Re: [R] \ escape sequence and windows path
Am 20.05.2014 19:03, schrieb Duncan Murdoch:
I have no idea what you mean by a callback from utils.
clipboard {utils}
readClipboard
function (format = 1L, raw = FALSE)
.Call(C_readClipboard, format, raw)
bytecode: 0x080d6c20
environment: namespace:utils
I do not know whether it is an real callback or a normal call to an C
subroutine
https://en.wikipedia.org/wiki/Callback_%28computer_programming%29Knut
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Re: [R] \ escape sequence and windows path
Sorry https://en.wikipedia.org/wiki/Callback_%28computer_programming%29 Knut __ 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] Unsubscribe mailing list
Hi, I would like to unsubscribe from the mailing list. Thanks -- -- Martina Marbà [[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] select randomly from a list
Hi, kindly I want to select randomly and item from list of items. the list generated in a looping process. I used sample(mylist,1) it works fine. BUTsome times the list have only one item. that should be chosen in this case since there is no other one. I found that sample return different item not included in the list thanks in advance RAE [[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] Combine factorial column intp a new column
Dear R-users, I'd be very greatful if you could help me with the following as after a few tests I haven't still been able to get the right outcome. I've got this data: dd_1 - data.frame(ID = c(1,2, 3, 4, 5), Class_a = c(a,NA, a, NA, NA), Class_b = c(NA, b, b, b, b)) And I'd like to produce a new column 'CLASS': dd_2 - data.frame(ID = c(1,2, 3, 4, 5), Class_a = c(a,NA, a, NA, NA), Class_b = c(NA, b, b, b, b), CLASS = c(a, b, a-b, b, b)) Thanks a lot! __ 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] Unsubscribe mailing list
Hi, Please use the link (https://stat.ethz.ch/mailman/listinfo/r-help) to unsubscribe. A.K. On Thursday, May 22, 2014 3:39 AM, Martina Marbà frees...@gmail.com wrote: Hi, I would like to unsubscribe from the mailing list. Thanks -- -- Martina Marbà [[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] Combine factorial column intp a new column
Hi, May be this helps: ddNew - transform(dd_1, CLASS= gsub(NA-|-NA,,paste(Class_a, Class_b, sep=-))) identical(ddNew, dd_2) #[1] TRUE A.K. On Thursday, May 22, 2014 4:07 AM, Beatriz R. Gonzalez Dominguez aguitatie...@hotmail.com wrote: Dear R-users, I'd be very greatful if you could help me with the following as after a few tests I haven't still been able to get the right outcome. I've got this data: dd_1 - data.frame(ID = c(1,2, 3, 4, 5), Class_a = c(a,NA, a, NA, NA), Class_b = c(NA, b, b, b, b)) And I'd like to produce a new column 'CLASS': dd_2 - data.frame(ID = c(1,2, 3, 4, 5), Class_a = c(a,NA, a, NA, NA), Class_b = c(NA, b, b, b, b), CLASS = c(a, b, a-b, b, b)) Thanks a lot! __ 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] select randomly from a list
Hi, I am not sure I understand the problem. Please provide a reproducible example using ?dput(). mylist - list(1:3, LETTERS[1:2], rnorm(4)) sample(mylist,1) sample(mylist,1) mylist1 - list(1:2) sample(mylist1,1) #[[1]] #[1] 1 2 sample(mylist1,1) #[[1]] #[1] 1 2 A.K. On Thursday, May 22, 2014 3:55 AM, Ragia Ibrahim ragi...@hotmail.com wrote: Hi, kindly I want to select randomly and item from list of items. the list generated in a looping process. I used sample(mylist,1) it works fine. BUTsome times the list have only one item. that should be chosen in this case since there is no other one. I found that sample return different item not included in the list thanks in advance RAE [[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] select randomly from a list
On Thu, 22 May 2014 09:54:13 AM Ragia Ibrahim wrote:
Hi,
kindly I want to select randomly and item from list of items. the list
generated in a looping process. I used sample(mylist,1) it works fine.
BUTsome times the list have only one item. that should be chosen in
this
case since there is no other one. I found that sample return different
item
not included in the list thanks in advance
RAE
Hi RAE,
This doesn't happen in an example like this:
for(i in 1:5) {
testlist-list()
for(j in 1:i) testlist[[j]]-sample(LETTERS[1:26],1)
cat(list is\n)
print(testlist)
cat(sample is\n)
print(sample(testlist,1))
}
How are you generating your lists?
Jim
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Re: [R] Mean of colMeans
Also, the ?colMeans() route seems to be slightly faster (in cases with no missing values). set.seed(398) x - matrix(rnorm(1e4*2e4), ncol=2e4) system.time(res1 - mean(colMeans(x))) # user system elapsed #0.227 0.000 0.227 system.time(res2 - mean(x)) # user system elapsed # 0.46 0.00 0.46 all.equal(res1,res2) #[1] TRUE A.K. On Wednesday, May 21, 2014 4:53 PM, Boris Steipe boris.ste...@utoronto.ca wrote: Not necessarily. Missing values may base the means on unequal numbers of observation: x - c(1,1,1,2,NA,2) dim(x) - c(3,2) cM - colMeans(x, na.rm = TRUE, dims = 1) mean(cM) mean(x, na.rm = TRUE) B. On 2014-05-21, at 3:38 PM, Frede Aakmann Tøgersen wrote: Hhhhmmhhhm, but is that not the same as mean(as.matrix(x)) ? Yours sincerely / Med venlig hilsen Frede Aakmann Tøgersen Specialist, M.Sc., Ph.D. Plant Performance Modeling Technology Service Solutions T +45 9730 5135 M +45 2547 6050 fr...@vestas.com http://www.vestas.com Company reg. name: Vestas Wind Systems A/S This e-mail is subject to our e-mail disclaimer statement. Please refer to www.vestas.com/legal/notice If you have received this e-mail in error please contact the sender. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Sarah Goslee Sent: 21. maj 2014 21:11 To: Kate Ignatius; r-help Subject: Re: [R] Mean of colMeans That would be because col is a function in base R, and thus a poor choice of names for user objects. Nonetheless, it worked when I ran it, but you didn't provide reproducible example so who knows. R set.seed(1) R x - data.frame(matrix(runif(150), ncol=10)) R # col is a function, so not a good name R col - colMeans(x) R mean(col) [1] 0.5119 It's polite to include the list on your reply. Sarah On Wed, May 21, 2014 at 2:50 PM, Kate Ignatius kate.ignat...@gmail.com wrote: That didn't work: gave me the error = [1] NA Warning message: In mean.default(col) : argument is not numeric or logical: returning NA But writing it like: mean(colMeans(x, na.rm = TRUE, dims = 1)), worked Thanks! On Wed, May 21, 2014 at 2:31 PM, Sarah Goslee sarah.gos...@gmail.com wrote: Is mean(col) not what you're looking for? Sarah On Wed, May 21, 2014 at 2:26 PM, Kate Ignatius kate.ignat...@gmail.com wrote: Hi All, I've successfully gotten out the colMeans for 60 columns using: col - colMeans(x, na.rm = TRUE, dims = 1) My next question is: is there a way of getting a mean of all the column means (ie a mean of a mean)? Thanks! -- Sarah Goslee http://www.functionaldiversity.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. __ 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. __ 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] Unsubscribe mailing list
Please do that by going to: https://stat.ethz.ch/mailman/listinfo/r-help and fill in the required information. Regards, Kuldeep Thanks and Regards, Kuldeep Singh On Wed, May 21, 2014 at 4:47 PM, Martina Marbà frees...@gmail.com wrote: Hi, I would like to unsubscribe from the mailing list. Thanks -- -- Martina Marbà [[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. [[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] HMM states simulation
Dear all,
I have 100 observations sampled from poisson and negative binomial
distributions and i want set a hidden states for the observations, class
with zero and one {0,1}. Larger observations to be one and smaller
observations to be zero using R.
Can someone please help?
Kind regards
Zakir
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Re: [R] select randomly from a list
It's a well known quirk of sample that it changes behavior when the x argument
has length 1:
replicate(10,sample(4:5, 1))
[1] 5 4 5 4 5 4 5 4 4 4
replicate(10,sample(5:5, 1))
[1] 5 3 1 1 1 2 5 3 2 2
One workaround is to zap the offending branch inside sample:
Sample - function (x, size, replace = FALSE, prob = NULL)
x[sample.int(length(x), size, replace, prob)]
replicate(10,Sample(5:5, 1))
[1] 5 5 5 5 5 5 5 5 5 5
(It's one of the cases of misguided user-friendliness that is probably long
regretted by its authors, but has been around for so long that it is painful to
change because code relies on the current behavior. A similar case is diag().)
-pd
On 22 May 2014, at 11:05 , Jim Lemon j...@bitwrit.com.au wrote:
On Thu, 22 May 2014 09:54:13 AM Ragia Ibrahim wrote:
Hi,
kindly I want to select randomly and item from list of items. the list
generated in a looping process. I used sample(mylist,1) it works fine.
BUTsome times the list have only one item. that should be chosen in
this
case since there is no other one. I found that sample return different
item
not included in the list thanks in advance
RAE
Hi RAE,
This doesn't happen in an example like this:
for(i in 1:5) {
testlist-list()
for(j in 1:i) testlist[[j]]-sample(LETTERS[1:26],1)
cat(list is\n)
print(testlist)
cat(sample is\n)
print(sample(testlist,1))
}
How are you generating your lists?
Jim
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and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
[R] Boxplot colors
Hi Im producing boxplots based on factors and rearranging them by median (This is for a Geochemistry element). Im giving each boxplot a unique color based on its level (factor) name. Im trying to produce a look up list to produce these colors as the order of the boxplots will change from element to element. Here is by color lool up and boxplot code: colors_list-list(Dalradian Appin Group quartzite=#00,Sperrins Dalradian (position uncertain)=#D7D79E, Namurian sandstone, shale=#C7D79E,Lr-Mid Ordovocian acid volcanics=#FFD37F, Tertiary granite, felsite=#FF5500,Caledonian appinite suite=#946379, Dalradian Argyll Gp quartzite=#00,Slishwood Division =#FFEDCC, Dalradian S Highland Gp volcanics=#00E6A9,LEITRIM GP; Visean mudstone, sandstone=#E6, Dalradian Argyll Group=#FFD9C7,Metadolerite or amphibolite=#4C7300, Caledonian granite=#FF7F7F,Dalradian Appin Group=#9EAAD7, Lr-Mid Ordovician basic volcanics=#448970,Ordovician Granite=#E6, Tertiary basic intrusion=#8400A8,Dalradian Argyll Gp volcanics=#00E6A9, VISEAN \\basal clastics\\=#73A6A6,Carboniferous volcs minor intrus=#AAFF00, Metagabbro, metadiorite (Tyrone Plu=#00A885,TYRONE GP; Visean mudstone, sandstone=#CC, Late Visean-Westphalian 'ORS'=#F5A27A,Dalradian S Highland Group=#FFD9C7, Mid Devonian ORS=#FFD37F,Mid-Up Ordovician slate=#C2753D, Interbasaltic formation laterite, b=#00,Visean shelf limestone, shale=#FF00C5, Tertiary rhyolite (volcintru)=#FFEABE,COURCEYAN \\basal clastics\\=#4D8099, Up Dev-Lr Carb ORS=#CDAA66,Tyrone CI (Corvanaghan=?Slishwood=#C9FFC9, Visean basinal limestone \\Calp\\=#CFD6EB,Mid-Up Ordovician g'wacke, sndst, shale=#C29ED7, Up Silurian - Lr Devonian ORS=#CD8966,Westphalian shale, sandstone=#B2B2B2, Upper Basalt Formation=#FAC2E0,Causeway Tholeiite Mbr=#C29ED7, Courceyan limestone=#73B2FF,Tertiary minor volcanics=#E600A9, Silurian sandstone, g'wacke, shale=#ADE6D1,Up Cretaceous limestone=#73FFDF, Lower Basalt Formation=#E0A8C7,Ballycastle succession=#CCB3B3, Navan Group=#408CBF,Devonian basic volcs, minor intrus=#A5F57A, Up. Ord-Sil \\Moffat shale\\ etc.=#DB,Oligocene clay, sand=#704489, Lr Jurassic mudstone=#704489,Lr Jurassic mudstone=#267300, Mid-Up Ordovician acid volcanics=#448970,Armagh Gp=#8073B3, Rathkenny Formation=#704489,Triassic sandstone=#F7DB5E, Waulsortian Limestones=#B2B2B2,Permian sandstone=#F5CA7A, Mid-Up Ordovician basic volcanics=#448970) boxplot(na.omit(C_pH$DATA.pH)~bymedian,axes=FALSE,horizontal=TRUE,col=unlist(colors_list),outwex=one,whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize) However it is not reading the colors based on the color_list given here. Anyone know how to solve this? Thanks for your help -- Shane [[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] K-mean Clustering URL's and ranking them according to there recent visit .
