[R] trouble installing SparseM
Dear R People: I am attempting to install SparseM on R 2.15.0 on a Linux 11.10 system. Here is the output install.packages(SparseM,depen=TRUE) Installing package(s) into ‘/home/erin/R/x86_64-pc-linux-gnu-library/2.15’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.at.r-project.org/src/contrib/SparseM_0.96.tar.gz' Content type 'application/x-gzip' length 749740 bytes (732 Kb) opened URL == downloaded 732 Kb * installing *source* package ‘SparseM’ ... ** package ‘SparseM’ successfully unpacked and MD5 sums checked ** libs gfortran -fpic -O3 -pipe -g -c bckslv.f -o bckslv.o gfortran -fpic -O3 -pipe -g -c chol.f -o chol.o gfortran -fpic -O3 -pipe -g -c chol2csr.f -o chol2csr.o gfortran -fpic -O3 -pipe -g -c cholesky.f -o cholesky.o gfortran -fpic -O3 -pipe -g -c csr.f -o csr.o gfortran -fpic -O3 -pipe -g -c extract.f -o extract.o gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c iohb.c -o iohb.o iohb.c: In function ‘readHB_info’: iohb.c:267: warning: cast from pointer to integer of different size iohb.c: In function ‘readHB_header’: iohb.c:309: warning: cast from pointer to integer of different size iohb.c:310: warning: cast from pointer to integer of different size iohb.c:328: warning: cast from pointer to integer of different size iohb.c:346: warning: cast from pointer to integer of different size iohb.c:347: warning: cast from pointer to integer of different size iohb.c:348: warning: cast from pointer to integer of different size iohb.c:349: warning: cast from pointer to integer of different size iohb.c:362: warning: cast from pointer to integer of different size iohb.c:305: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:313: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:323: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:336: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:354: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_double’: iohb.c:423: warning: cast from pointer to integer of different size iohb.c:446: warning: cast from pointer to integer of different size iohb.c:474: warning: cast from pointer to integer of different size iohb.c:427: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:450: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:478: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_double’: iohb.c:660: warning: cast from pointer to integer of different size iohb.c:627: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:638: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:645: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:665: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:693: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_char’: iohb.c:975: warning: cast from pointer to integer of different size iohb.c:998: warning: cast from pointer to integer of different size iohb.c:1026: warning: cast from pointer to integer of different size iohb.c:979: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1002: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1030: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_char’: iohb.c:1192: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1203: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1213: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1231: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1262: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘ParseRfmt’: iohb.c:1537: warning: cast from pointer to integer of different size iohb.c:1547: warning: cast from pointer to integer of different size iohb.c:1551: warning: cast from pointer to integer of different size iohb.c: In function ‘substr’: iohb.c:1585: warning: cast from pointer to integer of different size gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c readwrite.c -o readwrite.o readwrite.c: In function ‘read_HB1’: readwrite.c:33:
[R] Using apply() with a function involving ode()
Hello,
I am trying to get the output from the numerical simulation of a system
of ordinary differential equations for a range of values for three
parameters. I am using the ode() function (deSolve package) to run the
numerical simulation and apply() to run the simulation function for each
set of parameter values. I am having trouble getting the apply()
function to work.
Here is an example. Consider an epidemiology model of West Nile Virus:
wnv.model - function(Time, State, Pars){
with(as.list(c(State, Pars)), {
dSB - -(alphaB*betaB*SB*IM)/(SB + IB)
dIB - (alphaB*betaB*SB*IM)/(SB + IB) - deltaB*IB
dSM - bM*(SM + EM + IM) - (alphaM*betaB*IB*SM)/(SB
+ IB) - dM*SM
dEM - (alphaM*betaB*SM*IB)/(SB + IB) - kappaM*EM -
dM*EM
dIM - kappaM*EM - dM*IM
return(list(c(dSB, dIB, dSM, dEM, dIM)))
})
}
# I would like to run a numerical simulation of this model for each
combination of values for the parameters alphaB, betaB, # and kappaM:
av - seq(0, 1, by = 0.2) # vector of values for alphaB
bv - seq(0, 1, by = 0.2) # vector of values for betaB
kv - seq(0, 1, by = 0.2) # vector of values for kappaM
# Here is my function with ode() for the numerical simulation. The
function returns the last value of the simulation for the # IB state
variable (Infected Birds).
library(deSolve)
wnv.sim - function(x){
Pars - c(alphaB = x[1], betaB = x[2], deltaB = 0.2, bM =
0.03, dM = 0.03, alphaM = 0.69, kappaM = x[3])
State - c(SB = 100, IB = 0, SM = 1500, EM = 0, IM = 0.0001)
Time - seq(0, 60, by = 1)
model.out - as.data.frame(ode(func = wnv.incubation,
y = State,
parms = Pars,
times = Time))
model.out[nrow(model.out),]$IB
}
# Finally, here is my apply() function:
dat - apply(expand.grid(av, bv, kv), 1, wnv.sim)
However, the apply() function returns an error:
Error in eval(expr, envir, enclos) : object 'alphaB' not found
I am not sure what is wrong. Any help would be much appreciated.
Sincerely,
Adam
--
Adam Zeilinger
Post Doctoral Scholar
Department of Entomology
University of California Riverside
www.linkedin.com/in/adamzeilinger
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[R] Combining local data frame with external data
Hi, I have a list of IDs on a data frame and I want to extract information for those IDs from an external database. I did a workaround using the IN clause in SQL but when the number of IDs is big this method doesn't work. In other words, I want to create a join between a local and a external table using either RODBC() or any other package. Thanks a lot for your help! Orlando -- View this message in context: http://r.789695.n4.nabble.com/Combining-local-data-frame-with-external-data-tp4585355p4585355.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] Conditional average
Kebrab67-- It's difficult to answer in the absence of a small bit of representative data, and more information about it. How is age76 recorded, as numerical years or in categories (age ranges?) And what are years (you didn't mention them as a variable in your data.) That being said, perhaps by(), or summaryBy() in the doBy package, might help. --Chris Ryan SUNY Upstate Medical University Clincal Campus Binghamton, NY kebrab67 wrote: Hello, I have a set of data including age, wage and education level each called age76, wage76 and grade76 I want to know how i can calculate the average wage of people age 15 to 65 (each year separetly) , only for those who have an education level of 10 12 and 16... -- View this message in context: http://r.789695.n4.nabble.com/Conditional-average-tp4585313p4585313.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to reduce plot size on linux?
Hi, I am working on linux and i need to reduce plot size (bar plot) so that i can easily use in sweave. How can i implement it? Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-reduce-plot-size-on-linux-tp4585371p4585371.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] Removing the rows from dataset
Hi, I have data set where i have col1,col2,col3,col4 i want to write a condition where the rows has to removed from the dataset for col110 Please help, Thanks Santosh -- View this message in context: http://r.789695.n4.nabble.com/Removing-the-rows-from-dataset-tp4585710p4585710.html Sent from the R help mailing list archive at Nabble.com. [[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] trouble installing SparseM
On 25.04.2012 08:21, Erin Hodgess wrote: Dear R People: I am attempting to install SparseM on R 2.15.0 on a Linux 11.10 system. Here is the output install.packages(SparseM,depen=TRUE) Installing package(s) into ‘/home/erin/R/x86_64-pc-linux-gnu-library/2.15’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.at.r-project.org/src/contrib/SparseM_0.96.tar.gz' Content type 'application/x-gzip' length 749740 bytes (732 Kb) opened URL == downloaded 732 Kb * installing *source* package ‘SparseM’ ... ** package ‘SparseM’ successfully unpacked and MD5 sums checked ** libs gfortran -fpic -O3 -pipe -g -c bckslv.f -o bckslv.o gfortran -fpic -O3 -pipe -g -c chol.f -o chol.o gfortran -fpic -O3 -pipe -g -c chol2csr.f -o chol2csr.o gfortran -fpic -O3 -pipe -g -c cholesky.f -o cholesky.o gfortran -fpic -O3 -pipe -g -c csr.f -o csr.o gfortran -fpic -O3 -pipe -g -c extract.f -o extract.o gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c iohb.c -o iohb.o iohb.c: In function ‘readHB_info’: iohb.c:267: warning: cast from pointer to integer of different size iohb.c: In function ‘readHB_header’: iohb.c:309: warning: cast from pointer to integer of different size iohb.c:310: warning: cast from pointer to integer of different size iohb.c:328: warning: cast from pointer to integer of different size iohb.c:346: warning: cast from pointer to integer of different size iohb.c:347: warning: cast from pointer to integer of different size iohb.c:348: warning: cast from pointer to integer of different size iohb.c:349: warning: cast from pointer to integer of different size iohb.c:362: warning: cast from pointer to integer of different size iohb.c:305: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:313: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:323: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:336: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:354: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_double’: iohb.c:423: warning: cast from pointer to integer of different size iohb.c:446: warning: cast from pointer to integer of different size iohb.c:474: warning: cast from pointer to integer of different size iohb.c:427: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:450: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:478: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_double’: iohb.c:660: warning: cast from pointer to integer of different size iohb.c:627: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:638: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:645: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:665: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:693: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_char’: iohb.c:975: warning: cast from pointer to integer of different size iohb.c:998: warning: cast from pointer to integer of different size iohb.c:1026: warning: cast from pointer to integer of different size iohb.c:979: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1002: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1030: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_char’: iohb.c:1192: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1203: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1213: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1231: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1262: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘ParseRfmt’: iohb.c:1537: warning: cast from pointer to integer of different size iohb.c:1547: warning: cast from pointer to integer of different size iohb.c:1551: warning: cast from pointer to integer of different size iohb.c: In function ‘substr’: iohb.c:1585: warning: cast from pointer to integer of different size gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c readwrite.c -o readwrite.o readwrite.c:
Re: [R] Removing the rows from dataset
Seems you should re-read An Introduction to R that comes with it. Hint: look for subset. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. santoshdvn santosh...@gmail.com wrote: Hi, I have data set where i have col1,col2,col3,col4 i want to write a condition where the rows has to removed from the dataset for col110 Please help, Thanks Santosh -- View this message in context: http://r.789695.n4.nabble.com/Removing-the-rows-from-dataset-tp4585710p4585710.html Sent from the R help mailing list archive at Nabble.com. [[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] trouble installing SparseM
On 25/04/2012 08:05, Uwe Ligges wrote: On 25.04.2012 08:21, Erin Hodgess wrote: Dear R People: I am attempting to install SparseM on R 2.15.0 on a Linux 11.10 system. Here is the output install.packages(SparseM,depen=TRUE) Installing package(s) into ‘/home/erin/R/x86_64-pc-linux-gnu-library/2.15’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.at.r-project.org/src/contrib/SparseM_0.96.tar.gz' Content type 'application/x-gzip' length 749740 bytes (732 Kb) opened URL == downloaded 732 Kb * installing *source* package ‘SparseM’ ... ** package ‘SparseM’ successfully unpacked and MD5 sums checked ** libs gfortran -fpic -O3 -pipe -g -c bckslv.f -o bckslv.o gfortran -fpic -O3 -pipe -g -c chol.f -o chol.o gfortran -fpic -O3 -pipe -g -c chol2csr.f -o chol2csr.o gfortran -fpic -O3 -pipe -g -c cholesky.f -o cholesky.o gfortran -fpic -O3 -pipe -g -c csr.f -o csr.o gfortran -fpic -O3 -pipe -g -c extract.f -o extract.o gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c iohb.c -o iohb.o iohb.c: In function ‘readHB_info’: iohb.c:267: warning: cast from pointer to integer of different size iohb.c: In function ‘readHB_header’: iohb.c:309: warning: cast from pointer to integer of different size iohb.c:310: warning: cast from pointer to integer of different size iohb.c:328: warning: cast from pointer to integer of different size iohb.c:346: warning: cast from pointer to integer of different size iohb.c:347: warning: cast from pointer to integer of different size iohb.c:348: warning: cast from pointer to integer of different size iohb.c:349: warning: cast from pointer to integer of different size iohb.c:362: warning: cast from pointer to integer of different size iohb.c:305: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:313: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:323: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:336: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:354: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_double’: iohb.c:423: warning: cast from pointer to integer of different size iohb.c:446: warning: cast from pointer to integer of different size iohb.c:474: warning: cast from pointer to integer of different size iohb.c:427: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:450: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:478: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_double’: iohb.c:660: warning: cast from pointer to integer of different size iohb.c:627: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:638: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:645: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:665: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:693: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_mat_char’: iohb.c:975: warning: cast from pointer to integer of different size iohb.c:998: warning: cast from pointer to integer of different size iohb.c:1026: warning: cast from pointer to integer of different size iohb.c:979: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1002: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1030: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘readHB_aux_char’: iohb.c:1192: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1203: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1213: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1231: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c:1262: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result iohb.c: In function ‘ParseRfmt’: iohb.c:1537: warning: cast from pointer to integer of different size iohb.c:1547: warning: cast from pointer to integer of different size iohb.c:1551: warning: cast from pointer to integer of different size iohb.c: In function ‘substr’: iohb.c:1585: warning: cast from pointer to integer of different size gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fpic -O3 -pipe -g -c readwrite.c -o readwrite.o readwrite.c: In
[R] RCOM comInvoke
Hi, I have a quick question regarding RCOM and in particular the comInvoke function. When we use comCreateObject we have a way to check the object was properly created with comIsValidHandle; however, after calling comInvoke the only way to check it worked as expected is to use comGetProperty and see if we get a correct information. The issue is that comInvoke always returns NULL so I couldn't find a way to catch a problem with comInvoke. I was wondering whether there was a way to check whether comInvoke worked as expected before extracting data from the object, i.e. before using comGetProperty etc... As a side note I did try to use comInvoke with inaccurate object names, i.e. objects not created, and it also returned NULL, as it does when it works and is properly called. Thanks in advance for your help. Best Regards, Arnaud __ 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] [pROC] roc.test returns NA p-value...
