart
print(c(xMove, round(xMove, 3), yMove, round(yMove, 3)))
plot(rnorm(20), col=red)
#print(qplot(rnorm(100), geom=histogram))
pressed - FALSE
return(TRUE)
})
adddropsource(table) #, handler = defHandlerSource)
#defHandlerSource = function(h,...) print(source)
adddropmotion(g, handler=function(h, ...) {
print(motion)
})
adddroptarget(g, targetType=object, handler = defHandlerTarget, g)
defHandlerTarget = function(h, ...){
print(h$dropdata)
}
–––
Mark Heckmann
Dipl. Wirt.-Ing. cand. Psych.
Vorstraße 93 B01
28359 Bremen
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Rcpp 0.7.0: R/C++ integration now even easier
Rcpp version 0.7.0 went onto CRAN this weekend. The key new features are
o inline support, taken from Oleg Sklyar's neat inline package and
adapted/extented to support Rcpp as well as external libraries
(see below for an example); this even works on Windows (provided you
have Rtools installed and configured);
o addition of a new simple type RcppSexp for importing or exporting
simple types directly between R and C++
o addition of a number of new examples for both these features
o code reorganisation: every class has in its own header and source file
o last but not least relicensed from LGPL (=2.1) to GPL (= 2)
My blog (http://dirk.eddelbuettel.com/blog/) has two recent posts with a bit
more details (and colour highlighting of the code below) but let's just look
at one example of using GNU GSL functions for illustrative purposes (as you
wouldn't need this to access random-number generators as R has its own).
Consider this R code snippet:
## use Rcpp to pass down a parameter for the seed, and a vector size
gslrng - '
int seed = RcppSexp(s).asInt();
int len = RcppSexp(n).asInt();
gsl_rng *r;
gsl_rng_env_setup();
std::vectordouble v(len);
r = gsl_rng_alloc (gsl_rng_default);
gsl_rng_set (r, (unsigned long) seed);
for (int i=0; ilen; i++) {
v[i] = gsl_rng_get (r);
}
gsl_rng_free(r);
return RcppSexp(v).asSexp();
'
## turn into a function that R can call
## compileargs redundant on Debian/Ubuntu as gsl headers are found anyway
funx - cfunction(signature(s=numeric, n=numeric),
gslrng,
includes=#include gsl/gsl_rng.h,
Rcpp=TRUE,
cppargs=-I/usr/include,
libargs=-lgsl -lgslcblas)
print(funx(0, 5))
Brief notes on this:
1) The character variable gslrng contains valid C++ code.
2) The call to cfunction converts this C++ code into a function that
calls it -- and this function is compiled, linked and loaded
automagically simply using hints about GSL header files and libraries.
3) The interface is explit: cfunction() is told about a function signature
with numeric variables 's' and 'n'; this is what the C++ code convert
to (C++ local) variables 'seed' and 'len' using the RcppSexp class.
4) We return a single SEXP to R which is based on the std::vectordouble
v which is also converted on the fly.
5) The resulting vector is returned and simply printed at the R level.
More examples are in the source tarball and in the R-Forge SVN archive.
Cheers, Dirk
--
Three out of two people have difficulties with fractions.
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Re: [R] insert a dot to the numbers
Thanks a lot. 2009/12/20 Henrique Dallazuanna www...@gmail.com Try this also: sapply(s, function(x)format(x, big.interval = nchar(x) - 3, big.mark = .)) On Sun, Dec 20, 2009 at 7:50 PM, rusers.sh rusers...@gmail.com wrote: Hi, Anybody can give me some hints on the following problem? s-c(110,112321) I want to insert a dot . after the third number and get the following results. 110. 112.321 Thanks a lot. -- - Jane Chang Queen's [[alternative HTML version deleted]] __ R-help@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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- - Jane Chang Queen's [[alternative HTML version deleted]] __ R-help@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.