Hi all , I am having some URL's of facebook ,youtube , etc . I would like to applyk-mean clustering algorithm on those URL's which gives me the most recomended URL's which are visited frequently , I had seen some papers related to Web crawling using clustering ,but could not find outhow to proceed in R . is there any package to do so . Any idea would be much helpful . Thank You ASHIS DEB [[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] solve scalar linear equation
On 2014-05-20 10:00, r-help-requ...@r-project.org wrote: -- Message: 32 Date: Mon, 19 May 2014 23:04:27 +0100 From: Rui Barradas ruipbarra...@sapo.pt To: message let...@openmailbox.org, r-help@r-project.org uniroot(function(x) 5*x - 55, c(0, 20)) Why does this instruction fail if the interval is changed? uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign __ 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] Second axis on bottom of graph
Sorry, I don't know enough about R to understand what you said. But I don't think I need to understand it. And secondly, my reason for my stupid recent post about not finding a recent post was that I didn't notice that there was a little 1 2 in the lower right hand corner meaning that I had to select a second page. I think it is better to start a new thread when that happens so others will see my post too. Hurr -- View this message in context: http://r.789695.n4.nabble.com/Second-axis-on-bottom-of-graph-tp4690696p4691042.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] solve scalar linear equation
On 22-05-2014, at 12:28, message let...@openmailbox.org wrote: On 2014-05-20 10:00, r-help-requ...@r-project.org wrote: -- Message: 32 Date: Mon, 19 May 2014 23:04:27 +0100 From: Rui Barradas ruipbarra...@sapo.pt To: message let...@openmailbox.org, r-help@r-project.org uniroot(function(x) 5*x - 55, c(0, 20)) Why does this instruction fail if the interval is changed? uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign I don’t believe this. The error message says it all. 5*0-55 == -55 5*10-55 == -5 See the Details section of the uniroot documentation. Berend __ 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] solve scalar linear equation
Hello, Because there is no root of that function in the interval c(0, 10). Just like the error message says. Rui Barradas Em 22-05-2014 11:28, message escreveu: On 2014-05-20 10:00, r-help-requ...@r-project.org wrote: -- Message: 32 Date: Mon, 19 May 2014 23:04:27 +0100 From: Rui Barradas ruipbarra...@sapo.pt To: message let...@openmailbox.org, r-help@r-project.org uniroot(function(x) 5*x - 55, c(0, 20)) Why does this instruction fail if the interval is changed? uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign __ 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] select randomly from a list
You are probably encountering an annoying behaviour of sample(): when it is
given exactly one integer as an argument, it takes this as the upper limit of a
range.
a - c(3,5)
sample(a,10, replace=TRUE)
#[1] 5 5 3 3 3 3 3 3 3 5
a - c(5)
sample(a,10, replace=TRUE)
#[1] 2 1 3 1 1 3 4 5 1 4
#i.e. this is the same as
sample(1:5,10, replace=TRUE)
#An easy way to catch this is:
if (length(a) 1) sample(a,1) else a
Boris
On 2014-05-22, at 5:05 AM, Jim Lemon wrote:
On Thu, 22 May 2014 09:54:13 AM Ragia Ibrahim wrote:
Hi,
kindly I want to select randomly and item from list of items. the list
generated in a looping process. I used sample(mylist,1) it works fine.
BUTsome times the list have only one item. that should be chosen in
this
case since there is no other one. I found that sample return different
item
not included in the list thanks in advance
RAE
Hi RAE,
This doesn't happen in an example like this:
for(i in 1:5) {
testlist-list()
for(j in 1:i) testlist[[j]]-sample(LETTERS[1:26],1)
cat(list is\n)
print(testlist)
cat(sample is\n)
print(sample(testlist,1))
}
How are you generating your lists?
Jim
__
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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.
__
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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] Metafor: How to integrate effectsizes?
Hello,
thank you very much for your replies. I am almost done :-) but theres one
study left, where I only have sample size (not group size), mean values
and standarddeviations. Is there a way to compute cohens d from this data?
I thought it was correct to use measure=SMDH in the escalc () function to
compute cohens d? In your illustration, Wolfgang, you use SMD as measure -
now I am confused ;-)
Thank you very much in advance!
Best,
Verena
On Tue, May 6, 2014 at 7:14 PM, Michael Dewey i...@aghmed.fsnet.co.ukwrote:
At 14:23 06/05/2014, Viechtbauer Wolfgang (STAT) wrote:
Without the sample size of a study (i.e., either the group sizes or the
total sample size), you cannot convert the p-value to a t-value or a
t-value to a d-value. And for studies where you have the d-value but no
sample size, you cannot compute the corresponding sampling variance. So,
without additional information, you cannot include these studies. Maybe
studies where a d-value is directly reported also report a CI for the
d-value? Then the sampling variance can be back-calculated (since a 95% CI
for d is typically computed with d +- 1.96 sqrt(vi), where vi is the
sampling variance).
Verena,
What Wolfgang says is true of course but if you have _both_ the t value
and the p value you can backcalculate the number of degrees of freedom and
then if you are willing to assume equal arms you have the sample sizes.
finddf - function(t, pval) {
helper - function(df) {res - pval - pt(t, df, lower.tail = FALSE);
res}
res - uniroot(helper, interval = c(5, 1))
res
}
If you call finddf with the value of t and the _one-sided_ p-value (divide
by 2 if two-sided) it should give you a return value which, if you look at
the element of the list called root is its estimate of the degrees of
freedom. If you get errors from uniroot the interval supplied in the call
may need to be widened.
I would suggest that when you have your final dataset it would be a really
good idea to do some model checks using plot.influence to see whether the
studies for which you have imputed values are fundamentally different for
some reason. This will also check your calculations as a bonus.
Best,
Wolfgang
-Original Message-
From: Verena Weinbir [mailto:vwein...@gmail.com]
Sent: Tuesday, May 06, 2014 15:09
To: Michael Dewey
Cc: Viechtbauer Wolfgang (STAT); r-help@r-project.org
Subject: Re: [R] Metafor: How to integrate effectsizes?
Thank you very much for your illustration, Wolfgang! It helped me a
lot.Ã And also thank you for the package-hint, Michael!
Now, I have re-checked the respective studies, and there still are a
couple of studies left, only stating cohens d, and the respective
t-value
and p-value - sample and group sizes are not addressed (its data from an
older meta-analysis). Is there a way to embed these studies in my
sample?
Wolfgangs illustration addresses only cases in which group sizes are
stated, if I understand you correctly...
Many thanks in advance,
Verena
On Sat, Apr 26, 2014 at 1:38 PM, Michael Dewey i...@aghmed.fsnet.co.uk
wrote:
At 20:34 25/04/2014, Viechtbauer Wolfgang (STAT) wrote:
If you know the d-value and the corresponding group sizes for a study,
then it's possible to add that study to the rest of the dataset. Also,
if
you only know the test statistic from an independent samples t-test (or
only the p-value corresponding to that test), it's possible to back-
compute what the standardized mean difference is.
I added an illustration of this to the metafor package website:
http://www.metafor-project.org/doku.php/tips:assembling_data_smd
Verena might also like to look at the compute.es package available from
CRAN to see whether any of the conversions programmed there do the job.
Best,
Wolfgang
--
Wolfgang Viechtbauer, Ph.D., Statistician
Department of Psychiatry and Psychology
School for Mental Health and Neuroscience
Faculty of Health, Medicine, and Life Sciences
Maastricht University, P.O. Box 616 (VIJV1)
6200 MD Maastricht, The Netherlands
+31 (43) 388-4170Ã | http://www.wvbauer.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org]
On Behalf Of Michael Dewey
Sent: Friday, April 25, 2014 16:23
To: Verena Weinbir
Cc: r-help@r-project.org
Subject: Re: [R] Metafor: How to integrate effectsizes?
At 12:33 25/04/2014, you wrote:
Thank you very much for your reply and the book recommendation,
Michael.
Yes, I mean Cohen's d - sorry for the typo :-)
Just to make this sure for me: There is no
possibility to integrate stated Cohens' ds in an
R-Metaanalysis (or a MA at all), if there is no
further information traceable regarding SE or the like?
If there is really no other information like
sample sizes, significance level, value of some
significance test then you would have to
[R] Non-convergence in boot.stepAIC function with a logit model
Hi all I am getting warning when I try to perform a bootstrap selection procedure on variables (using boot.stepAIC function in the bootStepAIC package). I had previously established which variables were collinear and kept the one which had the lowest AIC following univariate regression on each predictor. I obtain a candidate list of variables that are not correlated at the end of this procedure. I then revisit those variables that were excluded at each step using bootstrapping. I have referred to other list questions (such as http://stackoverflow.com/questions/8596160/why-am-i-getting-algorithm-did-not-converge-and-fitted-prob-numerically-0-or) and I see that this is a common problem with logit models, but convergence fails only in a bootstrapping context. For the first set of previously excluded variables, I added individually each variable to candidate list of variables and then performed bootstrapping, and then added more than one variable to see if the algorithm would converge. Sometimes it did other times not (as indicated by dc). I suppose the presence of multicollinearity affects this process? From the next group on of excluded variables I only really considered adding variables separately one at a time and then checked if there was an improvement in AIC. If one of the previously excluded variables is in the candidate list, then i take that variable out and add the previously excluded one and see if there is an improvement in AIC. From this reasoning I end up adding two new variables to the list. They are not correlated with any of the variables in candidate list, nor are they correlated. My question is, is this a valid way to come up with my best set of predictors? Is there a way I can monitor more closely what is going on, i.e. if multicollinearity in is mathematically causing the algorithm not to converge for some variables? Here is my workflow using the boot.stepAIC function in the the forward stepwise direction (the forward direction seems to be more robust w.r.t convergence): (if reproducible code is required I can happily provide it - via dropbox for the data) #kept altitude (15454.23) (but not in candidate list) and excluded: # meanTemp (14422.72), minTemp (14435.72), bio1 (14767.88), bio6 (didn't converge (dc)), bio8 (15050.46), bio10 (14285.46), bio11 (14655.82), bio18 (15445.24), bio10+bio11 (dc), # bio10 + bio11 + # bio18 (dc), bio10+bio18 (dc), meanTemp + bio10+bio11 (14160.33), minTemp + bio10+bio11 (14204.41), meanTemp + minTemp + bio10+bio11 (14135.49), meanTemp+minTemp + bio10+bio11+bio1(dc), # bio10+bio11+bio1(dc) fit.1 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) bootGLM.1 - boot.stepAIC(fit.1, spatialVars, direction = backward, alpha = 0.05, B = 1000) #15445.24 # add bio10 (lowest AIC - 14285.46) # (backward drection dc) # kept bio2 and excluded: bio7 (dc), -bio2 + bio7 (14710.04), -bi02 - bio10 + bio7 (15676.62) fit.2 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) #15302.59 bootGLM.2 - boot.stepAIC(fit.2, spatialVars, direction = forward, alpha = 0.05, B = 1000) # keep bio2 in candidate list (+bio10) # kept bio5 and excluded: altitude (dc), -bio5 + altitude (15659.26), -bio5 +maxTemp (15637.91) fit.3 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) bootGLM.3 - boot.stepAIC(fit.3, spatialVars, direction = forward, alpha = 0.05, B = 1000) # keep bio5 in candidate list (+bio10) # kept bio17 (not in candidate list) (bio17 (14178.88)) and excluded: bio12 (dc), bio14 (14168.77), bio16 (14287.42), bio19 (14248.65), rain (14287.45), bio17+bio12(14162.3) fit.4 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10 + rain, weights = Examind, data = spatialVars, family=binomial) bootGLM.4 - boot.stepAIC(fit.4, spatialVars, direction = forward, alpha = 0.05, B = 1000) # add bio14 to candiate list (+bio10) # keptp bio15 (14168.77) (not included in candidate list) and excluded: bio17 (14161.03) fit.5 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10 + bio14, weights = Examind, data = spatialVars, family=binomial) bootGLM.5 - boot.stepAIC(fit.5, spatialVars, direction = forward, alpha = 0.05, B = 1000) Thanks very much (for any help, advice or thoughts) Justin [[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] solve scalar linear equation
On 2014-05-22 11:00, Berend Hasselman wrote: uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign I don’t believe this. The error message says it all. 5*0-55 == -55 5*10-55 == -5 The error states opposite sign, which suggests to a rudimentary novice that end points (0 and 10, or 0 and 20) must give results of the function such that one value is negative, another value is positive. Or is this interpretation wrong? See the Details section of the uniroot documentation. The documentation states that the upper end point (in this example 10, or 20) must be strictly larger than the lower point (0). What is being mis-understood please? __ 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] Second axis on bottom of graph