Hello, I am comparing two ROC curves with bootstraping. However, some runs return p-value = NA, and I have no clue why. Is this anyhow related to like sample size or no sufficient numbers of bootstraping? I used the default value (i.e. boot.n=2000), and the number of observations are quite big since I am comparing maps (e.g., the largest has more than 9 million observations). It'd would be great if someone could explain why this is happening. With thanks, OK -- View this message in context: http://r.789695.n4.nabble.com/pROC-roc-test-returns-NA-p-value-tp4585795p4585795.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] Using apply() with a function involving ode()
See inline comments.
On 25-04-2012, at 08:40, Adam Zeilinger wrote:
Hello,
I am trying to get the output from the numerical simulation of a system of
ordinary differential equations for a range of values for three parameters.
I am using the ode() function (deSolve package) to run the numerical
simulation and apply() to run the simulation function for each set of
parameter values. I am having trouble getting the apply() function to work.
Here is an example. Consider an epidemiology model of West Nile Virus:
wnv.model - function(Time, State, Pars){
with(as.list(c(State, Pars)), {
dSB - -(alphaB*betaB*SB*IM)/(SB + IB)
dIB - (alphaB*betaB*SB*IM)/(SB + IB) - deltaB*IB
dSM - bM*(SM + EM + IM) - (alphaM*betaB*IB*SM)/(SB + IB)
- dM*SM
dEM - (alphaM*betaB*SM*IB)/(SB + IB) - kappaM*EM - dM*EM
dIM - kappaM*EM - dM*IM
return(list(c(dSB, dIB, dSM, dEM, dIM)))
})
}
# I would like to run a numerical simulation of this model for each
combination of values for the parameters alphaB, betaB, # and kappaM:
av - seq(0, 1, by = 0.2) # vector of values for alphaB
bv - seq(0, 1, by = 0.2) # vector of values for betaB
kv - seq(0, 1, by = 0.2) # vector of values for kappaM
# Here is my function with ode() for the numerical simulation. The function
returns the last value of the simulation for the # IB state variable
(Infected Birds).
library(deSolve)
wnv.sim - function(x){
Pars - c(alphaB = x[1], betaB = x[2], deltaB = 0.2, bM = 0.03, dM =
0.03, alphaM = 0.69, kappaM = x[3])
State - c(SB = 100, IB = 0, SM = 1500, EM = 0, IM = 0.0001)
Time - seq(0, 60, by = 1)
model.out - as.data.frame(ode(func = wnv.incubation,
There is no function wnv.incubation.
Assuming you mean wnv.model
y = State,
parms = Pars,
times = Time))
model.out[nrow(model.out),]$IB
}
# Finally, here is my apply() function:
dat - apply(expand.grid(av, bv, kv), 1, wnv.sim)
However, the apply() function returns an error:
Error in eval(expr, envir, enclos) : object 'alphaB' not found
Have you inspected the object returned by expand.grid?
Do head(expand.grid(av,bv,kv))
and you will see that the columns have names which confuse the rest of your
code.
I was able to repair by doing
pars.grid - expand.grid(av, bv, kv)
names(pars.grid) - NULL
dat - apply(pars.grid, 1, wnv.sim)
HTH,
Berend
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Re: [R] How to reduce plot size on linux?
On 25.04.2012 04:02, Manish Gupta wrote: Hi, I am working on linux and i need to reduce plot size (bar plot) so that i can easily use in sweave. How can i implement it? You can control the size of the plots from Sweave, no need to change soemthing in the plts themselves (unless you want relative size corrections). But that would need some more specific question. uwe Ligges Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-reduce-plot-size-on-linux-tp4585371p4585371.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [pROC] roc.test returns NA p-value...
Probably you got NAs in some bootstrap results, or you got a 0 variance, or you hat 0 observations in a group you compared to another group? Check the data and the separate results! Uwe Ligges On 25.04.2012 09:28, O wrote: Hello, I am comparing two ROC curves with bootstraping. However, some runs return p-value = NA, and I have no clue why. Is this anyhow related to like sample size or no sufficient numbers of bootstraping? I used the default value (i.e. boot.n=2000), and the number of observations are quite big since I am comparing maps (e.g., the largest has more than 9 million observations). It'd would be great if someone could explain why this is happening. With thanks, OK -- View this message in context: http://r.789695.n4.nabble.com/pROC-roc-test-returns-NA-p-value-tp4585795p4585795.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Splitting data into test and train (80:20) kepping attributes similar
Hi, Could someone help me with this please , im trying to use Y = Attrition_data[,1] # extract labels from the data msk = sample.split (Y, SplitRatio=3/4) table(Y,msk) to do the splitting but it keeps throwing up and error Error: could not find function sample.split Could you please help Thanks in advance doy -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dwaipayan Dasgupta Sent: Tuesday, April 24, 2012 9:08 PM To: r-help@r-project.org Subject: [R] Splitting data into test and train (80:20) kepping attributes similar Hi, I am trying to do some predictive modeling around attrition and want to split the dataset into test and train (80:20) and keep the ratio of attritees:non attrites same. In my dataset the attrition indicator is coded as 0(for non-attritees) and 1 (for attritees) and I want to keep the ratio of 0's to 1 similar. I apologize for this trivial question but this is my second week with R. Thanks, Doy American Express made the following annotations on Tue Apr 24 2012 08:38:50 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Tue Apr 24 2012 08:38:50 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- [[alternative HTML version deleted]] American Express made the following annotations on Wed Apr 25 2012 03:39:08 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Wed Apr 25 2012 03:39:08 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- __ 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 reduce plot size on linux?
Cross posting http://stackoverflow.com/questions/10308955/how-to-reduce-image-size-in-sweave The code you have provided there is unusable, I assume your problems come from a lack of understanding how Sweave works (nothing to do with Linux). Always make sure that = and @ are always the first characters on the line, Comment character in R is # and % in Latex (never //). Have a look at Section 9 of this paper here http://www.stat.auckland.ac.nz/%7Eihaka/downloads/Sweave-customisation.pdf This should solve your problems with plot size in Sweave. Rgds, Rainer On Tuesday 24 April 2012 19:02:23 Manish Gupta wrote: Hi, I am working on linux and i need to reduce plot size (bar plot) so that i can easily use in sweave. How can i implement it? Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-reduce- plot-size-on-linux-tp4585371p4585371.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R shell script
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple example
explaining how to make this shell script.
Any help appreciated :)
Aoife
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Re: [R] Positioning main title
On 04/25/2012 12:28 AM, ramonovelar wrote: Hello, I have a barplot where each row has quite long texts and I have used par to make some room in the left: par(mar=c(0, 17, 3, 0), oma=c(0, 0, 0, 0)) barplot2(prueba, main = l, col=colores, horiz=TRUE,las=1, cex.names=.7) My problem is that main text appears justified to the plot. I want to put it in the middle of the image, and find a way to sort it out with mtex par(mar=c(0, 17, 3, 0), oma=c(0, 0, 0, 0)) barplot2(prueba, col=colores, horiz=TRUE,las=1, cex.names=.7) mtext(l,side = 3, at = -70, line = 0.8) But I will have to adjust this at =-70 to his plot, depending on the length of the title. I wonder if I can't find a way to place it in the center. Hi Ramon, Have a look at the getFigCtr function in the plotrix package. This returns the center of the figure region by default, but can be adjusted to return different proportions of the figure region. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding more files to list
Hi, I have searched over the last two days to try and sort this problem but unfortunately I cannot find the correct solution. I have a main directory - Spectra In this folder I have two subfolders Normal and Case I am using a package MALDIquant for processing of mass spectrometry data, The following command will open and list all my files from the Normal subfolder (70 files containing spectrums) Spectra - mqReadBrukerFlex(file.path(Spectra/, Normal)); How do I add the Case folder to the list (another 70 spectra), so that I can then process all the files in the same manner and yet maintain their origin so when I want to do stats analyse on them I can differentiate?? Any help would be great, I am very much a beginner, Brian [[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] take data from a file to another according to their correlation coefficient
Seems to work great! I have a last question (or 2) for you about it, and I will leave you alone afterwords, I promise :) I tested your function process.all for the automatization. It seems to be OK. It's just when I'd like to save the filled data files. If I name process.all, for example: test - process.all(lst, corr2008) and I save it: write.table(test, ...) and I check the test file, It has filled my data but all the files from lst are in one file (the columns are: ST001, ST001_time, ST002, ST002_time, . (with ST001 for station 1 for example)). How can I cut these files and save them automatically (one file for ST001, another for ST002, ...) according to these columns names? And it is possible in your script to take the second best correlated station data instead of the best one, if there are NAs in this best correlated station at the same lines with the NA gaps of the station to fill? Thanks again for all your help. If you come one day in France near the Alps or Chamonix (where I'm working), just tell me. I'll pay you some beers or a restaurant! You deserve it ^^ By the way, where do my rescuer come from? Are you a statistician? Geoffrey -- View this message in context: http://r.789695.n4.nabble.com/take-data-from-a-file-to-another-according-to-their-correlation-coefficient-tp4580054p4586079.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] Assignment problems
Thanks again! I would like to construct 14 new 'year' dummy variables. I have 14 years: 1992:2006 with 231 observations pr. year. The year dummies should assign a 1 if the observation is within the specific year and 0 otherwise. So for example: 1992dummyvariable=1 if year=0 and so on. P -- View this message in context: http://r.789695.n4.nabble.com/Assignment-problems-tp4578672p4586252.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] recommended way to group function calls in Sweave
Dear all
When using Sweave, I'm always hitting the same bump: I want to group
repetitive calls in a function, but I want both the results and the
function calls in the printed output. Let me explain myself.
Consider the following computation in an Sweave document:
summary(iris[,1:2])
cor(iris[,1:2])
When using these two calls directly, I obtain the following output:
summary(iris[,1:2])
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
cor(iris[,1:2])
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
However, if I try to group the calls in a function:
f - function(d, ind){
print(summary(d[ , ind]))
print(cor(d[ , ind]))
return(invisible(NULL))
}
Then I get a different output in the Sweave PDF:
f(iris, 1:2)
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
Of course I can use 'echo=F' to remove the ' f(iris, 1:2)' in the
above, but how can I do to keep the original calls 'summary(d[ ,
ind])' and 'cor(d[ , ind])'? Or even better, could the actual calls be
used, after the replacement of arguments by their values:
'summary(iris[,1:2])' and 'cor(iris[,1:2])'.
Or is the recommended way to use cat()-ed statements:
f - function(d, ind){
cat('Variable summary:\n')
print(summary(d[ , ind]))
cat('\nCorrelation table:\n')
print(cor(d[ , ind]))
return(invisible(NULL))
}
To obtain such output:
f(iris, 1:2)
Variable summary:
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
Correlation table:
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
What is the recommended way of grouping repetitive function calls?
What do you usually use in your own documents? Regards
Liviu
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Re: [R] Positioning main title
Many many thanks for the tip and for authoring this function! The final code for the plot goes like this: par(mar=c(0, 17, 3, 0), oma=c(0, 0, 0, 0)) barplot2(prueba, col=colores, horiz=TRUE,las=1, cex.names=.7) mtext(l,side = 3, at = getFigCtr()[1], line = 0.8, cex=1.5) On Wed, Apr 25, 2012 at 12:51 PM, Jim Lemon j...@bitwrit.com.au wrote: On 04/25/2012 12:28 AM, ramonovelar wrote: Hello, I have a barplot where each row has quite long texts and I have used par to make some room in the left: par(mar=c(0, 17, 3, 0), oma=c(0, 0, 0, 0)) barplot2(prueba, main = l, col=colores, horiz=TRUE,las=1, cex.names=.7) My problem is that main text appears justified to the plot. I want to put it in the middle of the image, and find a way to sort it out with mtex par(mar=c(0, 17, 3, 0), oma=c(0, 0, 0, 0)) barplot2(prueba, col=colores, horiz=TRUE,las=1, cex.names=.7) mtext(l,side = 3, at = -70, line = 0.8) But I will have to adjust this at =-70 to his plot, depending on the length of the title. I wonder if I can't find a way to place it in the center. Hi Ramon, Have a look at the getFigCtr function in the plotrix package. This returns the center of the figure region by default, but can be adjusted to return different proportions of the figure region. Jim -- == Ramón Ovelar [[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] Create a new Vector based on two columns
Hello,
I am trying to get a new vector 'x1' based on the not NA-values in
column 'a' and 'b'. I found a way but I am sure this is not the best
solution. So any ideas on how to optimize this would be great!
m - factor(c(a1, a1, a2, b1, b2, b3, d1, d1), ordered
= TRUE)
df - data.frame( a= m, b = m)
df[1,1] - NA
df[4,2] - NA
df[2,2] - NA
df[6,1] - NA
df
w - !apply(df, 2, is.na)
v - apply(w, 1, FUN=function(L) which(L == TRUE)[[1]])
for (i in 1:nrow(df) ) {
g[i] - df[i, v[i]]
}
df$x1 - g
Thanks for any help
Patrick
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Re: [R] trouble loading ggplot2 using R
I don't think I have touched at anything at all. I'm very newbie to R and to be honest I don't know what Ramdom.seed is. I will try to find out. I have seen other messages about restoring random.seed, but in order to check that the problem is really that I have used some Viewing data commands. The error says, in Spanish, that .Random.seed is not an integer vector but a list (.Random.seed no es un vector de números enteros pero es de tipo 'list') class(.Random.seed) [1] data.frame str(.Random.seed) 'data.frame': 626 obs. of 1 variable: $ .Random.seed: num 4.03e+02 1.00e+01 -1.28e+09 -1.40e+09 -1.06e+09 .. head(.Random.seed) .Random.seed 1 403 2 10 3 -1282779759 4 -1404015037 5 -1062445742 6665436644 tail(.Random.seed) .Random.seed 621 1369617214 622 -1673749493 623 -1883947891 624 1445895610 625 -903220232 626970996181 On Tue, Apr 24, 2012 at 4:26 PM, Hadley Wickham had...@rice.edu wrote: I have a similar error, running R in Snow Leopard too library(ggplot2) Error : .onAttach failed in attachNamespace() for 'ggplot2', details: call: stats::runif(1) error: .Random.seed no es un vector de números enteros pero es de tipo 'list' Error: package/namespace load failed for ggplot2 That's a completely different error. Are you setting .Random.seed to something non-standard? That's what the error message suggests. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ -- == Ramón Ovelar [[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] Create a new Vector based on two columns
On Wed, Apr 25, 2012 at 02:22:05PM +0200, Patrick Hausmann wrote:
Hello,
I am trying to get a new vector 'x1' based on the not NA-values in
column 'a' and 'b'. I found a way but I am sure this is not the best
solution. So any ideas on how to optimize this would be great!
m - factor(c(a1, a1, a2, b1, b2, b3, d1, d1), ordered
= TRUE)
df - data.frame( a= m, b = m)
df[1,1] - NA
df[4,2] - NA
df[2,2] - NA
df[6,1] - NA
df
w - !apply(df, 2, is.na)
v - apply(w, 1, FUN=function(L) which(L == TRUE)[[1]])
for (i in 1:nrow(df) ) {
g[i] - df[i, v[i]]
}
df$x1 - g
Hello.
