Re: [R] Extracting Data from SQL Server
Thanks for you help guys, your suggestion Ajay worked fine, cheers
Dyfed Thomas
Population Health Analyst
DDI: 07 858 5967
Mob: 021 409 800
Midlands Health Network
WEL House
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Hamilton 3240
Phone: 07 839 2888
Fax: 07 834 9242
From: Ajay Askoolum [via R] [mailto:ml-node+s789695n4281558...@n4.nabble.com]
Sent: Tuesday, 10 January 2012 10:40 p.m.
To: Dyfed Thomas
Subject: Re: Extracting Data from SQL Server
try:
SELECT a.UNIQUE_ID,
a.diag01
from LoadPUS a
left join CVD_ICD10 b
on a.diag01 = b.[ICD-10 Codes]
or a.diag02 = b.[ICD-10 Codes]
or a.diag03 = b.[ICD-10 Codes]
I am not sure why your table name CVD_ICD10 has a suffix $.
From: Jeff Newmiller <[hidden
email]>
To: dthomas <[hidden email]>;
[hidden email]
Sent: Tuesday, 10 January 2012, 8:00
Subject: Re: [R] Extracting Data from SQL Server
This is OT here. However, you might want to investigate the UNIQUE keyword in
the SQL Server documentation for SELECT.
---
Jeff NewmillerThe . . Go Live...
DCN:<[hidden email]>
Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
dthomas <[hidden email]> wrote:
>Hi,
>
>I am new to R (and rusty on SQL!) and I'm trying to extract records
>from a
>SQL server database. I have a table of patient records (LoadPUS) which
>have
>three code columns which i want to evaluate against a list of
>particular
>codes (CVD_ICD$ table). Given the size of the patient table I want to
>restrict the data I pull into R to the data I only want to analyse so I
>am
>using SQL to do this. The code i have is as follows:
>
>library(RODBC)
>channel<-odbcConnect("NatCollections")
>query<-"SELECT UNIQUE_ID, diag01 from LoadPUS
>WHERE (diag01 IN (SELECT [ICD-10 Codes] From CVD_ICD10$)) OR (diag02 IN
>(SELECT [ICD-10 Codes] From CVD_ICD10$))
>OR (diag03 IN (SELECT [ICD-10 Codes] From CVD_ICD10$))"
>
>This returns duplicate values, I don't want to hardcode the values
>because
>it is quite a long list. Running the "IN" function just for "diag01"
>returns
>the correct number of records, however when combining with another "IN"
>function it doesn't return the correct number of records. Can you see
>where
>my SQL is incorrect or is there another way of doing this?
>
>Much appreciated,
>D
>
>--
>View this message in context:
>http://r.789695.n4.nabble.com/Extracting-Data-from-SQL-Server-tp4281000p4281000.html
>Sent from the R help mailing list archive at Nabble.com.
>
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Re: [R] Finding percentile of a value from an empirical distribution
Thank you. That's easier than I thought. On Wed, Jan 11, 2012 at 12:06 AM, Jorge I Velez wrote: > Hi Robert, > > Try > > set.seed(123) > x <- seq(100) > x <- sample(x, 1000, replace = TRUE) > f <- ecdf(x) > f(10) > # [1] 0.099 > f(71) > # [1] 0.716 > > See ?ecdf for more information. > > HTH, > Jorge.- > > > On Tue, Jan 10, 2012 at 11:52 PM, Robert A'gata <> wrote: >> >> Hello, >> >> I am not sure how to do this in R. Any suggestion would be >> appreciated. I have a vector of values from where I build an empirical >> CDF. For example: >> >> > x <- seq(1,100) >> > x <- sample(x,1000,replace=T) >> > quantile(x,probs=seq(0,1,.05)) >> 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% >> 50% 55% >> 1.00 5.00 10.00 16.00 20.00 25.00 31.00 36.00 41.00 45.55 >> 50.00 56.00 >> 60% 65% 70% 75% 80% 85% 90% 95% 100% >> 60.00 65.00 70.00 74.00 80.00 85.00 91.00 95.05 100.00 >> >> I would like to write a function that takes in a number z and vector x >> (i.e. the raw vector).It returns percentile of z wrt x. E.g. >> >> > f(71,x) >> >> Should return something around 0.708 or 0.709. I am wondering if there >> is any pre-packaged functions that do this in R? If not, how can I >> write such a function. Any suggestion would be appreciated. >> >> Robert >> >> __ >> 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] Finding percentile of a value from an empirical distribution
Hi Robert, Try set.seed(123) x <- seq(100) x <- sample(x, 1000, replace = TRUE) f <- ecdf(x) f(10) # [1] 0.099 f(71) # [1] 0.716 See ?ecdf for more information. HTH, Jorge.- On Tue, Jan 10, 2012 at 11:52 PM, Robert A'gata <> wrote: > Hello, > > I am not sure how to do this in R. Any suggestion would be > appreciated. I have a vector of values from where I build an empirical > CDF. For example: > > > x <- seq(1,100) > > x <- sample(x,1000,replace=T) > > quantile(x,probs=seq(0,1,.05)) >0% 5%10%15%20%25%30%35%40%45% > 50%55% > 1.00 5.00 10.00 16.00 20.00 25.00 31.00 36.00 41.00 45.55 > 50.00 56.00 > 60%65%70%75%80%85%90%95% 100% > 60.00 65.00 70.00 74.00 80.00 85.00 91.00 95.05 100.00 > > I would like to write a function that takes in a number z and vector x > (i.e. the raw vector).It returns percentile of z wrt x. E.g. > > > f(71,x) > > Should return something around 0.708 or 0.709. I am wondering if there > is any pre-packaged functions that do this in R? If not, how can I > write such a function. Any suggestion would be appreciated. > > Robert > > __ > 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] Finding percentile of a value from an empirical distribution
Look at ?ecdf Michael On Jan 10, 2012, at 11:52 PM, "Robert A'gata" wrote: > Hello, > > I am not sure how to do this in R. Any suggestion would be > appreciated. I have a vector of values from where I build an empirical > CDF. For example: > >> x <- seq(1,100) >> x <- sample(x,1000,replace=T) >> quantile(x,probs=seq(0,1,.05)) >0% 5%10%15%20%25%30%35%40%45% > 50%55% > 1.00 5.00 10.00 16.00 20.00 25.00 31.00 36.00 41.00 45.55 > 50.00 56.00 > 60%65%70%75%80%85%90%95% 100% > 60.00 65.00 70.00 74.00 80.00 85.00 91.00 95.05 100.00 > > I would like to write a function that takes in a number z and vector x > (i.e. the raw vector).It returns percentile of z wrt x. E.g. > >> f(71,x) > > Should return something around 0.708 or 0.709. I am wondering if there > is any pre-packaged functions that do this in R? If not, how can I > write such a function. Any suggestion would be appreciated. > > Robert > > __ > 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] Finding percentile of a value from an empirical distribution
Hello, I am not sure how to do this in R. Any suggestion would be appreciated. I have a vector of values from where I build an empirical CDF. For example: > x <- seq(1,100) > x <- sample(x,1000,replace=T) > quantile(x,probs=seq(0,1,.05)) 0% 5%10%15%20%25%30%35%40%45% 50%55% 1.00 5.00 10.00 16.00 20.00 25.00 31.00 36.00 41.00 45.55 50.00 56.00 60%65%70%75%80%85%90%95% 100% 60.00 65.00 70.00 74.00 80.00 85.00 91.00 95.05 100.00 I would like to write a function that takes in a number z and vector x (i.e. the raw vector).It returns percentile of z wrt x. E.g. > f(71,x) Should return something around 0.708 or 0.709. I am wondering if there is any pre-packaged functions that do this in R? If not, how can I write such a function. Any suggestion would be appreciated. Robert __ 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] try() with silent=TRUE not preventing printing of error message
Hello, We're curious to know why, on builds lacking Tcl/Tk, try(library(tcltk),silent=TRUE) still allows the message to be printed: Error in firstlib(which.lib.loc, package) : Tcl/Tk support is not available on this system though it does succeed in trapping the error condition? Thanks, Ben __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help
That sort of name is allowed but not advised because it can lead to confusion
in certain non-standard evaluation functions like subset(). If you really want
the name like that add the check.names = FALSE argument to read.table()
Michael Weylandt
On Jan 10, 2012, at 5:57 PM, Anna Olofsson wrote:
> Thank you! The c was missing. I don't know if it's ok to continue on this
> thread, but I also had another question about reading data. I have this
> file containing 3 columns and 19 rows.
>
> 00.960.21
> 00.450.4
> 00.870.1
> 00.560.04
> 00.570.04
> 00.20.7
> 00.450.43
> 00.350.21
> 00.750.56
> 10.630.43
> 10.950.32
> 10.420.2
> 10.120.05
> 10.560.06
> 10.340.3
> 10.10.7
> 10.110.75
> 10.20.21
> 10.950.37
>
> I tried to read it into R, but I'm not exactly sure exactly what to use as
> input. This is my input line using read.table:
>
> data1 <- read.table(file = "filename.txt", header=FALSE, col.names =
> c("class", "P", "1G"))
>
> but in the output I get an X infront of "1G", which disappears when I run
> it with the name 'G' instead of '1G'. Am I not allowed to use numerical
> values?
>
> Best,
> Anna
>
>
>
>
> On Tue, 10 Jan 2012 23:02:04 +0100, Anna Olofsson
> wrote:
>> Hi,
>> I'm pretty new at programming and with the R language. I'm just trying
> to
>> get familiar with R and wrote a script in gedit (should I use emacs
>> instead?),
>>
>> x <- [10.4 5.6 3.1 6.4 21.7]
>> y <- [12,5.6, 7.2, 1.0, 9.3]
>> plot(x,y)
>>
>> then I went to the command window in the terminal (I'm using unix) to
> run
>> this with source("name_of_file"), but it doesn't work. Shouldn't a plot
>> come up automatically when I run it? What am I doing wrong? It knows
> what x
>> and y is, but I don't get an error of what might be wrong.
>>
>>> source("name_of_file")
>>> x
>> [1] 10.4 5.6 3.1 6.4 21.7
>>
>>
>> Best,
>> Anna
>
> __
> R-help@r-project.org mailing list
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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.
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Re: [R] help
Thank you! The c was missing. I don't know if it's ok to continue on this
thread, but I also had another question about reading data. I have this
file containing 3 columns and 19 rows.
0 0.960.21
0 0.450.4
0 0.870.1
0 0.560.04
0 0.570.04
0 0.2 0.7
0 0.450.43
0 0.350.21
0 0.750.56
1 0.630.43
1 0.950.32
1 0.420.2
1 0.120.05
1 0.560.06
1 0.340.3
1 0.1 0.7
1 0.110.75
1 0.2 0.21
1 0.950.37
I tried to read it into R, but I'm not exactly sure exactly what to use as
input. This is my input line using read.table:
data1 <- read.table(file = "filename.txt", header=FALSE, col.names =
c("class", "P", "1G"))
but in the output I get an X infront of "1G", which disappears when I run
it with the name 'G' instead of '1G'. Am I not allowed to use numerical
values?
Best,
Anna
On Tue, 10 Jan 2012 23:02:04 +0100, Anna Olofsson
wrote:
> Hi,
> I'm pretty new at programming and with the R language. I'm just trying
to
> get familiar with R and wrote a script in gedit (should I use emacs
> instead?),
>
> x <- [10.4 5.6 3.1 6.4 21.7]
> y <- [12,5.6, 7.2, 1.0, 9.3]
> plot(x,y)
>
> then I went to the command window in the terminal (I'm using unix) to
run
> this with source("name_of_file"), but it doesn't work. Shouldn't a plot
> come up automatically when I run it? What am I doing wrong? It knows
what x
> and y is, but I don't get an error of what might be wrong.
>
>> source("name_of_file")
>> x
> [1] 10.4 5.6 3.1 6.4 21.7
>
>
> Best,
> Anna
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[R] Problems with constrOptim
I am having problems with the constrOptim problem. I am trying to compute a function which must be concave. When I put the linear constraints in, constrOptim always returns the original values of the parameter entered after several iterations. However, if I set ui to the 0 matrix, it will optimize the function. Does this imply the constraints are restraining the algorithm from exploring the parameter space correctly? __ 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] 64bit R under 32bit winxp
Hi all: My OS is 32bit winxp,but I wanna install 64bit R2.14.1. >From the following website,it says "You can also go back and add 64-bit >components to a 32-bit install, or vice versa" http://cran.r-project.org/bin/windows/rw-FAQ.html#Can-both-32_002d-and-64_002dbit-R-be-installed-on-the-same-machine_003f Does it mean that I can install and run 64bit R2.14.1 under 32bit winxp?If so,how can I "add 64-bit components to a 32-bit install"? Many thanks for your help. My best [[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] different results from fligner.test
Yeah! it works, i know it because of the degrees of freedom. This is what i get now > g <- interaction(ano, edadysexo, zona, estacion) > fligner.test(rojos ~ g) Fligner-Killeen test of homogeneity of variances data: rojos by g Fligner-Killeen:med chi-squared = 249.7591, df = 87, p-value < 2.2e-16 I didn't repair on d.f. I would pay more attention next time. Thanks a lot, - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/different-results-from-fligner-test-tp4283312p4283753.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] short-hand to avoid use of length() in subsetting vectors?
On 12-01-10 6:04 PM, Eric Rupley wrote: Hi-- I suspect this is a frequently considered (and possibly asked) question, but I haven't thus far found an answer: For slicing a vector with x[…], is there a symbol for length(x)? No, there isn't. You don't need it much, but if you do you'll have to calculate it. You'll often save a noticeable amount of execution time if you store it in a local variable rather than calling length(x) repeatedly. Duncan Murdoch I'm seeking a short-hand for referring to the length of vector one is trying to index. E.g., for a data.frame, say, test.frame<-data.frame(matrix(c(1:100),ncol=10,nrow=10,byrow=T)) names(test.frame)<- names(islands)[1:10] is there a short-hand for subsetting test.frame$Baffin[1:(length(test.frame$Baffin)-3)] [1] 6 16 26 36 46 56 66 that would allow one to avoid the "(length(some.dataframe$variable)-offset)"? I was thinking something paralleling the use of negative indices in […] might exist with seq(from,to), e.g. for the above test.frame$Baffin[seq(,7)] [1] 6 16 26 36 46 56 66 works. But the fantasy test.frame$Baffin[seq(,-3)] obviously doesn't… Any suggestions will be gratefully appreciated… As always, many thanks to the patient list members who helps on these simple questions... Best, Eric -- Eric Rupley University of Michigan, Museum of Anthropology 1109 Geddes Ave, Rm. 4013 Ann Arbor, MI 48109-1079 erup...@umich.edu +1.734.276.8572 __ 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] runif with condition
AlanM said,
"Consider the pair {X, 1-X} where X is sampled from a uniform(0,1)
distribution. The quantity 1- X also comes from a uniform(0,1)
distribution and therefore is probabilistic and not deterministic. The
sum of independent random variables is itself a random variable. If X1,
X2 & X3 are uniformly distributed, then the distribution of Y = X1 + X2
+ X3 can be determined (i.e. Y is probabilistic and NOT deterministic).