Sarah's perhaps-too-clever comments had nothing to do with R, and everything to do with the fact that Nabble misleadingly suggests to you that this mailing list is hosted at their website. Very few of the participants on this EMAIL LIST use Nabble, so we don't have any idea about the troubles you may be having or the web page buttons you write about, and your mention of these issues is OFF TOPIC. Please read the Posting Guide. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On May 22, 2014 3:31:02 AM PDT, Hurr hill0...@umn.edu wrote: Sorry, I don't know enough about R to understand what you said. But I don't think I need to understand it. And secondly, my reason for my stupid recent post about not finding a recent post was that I didn't notice that there was a little 1 2 in the lower right hand corner meaning that I had to select a second page. I think it is better to start a new thread when that happens so others will see my post too. Hurr -- View this message in context: http://r.789695.n4.nabble.com/Second-axis-on-bottom-of-graph-tp4690696p4691042.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-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] multiple colour to symbols using brkdn.plot()
Dear all, I have been plotting response variable (defined as y in the example below) from 4 groups (defined as z variable ) under 5 conditions (defined by the x variable). The formula used is: y ~ z+x. I can differentiate the different z value using a call pch=c(symbol 1, symbol 2, ...) but it would be even better to have different colours for each of these symbols. In analogy to the pch=c(symbol 1, symbol 2, ...) i have tired to provide a bg(colour 1, colour 2, ...) but this give the colour to the x variables rather than the z ones. How can I give the colour to the z variables? best regards, Luigi # code: my.data-structure(list( row = 1:60, x = c( 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 3, 3, 3, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4), y = c( 2073.928223, 2131.830067, 2131.830067, 0.143912883, 2191.348468, 2073.928223, 2117.20479, 2017.59903, 1896.388977, 1976.358448, 2003.757427, 1883.378928, 2283.756186, 2363.732429, 2315.416732, 2206.485917, 2191.348468, 2176.314869, 1990.010783, 2059.700178, 1976.358448, 617.4528799, 613.2168858, 617.4528799, 1686.950197, 1819.655315, 1832.225173, 1480.122531, 1298.652866, 1212.260417, 495.3736815, 505.7106218, 538.0337432, 383.9842946, 365.919416, 330.0195927, 505.7106218, 541.7503854, 498.7956356, 512.7214729, 584.3675585, 564.5956413, 604.8318804, 604.8318804, 592.4688595, 1272.107849, 1298.652866, 1298.652866, 1935.96084, 2088.254554, 1962.799773, 4452.994159, 4422.444691, 4128.243033, 312.3359691, 316.6659968, 332.2993098, 1500.642011, 1531.95584, 1430.042989), z = c( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)), row.names = c(NA, -60L), class = data.frame) attach(my.data) ### *** PLOT DATA *** ### # open library library(plotrix) par(mfrow = c(1, 2)) brkdn.plot( y ~ z+x,data=my.data, mct=mean, md=sd, stagger=NA, dispbar=TRUE, type=p, pch=c(0, 1, 2, 5), main=z indicated by symbol , xlab=expression(bold(x)), ylab=expression(bold(y)) ) brkdn.plot( y ~ z+x,data=my.data, mct=mean, md=sd, stagger=NA, dispbar=TRUE, type=p, pch=c(21, 21, 21, 21), bg=c(blue, green, red, orange), main=z indicated by colour (???), xlab=expression(bold(x)), ylab=expression(bold(y)) ) __ 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] How to compute a P-value for a complex mixture of chi-squared distributions in R
On 01/06/2013, 12:26 AM, Tiago V. Pereira wrote: Hello, R users! I am struggling with the following problem: I need to compute a P-value for a mixture of two chi-squared distributions. My P-value is given by: P = 0.5*prob(sqrt(chi2(1)) = x) + 0.5*prob(sqrt(chi2(2)) = x) Isn't this simply 0.5*pchisq(x^2, df=1) + 0.5*pchisq(x^2, df=2) ? Duncan Murdoch In words, I need to compute the p-value for 50–50 mixture of the square root of a chi-squared random variable with 1 degree of freedom and the square root of a chi-squared with two degrees of freedom. Although I can quickly simulate data, the P-values I am looking for are at the tail of the distribution, that is, alpha levels below 10^-7. Hence, simulation is not efficient. Are you aware of smart approach? All the best, Tiago __ 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.
[R] manual installation errors; install.package, R CMD
Readers, As root user, from the command: install.packages() A dialogue window appears to select packages of interest. What to do if a specific packages exists on the cran page (http://cran.r-project.org/web/packages/available_packages_by_name.html), but not in the dialogue window of a repository? Sometimes, a dependent package is reported as being un-available, e.g.: install.packages(mclust) Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘mclust’ is not available Alternatively, tried the command 'R CMD': R CMD INSTALL http://cran.r-project.org/src/contrib/ChemoSpec_1.61-3.tar.gz Warning: invalid package ‘http://cran.r-project.org/src/contrib/ChemoSpec_1.61-3.tar.gz’ Error: ERROR: no packages specified The FAQ web page (http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-add_002don-packages-be-installed_003f) states that the above syntax should be accepted. What is my mistake please? __ 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] solve scalar linear equation
Hello, See inline. Em 22-05-2014 13:35, message escreveu: On 2014-05-22 11:00, Berend Hasselman wrote: uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign I don’t believe this. The error message says it all. 5*0-55 == -55 5*10-55 == -5 The error states opposite sign, which suggests to a rudimentary novice that end points (0 and 10, or 0 and 20) must give results of the function such that one value is negative, another value is positive. Or is this interpretation wrong? No, it's not. This is basic Math. See the Details section of the uniroot documentation. The documentation states that the upper end point (in this example 10, or 20) must be strictly larger than the lower point (0). What is being mis-understood please? I believe that Berend was trying to have you read that The function values at the endpoints must be of opposite signs (or zero), for extendInt=no, the default. __ 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] solve scalar linear equation
On 22-05-2014, at 14:35, message let...@openmailbox.org wrote: On 2014-05-22 11:00, Berend Hasselman wrote: uniroot(function(x) 5*x - 55, c(0, 10)) Error in uniroot(function(x) 5 * x - 55, c(0, 10)) : f() values at end points not of opposite sign I don’t believe this. The error message says it all. 5*0-55 == -55 5*10-55 == -5 The error states opposite sign, which suggests to a rudimentary novice that end points (0 and 10, or 0 and 20) must give results of the function such that one value is negative, another value is positive. Or is this interpretation wrong? No. It is exactly what the error message says. Both function values are negatieve i.e. not of opposite sign. Which is why you got the error message. See the Details section of the uniroot documentation. The documentation states that the upper end point (in this example 10, or 20) must be strictly larger than the lower point (0). What is being mis-understood please? Read the sentences after the first. Berend __ 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] manual installation errors; install.package, R CMD
Hi, On Thu, May 22, 2014 at 8:48 AM, message let...@openmailbox.org wrote: Readers, As root user, from the command: install.packages() A dialogue window appears to select packages of interest. What to do if a specific packages exists on the cran page (http://cran.r-project.org/web/packages/available_packages_by_name.html), but not in the dialogue window of a repository? Sometimes, a dependent package is reported as being un-available, e.g.: install.packages(mclust) Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘mclust’ is not available It's possible that a package is not available for your OS or version of R, but you don't tell us either. Alternatively, tried the command 'R CMD': R CMD INSTALL http://cran.r-project.org/src/contrib/ChemoSpec_1.61-3.tar.gz Warning: invalid package ‘http://cran.r-project.org/src/contrib/ChemoSpec_1.61-3.tar.gz’ Error: ERROR: no packages specified The FAQ web page (http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-add_002don-packages-be-installed_003f) states that the above syntax should be accepted. What is my mistake please? That FAQ doesn't say that R is capable of downloading it for you: you need to download the package manually, then use R CMD INSTALL to install the *local* package. Sarah -- Sarah Goslee http://www.functionaldiversity.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.
[R] Fwd: problem importing apparently common txt files
Dear list, I would like to import into R studio txt files coming from Adobe photoshop. These files contain different measures. The files are apparently common txt files with columns separate by tab. However with common function read.table the import process doesn't work and weird symbols appear once imported. An example file is attached. Do you have any idea what it is happening? Thank you in advance Zuzana Here just a view of a file. DocumentoLongitudÁnguloÁreaAlturaAnchura 6143066.17.4.12.DSC_0001.png8.832613-90.00 6143066.17.4.12.DSC_0001.png8.93791890.337288 6143066.17.4.12.DSC_0001.png551.10310916.736842 52.543860 6143066.17.4.12.DSC_0002.png8.942544-89.662891 6143066.17.4.12.DSC_0002.png8.994992-90.00 6143066.17.4.12.DSC_0002.png550.14650716.736842 52.491228 6106937.17.4.12.DSC_0004.png12.159715-89.008611 DocumentoLongitud ÁnguloÁrea Altura Anchura 6143066.17.4.12.DSC_0001.png 8.832613 -90.00 6143066.17.4.12.DSC_0001.png 8.937918 90.337288 6143066.17.4.12.DSC_0001.png 551.10310916.736842 52.543860 6143066.17.4.12.DSC_0002.png 8.942544 -89.662891 6143066.17.4.12.DSC_0002.png 8.994992 -90.00 6143066.17.4.12.DSC_0002.png 550.14650716.736842 52.491228 6106937.17.4.12.DSC_0004.png 12.159715 -89.008611 6106937.17.4.12.DSC_0004.png 12.054468 -88.54 6106937.17.4.12.DSC_0004.png 574.21083417.526316 52.192982 6106937.17.4.12.DSC_0005.png 12.210640 -89.753112 6106937.17.4.12.DSC_0005.png 12.316239 -89.510454 6106937.17.4.12.DSC_0005.png 573.66912917.561404 52.122807 6188609.17.4.12.DSC_0009.png 13.268267 -91.590146 6188609.17.4.12.DSC_0009.png 13.319529 -91.357689 6188609.17.4.12.DSC_0009.png 526.27516215.491228 52.438596 6188609.17.4.12.DSC_0010.png 13.196761 -91.371180 6188609.17.4.12.DSC_0010.png 13.231048 -91.215643 6188609.17.4.12.DSC_0010.png 526.78085615.491228 52.385965 6143090.15.4.12.DSC_0013.png 12.140351 -90.00 6143090.15.4.12.DSC_0013.png 12.00 -90.00 6143090.15.4.12.DSC_0013.png 610.21083418.421053 52.596491 6143090.15.4.12.DSC_0014.png 12.090731 -89.750728 6143090.15.4.12.DSC_0014.png 11.992736 -90.00 6143090.15.4.12.DSC_0014.png 610.64112018.245614 52.649123 6195159.14.5.12.DSC_0017.png 16.742131 -88.559723 6195159.14.5.12.DSC_0017.png 16.847362 -88.568721 6195159.14.5.12.DSC_0017.png 630.46506620.473684 55.684211 6195159.14.5.12.DSC_0018.png 16.797712 -88.205510 6195159.14.5.12.DSC_0018.png 16.846130 -88.747572 6195159.14.5.12.DSC_0018.png 630.21699020.491228 55.684211 6175701.15.4.12.DSC_0021.png 14.732429 -101.116408 6175701.15.4.12.DSC_0021.png 14.629281 -101.195797 6175701.15.4.12.DSC_0021.png 518.33271818.298246 53.614035 6175701.15.4.12.DSC_0022.png 14.780985 -101.703689 6175701.15.4.12.DSC_0022.png 14.801433 -101.064184 6175701.15.4.12.DSC_0022.png 517.61588218.245614 53.508772 6134530.15.4.12.DSC_0025.png 10.868923 -94.025757 6134530.15.4.12.DSC_0025.png 10.865356 -93.748940 6134530.15.4.12.DSC_0025.png 569.84641418.649123 54.526316 6134530.15.4.12.DSC_0026.png 10.902122 -93.875406 6134530.15.4.12.DSC_0026.png 10.899807 -93.691386 6134530.15.4.12.DSC_0026.png 570.21329618.67 54.456140 6140703.DSC_0029.png 13.057253 -94.469557 6140703.DSC_0029.png 13.019577 -94.327648 6140703.DSC_0029.png 507.53031716.157895 51.473684 6140703.DSC_0030.png 12.981982 -94.184916 6140703.DSC_0030.png 13.086967 -94.151285 6140703.DSC_0030.png 507.12526916.157895 51.736842 6140725.15.4.12.DSC_0033 - copia.png 14.174917 -95.682369 6140725.15.4.12.DSC_0033 - copia.png 14.213342 -95.809112 6140725.15.4.12.DSC_0033 - copia.png 583.036627 20.894737 54.017544
Re: [R] manual installation errors; install.package, R CMD
On 2014-05-22 13:08, Sarah Goslee wrote: It's possible that a package is not available for your OS or version of R, but you don't tell us either. For the record, GNU/Linux R2110 That FAQ doesn't say that R is capable of downloading it for you: you need to download the package manually, then use R CMD INSTALL to install the *local* package. For the benefit of other novices, that help page should state that the term path refers to _local_path. Thanks anyway. __ 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] Fwd: problem importing apparently common txt files
Hi, Check if this works: dat1 - read.table(folder2.txt, header=TRUE,stringsAsFactors=FALSE,sep=\t,fileEncoding=UTF-16) 'data.frame': 79 obs. of 6 variables: $ Documento: chr 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0002.png ... $ Longitud : num 8.83 8.94 NA 8.94 8.99 ... $ Ángulo : num -90 90.3 NA -89.7 -90 ... $ Área : num NA NA 551 NA NA ... $ Altura : num NA NA 16.7 NA NA ... $ Anchura : num NA NA 52.5 NA NA ... A.K. On Thursday, May 22, 2014 9:11 AM, zuzana zajkova zuzu...@gmail.com wrote: Dear list, I would like to import into R studio txt files coming from Adobe photoshop. These files contain different measures. The files are apparently common txt files with columns separate by tab. However with common function read.table the import process doesn't work and weird symbols appear once imported. An example file is attached. Do you have any idea what it is happening? Thank you in advance Zuzana Here just a view of a file. Documento Longitud Ángulo Área Altura Anchura 6143066.17.4.12.DSC_0001.png 8.832613 -90.00 6143066.17.4.12.DSC_0001.png 8.937918 90.337288 6143066.17.4.12.DSC_0001.png 551.103109 16.736842 52.543860 6143066.17.4.12.DSC_0002.png 8.942544 -89.662891 6143066.17.4.12.DSC_0002.png 8.994992 -90.00 6143066.17.4.12.DSC_0002.png 550.146507 16.736842 52.491228 6106937.17.4.12.DSC_0004.png 12.159715 -89.008611 __ 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.