The above code does not initialize g. Adding the command
g - rep(NA, times=nrow(df))
i get the same as using
df$x1 - ifelse(is.na(df$a), df$b, df$a)
df
ab x1
1 NA a1 1
2 a1 NA 1
3 a2 a2 2
4 b1 NA 3
5 b2 b2 4
6 NA b3 5
7 d1 d1 6
8 d1 d1 6
The codes are obtained, since the original data frame contains
factors. If the intention is to keep character values, then use
df$x1 - ifelse(is.na(df$a), as.character(df$b), as.character(df$a))
df
ab x1
1 NA a1 a1
2 a1 NA a1
3 a2 a2 a2
4 b1 NA b1
5 b2 b2 b2
6 NA b3 b3
7 d1 d1 d1
8 d1 d1 d1
Hope this helps.
Petr Savicky.
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[R] FW: Combined grouped and stacked bargraph
Dear R list, It appears that my request from yesterday got a bit garbled. I have tried to rephrase and retype it - please let me know if you can help me - I am having some trouble drawing a bar-graph with two groups, both of which are stacked. I would like to plot the conservation status according to two classifications (i.e. my groups - IUCN status and national status), and for each of those groups I would like data for the marine and terrestrial species to be stacked. My data look like this (where the names for the columns are conservation status': NE, LC, NT, VU, EN and CR; and the matrix name is cs.not.log.bp): IUCN.Terrestrial168 41 5 4 4 1 IUCN.Marine 69 6 4 2 2 0 National.CS.Terrestrial 16 148 7 7 4 2 National.CS.Marine 69 6 4 2 2 0 I have tried the following code, but it does not work: barplot(cs.not.log.bp[c(1:2),], xlab = Conservation status, ylab = Number of species, col = c(grey90,grey80),names = cs.names, ylim = c(0,250), space = 2) barplot(cs.not.log.bp[c(3:4),], col = c(grey60,grey30), beside = T,add = T,names.arg = NA) legend(topright,c(IUCN Terrestrial,IUCN Marine,National CS Terrestrial,National CS Marine), col = c(grey90,grey80,grey60,grey30), pch = 15) What happens is that some of the data in the second group stacks onto the first group and then the remainder forms a second group. I would like only like data (i.e. from the same database row) to stack within a group. There was one other similar post on the R-list (http://r.789695.n4.nabble.com/barplot-question-td3670861.html ) where the user had the same problem as I did, but it does not seem that this was resolved. Please let me know if you have any suggestions. Thanks and best wishes, Nicola -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Nicola Van Wilgen Sent: 24 April 2012 01:12 PM To: r-help@r-project.org Subject: [R] Combined grouped and stacked bargraph Dear R list, I am having some trouble drawing a bar-graph with two groups, both of which are stacked. A sample of my data (IUCN and national conservation status for marine and terrestrial species) look like this: Status IUCN Terrestrial IUCN Marine National CS Terrestrial National CS Marine NE 168 69 16 69 LC 41 6 148 6 NT 5 4 7 4 VU 4 2 7 2 EN 4 2 4 2 CR 1 0 2 0 I would like to plot the conservation status for two groups (IUCN status and national status), and for each of those groups I would like data for the marine and terrestrial species to be stacked. I needed to transpose the data to plot correctly, so my data in the code below look like this (where the names for data in the columns are NE, LC, NT, VU, EN and CR; and the matrix name is cs.not.log.bp): IUCN.Terrestrial168 41 5 4 4 1 IUCN.Marine 69 6 4 2 2 0 National.CS.Terrestrial 16 148 7 7 4 2 National.CS.Marine 69 6 4 2 2 0 I have tried the following code, but it does not work: barplot(cs.not.log.bp[c(1:2),], xlab = Conservation status, ylab = Number of species, col = c(grey90,grey80), names = cs.names, ylim = c(0,250), space = 2) barplot(cs.not.log.bp[c(3:4),], col = c(grey60,grey30), beside = T,add = T,names.arg = NA) legend(topright,c(IUCN Terrestrial,IUCN Marine,National CS Terrestrial,National CS Marine), col = c(grey90,grey80,grey60,grey30), pch = 15) What happens is that some of the data in the second group stacks onto the first group and then the remainder forms a second group. I would like only like data to stack within a group. There was one other similar post on the R-list (http://r.789695.n4.nabble.com/barplot-question-td3670861.html ) where the user had the same problem as I did, but it does not seem that this was resolved. Please let me know if you have any suggestions. Thanks and best wishes, Nicola This e-mail communication and any attachments are confidential and are intended only for the individual(s) or entity named above and others who have been specifically authorized to receive it. If you are not the intended recipient, please do not copy, use or disclose the contents of this communication to others. Please notify the sender that you have received this email in error by replying to the e-mail or by telephoning the sender. Please then delete the e-mail and any copies of it. This information may contain private, confidential or privileged material. Thank you. [[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
Re: [R] Use of optim to fit two curves at the same time ?
Dear list,
In order to find a solution to my problem, I created a third objective
function including both calculations done in the previous cases. This
function return a value (i.e. the value to be minimize by optim) equal to
the sum of the two sum of squares, but it does not work (see the code added
at the end of my previous script).
Any suggestion ?
Arnaud
Dear list,
Here is a small example code that use optim and optimize in order to fit
two functions.
Is it possible to fit two functions (like those two for example) at the
same time using optim ... or another function in R ?
Thanks
Arnaud
##
## function 1
x1 - 1:100
y1 - 5.468 * x + 3 # + rnorm(100,0, 10)
dfxy - cbind(x1,y1)
# Objective function
optfunc - function(x, dfxy){
a - x[1]
b - x[2]
xtest - dfxy[,1]
yobs - dfxy[,2]
ysim - a*xtest + b
sum((ysim - yobs)^2)
}
out- optim(par=c(0.2,5), fn=function(x){optfunc(x, dfxy)}, method =
Nelder-Mead, hessian = F)
## function 2
x2 - seq(0.01, 0.1, length=100)
y2 - exp(30*x2)
dfxy2 - cbind(x2,y2)
# objective function
optfunc2 - function(x, dfxy){
a - x[1]
xtest - dfxy[,1]
yobs - dfxy[,2]
ysim - exp(a*xtest)
sum((ysim - yobs)^2)
}
out- optimize(f=function(x){optfunc2(x, dfxy2)}, interval=c(0,500))
##
optfunc3 - function(x, dfxy, dfxy2){
a - x[1]
b - x[2]
xtest - dfxy[,1]
yobs - dfxy[,2]
ysim - a*xtest + b
obj1 - sum((ysim - yobs)^2)
c - x[3]
xtest2 - dfxy2[,1]
yobs2 - dfxy2[,2]
ysim2 - exp(c*xtest2)
obj2 - sum((ysim2 - yobs2)^2)
obj1 + obj2
}
out3- optim(par=c(0.2,5, 500), fn=function(x){optfunc3(x, dfxy, dfxy2)},
method = Nelder-Mead, hessian = F)
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Re: [R] FW: Combined grouped and stacked bargraph
Hi Nicola, You have provided the code and data as requested the data is in a very unfriendly format. If you would supply the data in an easily useable format so that the readers here can work with it it would help. Since it looks like you have a small data set the easiest thing is to use dput() [see ?dput for information] and then just paste the output into your email. I expect you would get faster and better answers. John Kane Kingston ON Canada -Original Message- From: nicola.vanwil...@sanparks.org Sent: Wed, 25 Apr 2012 14:42:49 +0200 To: r-help@r-project.org Subject: [R] FW: Combined grouped and stacked bargraph Dear R list, It appears that my request from yesterday got a bit garbled. I have tried to rephrase and retype it - please let me know if you can help me - I am having some trouble drawing a bar-graph with two groups, both of which are stacked. I would like to plot the conservation status according to two classifications (i.e. my groups - IUCN status and national status), and for each of those groups I would like data for the marine and terrestrial species to be stacked. My data look like this (where the names for the columns are conservation status': NE, LC, NT, VU, EN and CR; and the matrix name is cs.not.log.bp): IUCN.Terrestrial168 41 5 4 4 1 IUCN.Marine 69 6 4 2 2 0 National.CS.Terrestrial 16 148 7 7 4 2 National.CS.Marine 69 6 4 2 2 0 I have tried the following code, but it does not work: barplot(cs.not.log.bp[c(1:2),], xlab = Conservation status, ylab = Number of species, col = c(grey90,grey80),names = cs.names, ylim = c(0,250), space = 2) barplot(cs.not.log.bp[c(3:4),], col = c(grey60,grey30), beside = T,add = T,names.arg = NA) legend(topright,c(IUCN Terrestrial,IUCN Marine,National CS Terrestrial,National CS Marine), col = c(grey90,grey80,grey60,grey30), pch = 15) What happens is that some of the data in the second group stacks onto the first group and then the remainder forms a second group. I would like only like data (i.e. from the same database row) to stack within a group. There was one other similar post on the R-list (http://r.789695.n4.nabble.com/barplot-question-td3670861.html ) where the user had the same problem as I did, but it does not seem that this was resolved. Please let me know if you have any suggestions. Thanks and best wishes, Nicola -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Nicola Van Wilgen Sent: 24 April 2012 01:12 PM To: r-help@r-project.org Subject: [R] Combined grouped and stacked bargraph Dear R list, I am having some trouble drawing a bar-graph with two groups, both of which are stacked. A sample of my data (IUCN and national conservation status for marine and terrestrial species) look like this: Status IUCN Terrestrial IUCN Marine National CS Terrestrial National CS Marine NE 168 69 16 69 LC 41 6 148 6 NT 5 4 7 4 VU 4 2 7 2 EN 4 2 4 2 CR 1 0 2 0 I would like to plot the conservation status for two groups (IUCN status and national status), and for each of those groups I would like data for the marine and terrestrial species to be stacked. I needed to transpose the data to plot correctly, so my data in the code below look like this (where the names for data in the columns are NE, LC, NT, VU, EN and CR; and the matrix name is cs.not.log.bp): IUCN.Terrestrial168 41 5 4 4 1 IUCN.Marine 69 6 4 2 2 0 National.CS.Terrestrial 16 148 7 7 4 2 National.CS.Marine 69 6 4 2 2 0 I have tried the following code, but it does not work: barplot(cs.not.log.bp[c(1:2),], xlab = Conservation status, ylab = Number of species, col = c(grey90,grey80), names = cs.names, ylim = c(0,250), space = 2) barplot(cs.not.log.bp[c(3:4),], col = c(grey60,grey30), beside = T,add = T,names.arg = NA) legend(topright,c(IUCN Terrestrial,IUCN Marine,National CS Terrestrial,National CS Marine), col = c(grey90,grey80,grey60,grey30), pch = 15) What happens is that some of the data in the second group stacks onto the first group and then the remainder forms a second group. I would like only like data to stack within a group. There was one other similar post on the R-list (http://r.789695.n4.nabble.com/barplot-question-td3670861.html ) where the user had the same problem as I did, but it does not seem that this was resolved. Please let me know if you have any suggestions. Thanks and best wishes, Nicola FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing to find out more!
[R] fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
row - c(a,b,c,d,e,f,g) #rows from larger data frame row.1 - c(a,b,c,g) #rows of smaller data frame because d, e, and f don't contain any of the species, but the zeros are important x - data.frame(sp1=rnorm(4), sp2=rnorm(4), sp3=rnorm(4), sp4=rnorm(4)) rownames(x) - row.1 #I would like to make z as if I had y, but I only have the rownames of y y - data.frame(sp1=c(0,0,0), sp2=c(0,0,0), sp3=c(0,0,0), sp4=c(0,0,0)) rownames(y) - c(d, e, f) z - rbind(x,y) z - z[order(row.names(z)),] #I know I am missing something #many thanks, -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ 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] Where to find the p-value of a correlation test
Hey everyone, I hope this finds you in good cheer. I just have a quick question: What is the function that outputs the p-value for correlation? cor(x,y) only provides the R value. I would like the p-value associated with it. Thank you all for your help! [[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] trouble loading ggplot2 using R
On Wed, Apr 25, 2012 at 6:27 AM, Ramon Ovelar ramon.ove...@gmail.com wrote: I don't think I have touched at anything at all. I'm very newbie to R and to be honest I don't know what Ramdom.seed is. I will try to find out. I have seen other messages about restoring random.seed, but in order to check that the problem is really that I have used some Viewing data commands. The error says, in Spanish, that .Random.seed is not an integer vector but a list (.Random.seed no es un vector de números enteros pero es de tipo 'list') That's definitely not what it should look like - you might want to start from a fresh R workspace. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ 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] Where to find the p-value of a correlation test
Hi, Look at the See also section of ?cor Best, Ista On Wed, Apr 25, 2012 at 9:30 AM, Aaditya Nanduri aaditya.nand...@gmail.com wrote: Hey everyone, I hope this finds you in good cheer. I just have a quick question: What is the function that outputs the p-value for correlation? cor(x,y) only provides the R value. I would like the p-value associated with it. Thank you all for your help! [[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] recommended way to group function calls in Sweave
On 12-04-25 7:41 AM, Liviu Andronic wrote:
Dear all
When using Sweave, I'm always hitting the same bump: I want to group
repetitive calls in a function, but I want both the results and the
function calls in the printed output. Let me explain myself.