Y is a random variable, but it is correlated with X1, X2 and X3. The set
{X1, X2, X3, 100 - (X1 + X2 + X3) } contains 4 random variables, however
they are neither independent or identically distributed.
If you are curious, check this out.
Deriving the Probability Density for Sums of Uniform Random Variables
Edward J. Lusk and Haviland Wright
The American Statistician
Vol. 36, No. 2 (May, 1982), pp. 128-130"
(endquote)
Be *very* careful about how you state the problem. The quantity (1-X),
once you know the value of X, is deterministic. The confusion arises, I
think, in that if you do NOT know the value of X, then both X and (1-X)
are unknown and probabilistic. But, rather like entangled photons :-),
once you know one of the values, you immediately know the other.
I'll sign off with the reminder that, back when Marylin vos Savant first
published the Monte Hall problem, a number of people with PhDs in math,
some of whom claimed to be statisticians, angrily supported the
incorrect answer.
Carl
--
Sent from my Cray XK6
"Pendeo-navem mei anguillae plena est."
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[R] Vegan(ordistep) error: Error in if (aod[1, 5] <= Pin) { : missing value where TRUE/FALSE needed
I am getting the following erro rmessage in ordistep. I have a number of
similarly structured datasets using ordistep in a loop, and the message
only occurs for some of the datasets.
I cannot include a reproducible sample - the specific datasets where this
is occur ing are fairly large and there are several pcnm's in the rhs of
the formula.
thanks for any pointers that may allow me to track down the cause of the
error.
Error in if (aod[1, 5] <= Pin) { : missing value where TRUE/FALSE needed
In addition: There were 50 or more warnings (use warnings() to see the
first 50)
traceback()
9: ordistep(myrda0, scope = formula(myrda1), direction = "both",
Pin = 0.05, Pout = 0.1) at
RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
8: eval(expr, envir, enclos) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
7: eval(expr, pf) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
6: withVisible(eval(expr, pf)) at
RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
5: evalVis(expr) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
4: capture.output(ordistep(myrda0, scope = formula(myrda1), direction =
"both",
Pin = 0.05, Pout = 0.1)) at
RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
3: eval.with.vis(expr, envir, enclos)
2: eval.with.vis(ei, envir)
1:
source("~/Documents/Dropbox/thesis/CH3/Analysis/RDAPARTIALSexandAgeConnectandGEOGraphy2.R")
print(myrda1)
Call: rda(formula = mygenind@tab ~ pcnmTRE_25_100_CS25 + pcnmTRE_25_10_CS25
+ pcnmTRE_25_2_CS25 +
pcnmTRE_25_5_CS25 + mydata$TreeCov + mydata$Hab_Config +
pcnmEYR_EO_100_CS25 +
pcnmEYR_EO_5000_CS25 + pcnmEYR_TH_10_CS25 + pcnmEYR_TH_2_CS25 +
mydata$Site_No + mydata$Landscape
+ Condition(pcnmCS_NULL + mydata$LAT.x + mydata$LONG.x), na.action =
"na.omit")
Inertia Proportion Rank
Total 1.8110 1.
Conditional0.8681 0.4793 32
Constrained0. 0.0
Unconstrained 0.9429 0.5207 29
Inertia is variance
Some constraints were aliased because they were collinear (redundant)
Eigenvalues for unconstrained axes:
PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8
0.16008 0.14733 0.12183 0.09054 0.07380 0.06971 0.05578 0.04215
(Showed only 8 of all 29 unconstrained eigenvalues)
--
Nevil Amos
Molecular Ecology Research Group
Australian Centre for Biodiversity
Monash University
CLAYTON VIC 3800
Australia
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Re: [R] How to make this for() loop memory efficient?
Yeah -- just fired off an apology email before this landed in my inbox.
Sometimes I'm better off not trying to help at all -- this was one of
those cases ;-)
Whatever I was trying to do clearly was going down the wrong trail
Thankfully, you're on top of it though.
Sorry for the spam,
-steve
On Tue, Jan 10, 2012 at 7:33 PM, Ray Brownrigg
wrote:
> Steve:
>
> I don't understand why you couldn't get the original code working. You just
> have to
> notice that one comment overflows its line.
>
> However I couldn't get your code to match the output of the original -
> almost, but not
> quite!
>
> Ray
>
> On Wed, 11 Jan 2012, Steve Lianoglou wrote:
>> I'm having a really difficult time understanding what you're trying to
>> get -- copy and pasting your code is failing to run, and your question
>> isn't clear, ie:
>>
>> "For each phone call that BEGINS with the module which is denoted by 81
>> (i.e. of the form 81X,XXX), what is the expected number of modules in these
>> calls?"
>>
>> How does one calculate the expected number of "modules" in this
>> module? What does that even mean?
>>
>> Anyway, here's some using your `data` data.frame that calculates the
>> number of unique calls and other statistics on the "call id" within
>> each module prefix. I'm using both data.table and plyr ... there are
>> no for loops.
>>
>> You will want to do `whatever it is you really want to do` inside the
>> "blocks" below.
>>
>> ## R code
>> data <- transform(data, module.prefix=substring(modules, 1, 2))
>>
>> ## take a look at `data` now
>>
>> ## calulate "stuff" inside each module.prefix using data.table
>> xx <- data.table(data, key="module.prefix")
>>
>> ans <- xx[, {
>> ## the columns of the particular subset of your data.table
>> ## are "injected" into the scope for this expression block
>> ## which is where the `calls` variable below comes from
>> tabled <- table(as.character(calls))
>> list(unique.calls=length(tabled), min=min(tabled),
>> median=as.numeric(median(tabled)), max=max(tabled))
>> ## you will want to return your own list of "stuff"
>> }, by='module.prefix']
>>
>>
>> ## with plyr
>> library(plyr)
>> ans <- ddply(data, "module.prefix", function(x) {
>> ## `x` is a data.frame that all share the same module.prefix
>> ## do whatever you want with it here
>> tabled <- table(as.character(x$calls))
>> c(unique.calls=length(tabled), min=min(tabled),
>> median=median(tabled), max=max(tabled))
>> })
>>
>> You'll have to read up on the particulars of data.table and plyr. Both
>> are really powerful packages ... you should get familiar with at least
>> one.
>>
>> plyr is a bit more flexible in some ways.
>>
>> data.table is a bit more strict (cf. the need for
>> `as.numeric(median(tabled))`), but also tends to be (much) faster when
>> working over large datasets
>>
>> HTH,
>> -steve
>
> __
> 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.
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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Re: [R] How to make this for() loop memory efficient?
Let me just reply to myself.
Sorry, it's funny how much I don't get this, but it appears Ray is
following you and provides an answer -- scratch my email, it seems to
be way off
(you should still learn plyr and/or data.table if you haven't yet, tho ;-)
Apologies,
-steve
On Tue, Jan 10, 2012 at 7:18 PM, Steve Lianoglou
wrote:
> I'm having a really difficult time understanding what you're trying to
> get -- copy and pasting your code is failing to run, and your question
> isn't clear, ie:
>
> "For each phone call that BEGINS with the module which is denoted by 81
> (i.e. of the form 81X,XXX), what is the expected number of modules in these
> calls?"
>
> How does one calculate the expected number of "modules" in this
> module? What does that even mean?
>
> Anyway, here's some using your `data` data.frame that calculates the
> number of unique calls and other statistics on the "call id" within
> each module prefix. I'm using both data.table and plyr ... there are
> no for loops.
>
> You will want to do `whatever it is you really want to do` inside the
> "blocks" below.
>
> ## R code
> data <- transform(data, module.prefix=substring(modules, 1, 2))
>
> ## take a look at `data` now
>
> ## calulate "stuff" inside each module.prefix using data.table
> xx <- data.table(data, key="module.prefix")
>
> ans <- xx[, {
> ## the columns of the particular subset of your data.table
> ## are "injected" into the scope for this expression block
> ## which is where the `calls` variable below comes from
> tabled <- table(as.character(calls))
> list(unique.calls=length(tabled), min=min(tabled),
> median=as.numeric(median(tabled)), max=max(tabled))
> ## you will want to return your own list of "stuff"
> }, by='module.prefix']
>
>
> ## with plyr
> library(plyr)
> ans <- ddply(data, "module.prefix", function(x) {
> ## `x` is a data.frame that all share the same module.prefix
> ## do whatever you want with it here
> tabled <- table(as.character(x$calls))
> c(unique.calls=length(tabled), min=min(tabled),
> median=median(tabled), max=max(tabled))
> })
>
> You'll have to read up on the particulars of data.table and plyr. Both
> are really powerful packages ... you should get familiar with at least
> one.
>
> plyr is a bit more flexible in some ways.
>
> data.table is a bit more strict (cf. the need for
> `as.numeric(median(tabled))`), but also tends to be (much) faster when
> working over large datasets
>
> HTH,
> -steve
>
> --
> Steve Lianoglou
> Graduate Student: Computational Systems Biology
> | Memorial Sloan-Kettering Cancer Center
> | Weill Medical College of Cornell University
> Contact Info: http://cbio.mskcc.org/~lianos/contact
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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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] How to make this for() loop memory efficient?
Steve:
I don't understand why you couldn't get the original code working. You just
have to
notice that one comment overflows its line.
However I couldn't get your code to match the output of the original - almost,
but not
quite!
Ray
On Wed, 11 Jan 2012, Steve Lianoglou wrote:
> I'm having a really difficult time understanding what you're trying to
> get -- copy and pasting your code is failing to run, and your question
> isn't clear, ie:
>
> "For each phone call that BEGINS with the module which is denoted by 81
> (i.e. of the form 81X,XXX), what is the expected number of modules in these
> calls?"
>
> How does one calculate the expected number of "modules" in this
> module? What does that even mean?
>
> Anyway, here's some using your `data` data.frame that calculates the
> number of unique calls and other statistics on the "call id" within
> each module prefix. I'm using both data.table and plyr ... there are
> no for loops.
>
> You will want to do `whatever it is you really want to do` inside the
> "blocks" below.
>
> ## R code
> data <- transform(data, module.prefix=substring(modules, 1, 2))
>
> ## take a look at `data` now
>
> ## calulate "stuff" inside each module.prefix using data.table
> xx <- data.table(data, key="module.prefix")
>
> ans <- xx[, {
> ## the columns of the particular subset of your data.table
> ## are "injected" into the scope for this expression block
> ## which is where the `calls` variable below comes from
> tabled <- table(as.character(calls))
> list(unique.calls=length(tabled), min=min(tabled),
> median=as.numeric(median(tabled)), max=max(tabled))
> ## you will want to return your own list of "stuff"
> }, by='module.prefix']
>
>
> ## with plyr
> library(plyr)
> ans <- ddply(data, "module.prefix", function(x) {
> ## `x` is a data.frame that all share the same module.prefix
> ## do whatever you want with it here
> tabled <- table(as.character(x$calls))
> c(unique.calls=length(tabled), min=min(tabled),
> median=median(tabled), max=max(tabled))
> })
>
> You'll have to read up on the particulars of data.table and plyr. Both
> are really powerful packages ... you should get familiar with at least
> one.
>
> plyr is a bit more flexible in some ways.
>
> data.table is a bit more strict (cf. the need for
> `as.numeric(median(tabled))`), but also tends to be (much) faster when
> working over large datasets
>
> HTH,
> -steve
__
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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] How to make this for() loop memory efficient?
On Wed, 11 Jan 2012, Ray Brownrigg wrote:
> On Wed, 11 Jan 2012, iliketurtles wrote:
> > ##I have 2 columns of data. The first column is unique "event IDs" that
> > represent a phone call made to a customer.
> > ###So, if you see 3 entries together in the first column like follows:
> >
> > matrix(c("call1a","call1a","call1a") )
> >
> > ##then this means that this particular phone call (the first call that's
> > logged in the data set) was transferred
> > ##between 3 different "modules" before the call was terminated.
> >
> > ##The second column is a numerical description of the module the call
> > started with and then got transferred to prior to ##call termination.
> > Now, I'll construct a ##representative array of the type of data I'm
> > dealing with (the real data set goes ##on for X00,000s of rows):
> > ##(Ignore how I construct the following array, it’s completely unrelated
> > to how the actual data set was constructed).
> >
> >
> > a<-sapply(1:50,function(i){paste("call",i,sep="",collapse="")})
> > development.a<-seq(1,40,3)
> > development.a2<-seq(1,40,5)
> > a[development.a]<-a[development.a+1]
> > a[development.a2]<-a[development.a2+1]
> > a[1:2]<-"call2a";a[3]<-"call3a";a[4:5]<-"call5a";a[6:8]<-"call8a";a[9]<-"
> > ca ll9a"
> > b<-c(920010,960010,820009,920010,960500,970050,930010,920010,960500,97005
> > 0
> > ,930900,870010,840010,960500,920010,970050,930010,960500,920010,970050,9
> > 300
> > 10,960010,920010,940010,960010,970010,960500,920010,970050,930010,960500
> > ,92
> > 0010,970050,930010,960500,920010,970050,930010,920010,960500,970050,9300
> > 10, 920009,960500,970050,930009,940010,960500,960500,960500)
> > data<-as.data.frame(cbind(a,b))
> > colnames(data)<-c("phone calls","modules")
> > dim(data)
> > print(data[1:10,]) #sample of 10 rows
> >
> > # Note that in the real data set, data[,2] ranges from 810,000 to
> > 999,999. I've been tasked with the following:
> > # "For each phone call that BEGINS with the module which is denoted by 81
> > (i.e. of the form 81X,XXX), what is the expected number of modules in
> > these calls?"