[R] subsetting to exclude different values for each subject in study
Hi, I've written a code to determine the difference in score for a single subject and its non-neighbours o-(ao[,c(13,5)]) ##this is the table with the relevant information o-na.omit(o) ##omitted data with NA o-o[!o$NestkastNummer %in% c(176,140,162,713),] ##removed neighbours XO[7,1]-abs((XO[1,176]-(mean(o[,COR_LOC] #difference between that individual and average non-neighbours scores Since each subject has a different number of non-neighbours I was wondering if there is an efficient way of writing the code, instead of writing the same code again and again (76 subjects) for each subject and its non-neighbours. Best, Monaly. [[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] Boxplot colors
## Shane, ## This uses your color_list. Your other variables weren't included in the email ## so I invented some data. ## I recommend bwplot() using panel=panel.bwplot.superpose ## and I also show how to use boxplot() ## install.packages(HH) ## if necessary. library(HH) ## install.packages(reshape2) ## if necessary. library(reshape2) ## you didn't include data, so I am making up some data pH - data.frame(matrix(rnorm(200), 10, 20)) bwplot(value ~ variable, data=melt(pH), col=unlist(colors_list), panel=panel.bwplot.superpose, groups=variable) boxplot(value ~ variable, data=melt(pH), col=unlist(colors_list)) ## Rich On Thu, May 22, 2014 at 5:50 AM, Shane Carey careys...@gmail.com wrote: Hi Im producing boxplots based on factors and rearranging them by median (This is for a Geochemistry element). Im giving each boxplot a unique color based on its level (factor) name. Im trying to produce a look up list to produce these colors as the order of the boxplots will change from element to element. Here is by color lool up and boxplot code: colors_list-list(Dalradian Appin Group quartzite=#00,Sperrins Dalradian (position uncertain)=#D7D79E, Namurian sandstone, shale=#C7D79E,Lr-Mid Ordovocian acid volcanics=#FFD37F, Tertiary granite, felsite=#FF5500,Caledonian appinite suite=#946379, Dalradian Argyll Gp quartzite=#00,Slishwood Division =#FFEDCC, Dalradian S Highland Gp volcanics=#00E6A9,LEITRIM GP; Visean mudstone, sandstone=#E6, Dalradian Argyll Group=#FFD9C7,Metadolerite or amphibolite=#4C7300, Caledonian granite=#FF7F7F,Dalradian Appin Group=#9EAAD7, Lr-Mid Ordovician basic volcanics=#448970,Ordovician Granite=#E6, Tertiary basic intrusion=#8400A8,Dalradian Argyll Gp volcanics=#00E6A9, VISEAN \\basal clastics\\=#73A6A6,Carboniferous volcs minor intrus=#AAFF00, Metagabbro, metadiorite (Tyrone Plu=#00A885,TYRONE GP; Visean mudstone, sandstone=#CC, Late Visean-Westphalian 'ORS'=#F5A27A,Dalradian S Highland Group=#FFD9C7, Mid Devonian ORS=#FFD37F,Mid-Up Ordovician slate=#C2753D, Interbasaltic formation laterite, b=#00,Visean shelf limestone, shale=#FF00C5, Tertiary rhyolite (volcintru)=#FFEABE,COURCEYAN \\basal clastics\\=#4D8099, Up Dev-Lr Carb ORS=#CDAA66,Tyrone CI (Corvanaghan=?Slishwood=#C9FFC9, Visean basinal limestone \\Calp\\=#CFD6EB,Mid-Up Ordovician g'wacke, sndst, shale=#C29ED7, Up Silurian - Lr Devonian ORS=#CD8966,Westphalian shale, sandstone=#B2B2B2, Upper Basalt Formation=#FAC2E0,Causeway Tholeiite Mbr=#C29ED7, Courceyan limestone=#73B2FF,Tertiary minor volcanics=#E600A9, Silurian sandstone, g'wacke, shale=#ADE6D1,Up Cretaceous limestone=#73FFDF, Lower Basalt Formation=#E0A8C7,Ballycastle succession=#CCB3B3, Navan Group=#408CBF,Devonian basic volcs, minor intrus=#A5F57A, Up. Ord-Sil \\Moffat shale\\ etc.=#DB,Oligocene clay, sand=#704489, Lr Jurassic mudstone=#704489,Lr Jurassic mudstone=#267300, Mid-Up Ordovician acid volcanics=#448970,Armagh Gp=#8073B3, Rathkenny Formation=#704489,Triassic sandstone=#F7DB5E, Waulsortian Limestones=#B2B2B2,Permian sandstone=#F5CA7A, Mid-Up Ordovician basic volcanics=#448970) boxplot(na.omit(C_pH$DATA.pH)~bymedian,axes=FALSE,horizontal=TRUE,col=unlist(colors_list),outwex=one,whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize) However it is not reading the colors based on the color_list given here. Anyone know how to solve this? Thanks for your help -- Shane [[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] How to compute a P-value for a complex mixture of chi-squared distributions in R
On 22 May 2014, at 14:44 , Duncan Murdoch murdoch.dun...@gmail.com wrote: On 01/06/2013, 12:26 AM, Tiago V. Pereira wrote: Hello, R users! I am struggling with the following problem: I need to compute a P-value for a mixture of two chi-squared distributions. My P-value is given by: P = 0.5*prob(sqrt(chi2(1)) = x) + 0.5*prob(sqrt(chi2(2)) = x) Isn't this simply 0.5*pchisq(x^2, df=1) + 0.5*pchisq(x^2, df=2) ...as suggested by yours truly in 2011! ( :-o ???!!!) On Jun 12, 2011, at 03:36 , Thomas Lumley wrote: On Sun, Jun 12, 2011 at 12:44 PM, Tiago Pereira tiago.pere...@mbe.bio.br wrote: The test I am working on has an asymptotic 0.5*chi2(1)+0.5*chi2(2) distribution, where numbers inside parenthesis stand for the degrees of freedom. Is is possible to compute quickly in R the cumulative distribution of that distribution? There appear to be pchibar() functions in both the ibdreg and ic.infer packages that should do want you want. Simulation is also fairly efficient. Assuming that you mean a 50-50 mixture of the two, it should also work just to take the average of the two CDFs. The integral is a linear operator after all. -- Peter Dalgaard Center for Statistics, Copenhagen Business School ? Duncan Murdoch In words, I need to compute the p-value for 50–50 mixture of the square root of a chi-squared random variable with 1 degree of freedom and the square root of a chi-squared with two degrees of freedom. Although I can quickly simulate data, the P-values I am looking for are at the tail of the distribution, that is, alpha levels below 10^-7. Hence, simulation is not efficient. Are you aware of smart approach? All the best, Tiago __ 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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] subsetting to exclude different values for each subject in study
Hi, Sorry I'm fairly new to R and I don't really understand using dput(), when you say reproducible example do you mean the code with the output? Best, Monaly. On Thu, May 22, 2014 at 4:03 PM, arun smartpink...@yahoo.com wrote: Hi, It would be helpful if you provide a reproducible example using ?dput(). A.K. On Thursday, May 22, 2014 10:15 AM, Monaly Mistry monaly.mis...@gmail.com wrote: Hi, I've written a code to determine the difference in score for a single subject and its non-neighbours o-(ao[,c(13,5)]) ##this is the table with the relevant information o-na.omit(o) ##omitted data with NA o-o[!o$NestkastNummer %in% c(176,140,162,713),] ##removed neighbours XO[7,1]-abs((XO[1,176]-(mean(o[,COR_LOC] #difference between that individual and average non-neighbours scores Since each subject has a different number of non-neighbours I was wondering if there is an efficient way of writing the code, instead of writing the same code again and again (76 subjects) for each subject and its non-neighbours. Best, Monaly. [[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. [[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] Voronoi-Diagrams in R
Hello again, I have found further depths of confusion concerning delaunay triangulations to explore. Here is the code I'm using to create the confusing results: bm - getbm(x) # a data.frame with 2 columns and 800 rows, values are integers ranging from 1 to 82 del - deldir(bm) # creating an object of type deldir with the 800 coordinates from bm tri - triang.list(del) # creating a list of length n, holding the n triangles created by the delaunay triangulation with deldir(bm) as data.frames geodel - delaunayn(bm) # creating a matrix with 3 columns and m rows for the m triangles created by the delaunay triangulation using delaunayn(bm) from geometry-package now, the following is what I would expect: n and m should be the same, since both triangulations should give me the same number of triangles, right? but this is what I get: nrow(geodel) = 1584 length(tri) = 1186 This confuses me and I have the feeling I have left out some very important parameter either in delaunayn() or deldir() to create the different results. There is however more confusion to come, this time in the result of triang.list(del): Every element in that list is a data.frame holding information on one triangle. It has 3 columns and 3 rows with column headers 'ptNum', 'x' and 'y'. Now I would expect, that 'ptNum' would give me the index at which I can find the coordinates 'x' and 'y' as a row in my original data.frame of coordinates, bm. And for example bm[1,] gives me the same 'x' and 'y' that are listed in data.frame tri[1] as 'ptNum' = 1. But for some other points this does not work and for example the 'x' and 'y' of 'ptNum' = 129 do not match bm[129, ]. Have I totally misunderstood the meaning of 'ptNum' in this case or does my mistake lie somewhere else? Once again please excuse my slightly chaotic description of my problems and my faulty english and thanks for any help you can give! Raphael On 5/21/14, Rolf Turner r.tur...@auckland.ac.nz wrote: On 21/05/14 23:34, Raphael Päbst wrote: I believe you are right. A night of sleep has done wonders for my understanding of the problem. Oh sleep it is a gentle thing Beloved from pole to pole! Thank you for your patience and help! Patience? Moi? This must be some new use of the word patience with which I am not familiar! :-) Glad your tessellations are working out. cheers, Rolf __ 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] subsetting to exclude different values for each subject in study
Follow the link at the bottom of this message! -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Thu, May 22, 2014 at 8:31 AM, Monaly Mistry monaly.mis...@gmail.com wrote: Hi, Sorry I'm fairly new to R and I don't really understand using dput(), when you say reproducible example do you mean the code with the output? Best, Monaly. On Thu, May 22, 2014 at 4:03 PM, arun smartpink...@yahoo.com wrote: Hi, It would be helpful if you provide a reproducible example using ?dput(). A.K. On Thursday, May 22, 2014 10:15 AM, Monaly Mistry monaly.mis...@gmail.com wrote: Hi, I've written a code to determine the difference in score for a single subject and its non-neighbours o-(ao[,c(13,5)]) ##this is the table with the relevant information o-na.omit(o) ##omitted data with NA o-o[!o$NestkastNummer %in% c(176,140,162,713),] ##removed neighbours XO[7,1]-abs((XO[1,176]-(mean(o[,COR_LOC] #difference between that individual and average non-neighbours scores Since each subject has a different number of non-neighbours I was wondering if there is an efficient way of writing the code, instead of writing the same code again and again (76 subjects) for each subject and its non-neighbours. Best, Monaly. [[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. [[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] converting a matrix to array
Hi,
Try:
mat -
as.matrix(read.csv(R.csv,header=TRUE,stringsAsFactors=FALSE,sep=\t,row.names=1))
pdf(Hist_colwise.pdf)
lst1 - lapply(seq_len(ncol(mat)),function(i) {x- mat[,i,drop=FALSE]; hist(x,
main=paste(Histogram of, colnames(x)))})
dev.off()
##or may be this would be another way to look at the data
library(reshape2)
library(ggplot2)
dat - melt(mat)
pdf(Hist_all.pdf, width=14)
ggplot(dat, aes(x=Var2, y=value,
fill=Var1))+geom_histogram(stat=identity,position=dodge)+
theme(legend.position = none)+
xlab() +
theme(axis.line = element_line(colour = black),
panel.grid.major = element_blank(),
panel.grid.minor = element_blank(),
panel.border = element_blank(),
panel.background = element_blank())
dev.off()
A.K.