Consider the following computation in an Sweave document:
summary(iris[,1:2])
cor(iris[,1:2])
When using these two calls directly, I obtain the following output:
summary(iris[,1:2])
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
cor(iris[,1:2])
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
However, if I try to group the calls in a function:
f- function(d, ind){
print(summary(d[ , ind]))
print(cor(d[ , ind]))
return(invisible(NULL))
}
Then I get a different output in the Sweave PDF:
f(iris, 1:2)
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
Of course I can use 'echo=F' to remove the ' f(iris, 1:2)' in the
above, but how can I do to keep the original calls 'summary(d[ ,
ind])' and 'cor(d[ , ind])'? Or even better, could the actual calls be
used, after the replacement of arguments by their values:
'summary(iris[,1:2])' and 'cor(iris[,1:2])'.
Or is the recommended way to use cat()-ed statements:
f- function(d, ind){
cat('Variable summary:\n')
print(summary(d[ , ind]))
cat('\nCorrelation table:\n')
print(cor(d[ , ind]))
return(invisible(NULL))
}
To obtain such output:
f(iris, 1:2)
Variable summary:
Sepal.LengthSepal.Width
Min. :4.300 Min. :2.000
1st Qu.:5.100 1st Qu.:2.800
Median :5.800 Median :3.000
Mean :5.843 Mean :3.057
3rd Qu.:6.400 3rd Qu.:3.300
Max. :7.900 Max. :4.400
Correlation table:
Sepal.Length Sepal.Width
Sepal.Length1.000 -0.1175698
Sepal.Width-0.1175698 1.000
What is the recommended way of grouping repetitive function calls?
What do you usually use in your own documents? Regards
Liviu
I would use the last method, or if the calls were truly repetitive (i.e.
always identical, not just the same pattern), use a named chunk.
Duncan Murdoch
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Re: [R] trouble loading ggplot2 using R
On Apr 25, 2012, at 8:27 AM, Ramon Ovelar wrote: I don't think I have touched at anything at all. I'm very newbie to R and to be honest I don't know what Ramdom.seed is. I will try to find out. I have seen other messages about restoring random.seed, but in order to check that the problem is really that I have used some Viewing data commands. The error says, in Spanish, that .Random.seed is not an integer vector but a list (.Random.seed no es un vector de números enteros pero es de tipo 'list') class(.Random.seed) [1] data.frame That _is_ a problem. It _should_ be an integer atomic vector, although it appears to be the correct length (at least is the same as my .Random.seed which has the same leading entry, 403, as yours) and it appears to be all integers. Something you have done or some program has done has altered the default value. It is possible that this problem assignment has been saved in your (invisible) .Rdata file. It is generally a bad idea to do anything to .Random.seed, but I don't see any harm at this point in trying this: .Random.seed - unlist(.Random.seed) (... and then trying to install ggplot2) That should get rid of the 'data.frame' attribute. I would also track down your .Rdata file and maybe also your .Rhistory file and delete them. You will need to learn how to do this in a Terminal session or you will need to learn how to display invisible files in Finder.app. The archives of the SIG-Mac list will have instructions. -- David. str(.Random.seed) 'data.frame': 626 obs. of 1 variable: $ .Random.seed: num 4.03e+02 1.00e+01 -1.28e+09 -1.40e+09 -1.06e +09 .. head(.Random.seed) .Random.seed 1 403 2 10 3 -1282779759 4 -1404015037 5 -1062445742 6665436644 tail(.Random.seed) .Random.seed 621 1369617214 622 -1673749493 623 -1883947891 624 1445895610 625 -903220232 626970996181 On Tue, Apr 24, 2012 at 4:26 PM, Hadley Wickham had...@rice.edu wrote: I have a similar error, running R in Snow Leopard too library(ggplot2) Error : .onAttach failed in attachNamespace() for 'ggplot2', details: call: stats::runif(1) error: .Random.seed no es un vector de números enteros pero es de tipo 'list' Error: package/namespace load failed for ggplot2 That's a completely different error. Are you setting .Random.seed to something non-standard? That's what the error message suggests. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ -- == Ramón Ovelar [[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. 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.
Re: [R] Where to find the p-value of a correlation test
On 12-04-25 9:30 AM, Aaditya Nanduri wrote: Hey everyone, I hope this finds you in good cheer. I just have a quick question: What is the function that outputs the p-value for correlation? cor(x,y) only provides the R value. I would like the p-value associated with it. cor(x,y) calculates the correlation, it doesn't perform a test, so there is no p-value. cor.test() performs a test. 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] Splitting data into test and train (80:20) kepping attributes similar
Well, it throws an error, because there is no such function in default R. A bit of googling showed it might be the one in the caTools package. execute this: install.packages(caTools) library(caTools) before executing your code Am 25.04.2012 um 12:39 schrieb Dwaipayan Dasgupta: Hi, Could someone help me with this please , im trying to use Y = Attrition_data[,1] # extract labels from the data msk = sample.split (Y, SplitRatio=3/4) table(Y,msk) to do the splitting but it keeps throwing up and error Error: could not find function sample.split Could you please help Thanks in advance doy -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dwaipayan Dasgupta Sent: Tuesday, April 24, 2012 9:08 PM To: r-help@r-project.org Subject: [R] Splitting data into test and train (80:20) kepping attributes similar Hi, I am trying to do some predictive modeling around attrition and want to split the dataset into test and train (80:20) and keep the ratio of attritees:non attrites same. In my dataset the attrition indicator is coded as 0(for non-attritees) and 1 (for attritees) and I want to keep the ratio of 0's to 1 similar. I apologize for this trivial question but this is my second week with R. Thanks, Doy American Express made the following annotations on Tue Apr 24 2012 08:38:50 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Tue Apr 24 2012 08:38:50 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- [[alternative HTML version deleted]] American Express made the following annotations on Wed Apr 25 2012 03:39:08 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Wed Apr 25 2012 03:39:08 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do i read the source code of biplot?
I think this should help: require(pcaMethods) data(metaboliteDataComplete) mdc - scale(metaboliteDataComplete, center=TRUE, scale=FALSE) ## Now create 5% of outliers. cond - runif(length(mdc))0.05; mdcOut - mdc mdcOut[cond] - 10 ## Now we do a conventional PCA and robustPca on the original and the data ## with outliers. ## We use center=FALSE here because the large artificial outliers would ## affect the means and not allow to objectively compare the results. resSvd- pca(mdc, method = svd, nPcs = 10, center = FALSE) resSvdOut - pca(mdcOut, method = svd, nPcs = 10, center = FALSE) resRobPca - pca(mdcOut, method = robustPca, nPcs = 10, center = FALSE) biplot(resRobPca) # Draw the graph biplot # shows the generic methods(biplot) # which method is used? Depends on the class of the object class(resRobPca) # the class is 'pcaRes' biplot.pcaRes # show the actual code for drawing the object Kevin On Tue, Apr 24, 2012 at 1:44 PM, Michael comtech@gmail.com wrote: biplot standardGeneric for biplot defined from package stats function (x, ...) standardGeneric(biplot) environment: 0x0dd8 Methods may be defined for arguments: x Use showMethods(biplot) for currently available ones. showMethods(biplot) Function: biplot (package stats) x=ANY x=character (inherited from: x=ANY) x=Pca biplot(resRobPCA) But how do I get the source code of biplot? Thank you! [[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. -- 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] recommended way to group function calls in Sweave
On Wed, Apr 25, 2012 at 3:41 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
I would use the last method, or if the calls were truly repetitive (i.e.
always identical, not just the same pattern), use a named chunk.
Labeled chunks are indeed what I was looking for [1]. As far as I
understand, this is what Sweave functions (or are these macros?)
look like:
=
d - iris
ind - 1:2
@
sw=
summary(d[ , ind])
cor(d[ , ind])
@
=
d - iris
ind - 2:4
sw
@
Regards
Liviu
[1] vignette('Sweave', 'utils')
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[R] random effects in library mgcv
Hi, I am working with gam models in the mgcv library. My response variable (Y) is binary (0/1), and my dataset contains repeated measures over 110 individuals (same number of 0/1 within a given individual: e.g. 345-zero and 345-one for individual A, 226-zero and 226-one for individual B, etc.). The variable Factor is separating the individuals in three groups according to mass (group 0,1,2), Factor1 is a binary variable coding for individuals of group1, Factor2 is a binary variable for individuals of group 2 I use gam models of this sort with random effects coded using a s( ..., bs=re) term: gm-gam(Y~Factor+te(x1,x2,by=Factor) )+s(Individual,bs=re),dat=Data,family=binomial(link=logit),method=REML) gm1-gam(Y~Factor+te(x1,x2)+ te(x1,x2,by=Factor1)+ te(x1,x2,by=Factor2)+s(Individual,bs=re),dat=Data,family=binomial(link=logit),method=REML) 1)First question: is it OK to use gam() to model a binary variable with random effects coded as a bs=re term?? I have read that the gamm4() function gives better performance than gamm() to deal with binary variables when random effects are coded as: random=~(1|Individual) but does that mean that binary variables should not be used as response variable in gam() with random effects coded as bs=re??? 2)Second question: For some models, I obtain a p-value=NA and Chi-square=0 for the s(Individual) term, and for some other models a p-value=1 and high Chi-square. The difference between one model that can estimate a p-value and one that cannot is very slight: for example if I use a variable x3 instead of x2 in a model, it can change from p-value=NA to p-value=1. Does anyone know what can be happening? 3)Third question: Not linked to random effects but rather to what the two models gm and gm1 are actually testing. From my understanding, the first model creates a 2d-smooth for each level of my factor variable and test whether those smooth are significantly different from a straight line. The second model, also creates 3 smooth: one for the reference level of my Factor variable (group0), one showing the difference between the reference smooth and the smooth for group1, one showing the difference between the reference smooth and the smooth for group 2. The summary(gm1) gives p-values associated with each of those three smooths and which determine: if the reference smooth is different from 0, if the smooth for group1 is different from the reference smooth and if the smooth for group2 is different from the reference smooth. Do I understand well what the models are testing? The number of edf estimated for te(x1,x2):Factor2 in the gm1 model is 3,013 while it is 19,57 in the gm model. Does that mean that the difference between the reference smooth: te(x1,x2) and the smooth for group 2: te(x1,x2, by=Factor2) is small so it can be modeled with only 3 degrees of freedom? Still, the associated p-value is highly significant? When comparing AIC between the gm and gm1 models, I find sometimes that gm1 has a lower AIC than gm. How can that be interpreted?? Thanks a lot if anyone can help... Geraldine [[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] Create new Vector based on two colums
Hello,
I am trying to get a new vector 'x1' based on the not NA-values in
column 'a' and 'b'. I found a way but I am sure this is not the best
solution. So any ideas on how to optimize this would be great!
m - factor(c(a1, a1, a2, b1, b2, b3, d1, d1), ordered
= TRUE)
df - data.frame( a= m, b = m)
df[1,1] - NA
df[4,2] - NA
df[2,2] - NA
df[6,1] - NA
df
w - !apply(df, 2, is.na)
v - apply(w, 1, FUN=function(L) which(L == TRUE)[[1]])
for (i in 1:nrow(df) ) {
g[i] - df[i, v[i]]
}
df$x1 - g
Thanks for any help
Patrick
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Re: [R] Adding more files to list
Hi Brian,
On Wednesday 25 April 2012 13:15:03 Brian Flatley wrote:
I have a main directory - Spectra
In this folder I have two subfolders Normal and Case
The following should be enough:
library(readBrukerFlexData)
Spectra - mqReadBrukerFlex(Spectra/);
...
The following command will open and list all my files from the Normal
subfolder (70 files containing spectrums)
Spectra - mqReadBrukerFlex(file.path(Spectra/, Normal));
How do I add the Case folder to the list (another 70 spectra), so that I
can then process all the files in the same manner and yet maintain their
origin so when I want to do stats analyse on them I can differentiate??