> > #Then it's the same question for each module beginning with 82, 83,
> > 84. all the way until 99.
> > #I've created code that I think works for this, but I can't actually run
> > it on the whole data set. I left it for 30 minutes and it only had about
> > #5% of the task completed (I clicked "STOP" then checked my output to
> > see if I did it properly, and it seems correct).
> > #I know the apply() family specializes in vector operations, but I can't
> > figure out how to complete the above question in any way other than
> > #loops.
> >
> > L<-data
> >
> > A<-array(0,dim=c(19,2));rownames(A)<-seq(81,99,1)
> > A<-data.frame(A)
> >
> > for(i in 1:(nrow(L)-1))
> > {
> >
> > if(L[(i+1),1]!=L[i,1])
> > {
> >
> > A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse="")
> > ,1 ]<- {
> >
> > A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse="")
> > ,1 ]+length(grep(as.character(L[i+1,1]),L[,1],value=FALSE)) #aggregate
> > number of modules in the calls that begin with XX (not yet averaged).
> >
> > }
> >
> > A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse="")
> > ,2 ]<- {
> >
> > A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse="")
> > ,2 ]+1 }
> >
> > }
> >
> > }
> >
> > #If I can get this code to be more memory efficient such that I can do it
> > on a 400,000 row data set, I can do, for example,
> >
> > A[17,1]/A[17,2]
> >
> > #and I'll arrive at the mean number of modules per call where the call
> > starts with a module that starts with 97.
> >
> > A[17,1]
> > #is 10, which means that, out of every single call that started with a
> > module of 97X,XXX,
> > #they went through 10 modules in total.
> >
> > A[17,2]
> > #is 6, which means that there was 6 calls in total that began with a
> > 97X,XXX module.
> >
> > #Hence,
> >
> >
> > A[17,1]/A[17,2]
> >
> > #is the average number of modules that were executed in all the calls
> > that began with a 97X,XXX module.
> >
> >
> > -
> >
> >
> > Isaac
> > Research Assistant
> > Quantitative Finance Faculty, UTS
>
> I don't see any need for you to use data frames.
>
> If you make A and data (not a good use of a variable name) just matrices,
> you get the same answers at about 10 times the speed (using your example).
>
Further, you should calculate your rowname, namely:
paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse="")
only once each loop, instead of 4 times. this saves another 25-30% cputime.
And you can combine the two updates into a single assignment.
So using the code:
L <- as.matrix(data)
A <- array(0, dim=c(19, 2)); rownames(A) <- seq(81, 99, 1)
# A <- data.frame(A)
for(i in 1:(nrow(L)-1))
{
if(L[(i+1),1]!=L[i,1])
{
myrow <- paste(strsplit(as.character(L[i+1, 2]), "")[[1]][1:2], sep="",
collapse="")
A[myrow, ] <- A[myrow, ] +
c(length(g
Re: [R] How to make this for() loop memory efficient?
I'm having a really difficult time understanding what you're trying to
get -- copy and pasting your code is failing to run, and your question
isn't clear, ie:
"For each phone call that BEGINS with the module which is denoted by 81
(i.e. of the form 81X,XXX), what is the expected number of modules in these
calls?"
How does one calculate the expected number of "modules" in this
module? What does that even mean?
Anyway, here's some using your `data` data.frame that calculates the
number of unique calls and other statistics on the "call id" within
each module prefix. I'm using both data.table and plyr ... there are
no for loops.
You will want to do `whatever it is you really want to do` inside the
"blocks" below.
## R code
data <- transform(data, module.prefix=substring(modules, 1, 2))
## take a look at `data` now
## calulate "stuff" inside each module.prefix using data.table
xx <- data.table(data, key="module.prefix")
ans <- xx[, {
## the columns of the particular subset of your data.table
## are "injected" into the scope for this expression block
## which is where the `calls` variable below comes from
tabled <- table(as.character(calls))
list(unique.calls=length(tabled), min=min(tabled),
median=as.numeric(median(tabled)), max=max(tabled))
## you will want to return your own list of "stuff"
}, by='module.prefix']
## with plyr
library(plyr)
ans <- ddply(data, "module.prefix", function(x) {
## `x` is a data.frame that all share the same module.prefix
## do whatever you want with it here
tabled <- table(as.character(x$calls))
c(unique.calls=length(tabled), min=min(tabled),
median=median(tabled), max=max(tabled))
})
You'll have to read up on the particulars of data.table and plyr. Both
are really powerful packages ... you should get familiar with at least
one.
plyr is a bit more flexible in some ways.
data.table is a bit more strict (cf. the need for
`as.numeric(median(tabled))`), but also tends to be (much) faster when
working over large datasets
HTH,
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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[R] Rpad.org down? Searching latest R-Reference card
Hi, i'd like to update my R-Reference card and commit some edits, but i could not get the source from rpad.org Did it move? kind regards, -- Jonas Stein __ 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] short-hand to avoid use of length() in subsetting vectors?
> -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Steve Lianoglou > Sent: Tuesday, January 10, 2012 3:28 PM > To: Eric Rupley > Cc: r-help@r-project.org > Subject: Re: [R] short-hand to avoid use of length() in subsetting > vectors? > > Hi, > > On Tue, Jan 10, 2012 at 6:04 PM, Eric Rupley wrote: > > > > Hi-- > > > > I suspect this is a frequently considered (and possibly asked) > question, but I haven't thus far found an answer: > > > > For slicing a vector with x[…], is there a symbol for > length(x)? > > > > I'm seeking a short-hand for referring to the length of vector one is > trying to index. > > > > E.g., for a data.frame, say, > > > >> test.frame <-data.frame(matrix(c(1:100),ncol=10,nrow=10,byrow=T)) > >> names(test.frame) <- names(islands)[1:10] > > > > is there a short-hand for subsetting > > > >> test.frame$Baffin[1:(length(test.frame$Baffin)-3)] > > [1] 6 16 26 36 46 56 66 > >> > >> > > > > that would allow one to avoid the "(length(some.dataframe$variable)- > offset)"? > > Sadly there is no shorthand of the (exact) type you are looking for. > > You can access the data like so, however: > > R> head(test.frame$Baffin, -3) > [1] 6 16 26 36 46 56 66 > > I think that's as good as you're going to get, but let's see what > other suggestions pop up. > > -steve > Steve is probably right. But, here is one suggestion. Since every column of a data frame is the same length, You could try something like test.frame$Baffin[1:(nrow(test.frame)-3)] However, it is only a little bit shorter. Dan Daniel J. Nordlund Washington State Department of Social and Health Services Planning, Performance, and Accountability Research and Data Analysis Division Olympia, WA 98504-5204 __ 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 make this for() loop memory efficient?
On Wed, 11 Jan 2012, iliketurtles wrote:
> ##I have 2 columns of data. The first column is unique "event IDs" that
> represent a phone call made to a customer.
> ###So, if you see 3 entries together in the first column like follows:
>
> matrix(c("call1a","call1a","call1a") )
>
> ##then this means that this particular phone call (the first call that's
> logged in the data set) was transferred
> ##between 3 different "modules" before the call was terminated.
>
> ##The second column is a numerical description of the module the call
> started with and then got transferred to prior to ##call termination. Now,
> I'll construct a ##representative array of the type of data I'm dealing
> with (the real data set goes ##on for X00,000s of rows):
> ##(Ignore how I construct the following array, it’s completely unrelated to
> how the actual data set was constructed).
>
>
> a<-sapply(1:50,function(i){paste("call",i,sep="",collapse="")})
> development.a<-seq(1,40,3)
> development.a2<-seq(1,40,5)
> a[development.a]<-a[development.a+1]
> a[development.a2]<-a[development.a2+1]
> a[1:2]<-"call2a";a[3]<-"call3a";a[4:5]<-"call5a";a[6:8]<-"call8a";a[9]<-"ca
> ll9a"
> b<-c(920010,960010,820009,920010,960500,970050,930010,920010,960500,970050
> ,930900,870010,840010,960500,920010,970050,930010,960500,920010,970050,9300
> 10,960010,920010,940010,960010,970010,960500,920010,970050,930010,960500,92
> 0010,970050,930010,960500,920010,970050,930010,920010,960500,970050,930010,
> 920009,960500,970050,930009,940010,960500,960500,960500)
> data<-as.data.frame(cbind(a,b))
> colnames(data)<-c("phone calls","modules")
> dim(data)
> print(data[1:10,]) #sample of 10 rows
>
> # Note that in the real data set, data[,2] ranges from 810,000 to 999,999.
> I've been tasked with the following:
> # "For each phone call that BEGINS with the module which is denoted by 81
> (i.e. of the form 81X,XXX), what is the expected number of modules in these
> calls?"
> #Then it's the same question for each module beginning with 82, 83, 84.
> all the way until 99.
> #I've created code that I think works for this, but I can't actually run it
> on the whole data set. I left it for 30 minutes and it only had about #5%
> of the task completed (I clicked "STOP" then checked my output to see if I
> did it properly, and it seems correct).
> #I know the apply() family specializes in vector operations, but I can't
> figure out how to complete the above question in any way other than #loops.
>
> L<-data
>
> A<-array(0,dim=c(19,2));rownames(A)<-seq(81,99,1)
> A<-data.frame(A)
>
> for(i in 1:(nrow(L)-1))
> {
> if(L[(i+1),1]!=L[i,1])
> {
>
> A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),1
> ]<- {
>
> A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),1
> ]+length(grep(as.character(L[i+1,1]),L[,1],value=FALSE)) #aggregate number
> of modules in the calls that begin with XX (not yet averaged).
> }
>
> A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),2
> ]<- {
>
> A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),2
> ]+1 }
> }
>
> }
>
> #If I can get this code to be more memory efficient such that I can do it
> on a 400,000 row data set, I can do, for example,
>
> A[17,1]/A[17,2]
>
> #and I'll arrive at the mean number of modules per call where the call
> starts with a module that starts with 97.
>
> A[17,1]
> #is 10, which means that, out of every single call that started with a
> module of 97X,XXX,
> #they went through 10 modules in total.
>
> A[17,2]
> #is 6, which means that there was 6 calls in total that began with a
> 97X,XXX module.
>
> #Hence,
>
>
> A[17,1]/A[17,2]
>
> #is the average number of modules that were executed in all the calls that
> began with a 97X,XXX module.
>
>
> -
>
>
> Isaac
> Research Assistant
> Quantitative Finance Faculty, UTS
I don't see any need for you to use data frames.
If you make A and data (not a good use of a variable name) just matrices, you
get the same
answers at about 10 times the speed (using your example).
Hope this helps,
Ray Brownrigg
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] short-hand to avoid use of length() in subsetting vectors?
Hi, On Tue, Jan 10, 2012 at 6:04 PM, Eric Rupley wrote: > > Hi-- > > I suspect this is a frequently considered (and possibly asked) question, but > I haven't thus far found an answer: > > For slicing a vector with x[…], is there a symbol for length(x)? > > I'm seeking a short-hand for referring to the length of vector one is trying > to index. > > E.g., for a data.frame, say, > >> test.frame <-data.frame(matrix(c(1:100),ncol=10,nrow=10,byrow=T)) >> names(test.frame) <- names(islands)[1:10] > > is there a short-hand for subsetting > >> test.frame$Baffin[1:(length(test.frame$Baffin)-3)] > [1] 6 16 26 36 46 56 66 >> >> > > that would allow one to avoid the "(length(some.dataframe$variable)-offset)"? Sadly there is no shorthand of the (exact) type you are looking for. You can access the data like so, however: R> head(test.frame$Baffin, -3) [1] 6 16 26 36 46 56 66 I think that's as good as you're going to get, but let's see what other suggestions pop up. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can "prototype" and "initialize" coexist?
On 01/10/2012 11:47 AM, Keith Weintraub wrote:
Folks,
My object oriented background is in Java and C++. I am a novice to using
S4/object-oriented coding in R but not to R. Below is an example that I found
that I have expanded on.
I am not getting how "prototype" and "initialize" work together, if at all.
Here is output from a short "session"
setClass("xx",
+ representation(a="numeric", b ="numeric"),
+ prototype(a=33, b = 333)
+ )
[1] "xx"
#setMethod("initialize", "xx", function(.Object){.Object})
setMethod("initialize", "xx",
+ function(.Object, b) {
+ .Object@b<-b
+ .Object@a<-b^2
+ .Object
+ }
+ )
[1] "initialize"
new("xx", 3)
An object of class "xx"
Slot "a":
[1] 9
Slot "b":
[1] 3
new("xx")
Error in .local(.Object, ...) : argument "b" is missing, with no default
Hi Keith --
This might be a good place to start, setting aside prototype for the moment.
You'd like new("xx") to work. This means that all arguments to
initialize need to have a default value. You'd also like new("xx",
new("xx")) to work, because it's advertised as a copy constructor (a
slightly loose interpretation of ?new, which says for the ... argument
'Unnamed arguments must be objects from classes that this class
extends'). An appropriate initialize method is
setMethod(initialize, "xx", function(.Object, ..., b=2) {
callNextMethod(.Object, ..., b=b)
})
For instance,
> setClass("xx", representation(a="numeric", b="numeric"))
> new("xx")
An object of class "xx"
Slot "a":
numeric(0)
Slot "b":
[1] 2
> new("xx", new("xx", a=1))
An object of class "xx"
Slot "a":
[1] 1
Slot "b":
[1] 2
> new("xx", new("xx", a=1), a=2, b=3)
An object of class "xx"
Slot "a":
[1] 2
Slot "b":
[1] 3
with the basic initializer out of the way, you could introduce a prototype
setClass("xx",
representation(a="numeric", b="numeric"),
prototype(a=-1, b=-2))
> new("xx")
An object of class "xx"
Slot "a":
[1] -1
Slot "b":
[1] 2
which shows that the prototype and initializer are playing well together
(for slot 'a') and that the initialize method is over-riding the
prototype (for slot 'b').