On Thursday, May 22, 2014 10:48 AM, Effat Habibi habibi...@gmail.com wrote:
Thanks. I think columnwise histogram should be fine. Can you please help me
with the commands for making that?
On May 22, 2014 2:21 AM, arun smartpink...@yahoo.com wrote:
Hi Effat,
Sorry, I was a bit busy. I did read the data.
mat -
as.matrix(read.csv(R.csv,header=TRUE,stringsAsFactors=FALSE,sep=\t,row.names=1))
is.matrix(mat)
#[1] TRUE
head(mat,2)
COG0001H COG0002E COG0003P COG0004P COG0005F COG0006E COG0007H
NC_014218 NA NA NA NA 274 502 NA
NC_013501 437 345 NA 458 284 364 271
COG0008J COG0009J COG0010E COG0011S COG0012J COG0013J COG0014E
NC_014218 427 212 NA 103 365 890 NA
NC_013501 500 317 316 NA 366 953 NA
COG0015F
NC_014218 479
NC_013501 431
I am not sure I understand what you really wanted. In the matrix example I
showed, I did the table for all the elements of the matrix. Do you want a
columnwise histogram or a rowwise histogram??
A.K.
On Tuesday, May 20, 2014 1:29 PM, Effat Habibi habibi...@gmail.com wrote:
Thanks, it works. But this histogram doesn't show any information about my
data,
I want to make an informative histogram of my data which tells about length of
different COGs across
the genome.
__
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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] Post-hoc tests on linear mixed model give mixed results.
Dear all, I am quite new to R so apologies if I fail to ask properly. I have done a test comparing bat species richness in five habitats as assessed by three methods. I used a linear mixed model in lme4 and got habitat, method and the interaction between the two as significant, with the random effects explaining little variation. I then ran Tukey's post hoc tests as pairwise comparisons in three ways: Firstly in lsmeans: lsmeans(LMM.richness, pairwise~Habitat*Method, adjust=tukey) Then in agricolae: tx - with(diversity, interaction(Method, Habitat)) amod - aov(Richness ~ tx, data=diversity) library(agricolae) interaction -HSD.test(amod, tx, group=TRUE) interaction Then in ghlt 'multcomp': summary(glht(LMM.richness, linfct=mcp(Habitat=Tukey))) summary(glht(LMM.richness, linfct=mcp(Method=Tukey))) tuk - glht(amod, linfct = mcp(tx = Tukey)) summary(tuk) # standard display tuk.cld - cld(tuk) # letter-based display opar - par(mai=c(1,1,1.5,1)) par(mfrow=c(1,1)) plot(tuk.cld) par(opar) I got somewhat different levels of significance from each method, with ghlt giving me the greatest number of significant results and lsmeans the least. All the results from all packages make sense based on the graphs of the data. Can anyone tell me if there are underlying reasons why these tests might be more or less conservative, whether in any case I have failed to specify anything correctly or whether any of these post-hoc tests are not suitable for linear mixed models? Thankyou for your time, Claire [[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] Boxplot colors
Thanks very much, that worked superb! On Thu, May 22, 2014 at 3:41 PM, Richard M. Heiberger r...@temple.eduwrote: ## Shane, ## This uses your color_list. Your other variables weren't included in the email ## so I invented some data. ## I recommend bwplot() using panel=panel.bwplot.superpose ## and I also show how to use boxplot() ## install.packages(HH) ## if necessary. library(HH) ## install.packages(reshape2) ## if necessary. library(reshape2) ## you didn't include data, so I am making up some data pH - data.frame(matrix(rnorm(200), 10, 20)) bwplot(value ~ variable, data=melt(pH), col=unlist(colors_list), panel=panel.bwplot.superpose, groups=variable) boxplot(value ~ variable, data=melt(pH), col=unlist(colors_list)) ## Rich On Thu, May 22, 2014 at 5:50 AM, Shane Carey careys...@gmail.com wrote: Hi Im producing boxplots based on factors and rearranging them by median (This is for a Geochemistry element). Im giving each boxplot a unique color based on its level (factor) name. Im trying to produce a look up list to produce these colors as the order of the boxplots will change from element to element. Here is by color lool up and boxplot code: colors_list-list(Dalradian Appin Group quartzite=#00,Sperrins Dalradian (position uncertain)=#D7D79E, Namurian sandstone, shale=#C7D79E,Lr-Mid Ordovocian acid volcanics=#FFD37F, Tertiary granite, felsite=#FF5500,Caledonian appinite suite=#946379, Dalradian Argyll Gp quartzite=#00,Slishwood Division =#FFEDCC, Dalradian S Highland Gp volcanics=#00E6A9,LEITRIM GP; Visean mudstone, sandstone=#E6, Dalradian Argyll Group=#FFD9C7,Metadolerite or amphibolite=#4C7300, Caledonian granite=#FF7F7F,Dalradian Appin Group=#9EAAD7, Lr-Mid Ordovician basic volcanics=#448970,Ordovician Granite=#E6, Tertiary basic intrusion=#8400A8,Dalradian Argyll Gp volcanics=#00E6A9, VISEAN \\basal clastics\\=#73A6A6,Carboniferous volcs minor intrus=#AAFF00, Metagabbro, metadiorite (Tyrone Plu=#00A885,TYRONE GP; Visean mudstone, sandstone=#CC, Late Visean-Westphalian 'ORS'=#F5A27A,Dalradian S Highland Group=#FFD9C7, Mid Devonian ORS=#FFD37F,Mid-Up Ordovician slate=#C2753D, Interbasaltic formation laterite, b=#00,Visean shelf limestone, shale=#FF00C5, Tertiary rhyolite (volcintru)=#FFEABE,COURCEYAN \\basal clastics\\=#4D8099, Up Dev-Lr Carb ORS=#CDAA66,Tyrone CI (Corvanaghan=?Slishwood=#C9FFC9, Visean basinal limestone \\Calp\\=#CFD6EB,Mid-Up Ordovician g'wacke, sndst, shale=#C29ED7, Up Silurian - Lr Devonian ORS=#CD8966,Westphalian shale, sandstone=#B2B2B2, Upper Basalt Formation=#FAC2E0,Causeway Tholeiite Mbr=#C29ED7, Courceyan limestone=#73B2FF,Tertiary minor volcanics=#E600A9, Silurian sandstone, g'wacke, shale=#ADE6D1,Up Cretaceous limestone=#73FFDF, Lower Basalt Formation=#E0A8C7,Ballycastle succession=#CCB3B3, Navan Group=#408CBF,Devonian basic volcs, minor intrus=#A5F57A, Up. Ord-Sil \\Moffat shale\\ etc.=#DB,Oligocene clay, sand=#704489, Lr Jurassic mudstone=#704489,Lr Jurassic mudstone=#267300, Mid-Up Ordovician acid volcanics=#448970,Armagh Gp=#8073B3, Rathkenny Formation=#704489,Triassic sandstone=#F7DB5E, Waulsortian Limestones=#B2B2B2,Permian sandstone=#F5CA7A, Mid-Up Ordovician basic volcanics=#448970) boxplot(na.omit(C_pH$DATA.pH)~bymedian,axes=FALSE,horizontal=TRUE,col=unlist(colors_list),outwex=one,whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize) However it is not reading the colors based on the color_list given here. Anyone know how to solve this? Thanks for your help -- Shane [[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. -- Shane [[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] Post-hoc tests on linear mixed model give mixed results.
Wrong list! This does not concern R programming. Post on the r-sig-mixed-models list instead in **PLAIN TEXT** rather than html. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Thu, May 22, 2014 at 6:52 AM, Claire c.word...@live.com wrote: Dear all, I am quite new to R so apologies if I fail to ask properly. I have done a test comparing bat species richness in five habitats as assessed by three methods. I used a linear mixed model in lme4 and got habitat, method and the interaction between the two as significant, with the random effects explaining little variation. I then ran Tukey's post hoc tests as pairwise comparisons in three ways: Firstly in lsmeans: lsmeans(LMM.richness, pairwise~Habitat*Method, adjust=tukey) Then in ‘agricolae’: tx - with(diversity, interaction(Method, Habitat)) amod - aov(Richness ~ tx, data=diversity) library(agricolae) interaction -HSD.test(amod, tx, group=TRUE) interaction Then in ghlt 'multcomp': summary(glht(LMM.richness, linfct=mcp(Habitat=Tukey))) summary(glht(LMM.richness, linfct=mcp(Method=Tukey))) tuk - glht(amod, linfct = mcp(tx = Tukey)) summary(tuk) # standard display tuk.cld - cld(tuk) # letter-based display opar - par(mai=c(1,1,1.5,1)) par(mfrow=c(1,1)) plot(tuk.cld) par(opar) I got somewhat different levels of significance from each method, with ghlt giving me the greatest number of significant results and lsmeans the least. All the results from all packages make sense based on the graphs of the data. Can anyone tell me if there are underlying reasons why these tests might be more or less conservative, whether in any case I have failed to specify anything correctly or whether any of these post-hoc tests are not suitable for linear mixed models? Thankyou for your time, Claire [[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.
[R] How do I move the axis labels precisely
#trying for both period-labeled and frequency labeled horizontal axis
rm(list=ls(all=TRUE))
install.packages('plotrix')
library(plotrix)
staxlab-function(side=1,at,labels,nlines=2,top.line=0.5,
line.spacing=0.8,srt=NA,ticklen=0.03,adj=1,...) {
if(missing(labels)) labels-at
nlabels-length(labels)
if(missing(at)) at-1:nlabels
axislim-par(usr)[3:4-2*side%%2]
if(any(at axislim[1]) || any(at axislim[2]))
warning(Some axis labels are off the plot)
if(is.na(srt)) {
linepos-rep(top.line,nlines)
for(i in 2:nlines) linepos[i]-linepos[i-1]+line.spacing
linepos-rep(linepos,ceiling(nlabels/nlines))[1:nlabels]
axis(side=side,at=at,labels=rep(,nlabels))
mtext(text=labels,side=side,line=linepos,at=at,...)