To combine a list you could use c:
SpectraNormal - mqReadBrukerFlex(file.path(Spectra/, Normal));
SpectraCase - mqReadBrukerFlex(file.path(Spectra/, Case));
Spectra - c(SpectraNormal, SpectraCase);
To maintain the origin you could store the indices:
normal - 1:length(SpectraNormal);
cases - (normal+1):(normal+1+length(SpectraCase));
But I would prefer to fetch this information from the MassSpectrum metaData
slot:
filenames - sapply(Spectra, function(x) {return(metaData(x)$file); });
normal - grepl(pattern=/Normal/, x=filenames);
classes - as.factor(ifelse(normal, Normal, Case));
Now you could access the different cases in the following way:
Spectra[classes==Normal]
Spectra[classes==Case]
Bye,
Sebastian
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[R] Help on time series Hurst exponent
Hello, I'm an absolute beginner with R. I'm hoping to do some time-series analysis on my data. The data looks like #time value 18 153 20 426 70 7 83 130 84 7 and so on where time could be in seconds or hours or days (not all at the same time). How could I import such a file to R and do some simple stuff (say plot the values)? As per the tutorials on time series, I could use the ts() method to import the values (not timestamps). However, one problem with my data is that the *time intervals are not regular* -- i.e. I don't have observations at every delta_t. So, I possibly can't ignore the timestamps. I'm also interested to estimate the Hurst exponent for the above series. I've installed the fArma package. Again, I'm not sure how to use the above series there. Could someone please help me on this? -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[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] Conditional average
id age76 Wage76 Grade76 Black immigrt. ... 1 25 102456 12 1 0 2 27 15432 15 0 1 3 30 12340 16 1 1 Then I have lots of data variables 100 for 5000 individuals nearly. What I wanted is to discriminated by age and education level (age is years only and grade is years in educational system and wage is salary only no rational numbers here) I want to have for people of education level 10 12 16, the average wage for age 15, and then the same for age 16, and the same for age 17 and so on... Once i have these average I plot (age vs average salary) to see the distribution in average -- View this message in context: http://r.789695.n4.nabble.com/Conditional-average-tp4585313p4586691.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] fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
-- View this message in context: http://r.789695.n4.nabble.com/fill-a-dataframe-with-zeros-where-the-rows-are-a-smaller-subset-of-a-larger-dataframe-species-by-sit-tp4586534p4586711.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] take data from a file to another according to their correlation coefficient
Hello,
The first is easy.
How can I cut these files and save them automatically (one file for ST001,
another for ST002, ...) according to these columns names?
Similar to the way they were read, using lapply on the results list.
But first make a file names vector.
(I've used the file extension 'dat'.)
test - process.all(lst, m)
fl.names - paste(names(test), dat, sep=.)
lapply(seq_len(length(test)), function(i) write.table(test[[i]],
fl.names[i], ...))
The second is trickier.
And it is possible in your script to take the second best correlated
station data
instead of the best one, if there are NAs in this best correlated station
at the same
lines with the NA gaps of the station to fill?
In the function 'process.all', after the internal function 'f',
include the following.
g - function(station){
x - df.list[[station]]
if(any(is.na(x$data))){
mat[row(mat) == col(mat)] - -Inf
nas - which(is.na(x$data))
ord - order(mat[station, ], decreasing = TRUE)[-c(1,
ncol(mat))]
for(i in nas){
for(y in ord){
if(!is.na(df.list[[y]]$data[i])){
x$data[i] - df.list[[y]]$data[i]
break
}
}
}
}
x
}
Then, change the second pass to
# Note that the two passes are different
df.list - lapply(seq.int(n), f)
df.list - lapply(seq.int(n), g)
And I come from Portugal. I'm a mathematician (with 6 semesters of stats).
When I go to France, it's more to Charente-Maritime - Cognac, I have friends
there,
and I'll definitelly have a couple of cognacs on you.
Good luck with your assignment.
Rui Barradas
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Re: [R] Adding more files to list
As to combining, use c(). as to tracking, keep track of the directory that goes with the file. There are various ways to do this: a) keep a vector of filenames, and every time you combine vectors of filenames, do the same for the vectors of directory names. You will probably need to use rep() to turn one directory name into a vector of repeated values. b) an organized way to accomplish option a above is to keep the vectors in data frames, and use rbind() instead of c() to combine them. c) you will have to use paste() to combine the paths and filenames before you open them, so you might want to do so early on and keep track of which category each falls into using more presentable category names in the auxiliary vector(s) instead of tracking paths separately. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Brian Flatley b.flat...@pgr.reading.ac.uk wrote: Hi, I have searched over the last two days to try and sort this problem but unfortunately I cannot find the correct solution. I have a main directory - Spectra In this folder I have two subfolders Normal and Case I am using a package MALDIquant for processing of mass spectrometry data, The following command will open and list all my files from the Normal subfolder (70 files containing spectrums) Spectra - mqReadBrukerFlex(file.path(Spectra/, Normal)); How do I add the Case folder to the list (another 70 spectra), so that I can then process all the files in the same manner and yet maintain their origin so when I want to do stats analyse on them I can differentiate?? Any help would be great, I am very much a beginner, Brian [[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] trouble loading ggplot2 using R
Alternatively, the OP might just use set.seed(1) to get a fresh seed rather than playing with .Random.seed directly. To the OP: if you're on Mac, follow these instructions and we'll see if that fixes your problem: **) open Terminal **) type R -q --vanilla **) At the prompt type the following lines set.seed(1) install.packages(ggplot2) library(ggplot2) qplot(carat, price, data = head(diamonds, 200)) # Should make a nice ggplot graph with not so many dots **) If that works and you don't have anything important saved in your default start up, return to Terminal and type mv .RData .Trash (This just puts it in the trash so we can recover it if you realize later there actually is something important there) **) Then start R by typing R -q and enter library(ggplot2) qplot(carat, price, data = head(diamonds, 200)) and that should work. Hopefully that will isolate your problem. If anything fails along the way, let us know exactly where and how. Best, Michael On Wed, Apr 25, 2012 at 9:42 AM, David Winsemius dwinsem...@comcast.net wrote: On Apr 25, 2012, at 8:27 AM, Ramon Ovelar wrote: I don't think I have touched at anything at all. I'm very newbie to R and to be honest I don't know what Ramdom.seed is. I will try to find out. I have seen other messages about restoring random.seed, but in order to check that the problem is really that I have used some Viewing data commands. The error says, in Spanish, that .Random.seed is not an integer vector but a list (.Random.seed no es un vector de números enteros pero es de tipo 'list') class(.Random.seed) [1] data.frame That _is_ a problem. It _should_ be an integer atomic vector, although it appears to be the correct length (at least is the same as my .Random.seed which has the same leading entry, 403, as yours) and it appears to be all integers. Something you have done or some program has done has altered the default value. It is possible that this problem assignment has been saved in your (invisible) .Rdata file. It is generally a bad idea to do anything to .Random.seed, but I don't see any harm at this point in trying this: .Random.seed - unlist(.Random.seed) (... and then trying to install ggplot2) That should get rid of the 'data.frame' attribute. I would also track down your .Rdata file and maybe also your .Rhistory file and delete them. You will need to learn how to do this in a Terminal session or you will need to learn how to display invisible files in Finder.app. The archives of the SIG-Mac list will have instructions. -- David. str(.Random.seed) 'data.frame': 626 obs. of 1 variable: $ .Random.seed: num 4.03e+02 1.00e+01 -1.28e+09 -1.40e+09 -1.06e+09 .. head(.Random.seed) .Random.seed 1 403 2 10 3 -1282779759 4 -1404015037 5 -1062445742 6 665436644 tail(.Random.seed) .Random.seed 621 1369617214 622 -1673749493 623 -1883947891 624 1445895610 625 -903220232 626 970996181 On Tue, Apr 24, 2012 at 4:26 PM, Hadley Wickham had...@rice.edu wrote: I have a similar error, running R in Snow Leopard too library(ggplot2) Error : .onAttach failed in attachNamespace() for 'ggplot2', details: call: stats::runif(1) error: .Random.seed no es un vector de números enteros pero es de tipo 'list' Error: package/namespace load failed for ‘ggplot2’ That's a completely different error. Are you setting .Random.seed to something non-standard? That's what the error message suggests. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ -- == Ramón Ovelar [[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. 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-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] Selecting columns whose names contain mutated except when they also contain non or un
Hello Dr. Winsemius, There was a non-numeric column. Thanks for helping me to see the obvious. Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
I am subsetting a larger data frame that contains macroinverterate taxa. I am subsetting them at different levels of taxonomic resolution. Some of the sites do not have say Tipulidae present, so the rows are removed completely for this site. I would like to fill in the sites that were removed during the subsetting and fill these with 0. I have the non-subsetted dataframe which contains all of the sites. I would like to use the sites from this complete data set to expand the dataframe that is a subset of the sites in the original. Is that more clear. Stephen On Wed 25 Apr 2012 09:13:24 AM CDT, chuck.01 wrote: -- View this message in context: http://r.789695.n4.nabble.com/fill-a-dataframe-with-zeros-where-the-rows-are-a-smaller-subset-of-a-larger-dataframe-species-by-sit-tp4586534p4586711.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. -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ 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 on time series Hurst exponent
You really don't want to use ts() -- if you want to use the tools in fArma use a timeSeries (provided by the package of the same name) Michael On Wed, Apr 25, 2012 at 9:54 AM, Barun Saha barun.sah...@gmail.com wrote: Hello, I'm an absolute beginner with R. I'm hoping to do some time-series analysis on my data. The data looks like #time value 18 153 20 426 70 7 83 130 84 7 and so on where time could be in seconds or hours or days (not all at the same time). How could I import such a file to R and do some simple stuff (say plot the values)? As per the tutorials on time series, I could use the ts() method to import the values (not timestamps). However, one problem with my data is that the *time intervals are not regular* -- i.e. I don't have observations at every delta_t. So, I possibly can't ignore the timestamps. I'm also interested to estimate the Hurst exponent for the above series. I've installed the fArma package. Again, I'm not sure how to use the above series there. Could someone please help me on this? -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[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] fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
The clarification helps; the original description was rather terse. What about: row - c(a,b,c,d,e,f,g) #rows from larger data frame x - data.frame(sp1=rnorm(4), sp2=rnorm(4), sp3=rnorm(4), sp4=rnorm(4)) rownames(x) - row.1 merge(x, matrix(row, ncol=1), by.x=0, by.y=1, all=TRUE) Row.namessp1sp2sp3sp4 1 a 0.4964272 1.4989159 0.4302415 0.9648854 2 b 1.4137142 0.9430609 0.0728391 -0.6275084 3 c -0.8103023 -1.3375148 -0.3799518 0.4523287 4 d NA NA NA NA 5 e NA NA NA NA 6 f NA NA NA NA 7 g -0.1914184 0.5156566 0.5626614 0.8068154 Sarah On Wed, Apr 25, 2012 at 10:42 AM, Stephen Sefick sas0...@auburn.edu wrote: I am subsetting a larger data frame that contains macroinverterate taxa. I am subsetting them at different levels of taxonomic resolution. Some of the sites do not have say Tipulidae present, so the rows are removed completely for this site. I would like to fill in the sites that were removed during the subsetting and fill these with 0. I have the non-subsetted dataframe which contains all of the sites. I would like to use the sites from this complete data set to expand the dataframe that is a subset of the sites in the original. Is that more clear. Stephen -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] RODBC Error Code 202 on Mac OS X 10.6
Hello everybody out there using the RODBC package, On my linux (unixODBC) and windows machines, I can successfully use the RODBC package to connect to a PostgreSQL databse. On my Mac Book running Mac OS X 10.6 (Intel 64bit architecture), I get the error code 202 and message ? when I try to establish an connection by using con - odbcConnect(dsn=MyDataSource,UID=Me,pwd=secret) Could anyone please hint me towards additional diagnostics I can run to pin down the problem? I have the latest versions of unixODBC and psqlODBC installed from Macports. They seem to be okay, because $isql dsn uid pwd works fine to connect to the database. Thanks in advance for your help, Julia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R shell script
You can do this in bash but why not just do it in R directly? You probably need
list.files(pattern = .out)
to get started. Then just wrap your script in a function and pass it
to (s|l)apply something like:
sapply(list.files(pattern = *.out), function(x) wilcox.test ( ... ) )
Michael
On Wed, Apr 25, 2012 at 6:47 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple example
explaining how to make this shell script.
Any help appreciated :)
Aoife
[[alternative HTML version deleted]]
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[R] help with significance of the estimates when running the pgls function
Hello, I am running the pgls () function of the caper programme, the output has some estimates, some of them are negative and others are positive ( values can be less, equal, or more than 1 ). If I have an estimate of -0.12 for a coefficient (body size) on head size. How do I interpreted it? Thanks a lot for your help! :) -- BAS PhD Student Department of Animal Plant Sciences University of Sheffield Alfred Denny Building Western Bank Sheffield S10 2TN 0114 2220057 [[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] R shell script
Thanks for replying.
My problem is that i have say 50 input files, that i wanted to run a
particular command on, get 50 output files, and then when i close R, have
them in my directory?
so for example if i say:
R
library(MASS)
list.files(pattern = .out)
sapply(list.files(pattern = *.out), function(x) wilcox.test ( ... ) )
send each output to a different file, and save it in such a way that
when i close R the outputs are still there
i thought this might be easier in a shell way?
On Wed, Apr 25, 2012 at 4:03 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
You can do this in bash but why not just do it in R directly? You probably
need
list.files(pattern = .out)
to get started. Then just wrap your script in a function and pass it
to (s|l)apply something like:
sapply(list.files(pattern = *.out), function(x) wilcox.test ( ... ) )
Michael
On Wed, Apr 25, 2012 at 6:47 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple
example
explaining how to make this shell script.
Any help appreciated :)
Aoife
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R shell script
Check out the vignette for the optparse library:
http://cran.r-project.org/web/packages/optparse/vignettes/optparse.pdf
Super helpful library if you plan on making any semi-interesting
command line scripts w/ R.
-steve
On Wed, Apr 25, 2012 at 6:47 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple example
explaining how to make this shell script.
Any help appreciated :)
Aoife
[[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.
--
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] recommended way to group function calls in Sweave
On 25/04/2012 10:20 AM, Liviu Andronic wrote:
On Wed, Apr 25, 2012 at 3:41 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
I would use the last method, or if the calls were truly repetitive (i.e.
always identical, not just the same pattern), use a named chunk.