One could provide a default value to b in the initializer that is
derived from the prototype, which is used to populate .Object
setMethod(initialize, "xx", function(.Object, ..., b=.Object@b) {
callNextMethod(.Object, ..., b=b)
})
with
> new("xx", a=1)
An object of class "xx"
Slot "a":
[1] 1
Slot "b":
[1] -2
e.g., maybe you want to check user input and the check is expensive...
setMethod(initialize, "xx", function(.Object, ..., b=.Object@b) {
if (!missing(b))
## check user input, expensive
Sys.sleep(2)
callNextMethod(.Object, ..., b=b)
})
Maybe a slightly more common case is to provide a prototype for one slot
(e.g., for internal business, not the user) with initialize taking care
of others.
It is often the case that you'd like a constructor
xx <- function(a=-1, b=-2, ...)
new("xx", a=a, b=b, ...)
which is nicer for the end user, documents the necessary arguments,
frees one to separately implement the constructor vs. copy-constructor,
etc. Neither prototype nor initialize method would be implemented here,
leaving all the cleverness to the defaults.
One other thing is that the prototype must create a valid object; this
prototype allows R to produce invalid instances:
setClass("Abs", representation(x="numeric"),
prototype(x=-1),
validity=function(object) if (any(object@x < 0)) "oops" else TRUE)
> a = new("Abs")
An object of class "Abs"
Slot "x":
[1] -1
> validObject(a)
Error in validObject(a) : invalid class "Abs" object: oops
Not sure whether that helps, or is too much information...
Martin
Any help you can provide would be greatly appreciated,
Thanks,
KW
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[R] short-hand to avoid use of length() in subsetting vectors?
Hi-- I suspect this is a frequently considered (and possibly asked) question, but I haven't thus far found an answer: For slicing a vector with x[…], is there a symbol for length(x)? I'm seeking a short-hand for referring to the length of vector one is trying to index. E.g., for a data.frame, say, > test.frame <-data.frame(matrix(c(1:100),ncol=10,nrow=10,byrow=T)) > names(test.frame) <- names(islands)[1:10] is there a short-hand for subsetting > test.frame$Baffin[1:(length(test.frame$Baffin)-3)] [1] 6 16 26 36 46 56 66 > > that would allow one to avoid the "(length(some.dataframe$variable)-offset)"? I was thinking something paralleling the use of negative indices in […] might exist with seq(from,to), e.g. for the above > test.frame$Baffin[seq(,7)] [1] 6 16 26 36 46 56 66 > works. But the fantasy test.frame$Baffin[seq(,-3)] obviously doesn't… Any suggestions will be gratefully appreciated… As always, many thanks to the patient list members who helps on these simple questions... Best, Eric -- Eric Rupley University of Michigan, Museum of Anthropology 1109 Geddes Ave, Rm. 4013 Ann Arbor, MI 48109-1079 erup...@umich.edu +1.734.276.8572 __ 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] Aggregate by minimum
On Mon, Jan 9, 2012 at 8:00 PM, jim holtman wrote:
> try this:
>
>> x <- structure(list(speed = c(3,9,14,8,7,6), result = c(0.697, 0.011, 0.015,
>> 0.012, 0.018, 0.019), house = c(1,
> + 1, 1, 1, 1, 1), date = c(719, 1027, 1027, 1027, 1030, 1030),
> + id = c("1000", "1",
> + "10001", "10002", "10003", "10004")), .Names = c("speed",
> + "result", "house", "date", "id"), class = "data.frame", row.names =
> c("1000",
> + "1", "10001", "10002", "10003", "10004"))
>>
>> require(plyr)
>> ddply(x, .(date), .fun = function(a){
> + a[which.min(a$speed), ]
> + })
Or even more succinctly:
ddply(x, .(date), subset, speed == min(speed))
Hadley
--
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Department of Statistics / Rice University
http://had.co.nz/
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Re: [R] different results from fligner.test
On Jan 10, 2012, at 21:30 , gaiarrido wrote: > I've made fligner test with the same data, changing the orders of the > variables, and this what i get > >> fligner.test(rojos~edadysexo*zona*ano*estacion) > >Fligner-Killeen test of homogeneity of variances > > data: rojos by edadysexo by zona by ano by estacion > Fligner-Killeen:med chi-squared = 15.7651, df = 2, p-value = 0.0003773 > >> fligner.test(rojos~ano*edadysexo*zona*estacion) > >Fligner-Killeen test of homogeneity of variances > > data: rojos by ano by edadysexo by zona by estacion > Fligner-Killeen:med chi-squared = 86.5317, df = 3, p-value < 2.2e-16 > > Different results with the same variables!!! Why? i try to find an answer, > but i really surprised You are assuming that you can put interactions on the rhs of formulas, but nowhere in the documentation does it say that that should work. All examples have a single grouping variables. So the behavior is undefined. As far as I can see, only the _first_ variable is actually being used for grouping. Arguably, there is potential for better argument checking in fligner.test(), but in the meantime, I suspect that you need something like g <- interaction(ano, edadysexo, zona, estacion) fligner.test(rojos ~ g) > > - > Mario Garrido Escudero > PhD student > Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola > Universidad de Salamanca > -- > View this message in context: > http://r.789695.n4.nabble.com/different-results-from-fligner-test-tp4283312p4283312.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. -- 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] Correlograms
If I understand your question correctly, install the corrgram package from CRAN. Then, library(corrgram) cm <- cor(iris[ , 1:4]) corrgram(cm, type="corr") Also, see help for the "vote" data: ?vote Kevin Wright On Tue, Jan 10, 2012 at 2:06 PM, Natbyah wrote: > I would like to make a correlogram in which I also have a correlation > matrix > instead of one of the panels. > Is that possible? > > -- > View this message in context: > http://r.789695.n4.nabble.com/Correlograms-tp4283245p4283245.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. > -- Kevin Wright [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help
Tena koe Anna
[ is for subsetting, you need c():
x <- c(10.4, 5.6, 3.1, 6.4, 21.7)
y <- c(12, 5.6, 7.2, 1.0, 9.3)
plot(x, y)
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Anna Olofsson
Sent: Wednesday, 11 January 2012 11:02 a.m.
To: r-help@r-project.org
Subject: [R] help
Hi,
I'm pretty new at programming and with the R language. I'm just trying to
get familiar with R and wrote a script in gedit (should I use emacs
instead?),
x <- [10.4 5.6 3.1 6.4 21.7]
y <- [12,5.6, 7.2, 1.0, 9.3]
plot(x,y)
then I went to the command window in the terminal (I'm using unix) to run
this with source("name_of_file"), but it doesn't work. Shouldn't a plot
come up automatically when I run it? What am I doing wrong? It knows what x
and y is, but I don't get an error of what might be wrong.
> source("name_of_file")
> x
[1] 10.4 5.6 3.1 6.4 21.7
Best,
Anna
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Re: [R] help
There might be an x in your R session, but not from that script. Try
it by pasting
those three lines at the command line:
> x <- [10.4 5.6 3.1 6.4 21.7]
Error: unexpected '[' in "x <- ["
> y <- [12,5.6, 7.2, 1.0, 9.3]
Error: unexpected '[' in "y <- ["
> plot(x,y)
Error in plot(x, y) : object 'x' not found
You should get an error message when you try to source the script, I
would think.
Try instead:
x <- c(10.4, 5.6, 3.1, 6.4, 21.7)
y <- c(12, 5.6, 7.2, 1.0, 9.3)
plot(x, y)
and rereading your introductory materials.
If that still doesn't work then we'll need more information, like OS
and version of R.
Sarah
On Tue, Jan 10, 2012 at 5:02 PM, Anna Olofsson wrote:
> Hi,
> I'm pretty new at programming and with the R language. I'm just trying to
> get familiar with R and wrote a script in gedit (should I use emacs
> instead?),
>
> x <- [10.4 5.6 3.1 6.4 21.7]
> y <- [12,5.6, 7.2, 1.0, 9.3]
> plot(x,y)
>
> then I went to the command window in the terminal (I'm using unix) to run
> this with source("name_of_file"), but it doesn't work. Shouldn't a plot
> come up automatically when I run it? What am I doing wrong? It knows what x
> and y is, but I don't get an error of what might be wrong.
>
>> source("name_of_file")
>> x
> [1] 10.4 5.6 3.1 6.4 21.7
>
>
> Best,
> Anna
>
--
Sarah Goslee
http://www.functionaldiversity.org
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[R] Correlograms
I would like to make a correlogram in which I also have a correlation matrix instead of one of the panels. Is that possible? -- View this message in context: http://r.789695.n4.nabble.com/Correlograms-tp4283245p4283245.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] Can "prototype" and "initialize" coexist?
Folks,
My object oriented background is in Java and C++. I am a novice to using
S4/object-oriented coding in R but not to R. Below is an example that I found
that I have expanded on.
I am not getting how "prototype" and "initialize" work together, if at all.
Here is output from a short "session"
> setClass("xx",
+ representation(a="numeric", b ="numeric"),
+ prototype(a=33, b = 333)
+ )
[1] "xx"
>
> #setMethod("initialize", "xx", function(.Object){.Object})
>
> setMethod("initialize", "xx",
+ function(.Object, b) {
+ .Object@b<-b
+ .Object@a<-b^2
+ .Object
+ }
+ )
[1] "initialize"
>
> new("xx", 3)
An object of class "xx"
Slot "a":
[1] 9
Slot "b":
[1] 3
>
> new("xx")
Error in .local(.Object, ...) : argument "b" is missing, with no default
Any help you can provide would be greatly appreciated,
Thanks,
KW
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[R] help
Hi,
I'm pretty new at programming and with the R language. I'm just trying to
get familiar with R and wrote a script in gedit (should I use emacs
instead?),
x <- [10.4 5.6 3.1 6.4 21.7]
y <- [12,5.6, 7.2, 1.0, 9.3]
plot(x,y)
then I went to the command window in the terminal (I'm using unix) to run
this with source("name_of_file"), but it doesn't work. Shouldn't a plot
come up automatically when I run it? What am I doing wrong? It knows what x
and y is, but I don't get an error of what might be wrong.
> source("name_of_file")
> x
[1] 10.4 5.6 3.1 6.4 21.7
Best,
Anna
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[R] different results from fligner.test
I've made fligner test with the same data, changing the orders of the variables, and this what i get > fligner.test(rojos~edadysexo*zona*ano*estacion) Fligner-Killeen test of homogeneity of variances data: rojos by edadysexo by zona by ano by estacion Fligner-Killeen:med chi-squared = 15.7651, df = 2, p-value = 0.0003773 > fligner.test(rojos~ano*edadysexo*zona*estacion) Fligner-Killeen test of homogeneity of variances data: rojos by ano by edadysexo by zona by estacion Fligner-Killeen:med chi-squared = 86.5317, df = 3, p-value < 2.2e-16 Different results with the same variables!!! Why? i try to find an answer, but i really surprised - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/different-results-from-fligner-test-tp4283312p4283312.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] How to make this for() loop memory efficient?
##I have 2 columns of data. The first column is unique "event IDs" that
represent a phone call made to a customer.
###So, if you see 3 entries together in the first column like follows:
matrix(c("call1a","call1a","call1a") )
##then this means that this particular phone call (the first call that's
logged in the data set) was transferred
##between 3 different "modules" before the call was terminated.
##The second column is a numerical description of the module the call
started with and then got transferred to prior to ##call termination. Now,
I'll construct a ##representative array of the type of data I'm dealing with
(the real data set goes ##on for X00,000s of rows):
##(Ignore how I construct the following array, it’s completely unrelated to
how the actual data set was constructed).
a<-sapply(1:50,function(i){paste("call",i,sep="",collapse="")})
development.a<-seq(1,40,3)
development.a2<-seq(1,40,5)
a[development.a]<-a[development.a+1]
a[development.a2]<-a[development.a2+1]
a[1:2]<-"call2a";a[3]<-"call3a";a[4:5]<-"call5a";a[6:8]<-"call8a";a[9]<-"call9a"
b<-c(920010,960010,820009,920010,960500,970050,930010,920010,960500,970050,930900,870010,840010,960500,920010,970050,930010,960500,920010,970050,930010,960010,920010,940010,960010,970010,960500,920010,970050,930010,960500,920010,970050,930010,960500,920010,970050,930010,920010,960500,970050,930010,920009,960500,970050,930009,940010,960500,960500,960500)
data<-as.data.frame(cbind(a,b))
colnames(data)<-c("phone calls","modules")
dim(data)
print(data[1:10,]) #sample of 10 rows
# Note that in the real data set, data[,2] ranges from 810,000 to 999,999.
I've been tasked with the following:
# "For each phone call that BEGINS with the module which is denoted by 81
(i.e. of the form 81X,XXX), what is the expected number of modules in these
calls?"
#Then it's the same question for each module beginning with 82, 83, 84.
all the way until 99.
#I've created code that I think works for this, but I can't actually run it
on the whole data set. I left it for 30 minutes and it only had about #5% of
the task completed (I clicked "STOP" then checked my output to see if I did
it properly, and it seems correct).
#I know the apply() family specializes in vector operations, but I can't
figure out how to complete the above question in any way other than #loops.
L<-data
A<-array(0,dim=c(19,2));rownames(A)<-seq(81,99,1)
A<-data.frame(A)
for(i in 1:(nrow(L)-1))
{
if(L[(i+1),1]!=L[i,1])
{
A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),1]<-
{
A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),1]+length(grep(as.character(L[i+1,1]),L[,1],value=FALSE))
#aggregate number of modules in the calls that begin with XX (not yet
averaged).
}
A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),2]<-
{
A[paste(strsplit(as.character(L[i+1,2]),"")[[1]][1:2],sep="",collapse=""),2]+1
}
}
}
#If I can get this code to be more memory efficient such that I can do it on
a 400,000 row data set, I can do, for example,
A[17,1]/A[17,2]
#and I'll arrive at the mean number of modules per call where the call
starts with a module that starts with 97.
A[17,1]
#is 10, which means that, out of every single call that started with a
module of 97X,XXX,
#they went through 10 modules in total.
A[17,2]
#is 6, which means that there was 6 calls in total that began with a 97X,XXX
module.
#Hence,
A[17,1]/A[17,2]
#is the average number of modules that were executed in all the calls that
began with a 97X,XXX module.