}
else {
linewidth-strheight(M)
xylim-par(usr)
if(side == 1) {
xpos-at
if(par(ylog)) ypos-10^(xylim[3]-ticklen*(xylim[4]-xylim[3]))
else ypos-xylim[3]-ticklen*(xylim[4]-xylim[3])-top.line*linewidth
}
if(side == 3) {
xpos-at
if(par(ylog)) ypos-10^(xylim[4]+ticklen*(xylim[4]-xylim[3]))
else ypos-xylim[4]+ticklen*(xylim[4]-xylim[3])+top.line*linewidth
}
if(side == 2) {
ypos-at
if(par(xlog)) xpos-10^(xylim[1]-ticklen*(xylim[2]-xylim[1]))
else xpos-xylim[1]-ticklen*(xylim[2]-xylim[1])-top.line*linewidth
}
if(side == 4) {
ypos-at
if(par(xlog)) xpos-10^(xylim[2]+ticklen*(xylim[2]-xylim[1]))
else xpos-xylim[2]+ticklen*(xylim[2]-xylim[1])+top.line*linewidth
}
par(xpd=TRUE)
text(xpos,ypos,labels,srt=srt,adj=adj,...)
par(xpd=FALSE)
}
}
horAxisLims=c(0,7213); print(horAxisLims)
par(mar=c(8,4,4,2)+.1,lheight=0.7) #margins=c(bot,lef,top,rit);
default:c(5,4,4,2)+.1
#vertical text space = char height * character expansion * lheight
#set only by
par():ask,fig,fin,lheight,mai,mar,mex,mfcol,mfrow,mfg,new,oma,omd,omi,pin,plt,ps,pty,usr,xlog,ylog,ylbias
prdAxDistDown=0; frqAxDistDown=3.5;
verData=c(1,365,809,1252,1753,2191,2922,3409,3896,4383,4819,5255,5691,6128,6564,7000)
#not fussy
horData=c(1,300,800,1200,1700,2100,2900,3400,3800,4300,4800,5200,5600,6100,6500,7000)
#not at tics
prdTicLocs=c(1,365.,1252,1461,1753,2191,2922,4383,7000)
prdLabels=c(1Yr,1Da,7Hr,6Hr,5Hr,4Hr,3Hr,2Hr,1.25\nHr);
print(prdLabels)
nFrqTicInvls=9; frqTicLocs - vector(length=nFrqTicInvls+1); frqLabels -
vector(length=nFrqTicInvls+1);
frqInvl=(horAxisLims[2]-horAxisLims[1])/nFrqTicInvls
for(i in 1:(nFrqTicInvls+1))frqTicLocs[i]=horAxisLims[1]+(i-1)*frqInvl
for(i in
1:(nFrqTicInvls+1))frqLabels[i]=format(round(frqTicLocs[i]),digits=4,trim=TRUE,scientific=FALSE)
print(frqLabels)
#default mgp=c(3,1,0) margin line in mex units
c(axisTitle,axisLabel,axisLine)
las=2; #las=2 makes axis labels perpendicular to axis
plot(horData,verData,xaxt='n',xlim=horAxisLims,xlab=)
#tck if = 0.5 then fraction of relevant side; if =1 then gridlines
axis(1,tick=TRUE,line=prdAxDistDown,at=prdTicLocs,labels=rep(,length(prdLabels)),padj=0,tck=-.08)
staxlab(1,top.line=(prdAxDistDown),at=prdTicLocs,labels=prdLabels,srt=90,adj=c(1,0))
axis(1,tick=TRUE,line=frqAxDistDown,at=frqTicLocs,labels=rep(,length(frqLabels)),padj=0,tck=-.08)
staxlab(1,top.line=(frqAxDistDown),at=frqTicLocs,labels=frqLabels,srt=90,adj=c(1,0))
title(xlab=Cycles/Yr,line=6)
--
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Sent from the R help mailing list archive at Nabble.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] Running the cox model in R, through python or unix.
Hi guys, I have a piece of R code. Example code: R library(survival) ReadTable - read.table(TestFile,header=F) CoxModel -coxph(Surv(V1)~V2+V3+V4+V5,data=ReadTable) summary(CoxModel) plot(survfit(CoxModel)) I have to run this on 100 files, instead of 1 TestFile, and get outputs of 100 plots and summaries, instead of one summary and plot. Would anyone know how to run this set of R commands in a shell script on unix, or through a python script? I've been trying to understand Rpy2, PypeR and pyRserve, but I was wondering if someone had come across this before, and could show me a simple example of running R commands (such as those above) through unix or python. Many thanks Aoife -- View this message in context: http://r.789695.n4.nabble.com/Running-the-cox-model-in-R-through-python-or-unix-tp4691077.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] Questions in R about optim( ) and optimize( )
You need to provide reproducible examples if you want to get readers to
actually give you answers.
The two issues both come up from time to time and generally relate to
how the objective function is set up, though sometimes to options in the
call. However, generally really isn't good enough.
JN
Date: Wed, 21 May 2014 20:37:10 +
From: Gao, Vicky v...@panagora.com
To: 'r-help@r-project.org' r-help@r-project.org
Subject: [R] Questions in R about optim( ) and optimize( )
Message-ID:
e6231a6c907f1e4ebdc773df7f92327cdfe...@exmailbox1.panagora.com
Content-Type: text/plain
Hello,
I would be really thankful if I could bother you with two questions.
1. I ran into some problems when I used the code optim() in my work. I
used it to optimize a function-fn(x,y), where x is a two-dimensional
parameter. For y I want to pass the value later when I use optim(), because
it needs to be inside of a loop, which changes the value of y every time.
For (point in 100:500){
optim(c(0,0.5),fn,p=point)
}
But it always turned to be Error in fn(par, ...) : argument p is missing,
with no default. I'm confused about this, is that caused by optim() has
some specific requirements towards the objective function?
2. I'm also confused about optimize() actually. My code is like:
optimize(f,c(0,0.99),b=1.0,p=point). I have nothing wrong when I run it,
but the result is not accurate enough. Like I got the $minimum as 0.39 from
the interval as above (0,0.99), however, I got 0.87 when I change it to
(0.2,0.99), 0.96 for (0.3, 0.99). And the only thing I changed is the
interval value, so I'm not sure what I should do to improve it.
As above, could you help me with these two problems? I would really
appreciate that if you could give me some information.
Thank you very much! I look forward for your reply soon.
Vicky
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: problem importing apparently common txt files
Thank you! It seems to work. Zuzana On 22 May 2014 15:47, arun smartpink...@yahoo.com wrote: Hi, Check if this works: dat1 - read.table(folder2.txt, header=TRUE,stringsAsFactors=FALSE,sep=\t,fileEncoding=UTF-16) 'data.frame':79 obs. of 6 variables: $ Documento: chr 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0001.png 6143066.17.4.12.DSC_0002.png ... $ Longitud : num 8.83 8.94 NA 8.94 8.99 ... $ Ãngulo : num -90 90.3 NA -89.7 -90 ... $ Ãrea : num NA NA 551 NA NA ... $ Altura : num NA NA 16.7 NA NA ... $ Anchura : num NA NA 52.5 NA NA ... A.K. On Thursday, May 22, 2014 9:11 AM, zuzana zajkova zuzu...@gmail.com wrote: Dear list, I would like to import into R studio txt files coming from Adobe photoshop. These files contain different measures. The files are apparently common txt files with columns separate by tab. However with common function read.table the import process doesn't work and weird symbols appear once imported. An example file is attached. Do you have any idea what it is happening? Thank you in advance Zuzana Here just a view of a file. DocumentoLongitudÃnguloÃreaAlturaAnchura 6143066.17.4.12.DSC_0001.png8.832613-90.00 6143066.17.4.12.DSC_0001.png8.93791890.337288 6143066.17.4.12.DSC_0001.png551.10310916.736842 52.543860 6143066.17.4.12.DSC_0002.png8.942544-89.662891 6143066.17.4.12.DSC_0002.png8.994992-90.00 6143066.17.4.12.DSC_0002.png550.14650716.736842 52.491228 6106937.17.4.12.DSC_0004.png12.159715-89.008611 __ 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. [[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] Adding segments to a dot plot in ggplot2
I'm trying to plot a GWAS (in you will) with lined segments representing an overall p-value for each gene. Here is my code: skatg - ggplot(comm, aes(x = position,y = p, colour = grey)) + geom_point(size = 0.75) + geom_segment(data=rare, aes(x = txStart, y = -log10(p), xend=txEnd, yend = -log10(p), colour = darkgreen)) + labs(x = Position,y = -log10 P value) + facet_wrap(~ Chrom, scales = free, ncol = 4) Where comm is a file with 250k+ variants and genes.in.locus is a file with about 18k genes. When running this script, I get the error Don't know how to automatically pick scale for object of type function. Defaulting to continuous Error in data.frame(x = c(40840353L, 31902418L, 19468080L, 236748505L, : arguments imply differing number of rows: 79746, 0 Is this because there are different number of rows in each data frame I'm trying to plot? If so, what is a best way to overcome this error? Example of my data is as follows: comm: Namegene Chrom position p 1 rs1037FAM114A1 4 38924330 0.7513597 2 rs1250 CC2D2A 4 15482477 0.9202882 4 rs1911 USP38 4 144136193 0.8335902 5 rs10001 STXBP219 7711221 0.4709547 7 rs10001370 USP46 4 53463730 0.8759828 8 rs1000152 ZNF462 9 109687288 0.3451001 10 rs10002583POLN 4 2194953 0.7878575 12 rs10002971 EGF 4 110896050 0.5082255 15 rs10003873 SORBS2 4 186605868 0.2309855 16 rs10003909ARHGAP24 4 86915848 0.8714853 17 rs10003947 ANXA3 4 79512800 0.5141532 18rs10004SSR1 6 7310259 0.6851725 20 rs10004136 STX18 4 4463587 0.5296092 21 rs10004516 ENPEP 4 111398208 0.8564897 22 rs1000521 SLC8A314 70522484 0.6234326 23 rs10005849 DCHS2 4 155287317 0.8192577 24 rs10006362 RGS12 4 3319271 0.8061674 25 rs1000640WWP26 69905668 0.2682735 26 rs10006580 PCDH18 4 138449812 0.5178650 27 rs10006676 CYTL1 4 5021086 0.3531493 28 rs10006845 PCDH7 4 31116375 0.4817453 29 rs10007075 NEIL3 4 178274694 0.5433481 31 rs10008636 TMPRSS11BNL 4 69083563 0.8346434 32 rs10008910UBA6 4 68500171 0.5705853 33 rs10009228 CHRNA9 4 40356422 0.4223378 rare: geneName txStart txEnd Chromposition p 36131YTHDC16026 45746 4 6026 0.5009490 10898 FAM110C 38813 46588 19 38813 1.000 37306ZNF595 53178 88099 4 53178 0.1261045 16450 KIR2DL4 57208 6812319 57208 0.156 28406SCAND3 61610 77316 6 61610 0.2568 19926 MPG 127017 1358506 127017 00.000987456 34149TRIM27 174179 195169 6 174179 0.025698 I haven't included all information here. Any help will be greatly appreciated. Thanks! __ 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] Variation Inflation factor for GLS
Dear Laura, I've modified vif() in the development version of the car package on R-Forge so that it works with a wider variety of models, including gls models. Once the package is built on R-Forge, which usually takes about a day, you can install it via install.packages(car, repos=http://R-Forge.R-project.org;). Eventually, the development version of the car package will be moved to CRAN. Best, John -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Laura Riggi Sent: Tuesday, May 20, 2014 9:27 AM To: r-help@R-project.org Subject: [R] Variation Inflation factor for GLS Dear all, I am running a gls and I would like to check the vif of my model. It seems that the vif function in the car package and the vif.mer function available online do not work for gls. Would you know of a method to measure variance inflation factors for GLS? Thank you Laura [[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] Post-hoc tests on linear mixed model give mixed results.