Labeled chunks are indeed what I was looking for [1]. As far as I
understand, this is what Sweave functions (or are these macros?)
look like:
Yes. They're definitely macros, not functions: pure text substitution.
Duncan Murdoch
=
d- iris
ind- 1:2
@
sw=
summary(d[ , ind])
cor(d[ , ind])
@
=
d- iris
ind- 2:4
sw
@
Regards
Liviu
[1] vignette('Sweave', 'utils')
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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] Selecting columns whose names contain mutated except when they also contain non or un
On Apr 24, 2012, at 19:15 , Rui Barradas wrote:
Has anyone realized that both 'non' and 'un' end with the same letter? The
only one we really need to check?
(tmp - c('mutation','nonmutated','unmutated','verymutated','other'))
i1 - grepl(muta, tmp)
i2 - grepl(nmuta, tmp)
tmp[i1 !i2]
Yes, I was wondering why people were avoiding the obvious use of grepl(). I'm
not too happy about the nmuta technique though: What about deletionmutation
and such? Might as well do the safe(r) thing:
i2 - grepl(unmuta, tmp) | grepl(nonmuta, tmp)
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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Re: [R] Splitting data into test and train (80:20) kepping attributes similar
Thank you so much for replying. I tried what you said but it still throws the same error i.e could not find function sample.split Might be because of the version of R I am running (R version 2.12.2).i do not have admin rights to upgrade to the newest version. Is there anything else I can try? Im trying to split my data into 80:20 keeping the ratio of 0,1 in the Y variable(binary) constant. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jessica Streicher Sent: Wednesday, April 25, 2012 7:17 PM To: r-help@r-project.org Subject: Re: [R] Splitting data into test and train (80:20) kepping attributes similar Well, it throws an error, because there is no such function in default R. A bit of googling showed it might be the one in the caTools package. execute this: install.packages(caTools) library(caTools) before executing your code Am 25.04.2012 um 12:39 schrieb Dwaipayan Dasgupta: Hi, Could someone help me with this please , im trying to use Y = Attrition_data[,1] # extract labels from the data msk = sample.split (Y, SplitRatio=3/4) table(Y,msk) to do the splitting but it keeps throwing up and error Error: could not find function sample.split Could you please help Thanks in advance doy -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dwaipayan Dasgupta Sent: Tuesday, April 24, 2012 9:08 PM To: r-help@r-project.org Subject: [R] Splitting data into test and train (80:20) kepping attributes similar Hi, I am trying to do some predictive modeling around attrition and want to split the dataset into test and train (80:20) and keep the ratio of attritees:non attrites same. In my dataset the attrition indicator is coded as 0(for non-attritees) and 1 (for attritees) and I want to keep the ratio of 0's to 1 similar. I apologize for this trivial question but this is my second week with R. Thanks, Doy American Express made the following annotations on Tue Apr 24 2012 08:38:50 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Tue Apr 24 2012 08:38:50 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- [[alternative HTML version deleted]] American Express made the following annotations on Wed Apr 25 2012 03:39:08 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Wed Apr 25 2012 03:39:08 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire prévu, toute divulgation, duplication, utilisation ou distribution du courrier ou de toute pièce jointe est interdite. Si vous avez reçu cette communication par erreur, veuillez nous en aviser par courrier et détruire immédiatement le courrier et les pièces jointes. Merci. ** --- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] Create new Vector based on two colums
On Wed, Apr 25, 2012 at 6:14 AM, Patrick Hausmann
patrick.hausm...@covimo.de wrote:
Hello,
I am trying to get a new vector 'x1' based on the not NA-values in column
'a' and 'b'. I found a way but I am sure this is not the best solution. So
any ideas on how to optimize this would be great!
If by optimize you mean no loops, you could try
(df- structure(list(a = structure(c(NA, 1L, 2L, 3L, 4L, NA, 6L, 6L
), .Label = c(a1, a2, b1, b2, b3, d1), class = c(ordered,
factor)), b = structure(c(1L, NA, 2L, NA, 4L, 5L, 6L, 6L), .Label = c(a1,
a2, b1, b2, b3, d1), class = c(ordered, factor))),
.Names = c(a,
b), row.names = c(NA, -8L), class = data.frame))
x0 - factor(paste(ifelse(is.na(df[,1]),'',df[,1]),
ifelse(is.na(df[,2]),'',df[,2]), sep=''))
df$x1 - as.numeric(x0) # only needed for a numeric vector as in your solution
df
HTH
PS - In the future please avoid df and data as object names, as
they mask R functions.
m - factor(c(a1, a1, a2, b1, b2, b3, d1, d1), ordered =
TRUE)
df - data.frame( a= m, b = m)
df[1,1] - NA
df[4,2] - NA
df[2,2] - NA
df[6,1] - NA
df
w - !apply(df, 2, is.na)
v - apply(w, 1, FUN=function(L) which(L == TRUE)[[1]])
for (i in 1:nrow(df) ) {
g[i] - df[i, v[i]]
}
df$x1 - g
Thanks for any help
Patrick
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Re: [R] Help on time series Hurst exponent
www.rmetrics.org/ebooks There's a free book on time series classes in R here: I haven't used it myself, but it is by the fArma (RMetrics) folks so I presume it covers their own time series class. Michael On Apr 25, 2012, at 11:11 AM, Barun Saha barun.sah...@gmail.com wrote: Thanks, Michael! Could you plz point to some easy tutorials regarding this? On Wed, Apr 25, 2012 at 8:14 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: You really don't want to use ts() -- if you want to use the tools in fArma use a timeSeries (provided by the package of the same name) Michael On Wed, Apr 25, 2012 at 9:54 AM, Barun Saha barun.sah...@gmail.com wrote: Hello, I'm an absolute beginner with R. I'm hoping to do some time-series analysis on my data. The data looks like #time value 18 153 20 426 70 7 83 130 84 7 and so on where time could be in seconds or hours or days (not all at the same time). How could I import such a file to R and do some simple stuff (say plot the values)? As per the tutorials on time series, I could use the ts() method to import the values (not timestamps). However, one problem with my data is that the *time intervals are not regular* -- i.e. I don't have observations at every delta_t. So, I possibly can't ignore the timestamps. I'm also interested to estimate the Hurst exponent for the above series. I've installed the fArma package. Again, I'm not sure how to use the above series there. Could someone please help me on this? -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[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. -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[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] recommended way to group function calls in Sweave
What I usually do when I have to write a report with some functions I use
multiple times is that I put them in a separate file (call it setup.Rnw or
so).
The first chunk there loads the libraries, sets initial variable values etc:
echo=FALSE, results=hide=
library( xtable )
d - iris
ind - 1
@
Then I have a number of chunks that are not evaluated at this point:
chunk_name_1,eval=FALSE,echo=FALSE=
summary(d[ , ind])
cor(d[ , ind])
@
chunk_name_2,eval=FALSE,echo=FALSE=
# produce a nice table from some data
@
In my master file, I load these definitions first
SweaveInput( setup.Rnw )
and from here, I can suse the named chunks almost like function calls, as you
you describe below. The advantage (for me) is that I have only one place where
I maintain the functions code, and only one line in the real document,
rather than a lot of code, possibly distributed over the document..
Rgds,
Rainer
On Wednesday 25 April 2012 16:20:13 Liviu Andronic wrote:
On Wed, Apr 25, 2012 at 3:41 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
I would use the last method, or if the calls were truly repetitive (i.e.
always identical, not just the same pattern), use a named chunk.
Labeled chunks are indeed what I was looking for [1]. As far as I
understand, this is what Sweave functions (or are these macros?)
look like:
=
d - iris
ind - 1:2
@
sw=
summary(d[ , ind])
cor(d[ , ind])
@
=
d - iris
ind - 2:4
sw
@
Regards
Liviu
[1] vignette('Sweave', 'utils')
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Re: [R] Splitting data into test and train (80:20) kepping attributes similar
Don't know whats wrong there (except if you're using the eclipse R plugin on a mac like me and the window for choosing the download site doesn't pop up.. did it?^^) Anyway, you could just split all of your data into 2 datasets, one that has all the data labeled 0, the other for all labeled 1, then take a random 80% of both, put them back together into the 80% data, and put the rest back together to form the 20%. Since i don't know your data, heres an example: M-cbind(c(rep(0,10),rep(1,10)),1:20) M [,1] [,2] [1,]01 [2,]02 [3,]03 [4,]04 [5,]05 [6,]06 [7,]07 [8,]08 [9,]09 [10,]0 10 [11,]1 11 [12,]1 12 [13,]1 13 [14,]1 14 [15,]1 15 [16,]1 16 [17,]1 17 [18,]1 18 [19,]1 19 [20,]1 20 index1-which(M[,1]==1) index1 [1] 11 12 13 14 15 16 17 18 19 20 M1-M[index1,] M1 [,1] [,2] [1,]1 11 [2,]1 12 [3,]1 13 [4,]1 14 [5,]1 15 [6,]1 16 [7,]1 17 [8,]1 18 [9,]1 19 [10,]1 20 M0-M[-index1,] M0 [,1] [,2] [1,]01 [2,]02 [3,]03 [4,]04 [5,]05 [6,]06 [7,]07 [8,]08 [9,]09 [10,]0 10 s1-sample(1:dim(M1)[1],0.8*dim(M1)[1]) s1 [1] 10 3 5 9 2 6 8 7 s0-sample(1:dim(M0)[1],0.8*dim(M0)[1]) s0 [1] 8 10 9 3 7 4 2 1 data80-rbind(M1[s1,],M0[s0,]) data80 [,1] [,2] [1,]1 20 [2,]1 13 [3,]1 15 [4,]1 19 [5,]1 12 [6,]1 16 [7,]1 18 [8,]1 17 [9,]08 [10,]0 10 [11,]09 [12,]03 [13,]07 [14,]04 [15,]02 [16,]01 data20-rbind(M1[-s1,],M0[-s0,]) data20 [,1] [,2] [1,]1 11 [2,]1 14 [3,]05 [4,]06 which is probably not how you really do things efficiently, but it should work. greetings Jessi Am 25.04.2012 um 17:18 schrieb Dwaipayan Dasgupta: Thank you so much for replying. I tried what you said but it still throws the same error i.e could not find function sample.split Might be because of the version of R I am running (R version 2.12.2).i do not have admin rights to upgrade to the newest version. Is there anything else I can try? Im trying to split my data into 80:20 keeping the ratio of 0,1 in the Y variable(binary) constant. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jessica Streicher Sent: Wednesday, April 25, 2012 7:17 PM To: r-help@r-project.org Subject: Re: [R] Splitting data into test and train (80:20) kepping attributes similar Well, it throws an error, because there is no such function in default R. A bit of googling showed it might be the one in the caTools package. execute this: install.packages(caTools) library(caTools) before executing your code Am 25.04.2012 um 12:39 schrieb Dwaipayan Dasgupta: Hi, Could someone help me with this please , im trying to use Y = Attrition_data[,1] # extract labels from the data msk = sample.split (Y, SplitRatio=3/4) table(Y,msk) to do the splitting but it keeps throwing up and error Error: could not find function sample.split Could you please help Thanks in advance doy -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dwaipayan Dasgupta Sent: Tuesday, April 24, 2012 9:08 PM To: r-help@r-project.org Subject: [R] Splitting data into test and train (80:20) kepping attributes similar Hi, I am trying to do some predictive modeling around attrition and want to split the dataset into test and train (80:20) and keep the ratio of attritees:non attrites same. In my dataset the attrition indicator is coded as 0(for non-attritees) and 1 (for attritees) and I want to keep the ratio of 0's to 1 similar. I apologize for this trivial question but this is my second week with R. Thanks, Doy American Express made the following annotations on Tue Apr 24 2012 08:38:50 ** This message and any attachments are solely for the intended recipient and may contain confidential or privileged information. If you are not the intended recipient, any disclosure, copying, use, or distribution of the information included in this message and any attachments is prohibited. If you have received this communication in error, please notify us by reply e-mail and immediately and permanently delete this message and any attachments. Thank you. American Express a ajouté le commentaire suivant le Tue Apr 24 2012 08:38:50 Ce courrier et toute pièce jointe qu'il contient sont réservés au seul destinataire indiqué et peuvent renfermer des renseignements confidentiels et privilégiés. Si vous n'êtes pas le destinataire
[R] pca biplot.princomp has a bug?