-
Isaac
Research Assistant
Quantitative Finance Faculty, UTS
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem installing packages
What lists are you referring to when you state: "there are many packages that do not show up in the list of binaries. They do in the list of sources"? CRAN? To see all packages installed on your machine try rownames(installed.packages(()) I think available.packages() will give packages available from your local CRAN mirror. Just a hunch, but if you are seeing different behaviors between home and work it may well be that a work-firewall is blocking packages which contain pre-compiled DLLs/SOs while allowing the source code through. Also, if you aren't on Windows the vast majority of packages are quite easy to learn compile yourself Michael On Jan 10, 2012, at 4:01 PM, natalia norden wrote: > Thank you very much for your answers. I could do it by downloading the > package I needed manually and then installing it through the Terminal. Yet > the fundamental problem remains. I downloaded R 2.14.1 several times from > different mirrors and there are many packages that do not show up in the > list of binaries. They do in the list of sources, but then I have a > problem compiling the package... I did not have this problem from my home > computer when I installed R 2.14.0. > > Best, > Natalia > > > El 10/01/12 14:52, "Ken Hutchison" escribió: > >> Maybe check your proxy settings in your browser and make sure you're >> connecting to the mirror. >> Ken >> >> Sent from my iPhone >> >> On Jan 10, 2012, at 8:35 AM, natalia norden wrote: >> >>> Hello, >>> >>> I was using version 2.13.2 and I have just downloaded the latest version >>> 2.14.1. However, I'm trying to install the packages I was using and >>> when I >>> look for them in the packages list, I can´t find many in the CRAN >>> binaries >>> (e.g. "vegan"). I do find them in the CRAN sources but the installation >>> fails. I tried downloading the version 2.14.0 and I had the same >>> problem. I >>> re-installed the old version, and now it works again. Is this a problem >>> with >>> 2.14? >>> Thank you for your help. >>> Natalia Norden >>> >>> >>> >>> Natalia Norden >>> Profesor Asistente >>> Departamento de Ecología y Territorio >>> Facultad de Estudios Ambientales y Rurales >>> Pontificia Universidad Javeriana >>> Bogotá, Colombia >>> Tel: 320 83 20 Ext: 2448 >>> www.phylodiversity.net/nnorden/ >>> >>> >>> >>> [[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-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] problem installing packages
Thank you very much for your answers. I could do it by downloading the package I needed manually and then installing it through the Terminal. Yet the fundamental problem remains. I downloaded R 2.14.1 several times from different mirrors and there are many packages that do not show up in the list of binaries. They do in the list of sources, but then I have a problem compiling the package... I did not have this problem from my home computer when I installed R 2.14.0. Best, Natalia El 10/01/12 14:52, "Ken Hutchison" escribió: >Maybe check your proxy settings in your browser and make sure you're >connecting to the mirror. >Ken > >Sent from my iPhone > >On Jan 10, 2012, at 8:35 AM, natalia norden wrote: > >> Hello, >> >> I was using version 2.13.2 and I have just downloaded the latest version >> 2.14.1. However, I'm trying to install the packages I was using and >>when I >> look for them in the packages list, I can´t find many in the CRAN >>binaries >> (e.g. "vegan"). I do find them in the CRAN sources but the installation >> fails. I tried downloading the version 2.14.0 and I had the same >>problem. I >> re-installed the old version, and now it works again. Is this a problem >>with >> 2.14? >> Thank you for your help. >> Natalia Norden >> >> >> >> Natalia Norden >> Profesor Asistente >> Departamento de Ecología y Territorio >> Facultad de Estudios Ambientales y Rurales >> Pontificia Universidad Javeriana >> Bogotá, Colombia >> Tel: 320 83 20 Ext: 2448 >> www.phylodiversity.net/nnorden/ >> >> >> >>[[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 BY to a data.frame
On Jan 10, 2012, at 1:36 PM, Ramiro Barrantes wrote:
Hello,
I am trying to convert BY to a data frame, consider the following
example:
exampleDF<-data.frame(a=c(1,2),b=c(10,20),name=c("first","second"))
exampleBY<-by(exampleDF,with(exampleDF,paste(a,b,sep="_")),
function(x) {
data.frame(
name=as.character(x$name),
a=x$a,
b=x$b,
c=x$a + x$b)
}
)
It would be less confusing when the time comes around to coercing
vectors if you added stringsAsFactors =FALSE
exampleBY<-by(exampleDF,with(exampleDF,paste(a,b,sep="_")),
function(x) {
data.frame(
name=as.character(x$name),
a=x$a,
b=x$b,
c=x$a + x$b, stringsAsFactors=FALSE)
}
Then just:
do.call(rbind, exampleBY)
name a b c
1_10 first 1 10 11
2_20 second 2 20 22
I made this function:
convertByToDataFrame <- function( byObject ) {
data
.frame
(matrix
(unlist
(byObject
),nrow
=
length
(byObject
),ncol
=
length
(byObject[[1]]),byrow=TRUE,dimnames=list(NULL,names(byObject[[1]]
}
but when I run it:
convertByToDataFrame(exampleBY)
I either: (1) loose the types of the different values in the BY, or
2) get factors instead of the values
I tried the following but it doesn't scale:
exampleSUMRY <- data.frame(a =
as
.vector
(unlist(lapply(strsplit(names(exampleBY),split="_"),function(x)
x[1]))),
b =
as
.vector
(unlist(lapply(strsplit(names(exampleBY),split="_"),function(x)
x[2]))),
name=as.vector(unlist(lapply(exampleBY,function(x) x[[1]]
Any suggestions?
Thank you,
Ramiro
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David Winsemius, MD
West Hartford, CT
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[R] plotOHLC(alpha3): Error in plotOHLC(alpha3) : x is not a open/high/low/close time series
R version 2.12.0, 64 bit on Windows.
Here is a short script that illustrates the problem:
library(tseries)
library(xts)
setwd('C:\\cygwin\\home\\Ted\\New.Task\\NKs-01-08-12\\NKs\\tests')
x = read.table("quotes_h.2.dat", header = FALSE, sep="\t", skip=0)
str(x)
y <- data.frame(as.POSIXlt(paste(x$V2,substr(x$V4,4,8),sep="
"),format='%Y-%m-%d %H:%M'),x$V5)
colnames(y) <- c("tickdate","price")
str(y)
plot(y)
z <- as.irts(y)
str(z)
plot(z)
str(alpha3)
List of 2
$ time : POSIXt[1:98865], format: "2010-06-30 15:47:00" "2010-06-30
15:53:00" "2010-06-30 17:36:00" ...
$ value: num [1:98865, 1:4] 9215 9220 9205 9195 9195 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:4] "z.Open" "z.High" "z.Low" "z.Close"
- attr(*, "class")= chr "ts"
- attr(*, "tsp")= num [1:3] 1 2 1
alpha3 <- as.xts(to.minutes3(z,OHLC = TRUE))
plotOHLC(alpha3)
Error in plotOHLC(alpha3) : x is not a open/high/low/close time series
The file quotes_h.2.dat contains real time tick data for futures contracts,
so the above manipulation is my attempt to just get a time series with one
column being a date/time and the other being tick price. I believe I have
to use read.table to make a data frame, and then the manipulations to
combine the date and time fields from that feed, along with the price.
My first attempt at using to.minutes3 (and I am interested in the other
'to.period' functions too), is to get a regular time series to which I can
apply rollapply, along with a function in which I use various autoregression
methods, along with forecasting for as long as the 95% confidence intervals
is reasonably close - I want to know how far into the future the forecast
contains useful information. And then, I want to create a plot in which I
do the autoregression, and then plot the actual and forecast prices (along
with the confidence interval), as a function of time, embed that in a
function, which rollappply works with, so I can have a plot comprised of all
those individual plots (plotting only the comparison of actual and forecast
values).
It seems everything works adequately until I try the plotOHLC function
itself, which gives me the error in the subject line.
I would ask for two things:
1) what the fix is to get rid of that error plotOHLC gives me
2) some tips on the 'walk-forward' method I am looking at using.
Thanks
Ted
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Re: [R] glmmPQL and predict
The whole of idea of 'level' in mixed models is confusing to some. Professor Snijders (who teaches our students) and Professor Bates label from opposite ends. But, assuming this is my work in package MASS (Master Harwood: it is childish, to put it mildly, to fail to give due credit), it follows lme in package nlme: glmmPQL's predict method just passes this on to nlme, and the documentation is identical. So not only was it unfair to fail to mention which package and whose work this was, it was even more unfair to attribute personal lack of understanding to my work and not package nlme. Just because R and many contributed packages are free does not entitle you to treat them as zero-value: very much to the contrary. On 10/01/2012 16:38, Ben Bolker wrote: Mike Harwood gmail.com> writes: Is the labeling/naming of levels in the documentation for the predict.glmmPQL function "backwards"? The documentation states "Level values increase from outermost to innermost grouping, with level zero corresponding to the population predictions". Taking the sample in the documentation: fit<- glmmPQL(y ~ trt + I(week> 2), random = ~1 | ID, family = binomial, data = bacteria) head(predict(fit, bacteria, level = 0, type="response")) [1] 0.9680779 0.9680779 0.8587270 0.8587270 0.9344832 0.9344832 head(predict(fit, bacteria, level = 1, type="response")) X01 X01 X01 X01 X02 X02 0.9828449 0.9828449 0.9198935 0.9198935 0.9050782 0.9050782 head(predict(fit, bacteria, type="response")) ## population prediction X01 X01 X01 X01 X02 X02 0.9828449 0.9828449 0.9198935 0.9198935 0.9050782 0.9050782 The returned values for level=1 and level= match, which is not what I expected based upon the documentation. Well, the documentation says: "Defaults to the highest or innermost level of grouping", which is level 1 in this case -- right? Exponentiating the intercept coefficients from the fitted regression, the level=0 values match when the random effect intercept is included Do you mean "is NOT included" here? 0.9680779 (no random effect, below) matches the level=0 prediction above 0.9828449 (include random effect, below) matches the level=1 prediction, which is also the default prediction, above. 1/(1+exp(-3.412014)) ## only the fixed effect [1] 0.9680779 1/(1+exp(-1*(3.412014+0.63614382))) ## fixed and random effect intercepts [1] 0.9828449 This all matches my expectations. If your expectations still go in the other direction, could you explain in more detail? By the way, I recommend r-sig-mixed-mod...@r-project.org for mixed model questions in general ... Ben Bolker __ 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. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] grplasso
I want to use the grplasso package on a data set where I want to fit a linear model. My interest is in identifying significant beta coefficients. The documentation is a bit cryptic so I'd appreciate some help. I know this is a strategy for large numbers of variables but consider a simple case for pedagogical puposes. Say I have two 3 category predictors (2 dummies each), a binary predictor and a continuous predictor with a continuous outcome: y x1 x2 x3 x4 x5 x6 rows of data here .. .. Naturally, I want to select x1 and x2 as a group and x3 and x4 as another group. The documentation has a couple of examples but it's not clear how they translate to the current problem. How do I specify my groups and run the lasso regression? Looks like this is the grouping part: index<-c(NA,) but I'm not sure how to specify the df for the variables past the NA for the intercept. Once that's defined the penalty can be specified: lambda <- lambdamax(x, y = y, index = index, penscale = sqrt, model = LogReg()) * 0.5^(0:5) In my case I'd use LinReg for the model. Then the model: fit <- grplasso(x, y = y, index = index, lambda = lambda, model = LogReg(), penscale = sqrt, control = grpl.control(update.hess = "lambda", trace = 0)) again using LinReg for the model. This can be plotted against lambda, but when I do lasso regression in other software I end up with a plot of the coefficients against the tuning parameter with a cutpoint or a table and graph that tells me what to include in the model based on some selected criterion. It's not clear from the example if there's a cross-validation or some other procedure to determine what variables to include. Plot(fit) produces a graph of coefficients against lambda but nothig to indicate what to include. What is used in the package, if anything, to make that determination? __ 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] Restricting R session
Hello, Is it possible to use R on public server where each user has its own restricted R session? In particular, how to prohibit some set of functions, for example, from "base" package? How to limit session operating memory and CPU time? What additional security considerations must be taken care of? -- Kind regards, Antonio Rodriges __ 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: Sum of a couple of variables of which a few have NA values
Dear Ivan, Thank you very much for your help. How do I use rowSums if I need to "skip" a variable from summing? (example: sum var1, var2, var3, var5, var34 only). Thanks in advance, Petra Opic -- View this message in context: http://r.789695.n4.nabble.com/Sum-of-a-couple-of-variables-of-which-a-few-have-NA-values-tp4282448p4282969.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] rpart vs. tree and deviance calculations
Hi Everyone, I'm working on building some classification trees, and up to this point I've been using rpart. However, I recently discovered the tree package, and found that it had some useful functions (in particular deviance(), which I would really like to use for my project). I can't seem to find an equivalent function for rpart. I've considered using tree() in place of rpart(), but I read an old post on this list that recommended using rpart() as tree() was built for bug-checking in S. So, to summarize: Is there a way to compute deviance for an rpart object? If not, would it be a bad idea to use tree() in place of rpart()? My rpart models are currently quite large and take ~ 15 minutes to run... Thanks everyone! [[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] Error when using foreach package for parralelization
On 01/03/2012 03:19 PM, Julien Textoris wrote:
> I'm trying to parallelize the following R code :
>
> pk2test =
> c(1:16,(12*16+1):(12*16+16),(16*16+1):(16*16+16),(20*16+1):(20*16+16))
> score.mat = matrix(nc=16*4,nr=16*4)
> for(i in 1:(16*4)) {
>
> for(j in i:(16*4)) {
> score.mat[i,j] = score.mat[j,i] =
> computeScore(pk[[pk2test[i]]],pk[[pk2test[j]]],10,5)$score
> }
> }
>
> pk is a list of Object of type MassPeak from MALDIquant library. Each
> object is composed with a mass vector (@mass) an intensity vector
> (@intensity) and a metaData field (another list)
>
> score.mat is a matrix with scores (reals)
> pk2test is just a vector to know which objects in pk i want to deal with
> computeScore is the function i wrote to compute the score, it calls
> another function called filterSpectra
>
>
> I write the function like that, and i got the error below, and i can't
> figure out why ?
>
> pk2test =
> c(1:16,(12*16+1):(12*16+16),(16*16+1):(16*16+16),(21*16+1):(21*16+16))
> score.mat = matrix(nc=16*4,nr=16*4)
> for(i in 1:4) {
> score.mat[i,i:4] =
> foreach(filterSpectra=filterSpectra,
> computeScore=computeScore,
> pk=pk,pk2test=pk2test,i=i,j=c(i:4),
> .combine="c", .packages="MALDIquant" ) %dopar% {
> computeScore(pk[[pk2test[i]]],pk[[pk2test[j]]],10,5)$score
> }
> }
>
> Error: trying to get slot "mass" from an object of a basic class
> ("integer") with no slots.