Thanks Bert, Will post on r-sig-mixed-models list. Can't help it being in html though as i sent the query via -email. Cheers Claire Date: Thu, 22 May 2014 09:29:44 -0700 Subject: Re: [R] Post-hoc tests on linear mixed model give mixed results. From: gunter.ber...@gene.com To: c.word...@live.com CC: r-help@r-project.org Wrong list! This does not concern R programming. Post on the r-sig-mixed-models list instead in **PLAIN TEXT** rather than html. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Thu, May 22, 2014 at 6:52 AM, Claire c.word...@live.com wrote: Dear all, I am quite new to R so apologies if I fail to ask properly. I have done a test comparing bat species richness in five habitats as assessed by three methods. I used a linear mixed model in lme4 and got habitat, method and the interaction between the two as significant, with the random effects explaining little variation. I then ran Tukey's post hoc tests as pairwise comparisons in three ways: Firstly in lsmeans: lsmeans(LMM.richness, pairwise~Habitat*Method, adjust=tukey) Then in agricolae: tx - with(diversity, interaction(Method, Habitat)) amod - aov(Richness ~ tx, data=diversity) library(agricolae) interaction -HSD.test(amod, tx, group=TRUE) interaction Then in ghlt 'multcomp': summary(glht(LMM.richness, linfct=mcp(Habitat=Tukey))) summary(glht(LMM.richness, linfct=mcp(Method=Tukey))) tuk - glht(amod, linfct = mcp(tx = Tukey)) summary(tuk) # standard display tuk.cld - cld(tuk) # letter-based display opar - par(mai=c(1,1,1.5,1)) par(mfrow=c(1,1)) plot(tuk.cld) par(opar) I got somewhat different levels of significance from each method, with ghlt giving me the greatest number of significant results and lsmeans the least. All the results from all packages make sense based on the graphs of the data. Can anyone tell me if there are underlying reasons why these tests might be more or less conservative, whether in any case I have failed to specify anything correctly or whether any of these post-hoc tests are not suitable for linear mixed models? Thankyou for your time, Claire [[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. [[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] Random forest proximity measure
Hi all, I've been using the randomForest package on a dataset (described later) and my problem is: even though I specify proximity= TRUE in the call I get a NULL proximity matrix. Any thoughts on why that may happen? Unfortunately I can't post my dataset, which is particularly problematic here since i believe that's where the problem is. So I'll try to give as detailed of an account as i can. The outcome is binary, highly skewed with the positive outcome being 1.5% of the data. The dataset has ~7000 observations and 200 predictors. these are either 2 level factors or continuous variables. Extremely sparse. Here is my call: #i pass a balanced dataset for each tree, to deal with the skewed outcome. rf-randomForest(y~. ,data=train, ntree=800,replace=TRUE,sampsize = c(112, 112), proximilty=TRUE) Any ideas on why im getting a null proximity measure/ solutions? Thanks! [[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] Zip multiple files with same prefix?
I have thousands of files that need to be zipped. The files that need to be zipped together have the same seven digit prefix e.g. 40111h1. I am a novice at R-Studio and would appreciate some help on this matter. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Zip-multiple-files-with-same-prefix-tp4691088.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] equivalent of R CMD BATCH --vanilla for those who can't do anything non-GUI?
For a Google Group about aster models, I want to say that people wanting help are best advised to provide an example that works as R CMD BATCH --vanilla foo.R but I realize that many R users have zero idea of how to start R in any way other than clicking on an icon. Is there a way to start up the standard mac and windows GUIs or Rstudio with no loaded saved global environment? (Without making it impossible to go back to what they were doing before?) Is there a way to make the source function do the job (ignore everything in the global environment)? What do you tell users about how to make an example that doesn't assume there is huge amounts of crap that the user doesn't even remember what it is that is involved? Do I just have to explain the command line to all the GUI fans? -- Charles Geyer Professor, School of Statistics University of Minnesota char...@stat.umn.edu [[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] equivalent of R CMD BATCH --vanilla for those who can't do anything non-GUI?
You could have them spawn a vanilla R session using system instead of
the command line:
system('R CMD BATH --vanilla foo.R')
Or you could use the local argument to source to evaluate in a new
environment that does not inherit from the global environment:
source('foo.R', local=new.env(parent=parent.env(.GlobalEnv)))
this will prevent the code seeing anything in the global environment,
but it could still use functions/objects from any loaded packages.
On Thu, May 22, 2014 at 3:00 PM, Charles Geyer char...@stat.umn.edu wrote:
For a Google Group about aster models, I want to say that people wanting
help are best advised to provide an example that works as
R CMD BATCH --vanilla foo.R
but I realize that many R users have zero idea of how to start R in any way
other than clicking on an icon. Is there a way to start up the standard
mac and windows GUIs or Rstudio with no loaded saved global environment?
(Without making it impossible to go back to what they were doing before?)
Is there a way to make the source function do the job (ignore everything in
the global environment)?
What do you tell users about how to make an example that doesn't assume
there is huge amounts of crap that the user doesn't even remember what it
is that is involved?
Do I just have to explain the command line to all the GUI fans?
--
Charles Geyer
Professor, School of Statistics
University of Minnesota
char...@stat.umn.edu
[[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.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.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.
Re: [R] subsetting to exclude different values for each subject in study
Hi Everyone, I hope I did this correctly (I called my data frame ao) and Thank you very much for the info about using dput(), I'm starting to understand all the different things that can be done in R and I appreciate all the advice. I must appologize in advance since my coding is quite long but hopefully it makes sense. and there is a efficient way to do this. structure(list(num = 1:99, FORM_CHK = c(20870L, 22018L, 30737L, 22010L, 22028L, 36059L, 36063L, 36066L, 30587L, 30612L, 36056L, 30376L, 35153L, 30435L, 30536L, 30486L, 30475L, 36053L, 36048L, 36076L, 36045L, 36065L, 35772L, 36949L, 35702L, 36894L, 36080L, 35542L, 35457L, 35533L, 36042L, 36925L, 36827L, 36008L, 35817L, 36350L, 35985L, 35973L, 35801L, 36639L, 35810L, 35812L, 35807L, 36351L, 35967L, 35944L, 37006L, 36345L, 36062L, 36077L, 35802L, 35984L, 36043L, 35769L, 36360L, 36082L, 36071L, 36354L, 35771L, 35754L, 36295L, 35746L, 36064L, 35779L, 35751L, 35752L, 35785L, 35792L, 37011L, 36003L, 36040L, 36831L, 36031L, 36652L, 36992L, 36965L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), RingNummerMan = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 19L, 22L, 23L, 24L, 25L, 26L, 27L, 29L, 30L, 31L, 34L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 43L, 44L, 46L, 47L, 48L, 49L, 50L, 51L, 52L, 54L, 55L, 56L, 57L, 58L, 59L, 60L, 61L, 63L, 65L, 67L, 69L, 70L, 73L, 74L, 75L, 76L, 78L, 79L, 80L, 81L, 82L, 83L, 85L, 86L, 87L, 88L, 89L, 93L, 96L, 97L, 18L, 20L, 21L, 28L, 32L, 33L, 45L, 53L, 62L, 64L, 66L, 68L, 71L, 72L, 77L, 84L, 90L, 91L, 92L, 94L, 95L, 98L, 99L), .Label = c(AJ...75425, AL...62371, AR...11060, AR...29297, AR...29307, AR...29502, AR...29504, AR...29507, AR...30039, AR...30085, AR...30165, AR...30491, AR...30563, AR...30616, AR...30652, AR...30687, AR...30701, AR...30927, AR...30959, AR...30963, AR...30964, AR...30965, AR...30966, AR...30985, AR...30988, AR...40917, AR...40996, AR...45735, AR...45904, AR...45928, AR...47609, AR...65387, AR...65479, AR...65550, AR...65629, AR...65948, AR...86074, AR...86521, AR...86527, AR...90061, AR...90064, AR...90067, AR...90077, AR...90081, AR...90098, AR...90101, AR...90106, AR...90112, AR...90133, AR...90155, AR...90176, AR...90178, AR...90180, AR...90187, AR...90212, AR...90247, AR...90252, AR...90256, AR...90258, AR...90269, AR...90272, AR...90275, AR...90294, AR...90298, AR...90300, AR...90337, AR...90338, AR...90367, AR...90397, AR...90410, AR...90463, AR...90520, AR...90544, AR...90556, AR...90678, AR...90712, AR...90737, AR...90744, AR...90829, AR...90862, AR...90863, AR...90873, AR...90880, AR...90892, AR...90898, AR...90945, AR...90951, AR...90965, AR...90970, AR...90972, AU...15008, AU...15009, AU...15027, AU...15032, AU...15036, AU...15038, AU...15046, AU...15049, AU...15505), class = factor), year_score_taken = c(2006L, 2008L, 2009L, 2008L, 2008L, 2011L, 2011L, 2011L, 2009L, 2009L, 2011L, 2009L, 2010L, 2009L, 2009L, 2009L, 2009L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2012L, 2011L, 2012L, 2011L, 2010L, 2010L, 2010L, 2011L, 2012L, 2012L, 2011L, 2011L, 2012L, 2011L, 2011L, 2011L, 2012L, 2011L, 2011L, 2011L, 2012L, 2011L, 2011L, 2013L, 2012L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2012L, 2012L, 2011L, 2012L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2013L, 2011L, 2011L, 2012L, 2011L, 2012L, 2012L, 2012L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), COR_LOC = c(15.13404, 13.88054, 30.0969, 19.09152, 16.88054, 14.15718, 39.15718, 16.15718, 16.13566, 23.07538, 39.15718, 24.56838, 12.13942, 21.4123, 19.06945, 12.33264, 32.48872, 30.15718, 37.15718, 37.15718, 49.15718, 22.15718, 18.50272, 23.69432, 24.9322, 47.29712, 41.15718, 21.47903, 38.6588, 34.99572, 28.15718, 13.08614, 16.71908, 22.68894, 19.2616, 15.96234, 22.83964, 13.89992, 14.2616, 18.17118, 24.2616, 22.2616, 13.2616, 23.96234, 24.89992, 24.05062, 10.20884, 6.96234, 13.15718, 17.15718, 40.2616, 21.83964, 20.15718, 39.50272, 26.81164, 20.3843, 14.15718, 7.96234, 19.50272, 40.74384, 5.7675, 42.95482, 29.15718, 18.32188, 28.74384, 37.74384, 22.32188, 25.32188, 18.20884, 14.68894, 22.15718, 39.71908, 18.2067, 15.1109, 15.61466, 47.4532, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), IndividuID = c(11394L, 15676L, 342518L, 344902L, 344909L, 377497L, 377499L, 377504L, 352003L, 351986L, 352260L, 352392L, 353800L, 353892L, 353949L, 354060L, 354074L, 377487L, 377490L, 377511L, 377513L, 377495L, 377297L, 357796L, 366326L, 378446L, 377518L, 358157L, 358730L, 366215L, 377519L, 378407L, 378453L, 377443L, 377358L, 377726L, 377422L, 377402L, 377341L, 378354L, 377350L, 377352L, 377347L, 378408L, 377396L, 377374L, 34L, 377743L, 377500L, 377510L, 377342L, 377421L, 377786L, 377294L, 377836L, 378291L, 377508L, 378199L, 377296L, 377280L, 373000L, 373020L, 377496L, 377306L, 373025L, 377278L, 377310L, 377317L, 377337L, 377439L, 377450L, 377464L,
Re: [R] How do I move the axis labels precisely
I included runnable code to help demonstrate what I want to do but Sorry in my rush I forgot to include this request: I want to move the x axis labels to make the right end close to the axis with a very small gap, and the bottom close to the tick with an even smaller gap. The staxlab function to be standard soon was sent to me by the author to make staxlab work on the second x axis. Hurr -- View this message in context: http://r.789695.n4.nabble.com/How-do-I-move-the-axis-labels-precisely-tp4691082p4691095.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] equivalent of R CMD BATCH --vanilla for those who can't do anything non-GUI?
On 22/05/2014, 5:00 PM, Charles Geyer wrote: For a Google Group about aster models, I want to say that people wanting help are best advised to provide an example that works as R CMD BATCH --vanilla foo.R but I realize that many R users have zero idea of how to start R in any way other than clicking on an icon. Is there a way to start up the standard mac and windows GUIs or Rstudio with no loaded saved global environment? (Without making it impossible to go back to what they were doing before?) On Windows, you can copy the shortcut, and edit it to add a --vanilla option to the command line. Is there a way to make the source function do the job (ignore everything in the global environment)? Only in the way Greg said: start a new R process to run it. What do you tell users about how to make an example that doesn't assume there is huge amounts of crap that the user doesn't even remember what it is that is involved? I just ask for a minimal, self-contained example. You won't always get that, but I suspect people won't always follow your --vanilla instructions, either. Duncan Murdoch Do I just have to explain the command line to all the GUI fans? __ 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] Post-hoc tests on linear mixed model give mixed results.