x=rmvnorm(2000, rep(0, 6), diag(c(5, rep(1,5 x=scale(x, center=T, scale=F) pc - princomp(x) biplot(pc) There are a bunch of red arrows plotted, what do they mean? I knew that the first arrow labelled with Var1 should be pointing the most varying direction of the data-set (if we think them as 2000 data points, each being a vector of size 6). I also read from somewhere, the most varying direction should be the direction of the 1st eigen vector. However, reading into the code of biplot in R. The line about the arrows is: if(var.axes) arrows(0, 0, y[,1L] * 0.8, y[,2L] * 0.8, col = col[2L], Where `y` is the actually the loadings matrix, which is the eigenvector matrix. So it looks like the 1st arrow is actually pointing from `(0, 0)` to `(y[1, 1], y[1, 2])`. I understand that we are trying to plot a high dimensional arrow onto a 2D plane. That's why we are taking the 1st and 2nd element of the `y[1, ]` vector. However what I don't understand is: Shouldn't the 1st eigenvector direction be the vector denoted by `y[, 1]`, instead of `y[1, ]`? (Again, here `y` is the eigenvector matrix, obtained by PCA or by eigendecomposition of `t(x) %*% x`.) i.e. the eigenvectors should be column vectors, not those horizontal vectors. Even though we are plotting them on 2D plane, we should draw the 1st direction to be from `(0, 0)` pointing to `(y[1, 1], y[2, 1])`? I think the R code biplot.princomp has a bug: the loading matrix (eigenvector matrix) should be transposed before being sent into biplot.princomp... Any thoughts? [[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] RODBC Error Code 202 on Mac OS X 10.6
On Apr 25, 2012, at 10:03 AM, julia.jacob...@arcor.de wrote: Hello everybody out there using the RODBC package, On my linux (unixODBC) and windows machines, I can successfully use the RODBC package to connect to a PostgreSQL databse. On my Mac Book running Mac OS X 10.6 (Intel 64bit architecture), I get the error code 202 and message ? when I try to establish an connection by using con - odbcConnect(dsn=MyDataSource,UID=Me,pwd=secret) Could anyone please hint me towards additional diagnostics I can run to pin down the problem? I have the latest versions of unixODBC and psqlODBC installed from Macports. They seem to be okay, because $isql dsn uid pwd works fine to connect to the database. Thanks in advance for your help, Julia Anytime Fink or MacPorts comes into the picture, I get uncontrollable shakes...never had good luck with them when I moved to OSX from Linux three years ago and they really screwed up my MacBook Pro, conflicting with all kinds of stuff. If I needed something, I always go to the upstream source. I can't seem to locate an error code 202 in any ODBC documentation. Is the above a direct copy and paste of the _complete_ error message? That being said, do you know if the MacPorts installs are 32 bit, 64 bit or both? The entire tool chain, including unixODBC, the ODBC driver and R must all be of the same architecture, or you get conflicts. Check to see which version of R you are running by using: .Machine$sizeof.pointer If it comes back with 8, you are running 64 bit R, if 4, then 32 bit R. Runnning: vignette(RODBC) will bring up additional installation and other relevant details on using RODBC, including pointers to Actual Technologies: http://www.actualtech.com/products.php which provides relatively inexpensive commercial ODBC drivers for OSX and which is what I use to access Oracle on OSX, using RODBC. You would want: http://www.actualtech.com/product_opensourcedatabases.php They are easy to install and configure and also provide the ODBC Manager app, which replaces the no longer included Apple ODBC Administrator app. Also, you should post this query (now follow ups) to R-SIG-DB, as that list is focused on DB access related issues: https://stat.ethz.ch/mailman/listinfo/r-sig-db Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R shell script
On 25.04.2012 17:12, Steve Lianoglou wrote:
Check out the vignette for the optparse library:
http://cran.r-project.org/web/packages/optparse/vignettes/optparse.pdf
Super helpful library if you plan on making any semi-interesting
command line scripts w/ R.
-steve
On Wed, Apr 25, 2012 at 6:47 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple example
explaining how to make this shell script.
Any help appreciated :)
Aoife
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Hi Aoife
you can use the capture.output function if you want so sent R output to
a file
Regards!
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[R] Average for Huge file
Hi, I am a PhD student. Basic knowing R. I have 2 datw files 1st 5min intervals the 2nd 1min for a month. My taske is to have 15min average for all. This how data look; Date Pm FF LL KK HH NNWwDD 01/01/2012 00:00:00 349 120 10 8 1178 12922005 762 01/01/01/2012 00:00:05 356 119 12 7 1167 1289 1992778 01/01/2012 00:00:10 360 115 15 8 189 1302 2010 770 -- View this message in context: http://r.789695.n4.nabble.com/Average-for-Huge-file-tp4586926p4586926.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] transforming data based on factors in a dataframe
Hello R-help list, I would really appreciate help with my factoring problem. My generated data is this: df - expand.grid(T=seq(10,80, by=5), conc=rep(c(1, 3, 7), 2)) df$curve - as.factor(rep(1:6, each=length(seq(10,80, by=5 df$counts - 3*df$T/df$conc + rnorm(df$T,0,2) plot(counts~T, df) What I would like to do add a new column to the dataframe of zeroed data (say df$counts.zeroed). For each curve (designated by factor df$curve) I want to take the value in counts and subtract the minimum value for that curve. However, I don't really have an idea of how to approach this problem and haven't found anything in my searches. And just as a second question, my second line of code assigns the factors, but if there is a nicer way of doing this I would really appreciate knowing how. Thanks for any help! Carly __ 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 confidence interval graphics
Hi everybody. I'm making a confidence interval plot using the function plotCI. I would like to decrease the space between the lines. How can I do that? Thanks alot -- View this message in context: http://r.789695.n4.nabble.com/Help-confidence-interval-graphics-tp4587052p4587052.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] Help on time series Hurst exponent
Thanks, Michael! Could you plz point to some easy tutorials regarding this? On Wed, Apr 25, 2012 at 8:14 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: You really don't want to use ts() -- if you want to use the tools in fArma use a timeSeries (provided by the package of the same name) Michael On Wed, Apr 25, 2012 at 9:54 AM, Barun Saha barun.sah...@gmail.com wrote: Hello, I'm an absolute beginner with R. I'm hoping to do some time-series analysis on my data. The data looks like #time value 18 153 20 426 70 7 83 130 84 7 and so on where time could be in seconds or hours or days (not all at the same time). How could I import such a file to R and do some simple stuff (say plot the values)? As per the tutorials on time series, I could use the ts() method to import the values (not timestamps). However, one problem with my data is that the *time intervals are not regular* -- i.e. I don't have observations at every delta_t. So, I possibly can't ignore the timestamps. I'm also interested to estimate the Hurst exponent for the above series. I've installed the fArma package. Again, I'm not sure how to use the above series there. Could someone please help me on this? -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[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. -- Thanks, Barun Saha JPA IIT, Kharagpur http://pothi.com/pothi/book/barun-saha-swapner-kheya http://delay-tolerant-networks.blogspot.com/p/one-tutorial.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] r-square for non-linear regression
Hi, I saw you discussed about the meaning of the R squared in a nls regression. Do you have a source or a quotation please? kind regards, Pierre Grison Tel: 06 01 79 74 22 Mail: pgri...@hotmail.fr [[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] R shell script
Hi,
On Wed, Apr 25, 2012 at 11:11 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Thanks for replying.
My problem is that i have say 50 input files, that i wanted to run a
particular command on, get 50 output files, and then when i close R, have
them in my directory?
so for example if i say:
R
library(MASS)
list.files(pattern = .out)
sapply(list.files(pattern = *.out), function(x) wilcox.test ( ... ) )
send each output to a different file, and save it in such a way that
when i close R the outputs are still there
i thought this might be easier in a shell way?
In this case, just make your function write a text file -- you have to
figure out what you want to save and serialize it to text. Or you can
write as many output rds (or rda) files as you do tests, for instance:
filez - list.files(pattern=*.out)
for (f in filez) {
## something to load the data in file `f` I presume
w - wilcox.test(... on the data you loaded ...)
saveRDS(w, gsub('.out', '.rds', f) ## if you want to save the object
}
or
info - lapply(filez, function(x) {
## load the file
w - wilcox.test(... on the data you loaded ...)
data.frame(file.name=x, statistic=w$statistic, p.value=w$p.value,
... anything else you want?)
})
result - do.call(rbind, info)
write.table(result, 'wilcox.results.txt', ...)
HTH,
-steve
On Wed, Apr 25, 2012 at 4:03 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
You can do this in bash but why not just do it in R directly? You probably
need
list.files(pattern = .out)
to get started. Then just wrap your script in a function and pass it
to (s|l)apply something like:
sapply(list.files(pattern = *.out), function(x) wilcox.test ( ... ) )
Michael
On Wed, Apr 25, 2012 at 6:47 AM, aoife doherty
aoife.m.dohe...@gmail.com wrote:
Hey guys,
Does anyone have an example of a REALLY simple shell script in R.
Basically i want to run this command:
library(MASS)
wilcox.test(list1,list2,paired=TRUE,alternative=c(greater),correct=TRUE,exact=FALSE)
in a shell script something like this:
#!/bin/bash
R
library(MASS)
for i in *.out
do
wilcox.test($i,${i/out}.out2,paired=TRUE) $i.out
done
that i can run on a command line this this:
sh R.sh
because i've SO many files to run this command on.
I've been googling, but i'm having trouble of just finding a simple
example
explaining how to make this shell script.
Any help appreciated :)
Aoife
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--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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Re: [R] How to insert filename as column in a file
Thanks Jeff. I had tried the 'list' approach as well but got stuck with the
below error:
Error in `$-.data.frame`(`*tmp*`, date, value = 20100701) :
replacement has 1 rows, data has 0
Couldnt find a work around to this, hence resorted to the multiple
dataframes approach. Any insights into this?
On Tue, Apr 24, 2012 at 9:51 PM, Jeff Newmiller jdnew...@dcn.davis.ca.uswrote:
Programatically dealing with large numbers of separately-named objects
leads to syntactically complicated code that is hard to read and maintain.
Load the data frames into a list so you can access them by numeric or
named index, and then getting at the loaded data will be much easier.
fnames = list.files(path = getwd())
# preallocating the list for efficiency (execution speed)
dtalist - vector( list, length(fnames) )
for (i in seq_len(length(fnames))){
dtalist[[i]] - read.csv.sql(fnames[i], sql = select * from file where
V3 == 'XXX' and V5=='YYY',header = FALSE, sep= '|', eol =\n))
dtalist[[i]]$date - substr(fnames[i],1,8))
}
names(dtalist) - fnames
# now you can optionally refer to dtalist$file20120424.csv or
dtalist[[file20120424]] if you wish.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
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---
Sent from my phone. Please excuse my brevity.
Shivam shivamsi...@gmail.com wrote:
Reposting in hope of a reply.
On Tue, Apr 24, 2012 at 1:12 AM, Shivam shivamsi...@gmail.com wrote:
Thanks for the quick response. It works for an individual dataframe,
but I
have many dataframes. This is the code so far
fnames = list.files(path = getwd())
for (i in 1:length(fnames)){
assign(paste(file,i,sep=),read.csv.sql(fnames[i], sql = select *
from
file where V3 == 'XXX' and V5=='YYY',header = FALSE, sep= '|', eol =
\n))
}
This generates dataframes named as as file1,file2,...,file250. Is
there a
way to do something like below within the same loop?
file1$date = substr(fnames[1],1,8))
file2$date = substr(fnames[2],1,8))
.
.
file250$date = substr(fnames[250],1,8))
assign(paste(file,i,sep=)$date doesnt work.
Any help?
On Tue, Apr 24, 2012 at 12:01 AM, MacQueen, Don
macque...@llnl.govwrote:
This little example might help.
foo - data.frame(a=1:10, b=letters[1:0])
foo
a b
1 1 a
2 2 a
3 3 a
4 4 a
5 5 a
6 6 a
7 7 a
8 8 a
9 9 a
10 10 a
foo$date - '20120423'
foo
a b date
1 1 a 20120423
2 2 a 20120423
3 3 a 20120423
4 4 a 20120423
5 5 a 20120423
6 6 a 20120423
7 7 a 20120423
8 8 a 20120423
9 9 a 20120423
10 10 a 20120423
In other words, immediately after reading the data into a data
frame, add
a date column as in the example. You'll have to extract the date
from the
filename, of course.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 4/23/12 9:29 AM, Shivam shivamsi...@gmail.com wrote:
Hi,
I am relatively new to R. Have scourged the help files and the www
but
havent been able to get a solution.
I have around 250 csv files, one file for each date. They have
columns of
all types, numeric, string etc. The name of each file is the date
in the
form of 'mmdd'. There is no column within the file which helps
me
identify the date on which the file was generated, only the
filename has
that info.
I am selecting some data (using read.csv.sql) from each file and
creating
a
dataset for each day. Ultimately I will combine all the datasets. I
can
accomplish the select and combine part, but after combining I wont
have a
record as to the date corresponding to the data.
Hence I want to insert the filename as a column in the respective
file to
help me in identifying to what date each data row belongs to.
Sorry for the long mail, but wanted to make myself clear. Any help
would
be
greatly appreciated.
Thanks in advance,
Shivam
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--
*Victoria Concordia Crescit*
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Re: [R] r-square for non-linear regression
This question has been asked many times before. Please search the archives. The short answer is: R-squared is more or less meaningless in nonlinear regression. The archives provide elaboration and caveats to this claim. -- Bert On Wed, Apr 25, 2012 at 8:12 AM, Pierre Grison pgri...@hotmail.fr wrote: Hi, I saw you discussed about the meaning of the R squared in a nls regression. Do you have a source or a quotation please? kind regards, Pierre Grison Tel: 06 01 79 74 22 Mail: pgri...@hotmail.fr [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] transforming data based on factors in a dataframe
try this: (uses 'ave') df - expand.grid(T=seq(10,80, by=5), conc=rep(c(1, 3, 7), 2)) df$curve - as.factor(rep(1:6, each=length(seq(10,80, by=5 df$counts - 3*df$T/df$conc + rnorm(df$T,0,2) plot(counts~T, df) df$zero - ave(df$counts, df$curve, FUN = function(x) x - min(x)) df T conc curve counts zero 1 101 1 30.4148210 0.00 2 151 1 43.2169594 12.802138 3 201 1 64.3876491 33.972828 4 251 1 75.2957249 44.880904 5 301 1 95.4479888 65.033168 6 351 1 103.5814348 73.166614 7 401 1 121.0131061 90.598285 8 451 1 135.8270827 105.412262 9 501 1 152.7179565 122.303136 10 551 1 165.2928662 134.878045 11 601 1 181.1899238 150.775103 12 651 1 197.8600462 167.445225 13 701 1 210.0112723 179.596451 14 751 1 224.1807064 193.765885 15 801 1 237.5717249 207.156904 16 103 2 8.2411801 0.00 17 153 2 14.7415660 6.500386 18 203 2 20.9944483 12.753268 19 253 2 23.4369433 15.195763 On Wed, Apr 25, 2012 at 10:57 AM, Carly Huitema carly.huit...@gmail.com wrote: Hello R-help list, I would really appreciate help with my factoring problem. My generated data is this: df - expand.grid(T=seq(10,80, by=5), conc=rep(c(1, 3, 7), 2)) df$curve - as.factor(rep(1:6, each=length(seq(10,80, by=5 df$counts - 3*df$T/df$conc + rnorm(df$T,0,2) plot(counts~T, df) What I would like to do add a new column to the dataframe of zeroed data (say df$counts.zeroed). For each curve (designated by factor df$curve) I want to take the value in counts and subtract the minimum value for that curve. However, I don't really have an idea of how to approach this problem and haven't found anything in my searches. And just as a second question, my second line of code assigns the factors, but if there is a nicer way of doing this I would really appreciate knowing how. Thanks for any help! Carly __ 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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] transforming data based on factors in a dataframe
On Apr 25, 2012, at 10:57 AM, Carly Huitema wrote:
Hello R-help list,
I would really appreciate help with my factoring problem.