Hi Julien!
Are these two code sequences supposed to produce the same result? The
two definitions of pk2test are slightly different. Also, in the
attempted parallelized version, you are only assigning to a small part
of score.mat. Is that intentional?
The real error in this case seems to be that you mistakenly redefine
some variables in the foreach() call. As far as I can tell, you should
not redefine the variables 'filterSpectra', 'computeScore', 'pk',
'pk2test' or 'i'. In the foreach() call, you should only define
iteration variables, i.e. variables whose value changes from one
iteration to another (like 'j'). Now you actually accidentally iterate
over some data structures. For example, 'pk' inside the %dopar% loop is
a single element of the original 'pk' list (which may get overwritten,
depending on whether the loop is actually run in parallel). This is
probably not what you want.
- Mikko Korpela
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[R] Converting BY to a data.frame
Hello,
I am trying to convert BY to a data frame, consider the following example:
exampleDF<-data.frame(a=c(1,2),b=c(10,20),name=c("first","second"))
exampleBY<-by(exampleDF,with(exampleDF,paste(a,b,sep="_")),
function(x) {
data.frame(
name=as.character(x$name),
a=x$a,
b=x$b,
c=x$a + x$b)
}
)
I made this function:
convertByToDataFrame <- function( byObject ) {
data.frame(matrix(unlist(byObject),nrow=length(byObject),ncol=length(byObject[[1]]),byrow=TRUE,dimnames=list(NULL,names(byObject[[1]]
}
but when I run it:
convertByToDataFrame(exampleBY)
I either: (1) loose the types of the different values in the BY, or 2) get
factors instead of the values
I tried the following but it doesn't scale:
exampleSUMRY <- data.frame(a =
as.vector(unlist(lapply(strsplit(names(exampleBY),split="_"),function(x)
x[1]))),
b =
as.vector(unlist(lapply(strsplit(names(exampleBY),split="_"),function(x)
x[2]))),
name=as.vector(unlist(lapply(exampleBY,function(x)
x[[1]]
Any suggestions?
Thank you,
Ramiro
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Re: [R] Calculating rolling mean by group
Thanks for getting me on the right path Gabor! I have one outstanding
issue though.
On Mon, Jan 9, 2012 at 4:21 PM, Gabor Grothendieck
wrote:
> On Mon, Jan 9, 2012 at 6:39 PM, Sam Albers wrote:
>> Hello all,
>>
>> I am trying to determine how to calculate rolling means in R using a
>> grouping variable. Say I have a dataframe like so:
>>
>> dat1 <- data.frame(x = runif(2190, 0, 125), year=rep(1995:2000,
>> each=365), jday=1:365, site="here")
>> dat2 <- data.frame(x = runif(2190, 0, 200), year=rep(1995:2000,
>> each=365), jday=1:365, site="there")
>> dat <- rbind(dat1,dat2)
>>
>> ## What I would like to do is calculate a rolling 7 day mean
>> separately for each site. I have looked at both
>> ## rollmean() in the zoo package and running.mean() in the igraph
>> package but neither seem to have led
>> ## me to calculating a rolling mean by group. My first thought was to
>> use the plyr package but I am confused
>> ## by this output:
>>
>> library(plyr)
>> library(zoo)
>>
>> ddply(dat, c("site"), function(df) return(c(roll=rollmean(df$x, 7
>>
>> ## Can anyone recommend a better way to do this or shed some light on
>> this output?
>>
>
> Using dat in the question, try this:
>
> library(zoo)
> z <- read.zoo(dat, index = 2:3, split = 4, format = "%Y %j")
> zz <- rollmean(z, 7)
>
> The result, zz, is a multivariate zoo series with one column per group.
Using the zoo approach works well except that an wrinkle in my dataset
not reflected in the sample data caused some problems. I am actually
dealing with a situation where there is an unequal number of
observations in each group like the below data set
library(zoo)
dat1 <- data.frame(x = runif(2190, 0, 125), year=rep(1995:2000,
each=365), jday=1:365, site="here")
dat2 <- data.frame(x = runif(4380, 0, 200), year=rep(1989:2000,
each=365), jday=1:365, site="there")
dat <- rbind(dat1,dat2)
## When I use read.zoo everything is read in fine
z <- read.zoo(dat, index = 2:3, split = 4, format = "%Y %j")
## But when I use rollmean to get a 7 day average for both the 'here'
and 'there' columns only the 'there' column 7 day
## average is calculated
zz <- rollmean(z, 7)
Any thoughts on how I can then calculate a rolling mean on groups
where there is an unequal number of observations in each group?
Thanks for the previous post and in advance.
Sam
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
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Re: [R] 2 sample wilcox.test != kruskal.test
2012/1/10 syrvn > And why does kruskal.test(x~y) differ from kruskal.test(f~d)?? > Your formula is wrong, but function doesn't see errors. "formula a formula of the form lhs ~ rhs where lhs gives the data values and rhs the corresponding groups." And that leads to kruskal.test(d~as.factor(f)) which is fine. -- Mi³ego dnia [[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] problem installing packages
On 10.01.2012 14:35, natalia norden wrote: Hello, I was using version 2.13.2 and I have just downloaded the latest version 2.14.1. However, I'm trying to install the packages I was using and when I look for them in the packages list, I can´t find many in the CRAN binaries (e.g. "vegan"). I do find them in the CRAN sources but the installation fails. I tried downloading the version 2.14.0 and I had the same problem. I re-installed the old version, and now it works again. Is this a problem with 2.14? No. According to http://cran.r-project.org/web/checks/check_results_vegan.html the package works fine with R-2.14.1 (aka R-release). Uwe Ligges Thank you for your help. Natalia Norden Natalia Norden Profesor Asistente Departamento de Ecología y Territorio Facultad de Estudios Ambientales y Rurales Pontificia Universidad Javeriana Bogotá, Colombia Tel: 320 83 20 Ext: 2448 www.phylodiversity.net/nnorden/ [[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] 2 sample wilcox.test != kruskal.test
Hello,
I think I am right in saying that a 2 sample wilcox.test is equal to a 2
sample kruskal.test and
a 2 sample t.test is equal to a 2 sample anova. This is also stated in the
?kruskal.test man page:
The Wilcoxon rank sum test (wilcox.test) as the special case for two
samples; lm together with anova for performing one-way location analysis
under normality assumptions; with Student's t test (t.test) as the special
case for two samples.
>From this example it seems like it doesn't but I cannot figure out what I am
doing wrong.
x <- c(10,11,15,8,16,12,20)
y <- c(10,14,18,25,28,30,35)
f <- c(rep("a",7), rep("b",7))
d <- c(x,y)
wilcox.test(x,y)
kruskal.test(x,y)
kruskal.test(x~y)
kruskal.test(f~d)
t.test(x,y)
anova(lm(x~y))
summary(aov(lm(x~y)))
And why does kruskal.test(x~y) differ from kruskal.test(f~d)??
Cheers
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Re: [R] Adding Institution-Affiliation to Description File of R Package
On 10/01/2012 1:02 PM, Ben Ganzfried wrote: Hi, I'm just finishing up an R package and I was wondering if anyone knows how to include institution name in the "Description File." That is, my current Description File looks like: Package: curatedCancerData Type: Package Title: Cancer Gene Expression Analysis Version: 1.0 Date: 2011-12-24 Author: Benjamin F. Ganzfried, et al. Maintainer: Benjamin F. Ganzfried Description: The curatedCancerData package provides relevant functions and data for gene expression analysis. License: GPL (>= 3) Ideally I would like the final PDF to have footnotes showing the university affiliations of all the authors (more will be listed than currently are above). I would greatly appreciate any suggestions! Write a curatedCancerData-package.Rd topic, and put the information in there. It won't show up on the title page of the reference manual, but it will be the first topic. See "Documenting Packages" (section 2.1.4, I think) in the "Writing R Extensions" manual. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] colored outliers
woops! see inline.
Hope that helps, and enjoy R.
Justin
On Tue, Jan 10, 2012 at 8:40 AM, Geophagus
wrote:
> Hi Justin,
> thanks a lot for your quick answer.
> If I use your code, all points become red.
> How do you include the sorted and separated four values into the "points"
> argument?
> The variable in your script is called "circ" but this is not fronted up
> anymore.
> Here the script again:
>
>
> TOC_NI<-read.csv2("C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv",
> sep=";", dec=",", encoding="UTF-8")
>
this line just needs trimming. not sure how i missed that on my copy...
anyway, order puts the data.frame in order of the given vector, default
behavior sorts in ascending order unless you specify decreasing=TRUE.
circ<-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,]
>
and it should work
> plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
> abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
> points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3)
>
> Thanks a lot for your help!
> GeO
>
>
>
> --
> View this message in context:
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> Sent from the R help mailing list archive at Nabble.com.
>
>
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>
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[R] Adding Institution-Affiliation to Description File of R Package
Hi, I'm just finishing up an R package and I was wondering if anyone knows how to include institution name in the "Description File." That is, my current Description File looks like: Package: curatedCancerData Type: Package Title: Cancer Gene Expression Analysis Version: 1.0 Date: 2011-12-24 Author: Benjamin F. Ganzfried, et al. Maintainer: Benjamin F. Ganzfried Description: The curatedCancerData package provides relevant functions and data for gene expression analysis. License: GPL (>= 3) Ideally I would like the final PDF to have footnotes showing the university affiliations of all the authors (more will be listed than currently are above). I would greatly appreciate any suggestions! Thanks! Ben [[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] Online 'Beginner's Guide to R' course (with video)
Apologies for cross-posting We would like to announce an on-line 'Beginner's Guide to R' course With video presentations of theory and solutions For details: http://www.highstat.com/statscourse.htm Kind regards, Alain Zuur __ 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] runif with condition
I have to disagree with what's been posted, but I think some very interesting
points have been addressed. I'd like to add my two cents.
Consider the pair {X, 1-X} where X is sampled from a uniform(0,1)
distribution. The quantity 1- X also comes from a uniform(0,1) distribution
and therefore is probabilistic and not deterministic.
The sum of independent random variables is itself a random variable. If X1,
X2 & X3 are uniformly distributed, then the distribution of Y = X1 + X2 + X3
can be determined (i.e. Y is probabilistic and NOT deterministic). Y is a
random variable, but it is correlated with X1, X2 and X3. The set {X1, X2,
X3, 100 - (X1 + X2 + X3) } contains 4 random variables, however they are
neither independent or identically distributed.
If you are curious, check this out.
Deriving the Probability Density for Sums of Uniform Random Variables
Edward J. Lusk and Haviland Wright
The American Statistician
Vol. 36, No. 2 (May, 1982), pp. 128-130
Thanks to the OP. This has become an interesting thread.
-Alan Mitchell
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Re: [R] Sum of a couple of variables of which a few have NA values
x = runif(10) x[4] = NA sum(x, na.rm = T) -- View this message in context: http://r.789695.n4.nabble.com/Sum-of-a-couple-of-variables-of-which-a-few-have-NA-values-tp4282448p4282483.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] colored outliers
Hi Justin,
thanks a lot for your quick answer.
If I use your code, all points become red.
How do you include the sorted and separated four values into the "points"
argument?
The variable in your script is called "circ" but this is not fronted up
anymore.
Here the script again:
TOC_NI<-read.csv2("C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv",
sep=";", dec=",", encoding="UTF-8")
circ<-TOC_NI[order(TOC_NI$NI,decreasing=T),]
plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3)
Thanks a lot for your help!