On 23/05/14 06:59, Claire wrote: Thanks Bert, Will post on r-sig-mixed-models list. Can't help it being in html though as i sent the query via -email. Huh? What do you mean by -email? Do you really mean email (without the minus sign)? We *all* send our queries via email. This *is* email. Set your expletive deleted email *not* to post in html!!! cheers, Rolf __ 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] HMM states simulation
On 22/05/14 21:38, Baba Bukar wrote:
Dear all,
I have 100 observations sampled from poisson and negative binomial
distributions and i want set a hidden states for the observations, class
with zero and one {0,1}. Larger observations to be one and smaller
observations to be zero using R.
Can someone please help?
No, I don't think anyone can help you. You question is too ill-posed,
incoherent and garbled.
If you can actually explain what you want to do, in meaningful terms,
then some *might* be able to help. However I suspect that you really
have no clear idea of what you want to do, and/or what you want to do
makes no sense at all.
cheers,
Rolf Turner
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Re: [R] Voronoi-Diagrams in R
I just tried the following: require(deldir) require(geometry) set.seed(42) bm - data.frame(x=sample(1:82,800,TRUE),y=sample(1:82,800,TRUE)) del - deldir(bm) tri - triang.list(del) geodel - delaunayn(bm) length(tri) # Got [1] 1481 dim(geodel) # Got [1] 14813 So all seems to be in harmony in the universe. Something is weird about your bm; hard to say what without seeing it. One thought: Are there any duplicated points in your bm? (Upon re-reading your message and noting the problem with ptNum, I think this might be the case.) Try doing something like: newbm - bm[!duplicated(bm),] and then hit newbm with both deldir() and delaunayn(). cheers, Rolf Turner On 23/05/14 03:36, Raphael Päbst wrote: Hello again, I have found further depths of confusion concerning delaunay triangulations to explore. Here is the code I'm using to create the confusing results: bm - getbm(x) # a data.frame with 2 columns and 800 rows, values are integers ranging from 1 to 82 del - deldir(bm) # creating an object of type deldir with the 800 coordinates from bm tri - triang.list(del) # creating a list of length n, holding the n triangles created by the delaunay triangulation with deldir(bm) as data.frames geodel - delaunayn(bm) # creating a matrix with 3 columns and m rows for the m triangles created by the delaunay triangulation using delaunayn(bm) from geometry-package now, the following is what I would expect: n and m should be the same, since both triangulations should give me the same number of triangles, right? but this is what I get: nrow(geodel) = 1584 length(tri) = 1186 This confuses me and I have the feeling I have left out some very important parameter either in delaunayn() or deldir() to create the different results. There is however more confusion to come, this time in the result of triang.list(del): Every element in that list is a data.frame holding information on one triangle. It has 3 columns and 3 rows with column headers 'ptNum', 'x' and 'y'. Now I would expect, that 'ptNum' would give me the index at which I can find the coordinates 'x' and 'y' as a row in my original data.frame of coordinates, bm. And for example bm[1,] gives me the same 'x' and 'y' that are listed in data.frame tri[1] as 'ptNum' = 1. But for some other points this does not work and for example the 'x' and 'y' of 'ptNum' = 129 do not match bm[129, ]. Have I totally misunderstood the meaning of 'ptNum' in this case or does my mistake lie somewhere else? Once again please excuse my slightly chaotic description of my problems and my faulty english and thanks for any help you can give! __ 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] multiple colour to symbols using brkdn.plot()
On Thu, 22 May 2014 01:43:23 PM Luigi Marongiu wrote: Dear all, I have been plotting response variable (defined as y in the example below) from 4 groups (defined as z variable ) under 5 conditions (defined by the x variable). The formula used is: y ~ z+x. I can differentiate the different z value using a call pch=c(symbol 1, symbol 2, ...) but it would be even better to have different colours for each of these symbols. In analogy to the pch=c(symbol 1, symbol 2, ...) i have tired to provide a bg(colour 1, colour 2, ...) but this give the colour to the x variables rather than the z ones. How can I give the colour to the z variables? Hi Luigi, Very good example. The brkdn.plot function assigns colors to the group variable based upon the col argument. Try using this: brkdn.plot(y~z+x, data=my.data, mct=mean, md=sd, stagger=NA, dispbar=TRUE, type=p, pch=20, col=2:6, main=z indicated by colour (???), xlab=x,ylab=y,cex=2) Jim __ 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] subsetting to exclude different values for each subject in study
Hi, Neighbours in this case were selected if they shared a boundary in the voroni tesellation. Best, Monaly On May 23, 2014 3:19 AM, arun smartpink...@yahoo.com wrote: HI Monaly, Thanks for the code and dput. But, I have a doubt about how you are selecting the neigbours. Is there another dataset with the information? Sorry, if I have missed something For e.g. ### average difference b/n neighbours for each individual XO[avg, 176]- mean(abs((XO[1,176])-XO[1,c(140,162,713)])) A.K. On Thursday, May 22, 2014 5:21 PM, Monaly Mistry monaly.mis...@gmail.com wrote: Hi Everyone, I hope I did this correctly (I called my data frame ao) and Thank you very much for the info about using dput(), I'm starting to understand all the different things that can be done in R and I appreciate all the advice. I must appologize in advance since my coding is quite long but hopefully it makes sense. and there is a efficient way to do this. structure(list(num = 1:99, FORM_CHK = c(20870L, 22018L, 30737L, 22010L, 22028L, 36059L, 36063L, 36066L, 30587L, 30612L, 36056L, 30376L, 35153L, 30435L, 30536L, 30486L, 30475L, 36053L, 36048L, 36076L, 36045L, 36065L, 35772L, 36949L, 35702L, 36894L, 36080L, 35542L, 35457L, 35533L, 36042L, 36925L, 36827L, 36008L, 35817L, 36350L, 35985L, 35973L, 35801L, 36639L, 35810L, 35812L, 35807L, 36351L, 35967L, 35944L, 37006L, 36345L, 36062L, 36077L, 35802L, 35984L, 36043L, 35769L, 36360L, 36082L, 36071L, 36354L, 35771L, 35754L, 36295L, 35746L, 36064L, 35779L, 35751L, 35752L, 35785L, 35792L, 37011L, 36003L, 36040L, 36831L, 36031L, 36652L, 36992L, 36965L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), RingNummerMan = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 19L, 22L, 23L, 24L, 25L, 26L, 27L, 29L, 30L, 31L, 34L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 43L, 44L, 46L, 47L, 48L, 49L, 50L, 51L, 52L, 54L, 55L, 56L, 57L, 58L, 59L, 60L, 61L, 63L, 65L, 67L, 69L, 70L, 73L, 74L, 75L, 76L, 78L, 79L, 80L, 81L, 82L, 83L, 85L, 86L, 87L, 88L, 89L, 93L, 96L, 97L, 18L, 20L, 21L, 28L, 32L, 33L, 45L, 53L, 62L, 64L, 66L, 68L, 71L, 72L, 77L, 84L, 90L, 91L, 92L, 94L, 95L, 98L, 99L), .Label = c(AJ...75425, AL...62371, AR...11060, AR...29297, AR...29307, AR...29502, AR...29504, AR...29507, AR...30039, AR...30085, AR...30165, AR...30491, AR...30563, AR...30616, AR...30652, AR...30687, AR...30701, AR...30927, AR...30959, AR...30963, AR...30964, AR...30965, AR...30966, AR...30985, AR...30988, AR...40917, AR...40996, AR...45735, AR...45904, AR...45928, AR...47609, AR...65387, AR...65479, AR...65550, AR...65629, AR...65948, AR...86074, AR...86521, AR...86527, AR...90061, AR...90064, AR...90067, AR...90077, AR...90081, AR...90098, AR...90101, AR...90106, AR...90112, AR...90133, AR...90155, AR...90176, AR...90178, AR...90180, AR...90187, AR...90212, AR...90247, AR...90252, AR...90256, AR...90258, AR...90269, AR...90272, AR...90275, AR...90294, AR...90298, AR...90300, AR...90337, AR...90338, AR...90367, AR...90397, AR...90410, AR...90463, AR...90520, AR...90544, AR...90556, AR...90678, AR...90712, AR...90737, AR...90744, AR...90829, AR...90862, AR...90863, AR...90873, AR...90880, AR...90892, AR...90898, AR...90945, AR...90951, AR...90965, AR...90970, AR...90972, AU...15008, AU...15009, AU...15027, AU...15032, AU...15036, AU...15038, AU...15046, AU...15049, AU...15505), class = factor), year_score_taken = c(2006L, 2008L, 2009L, 2008L, 2008L, 2011L, 2011L, 2011L, 2009L, 2009L, 2011L, 2009L, 2010L, 2009L, 2009L, 2009L, 2009L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2012L, 2011L, 2012L, 2011L, 2010L, 2010L, 2010L, 2011L, 2012L, 2012L, 2011L, 2011L, 2012L, 2011L, 2011L, 2011L, 2012L, 2011L, 2011L, 2011L, 2012L, 2011L, 2011L, 2013L, 2012L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2012L, 2012L, 2011L, 2012L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2013L, 2011L, 2011L, 2012L, 2011L, 2012L, 2012L, 2012L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), COR_LOC = c(15.13404, 13.88054, 30.0969, 19.09152, 16.88054, 14.15718, 39.15718, 16.15718, 16.13566, 23.07538, 39.15718, 24.56838, 12.13942, 21.4123, 19.06945, 12.33264, 32.48872, 30.15718, 37.15718, 37.15718, 49.15718, 22.15718, 18.50272, 23.69432, 24.9322, 47.29712, 41.15718, 21.47903, 38.6588, 34.99572, 28.15718, 13.08614, 16.71908, 22.68894, 19.2616, 15.96234, 22.83964, 13.89992, 14.2616, 18.17118, 24.2616, 22.2616, 13.2616, 23.96234, 24.89992, 24.05062, 10.20884, 6.96234, 13.15718, 17.15718, 40.2616, 21.83964, 20.15718, 39.50272, 26.81164, 20.3843, 14.15718, 7.96234, 19.50272, 40.74384, 5.7675, 42.95482, 29.15718, 18.32188, 28.74384, 37.74384, 22.32188, 25.32188, 18.20884, 14.68894, 22.15718, 39.71908, 18.2067, 15.1109, 15.61466, 47.4532, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), IndividuID =
[R] Loop Issue
Hi everybody.
Consider the following exampling code:
x=numeric()
for(i in 1:10){
u=runif(1,-1,2)
x[i]=log(u)
}
This code, in each interation, generates a random value in the (-1,2) interval
and then calculates the log of the value.
When the generated value is less than 0 the log produces a NaN, which gives a
warning.
What I want is to make it start over when a warning is produced, in order to
repeat it until a positive value is generated and therefore the log is
calculated. Logically, would be like: if there's a warning here, go back at
the beginning and start over, without changing the iteration.
Could someone help me with some directions?
Thanks a lot,
Ricardo
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Re: [R] Loop Issue
Hi Ricardo
Assuming you have a good reason to use such approach (what are you
trying to do ultimately?), you can just increment your counter when
you get a good value, i.e.:
x - numeric()
n - 0
while (n 10) {
u - log(runif(1, -1, 2))
if (is.finite(u)) {
n - n+1
x[n] - u
}
}
x
On 23 May 2014 12:11, Ricardo Rocha ricardoroch...@hotmail.com wrote:
Hi everybody.
Consider the following exampling code:
x=numeric()
for(i in 1:10){
u=runif(1,-1,2)
x[i]=log(u)
}
This code, in each interation, generates a random value in the (-1,2)
interval and then calculates the log of the value.
When the generated value is less than 0 the log produces a NaN, which gives a
warning.
What I want is to make it start over when a warning is produced, in order to
repeat it until a positive value is generated and therefore the log is
calculated. Logically, would be like: if there's a warning here, go back at
the beginning and start over, without changing the iteration.
Could someone help me with some directions?
Thanks a lot,
Ricardo
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