GeO
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Re: [R] question about R 2.15.0
On 10/01/2012 11:31 AM, Wang, Jing wrote: Dear Sir/Madam, I want to download R development version 2.15.0 source code. But I just found the version for windows and MacOS. So would you please give me some instruction about how can I download the R 2.15.0 source code? Thank you very much for you help! Go to your local mirror of CRAN, and look in the second box "Source Code for all Platforms" on the main page. There is no 2.15.0 (it hasn't been released yet); the closest we have is the daily snapshot of the development version (R-devel). You might instead want R-patched, which contains bug fixes to 2.14.1, and which will eventually be released as 2.14.2. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum of a couple of variables of which a few have NA values
On Jan 10, 2012, at 11:25 AM, Petra Opic wrote: Dear everyone, I have looked all over the internet but I cannot find a way to solve my problem. ? rowSums # has an na.rm argument In my data I want to sum a couple of variables. Some of these variables have NA values, and when I add them together, the result is NA snip attach(dat) You would be well-advised to forget `attach`. Use with(dat, ...) instead. It will prevent frustration and embarrassing postings to rhelp. dat$sum <- var2 + var3 + var4 The plus infix operator does not have an na.rm argment dat$sum <- rowSums(dat[ , 2:4] , na.rm=TRUE) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: Sum of a couple of variables of which a few have NA values
Hi Petra, Try this: dat$sums <- rowSums(dat[3:5], na.rm=TRUE) I think this should do what you're looking for HTH, Ivan Message original Sujet: [R] Sum of a couple of variables of which a few have NA values Date : Tue, 10 Jan 2012 17:25:21 +0100 De :Petra Opic Pour : r-help@r-project.org Dear everyone, I have looked all over the internet but I cannot find a way to solve my problem. In my data I want to sum a couple of variables. Some of these variables have NA values, and when I add them together, the result is NA dat<- data.frame( id = gl(5,1), var1 = rnorm(5, 10), var2 = rnorm(5, 7), var3 = rnorm(5, 6), var4 = rnorm(5, 3), var5 = rnorm(5, 8) ) dat[3,3]<- NA dat[4,5]<- NA dat id var1 var2 var3 var4 var5 1 1 9.371328 7.830814 5.032541 3.491053 7.626418 2 2 10.413516 7.333630 6.557178 1.465597 8.591770 3 3 10.967073 NA 6.674079 3.946451 7.251263 4 4 9.900380 7.727111 5.059698 NA 6.632962 5 5 9.191068 7.901271 6.652410 2.734856 8.484757 attach(dat) dat$sum<- var2 + var3 + var4 # I think I'm doing this wrong, but I don't know what command to use dat id var1 var2 var3 var4 var5 sum 1 1 9.371328 7.830814 5.032541 3.491053 7.626418 16.35441 2 2 10.413516 7.333630 6.557178 1.465597 8.591770 15.35640 3 3 10.967073 NA 6.674079 3.946451 7.251263 NA 4 4 9.900380 7.727111 5.059698 NA 6.632962 NA 5 5 9.191068 7.901271 6.652410 2.734856 8.484757 17.28854 I would like to omit the values of NA and just sum the rest. I tried to use rowSums() but that sums an entire row and I only need a few variables. Does anyone know how to do this? Thanks in advance, Petra __ 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. -- Ivan CALANDRA Université de Bourgogne UMR CNRS 5561 Biogéosciences 6 Boulevard Gabriel 21000 Dijon, FRANCE ivan.calan...@u-bourgogne.fr __ 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] glmmPQL and predict
Mike Harwood gmail.com> writes: > Is the labeling/naming of levels in the documentation for the > predict.glmmPQL function "backwards"? The documentation states "Level > values increase from outermost to innermost grouping, with level zero > corresponding to the population predictions". Taking the sample in > the documentation: > > fit <- glmmPQL(y ~ trt + I(week > 2), random = ~1 | ID, >family = binomial, data = bacteria) > > > head(predict(fit, bacteria, level = 0, type="response")) > [1] 0.9680779 0.9680779 0.8587270 0.8587270 0.9344832 0.9344832 > > head(predict(fit, bacteria, level = 1, type="response")) > X01 X01 X01 X01 X02 X02 > 0.9828449 0.9828449 0.9198935 0.9198935 0.9050782 0.9050782 > > head(predict(fit, bacteria, type="response")) ## population prediction > X01 X01 X01 X01 X02 X02 > 0.9828449 0.9828449 0.9198935 0.9198935 0.9050782 0.9050782 > > The returned values for level=1 and level= match, which > is not what I expected based upon the documentation. Well, the documentation says: "Defaults to the highest or innermost level of grouping", which is level 1 in this case -- right? > Exponentiating > the intercept coefficients from the fitted regression, the level=0 > values match when the random effect intercept is included Do you mean "is NOT included" here? 0.9680779 (no random effect, below) matches the level=0 prediction above 0.9828449 (include random effect, below) matches the level=1 prediction, which is also the default prediction, above. > > > 1/(1+exp(-3.412014)) ## only the fixed effect > [1] 0.9680779 > > 1/(1+exp(-1*(3.412014+0.63614382))) ## fixed and random effect intercepts > [1] 0.9828449 This all matches my expectations. If your expectations still go in the other direction, could you explain in more detail? By the way, I recommend r-sig-mixed-mod...@r-project.org for mixed model questions in general ... Ben Bolker __ 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] Problem with segmented
Hi everyone. I'm trying to use the segmented function with the following data: For instance, I use segmented package as follow: myreg2 = lm(xy$y ~ xy$x) mysegmented = segmented(myreg2, seg.Z=~x, psi=c(245000), control = seg.control(display=FALSE)) Which get me to the following error : As a break point, a starting guess of 245000 seems fair. Anyone has an idea why I'm getting such error? Regards, Phil -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-segmented-tp4282398p4282398.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] question about R 2.15.0
Dear Sir/Madam, I want to download R development version 2.15.0 source code. But I just found the version for windows and MacOS. So would you please give me some instruction about how can I download the R 2.15.0 source code? Thank you very much for you help! Best, Jing Wang [[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] Sum of a couple of variables of which a few have NA values
Dear everyone, I have looked all over the internet but I cannot find a way to solve my problem. In my data I want to sum a couple of variables. Some of these variables have NA values, and when I add them together, the result is NA dat <- data.frame( id = gl(5,1), var1 = rnorm(5, 10), var2 = rnorm(5, 7), var3 = rnorm(5, 6), var4 = rnorm(5, 3), var5 = rnorm(5, 8) ) dat[3,3] <- NA dat[4,5] <- NA > dat id var1 var2 var3 var4 var5 1 1 9.371328 7.830814 5.032541 3.491053 7.626418 2 2 10.413516 7.333630 6.557178 1.465597 8.591770 3 3 10.967073 NA 6.674079 3.946451 7.251263 4 4 9.900380 7.727111 5.059698 NA 6.632962 5 5 9.191068 7.901271 6.652410 2.734856 8.484757 attach(dat) dat$sum <- var2 + var3 + var4 # I think I'm doing this wrong, but I don't know what command to use > dat id var1 var2 var3 var4 var5 sum 1 1 9.371328 7.830814 5.032541 3.491053 7.626418 16.35441 2 2 10.413516 7.333630 6.557178 1.465597 8.591770 15.35640 3 3 10.967073 NA 6.674079 3.946451 7.251263 NA 4 4 9.900380 7.727111 5.059698 NA 6.632962 NA 5 5 9.191068 7.901271 6.652410 2.734856 8.484757 17.28854 I would like to omit the values of NA and just sum the rest. I tried to use rowSums() but that sums an entire row and I only need a few variables. Does anyone know how to do this? Thanks in advance, Petra __ 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] Error message in vegan ordistep
I am getting the following erro rmessage in ordistep. I have a number of
similarly structured datasets using ordistep in a loop, and the message only
occurs for some of the datasets.
I cannot include a reproducible sample - the specific datasets where this is
occur ing are fairly large and there are several pcnm's in the rhs of the
formula.
thanks for any pointers that may allow me to track down the cause of the error.
Nevil Amos
Error in if (aod[1, 5] <= Pin) { : missing value where TRUE/FALSE needed
In addition: There were 50 or more warnings (use warnings() to see the first 50)
> traceback()
9: ordistep(myrda0, scope = formula(myrda1), direction = "both",
Pin = 0.05, Pout = 0.1) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
8: eval(expr, envir, enclos) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
7: eval(expr, pf) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
6: withVisible(eval(expr, pf)) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
5: evalVis(expr) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
4: capture.output(ordistep(myrda0, scope = formula(myrda1), direction = "both",
Pin = 0.05, Pout = 0.1)) at RDAPARTIALSexandAgeConnectandGEOGraphy2.R#86
3: eval.with.vis(expr, envir, enclos)
2: eval.with.vis(ei, envir)
1:
source("~/Documents/Dropbox/thesis/CH3/Analysis/RDAPARTIALSexandAgeConnectandGEOGraphy2.R")
>
> print(myrda1)
Call: rda(formula = mygenind@tab ~ pcnmTRE_25_100_CS25 + pcnmTRE_25_10_CS25 +
pcnmTRE_25_2_CS25 +
pcnmTRE_25_5_CS25 + mydata$TreeCov + mydata$Hab_Config + pcnmEYR_EO_100_CS25 +
pcnmEYR_EO_5000_CS25 + pcnmEYR_TH_10_CS25 + pcnmEYR_TH_2_CS25 + mydata$Site_No
+ mydata$Landscape
+ Condition(pcnmCS_NULL + mydata$LAT.x + mydata$LONG.x), na.action = "na.omit")
Inertia Proportion Rank
Total 1.8110 1.
Conditional0.8681 0.4793 32
Constrained0. 0.0
Unconstrained 0.9429 0.5207 29
Inertia is variance
Some constraints were aliased because they were collinear (redundant)
Eigenvalues for unconstrained axes:
PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8
0.16008 0.14733 0.12183 0.09054 0.07380 0.06971 0.05578 0.04215
(Showed only 8 of all 29 unconstrained eigenvalues)
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Re: [R] colored outliers
# find top 4 points
circ
<-
TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,]TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,]
# add them to your plot!
plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3)
Justin
On Tue, Jan 10, 2012 at 7:11 AM, Geophagus
wrote:
> Hi @ all,
> I have question how to mark significant outliers in R.
> This is my very simple script to plot a regression:
>
> TOC_NI<-read.csv2("C:/Users/XYZ/Desktop/Master/Daten/Statistik/TOC-NI.csv",
> sep=";", dec=",", encoding="UTF-8")
> plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
> abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
> summary(lm(NI~TOC,data=TOC_NI))
>
> The result is the following pic:
> http://r.789695.n4.nabble.com/file/n4282207/nickel_TOC_5f.png
> nickel_TOC_5f.png
>
> Now I want to make small red circles around the four highest values of Ni.
> Does anyone has an idea how to do that?
> Thanks a lot!
>
> Best Regards
> Geophagus
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/colored-outliers-tp4282207p4282207.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> and provide commented, minimal, self-contained, reproducible code.
>
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[R] rjags installation trouble
Trying to install latest rjags (3-5) from CRAN with JAGS 3.2.0
installed on Ubuntu 10.04, with r-devel ... the bottom line is that it
fails while loading with
/libs/rjags.so: undefined symbol: _ZN7Console15checkAdaptationERb
Has anyone else seen this or is it a glitch somewhere in my system?
thanks
Ben Bolker
==
bolker@ubuntu-10-new:~/R/pkgs/rjags$ jags
Welcome to JAGS 3.2.0 on Tue Jan 10 10:38:31 2012
JAGS is free software and comes with ABSOLUTELY NO WARRANTY
Loading module: basemod: ok
Loading module: bugs: ok
.
===
install.packages("rjags") starts out fine:
Installing package(s) into
‘/mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library’
(as ‘lib’ is unspecified)
trying URL 'http://probability.ca/cran/src/contrib/rjags_3-5.tar.gz'
Content type 'application/x-gzip' length 66429 bytes (64 Kb)
opened URL
==
downloaded 64 Kb
* installing *source* package ‘rjags’ ...
** package ‘rjags’ successfully unpacked and MD5 sums checked
checking for prefix by checking for jags... /usr/bin/jags
[snip snip]
** libs
g++ -I/usr/local/lib/R/include -DNDEBUG -I/usr/include/JAGS
-I/usr/local/include-fpic -g -O2 -c jags.cc -o jags.o
g++ -I/usr/local/lib/R/include -DNDEBUG -I/usr/include/JAGS
-I/usr/local/include-fpic -g -O2 -c parallel.cc -o parallel.o
g++ -shared -L/usr/local/lib -o rjags.so jags.o parallel.o -L/usr/lib -ljags
installing to /mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library/rjags/libs
** R
[snip snip]
but fails at:
** building package indices
Error : .onLoad failed in loadNamespace() for 'rjags', details:
call: dyn.load(file, DLLpath = DLLpath, ...)
error: unable to load shared object
'/mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library/rjags/libs/rjags.so':
/mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library/rjags/libs/rjags.so:
undefined symbol: _ZN7Console15checkAdaptationERb
ERROR: installing package indices failed
* removing ‘/mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library/rjags’
* restoring previous
‘/mnt/hgfs/bolker/Documents/LOCAL/lib/R/site-library/rjags’
The downloaded source packages are in
‘/tmp/RtmpSKHIz5/downloaded_packages’
Warning message:
In install.packages("rjags") :
installation of package ‘rjags’ had non-zero exit status
> sessionInfo()
R Under development (unstable) (2012-01-01 r58032)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_CA.utf8 LC_NUMERIC=C
[3] LC_TIME=en_CA.utf8LC_COLLATE=en_CA.utf8
[5] LC_MONETARY=en_CA.utf8LC_MESSAGES=en_CA.utf8
[7] LC_PAPER=CLC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_CA.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] tools_2.15.0
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[R] colored outliers
Hi @ all,
I have question how to mark significant outliers in R.
This is my very simple script to plot a regression:
TOC_NI<-read.csv2("C:/Users/XYZ/Desktop/Master/Daten/Statistik/TOC-NI.csv",
sep=";", dec=",", encoding="UTF-8")
plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
summary(lm(NI~TOC,data=TOC_NI))
The result is the following pic:
http://r.789695.n4.nabble.com/file/n4282207/nickel_TOC_5f.png
nickel_TOC_5f.png
Now I want to make small red circles around the four highest values of Ni.
Does anyone has an idea how to do that?
Thanks a lot!
Best Regards
Geophagus
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Re: [R] problem installing packages
On Jan 10, 2012, at 8:45 AM, Gavin Blackburn wrote: The packages you require might not have been updated yet. You could contact the package admin. That would not be the first option. Checking to see if your mirror is deficient by looking at another mirror would be the first option. You should also look at at the Package Checks page; http://cran.r-project.org/web/checks/check_summary.html There is a version of vegan on the mirror I am using and I have a relatively recent version of R 2.14.1 Patched (with a Mac). And there are no errors reported for current versions. So leave the package maintainer in peace, since he has been doing his job. Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org ] On Behalf Of natalia norden Sent: 10 January 2012 13:35 To: r-help@r-project.org Subject: [R] problem installing packages Hello, I was using version 2.13.2 and I have just downloaded the latest version 2.14.1. However, I'm trying to install the packages I was using and when I look for them in the packages list, I can´t find many in the CRAN binaries (e.g. "vegan"). I do find them in the CRAN sources but the installation fails. I tried downloading the version 2.14.0 and I had the same problem. I re-installed the old version, and now it works again. Is this a problem with 2.14? Thank you for your help. Natalia Norden -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lapack routine dgesv: system is exactly singular
I was sent a copy of the data on request. A quick look shows that
> range(days.alive[censored==0])
[1]0 1825
> range(days.alive[censored==1])
[1] 1826 1826
The original call of survdiff(Surv(days.alive, censored) ~ group) will
assume that censored=1 corresponds to deaths and 0 to alive; from the
above this is almost certainly backwards. If instead we force 0 to be
the "yes they're dead" group the results look better.
> survdiff(Surv(days.alive, censored==0) ~ group, f2)
Call:
survdiff(formula = Surv(days.alive, censored == 0) ~ group, data = f2)
N Observed Expected (O-E)^2/E (O-E)^2/V
group=PRI_CAS_5_NODU 3326 1129 1745 217.7 281
group=SEC_CAS_5_NODUP 13469 6731 6115 62.1 281
Chisq= 281 on 1 degrees of freedom, p= 0
In the original setup the test statistic had value 0 + roundoff error
and std = 0 + roundoff error due to all the "deaths" being tied on
exactly the same day, and you were getting the matrix version of a 0/0
error.
Terry T
--- begin included message
I have a problem with this error, I have searched the archives and found
previous discussion about this, can I cannot understand how the
explanations
apply to what I am trying to do.
I am trying to do Log_rank Survival analysis, I have included tables and
str command, is it a factor/integer problem? If so how do I correct
this, as all my attempt to recode the data have failed.
> survdiff(Surv(f2$days.alive , f2$censored)~group, data=f2)
Error in drop(.Call("La_dgesv", a, as.matrix(b), tol, PACKAGE =
"base")) :
Lapack routine dgesv: system is exactly singular
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Re: [R] strange Sys.Date() side effect
On 12-01-10 8:04 AM, Czerminski, Ryszard wrote:
Any ideas what is the problem with this code?
N<- 2; c(Sys.Date(), sprintf('N = %d', N))
[1] "2012-01-10" NA
Warning message:
In as.POSIXlt.Date(x) : NAs introduced by coercion
You are trying to create a vector combining a Date object and a
character object. R is trying to coerce both objects to dates, and that
fails.
You probably want two strings; so convert the date explicitly:
c(as.character(Sys.Date()), sprintf('N = %d', N))
(or use format or some other function to convert the date to a string.)
Duncan Murdoch
Best regards,
Ryszard
Ryszard Czerminski
AstraZeneca Pharmaceuticals LP
35 Gatehouse Drive
Waltham, MA 02451
USA
781-839-4304
ryszard.czermin...@astrazeneca.com
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Re: [R] problem installing packages
The packages you require might not have been updated yet. You could contact the package admin. Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of natalia norden Sent: 10 January 2012 13:35 To: r-help@r-project.org Subject: [R] problem installing packages Hello, I was using version 2.13.2 and I have just downloaded the latest version 2.14.1. However, I'm trying to install the packages I was using and when I look for them in the packages list, I can´t find many in the CRAN binaries (e.g. "vegan"). I do find them in the CRAN sources but the installation fails. I tried downloading the version 2.14.0 and I had the same problem. I re-installed the old version, and now it works again. Is this a problem with 2.14? Thank you for your help. Natalia Norden Natalia Norden Profesor Asistente Departamento de Ecología y Territorio Facultad de Estudios Ambientales y Rurales Pontificia Universidad Javeriana Bogotá, Colombia Tel: 320 83 20 Ext: 2448 www.phylodiversity.net/nnorden/ [[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] Propensity score matching in R using Classification tree method
A single tree will undermatch subjects.
Frank
ardsiiitmg wrote
>
> I can able to calculate the propensity score using classification tree
> method.
> And if i am trying to find the PS matching i am getting error.(Error in
> Match(Y = Y, Tr = Tr, X = ps0) : length(Y) != length(Tr))
>
> Propensity score matching:
> library("rgenoud")
> library("Matching")
> data("Passport")
> attach("Passport")
>
> Y<-Passport$A3
> Tr<-Passport$fserv_cd
> rr1<-Match(Y=Y, Tr = Tr, X= glm1$fitted)
> [in above command i am getting Error in Match(Y = Y, Tr = Tr, X = ps0) :
> length(Y) != length(Tr)]
>
> Matchbalance(fserv_cd~ A4,match.out=rr1,nboots=1000,data=Passport)
>
> Please suggest me right approach for Propensity score matching using
> classification trees(if possible code me) .
> ps0-propensity score values
>
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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[R] Propensity score matching in R using Classification tree method
I can able to calculate the propensity score using classification tree
method.
And if i am trying to find the PS matching i am getting error.(Error in
Match(Y = Y, Tr = Tr, X = ps0) : length(Y) != length(Tr))
Propensity score matching:
library("rgenoud")
library("Matching")
data("Passport")
attach("Passport")
Y<-Passport$A3
Tr<-Passport$fserv_cd
rr1<-Match(Y=Y, Tr = Tr, X= glm1$fitted)
[in above command i am getting Error in Match(Y = Y, Tr = Tr, X = ps0) :
length(Y) != length(Tr)]
Matchbalance(fserv_cd~ A4,match.out=rr1,nboots=1000,data=Passport)
Please suggest me right approach for Propensity score matching using
classification trees(if possible code me) .
ps0-propensity score values
--
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[R] problem installing packages
Hello, I was using version 2.13.2 and I have just downloaded the latest version 2.14.1. However, I'm trying to install the packages I was using and when I look for them in the packages list, I can´t find many in the CRAN binaries (e.g. "vegan"). I do find them in the CRAN sources but the installation fails. I tried downloading the version 2.14.0 and I had the same problem. I re-installed the old version, and now it works again. Is this a problem with 2.14? Thank you for your help. Natalia Norden Natalia Norden Profesor Asistente Departamento de Ecología y Territorio Facultad de Estudios Ambientales y Rurales Pontificia Universidad Javeriana Bogotá, Colombia Tel: 320 83 20 Ext: 2448 www.phylodiversity.net/nnorden/ [[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] strange Sys.Date() side effect
Any ideas what is the problem with this code?
> N <- 2; c(Sys.Date(), sprintf('N = %d', N))
[1] "2012-01-10" NA
Warning message:
In as.POSIXlt.Date(x) : NAs introduced by coercion
Best regards,
Ryszard
Ryszard Czerminski
AstraZeneca Pharmaceuticals LP
35 Gatehouse Drive
Waltham, MA 02451
USA
781-839-4304
ryszard.czermin...@astrazeneca.com
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[R] rworldmap: xlim, ylim do not change plotting region
Specifying xlim or ylim in the mapCountryData function of the rworldmap library do not alter the plotting region on my system. #Using the example from rworldmap library(rworldmap) mapCountryData() #uses the sample dataset mapCountryData(ylim = c(-45,45)) #makes no difference to the plot R 2.14.1 on Linux Mint 11 64bit. Thanks, Dan Dr Dan Bebber Head of Climate Change Research, Earthwatch Institute, 256 Banbury Road, Oxford OX2 7DE, UK Research Fellow in Biology, St. Peter's College, University of Oxford Tel. +44 (0)1865 318842, Mob. +44 (0)7729 167502, Fax. +44 (0)1865 318824, skype danbebber, email dbeb...@earthwatch.org.uk, web http://www.earthwatch.org/europe This e-mail (and any attachments) is confidential and may contain personal views, which are not the views of Earthwatch Institute Europe unless specifically stated. If you have received it in error, please delete it from your system, do not use, copy or disclose the information in any way nor act in reliance on it and notify the sender immediately. Please note that Earthwatch Institute (Europe) monitors e-mails sent or received. Further communication will signify your consent to this.Conservation Education & Research Trust also known as Earthwatch Institute (Europe) is a company limited by guarantee and registered in England and Wales under company number 4373313 and charity number 1094467. The registered address is, Mayfield House, 256 Banbury Road, Oxford, OX2 7DE England. __ 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] "tau + h > 1: error in summary.rq"
Dear all,
I am doing a simulation for my model that works when I use only the rq()
command. However, since I need to use the varcov matrix for my Wald test, I
need to compute summary(rq(), cov=TRUE). But the simulation does not work
because of the error: tau + h > 1: error in summary.rq
I tried to use:
if (tau + h > 1)
stop("tau + h > 1: error in summary.rq")
But the Hall-Sheather bandwidth is very high because I also vary the number of
observations from 40 to 300.
Is there anyone that could help me?
Thanks in advance!
All the best,
Julia
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Re: [R] [R-sig-Geo] Spatial data, rpoispp, using window with fixed radius?
Hi Adrian, I can not see any reference about scanmeasure and deviation functions in the current spatstat manual: are these included in the newest version of spatstat? Mathieu 2012/1/9 > > The following message appeared on R-help but this discussion should be > moved to R-sig-geo > > On 07/01/12 02:17, herbert8...@gmx.de wrote: > > > I was searching through the spatstat manual in order to find a function > > to simulate a Poisson pattern only within a fixed radius (circular > moving window) > > around individual points. If points are distributed heterogeneously > over a > > large area this may help to only assess deviation from CSR within the > window > > and thus does not require additional information on a covariate. > > I could not find such a function in spatstat. Can please anyone help? > > What do you want to happen if two of the circles overlap? Should the > density of random points be twice as high? > > If the answer is 'yes' then do the following (where X is your original > point pattern of centres, and 'r' is the radius of the circles, and > 'lambda' is the intensity of random points per unit area in each circle) > > V <- scanmeasure(X, r) > V <- eval.im(lambda * V) > Y <- rpoispp(V) > > If the answer is 'no' then do > W <- dilation(X, r) > Y <- rpoispp(lambda, win=W) > > Adrian Baddeley > > ___ > R-sig-Geo mailing list > r-sig-...@r-project.org > https://stat.ethz.ch/mailman/listinfo/r-sig-geo > [[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] Extracting Data from SQL Server
try:
SELECT a.UNIQUE_ID,
a.diag01
from LoadPUS a
left join CVD_ICD10 b
on a.diag01 = b.[ICD-10 Codes]
or a.diag02 = b.[ICD-10 Codes]
or a.diag03 = b.[ICD-10 Codes]
I am not sure why your table name CVD_ICD10 has a suffix $.
From: Jeff Newmiller
To: dthomas ; r-help@r-project.org
Sent: Tuesday, 10 January 2012, 8:00
Subject: Re: [R] Extracting Data from SQL Server
This is OT here. However, you might want to investigate the UNIQUE keyword in
the SQL Server documentation for SELECT.
---
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dthomas wrote:
>Hi,
>
>I am new to R (and rusty on SQL!) and I'm trying to extract records
>from a
>SQL server database. I have a table of patient records (LoadPUS) which
>have
>three code columns which i want to evaluate against a list of
>particular
>codes (CVD_ICD$ table). Given the size of the patient table I want to
>restrict the data I pull into R to the data I only want to analyse so I
>am
>using SQL to do this. The code i have is as follows:
>
>library(RODBC)
>channel<-odbcConnect("NatCollections")
>query<-"SELECT UNIQUE_ID, diag01 from LoadPUS
>WHERE (diag01 IN (SELECT [ICD-10 Codes] From CVD_ICD10$)) OR (diag02 IN
>(SELECT [ICD-10 Codes] From CVD_ICD10$))
>OR (diag03 IN (SELECT [ICD-10 Codes] From CVD_ICD10$))"
>
>This returns duplicate values, I don't want to hardcode the values
>because
>it is quite a long list. Running the "IN" function just for "diag01"
>returns
>the correct number of records, however when combining with another "IN"
>function it doesn't return the correct number of records. Can you see
>where
>my SQL is incorrect or is there another way of doing this?
>
>Much appreciated,
>D
>
>--
>View this message in context:
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>Sent from the R help mailing list archive at Nabble.com.
>
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Unexpected results using the oneway_test in the coin package
On Jan 09, 2012 at 11:48am Christoph Liedtke wrote: > I should be detecting some non-significance between groups I and III at > least, but the test comes back with > extremely low p-values. Where am I going wrong? Nowhere, I think. This does seem to be an error in coin. You should send your example to the maintainers of the package. Apart from the visual you have provided, other reasons for thinking that this is an error are the following. First, if you redo the analysis excluding habitat II, then the contrast is not significant, as expected. Secondly, if you repeat the full analysis using package nparcomp then you get the results you are expecting, based on the graphical representation of the data. See the examples below. ## drop habitat == II NDWD <- oneway_test(breeding ~ habitat, data = droplevels(subset(mydata, habitat != "II")), ytrafo = function(data) trafo(data, numeric_trafo = rank), xtrafo = function(data) trafo(data, factor_trafo = function(x) model.matrix(~x - 1) %*% t(contrMat(table(x), "Tukey"))), teststat = "max", distribution = approximate(B = 90)) print(NDWD) print(pvalue(NDWD, method = "single-step")) ## use nparcomp library(nparcomp) npar <- nparcomp(breeding ~ habitat, data = mydata, type = "Tukey") npar Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Unexpected-results-using-the-oneway-test-in-the-coin-package-tp4278371p4281329.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] error in Recursive
arunkumar wrote
>
> Hi
>
> I need help in the recursive problem. this is my code
>
> #Generate two random Numbers
> minval=20
> maxval=100
> cutoffValue=50
>
> optVal<- function(cutoffValue,minval,maxval)
> {
>x=runif(2)
>x=x*cutoffValue
>for( i in 1:2)
> {
> if(x[i] < 30 || x[i] >60) # checking it falls between the range
> {
> optVal(cutoffValue,minval,maxval)
> }
>}
>return(x)
> }
>
> optVal(40,20,60)
>
> I'm getting Error like this
> *Error: evaluation nested too deeply: infinite recursion /
> options(expressions=)?*
>
You can easily find out what is wrong by inserting a print(x) just before
the if(...).
Why are you not using the minval and maxval arguments in your function
optVal?
Change the line with the if to
if(x[i] < minval || x[i] > maxval) # checking it falls between the
range
and call optVal with a smaller value for minval
optVal(40,20,60)
and you will avoid the error message. Your minval of 30 is simply too high
i.e. you are rejecting an x too quickly.
Berend
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Re: [R] Extracting Data from SQL Server
This is OT here. However, you might want to investigate the UNIQUE keyword in
the SQL Server documentation for SELECT.
---
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/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
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dthomas wrote:
>Hi,
>
>I am new to R (and rusty on SQL!) and I'm trying to extract records
>from a
>SQL server database. I have a table of patient records (LoadPUS) which
>have
>three code columns which i want to evaluate against a list of
>particular
>codes (CVD_ICD$ table). Given the size of the patient table I want to
>restrict the data I pull into R to the data I only want to analyse so I
>am
>using SQL to do this. The code i have is as follows:
>
>library(RODBC)
>channel<-odbcConnect("NatCollections")
>query<-"SELECT UNIQUE_ID, diag01 from LoadPUS
>WHERE (diag01 IN (SELECT [ICD-10 Codes] From CVD_ICD10$)) OR (diag02 IN
>(SELECT [ICD-10 Codes] From CVD_ICD10$))
>OR (diag03 IN (SELECT [ICD-10 Codes] From CVD_ICD10$))"
>
>This returns duplicate values, I don't want to hardcode the values
>because
>it is quite a long list. Running the "IN" function just for "diag01"
>returns
>the correct number of records, however when combining with another "IN"
>function it doesn't return the correct number of records. Can you see
>where
>my SQL is incorrect or is there another way of doing this?
>
>Much appreciated,
>D
>
>--
>View this message in context:
>http://r.789695.n4.nabble.com/Extracting-Data-from-SQL-Server-tp4281000p4281000.html
>Sent from the R help mailing list archive at Nabble.com.
>
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