My generated data is this:
df - expand.grid(T=seq(10,80, by=5), conc=rep(c(1, 3, 7), 2))
df$curve - as.factor(rep(1:6, each=length(seq(10,80, by=5
df$counts - 3*df$T/df$conc + rnorm(df$T,0,2)
plot(counts~T, df)
What I would like to do add a new column to the dataframe of zeroed
data (say df$counts.zeroed). For each curve (designated by factor
df$curve) I want to take the value in counts and subtract the minimum
value for that curve.
translated to R that request would be something along these minimally
tested lines:
df$counts.zeroed - with( df, ave(counts, curve, FUN=function(x){ x-
min(x) }) )
However, I don't really have an idea of how to
approach this problem and haven't found anything in my searches.
And just as a second question, my second line of code assigns the
factors, but if there is a nicer way of doing this I would really
appreciate knowing how.
There is a `gl` function that I think creates grouping factors, but I
generally use `rep` because I understand it better.
Thanks for any help!
Carly
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David Winsemius, MD
West Hartford, CT
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Re: [R] fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
Thank you all very much, that did the trick. Stephen On Wed 25 Apr 2012 09:59:38 AM CDT, Sarah Goslee wrote: The clarification helps; the original description was rather terse. What about: row- c(a,b,c,d,e,f,g) #rows from larger data frame x- data.frame(sp1=rnorm(4), sp2=rnorm(4), sp3=rnorm(4), sp4=rnorm(4)) rownames(x)- row.1 merge(x, matrix(row, ncol=1), by.x=0, by.y=1, all=TRUE) Row.namessp1sp2sp3sp4 1 a 0.4964272 1.4989159 0.4302415 0.9648854 2 b 1.4137142 0.9430609 0.0728391 -0.6275084 3 c -0.8103023 -1.3375148 -0.3799518 0.4523287 4 d NA NA NA NA 5 e NA NA NA NA 6 f NA NA NA NA 7 g -0.1914184 0.5156566 0.5626614 0.8068154 Sarah On Wed, Apr 25, 2012 at 10:42 AM, Stephen Seficksas0...@auburn.edu wrote: I am subsetting a larger data frame that contains macroinverterate taxa. I am subsetting them at different levels of taxonomic resolution. Some of the sites do not have say Tipulidae present, so the rows are removed completely for this site. I would like to fill in the sites that were removed during the subsetting and fill these with 0. I have the non-subsetted dataframe which contains all of the sites. I would like to use the sites from this complete data set to expand the dataframe that is a subset of the sites in the original. Is that more clear. Stephen -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ 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] On the Design of the R Language
http://www.cs.purdue.edu/homes/jv/pubs/ecoop12.pdf A new paper out on R the language -- I'm not all the way through it but it's been an interesting read so far. Thought it might be of interest to the list. Michael Weylandt [[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] Using apply() with a function involving ode()
Dear Berend,
Yes, the wnv.sim function should have included wnv.model instead of
wnv.incubation. Sorry for the typo.
Although I had inspected the expand.grid() object, I had not connected
the error with a naming problem. Thanks so much for the help! The
function works perfectly.
Sincerely,
Adam
On 4/25/2012 2:02 AM, Berend Hasselman wrote:
See inline comments.
On 25-04-2012, at 08:40, Adam Zeilinger wrote:
Hello,
I am trying to get the output from the numerical simulation of a system of
ordinary differential equations for a range of values for three parameters. I
am using the ode() function (deSolve package) to run the numerical simulation
and apply() to run the simulation function for each set of parameter values. I
am having trouble getting the apply() function to work.
Here is an example. Consider an epidemiology model of West Nile Virus:
wnv.model- function(Time, State, Pars){
with(as.list(c(State, Pars)), {
dSB- -(alphaB*betaB*SB*IM)/(SB + IB)
dIB- (alphaB*betaB*SB*IM)/(SB + IB) - deltaB*IB
dSM- bM*(SM + EM + IM) - (alphaM*betaB*IB*SM)/(SB + IB) -
dM*SM
dEM- (alphaM*betaB*SM*IB)/(SB + IB) - kappaM*EM - dM*EM
dIM- kappaM*EM - dM*IM
return(list(c(dSB, dIB, dSM, dEM, dIM)))
})
}
# I would like to run a numerical simulation of this model for each combination
of values for the parameters alphaB, betaB, # and kappaM:
av- seq(0, 1, by = 0.2) # vector of values for alphaB
bv- seq(0, 1, by = 0.2) # vector of values for betaB
kv- seq(0, 1, by = 0.2) # vector of values for kappaM
# Here is my function with ode() for the numerical simulation. The function returns the last value
of the simulation for the # IB state variable (Infected Birds).
library(deSolve)
wnv.sim- function(x){
Pars- c(alphaB = x[1], betaB = x[2], deltaB = 0.2, bM = 0.03, dM =
0.03, alphaM = 0.69, kappaM = x[3])
State- c(SB = 100, IB = 0, SM = 1500, EM = 0, IM = 0.0001)
Time- seq(0, 60, by = 1)
model.out- as.data.frame(ode(func = wnv.incubation,
There is no function wnv.incubation.
Assuming you mean wnv.model
y = State,
parms = Pars,
times = Time))
model.out[nrow(model.out),]$IB
}
# Finally, here is my apply() function:
dat- apply(expand.grid(av, bv, kv), 1, wnv.sim)
However, the apply() function returns an error:
Error in eval(expr, envir, enclos) : object 'alphaB' not found
Have you inspected the object returned by expand.grid?
Do head(expand.grid(av,bv,kv))
and you will see that the columns have names which confuse the rest of your
code.
I was able to repair by doing
pars.grid- expand.grid(av, bv, kv)
names(pars.grid)- NULL
dat- apply(pars.grid, 1, wnv.sim)
HTH,
Berend
--
Adam Zeilinger
Post Doctoral Scholar
Department of Entomology
University of California Riverside
www.linkedin.com/in/adamzeilinger
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Re: [R] pca biplot.princomp has a bug?
The arrows are not pointing in the most-varying direction of the data. The principal components are pointing in the most-varying direction of the data. But you are not plotting the data on the original scale, you are plotting the data on the rotated scale, and thus the horizontal axis is the most-varying direction of the data. The arrows are pointing in the direction of the variables, as projected into the 2-d plane of the biplot. There is no bug. Kevin Wright On Wed, Apr 25, 2012 at 11:29 AM, Michael comtech@gmail.com wrote: x=rmvnorm(2000, rep(0, 6), diag(c(5, rep(1,5 x=scale(x, center=T, scale=F) pc - princomp(x) biplot(pc) There are a bunch of red arrows plotted, what do they mean? I knew that the first arrow labelled with Var1 should be pointing the most varying direction of the data-set (if we think them as 2000 data points, each being a vector of size 6). I also read from somewhere, the most varying direction should be the direction of the 1st eigen vector. However, reading into the code of biplot in R. The line about the arrows is: if(var.axes) arrows(0, 0, y[,1L] * 0.8, y[,2L] * 0.8, col = col[2L], Where `y` is the actually the loadings matrix, which is the eigenvector matrix. So it looks like the 1st arrow is actually pointing from `(0, 0)` to `(y[1, 1], y[1, 2])`. I understand that we are trying to plot a high dimensional arrow onto a 2D plane. That's why we are taking the 1st and 2nd element of the `y[1, ]` vector. However what I don't understand is: Shouldn't the 1st eigenvector direction be the vector denoted by `y[, 1]`, instead of `y[1, ]`? (Again, here `y` is the eigenvector matrix, obtained by PCA or by eigendecomposition of `t(x) %*% x`.) i.e. the eigenvectors should be column vectors, not those horizontal vectors. Even though we are plotting them on 2D plane, we should draw the 1st direction to be from `(0, 0)` pointing to `(y[1, 1], y[2, 1])`? I think the R code biplot.princomp has a bug: the loading matrix (eigenvector matrix) should be transposed before being sent into biplot.princomp... Any thoughts? [[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. -- 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] On the Design of the R Language
Hmmm an 'objective' assessment? Maybe. But it looks to me that some commercial enterprise paid for this study as a means to argue against the use of R in favor of a commercial package On Wed, Apr 25, 2012 at 1:48 PM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: http://www.cs.purdue.edu/homes/jv/pubs/ecoop12.pdf A new paper out on R the language -- I'm not all the way through it but it's been an interesting read so far. Thought it might be of interest to the list. Michael Weylandt [[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. -- Thomas E Adams National Weather Service Ohio River Forecast Center 1901 South State Route 134 Wilmington, OH 45177 EMAIL: thomas.ad...@noaa.gov VOICE: 937-383-0528 FAX:937-383-0033 [[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] locate values to their positions based on their pixel and line in other files
I have 3 files A ,B ,C as binary files. C represents the values of tepm
measured every 3 hours for one month so it has 674200 columns and 248
rows.values of A represent the correspond lines(these values could be within
1to360) and , while values of B represent the correspond pixel(1to720) for
each point in file C(so both they have 1 row and 67420rows).Now I will
create a new matrix ,M ,which has this dim(360,720).for example the first
value in C it will be represented in the creatd matrix based on the
corspding pixl and lines.assume the first value is 54(row 1 and column 1)in
C so in order to locate it in my new matrix we need to know what its pixl
and line that this point shoud be assocated with ,therfore I need to tell R
that take its pixl value from A and its line from B and locate it in my new
matrix. this is to be done for first row out of 248 rows and write it to new
file then do the same for the second row and write to a new file and so on
for all 248 rows.
library(Matrix)
M - Matrix(-, 360, 720)## creat matrix with 720 columns and 360 rows
with valus of -
long - file(C:\\New folder (5)\\inra.bin, rb)
A=readBin(long, integer(), size=2,n=1*67420, signed=TRUE)
lot - file(C:\\New folder (5)\\lat.img, rb)
B=readBin(lot, integer(), size=2,n=1*67420, signed=TRUE)
for (i in c(1:67420)) {
wind - file(C:\\Wind_WFD_200201.bin, rb)
C=readBin(wind, integer(), size=2,n=248*67420, signed=TRUE)##it has 67420
columns and 248 rows
M(A(i),B(i))= C(i)}
I wrote the above code but didn't work.I am so sorry for this long
description and thanks in advance
--
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Re: [R] On the Design of the R Language
I have not read it yet, but the acknowledgements at the end of the paper note that: 1. It was supported by an NSF grant. 2. At least two members of R Core are recognized amongst the list of names. Regards, Marc Schwartz On Apr 25, 2012, at 1:07 PM, Thomas Adams wrote: Hmmm∑ an 'objective' assessment? Maybe. But it looks to me that some commercial enterprise paid for this study as a means to argue against the use of R in favor of a commercial package∑ On Wed, Apr 25, 2012 at 1:48 PM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: http://www.cs.purdue.edu/homes/jv/pubs/ecoop12.pdf A new paper out on R the language -- I'm not all the way through it but it's been an interesting read so far. Thought it might be of interest to the list. Michael Weylandt [[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] Equate two (or more) functions
Dear List, I was wondering if there is a simple way to solve (equate) two functions in R. For example: (1) 5x + 3y = 30 (2) 12x - 2y = 26 I would like R to finde x = 3 and y = 5 Is there an implemented function in R or would I have to write an own function? Thank you very much for your help. -- View this message in context: http://r.789695.n4.nabble.com/Equate-two-or-more-functions-tp4587302p4587302.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] Average for Huge file
Hi. I would use the function aggregate, but first you will have to tag each row with a special code so R can recognize the group of data and apply the function you desire. For example, with your data I would do this: Date Pm FF LL KKHH NNWwDD code 01/01/2012 00:00:00 349 120 10 8 1178 12922005 762 01_01_2012_1 01/01/01/2012 00:00:05 356 119 127 1167 1289 1992778 01_01_2012_1 01/01/2012 00:00:10 360 115 15 8 189 1302 2010 770 01_01_2012_1 01/01/2012 00:00:15 349 120 10 8 1178 12922005 762 01_01_2012_1 01/01/01/2012 00:00:20 356 119 127 1167 1289 1992778 01_01_2012_2 01/01/2012 00:00:25 360 115 15 8 189 1302 2010 770 01_01_2012_2 01/01/2012 00:00:30 360 115 15 8 189 1302 2010 770 01_01_2012_2 ##Then apply the function aggregate aggregate(name of the variable you want to obtain the mean, by=list(variable used for grouping),FUN=mean) For example, if you want to aggregate Pm by groups of 15 min, you write it like this aggregate(Pm,by=list(code),FUN=sum) and you'll obtain the mean of the rows that have the same code . In this example, you'll obtain the mean of two groups: the Pm measurments wich their labels are 01_01_2012_1 and 01_01_2012_2. Hope it works M.C. Luis Antonio Arias Medellín National Institute of Public Health Cuernavaca, Morelos, Mexico -- View this message in context: http://r.789695.n4.nabble.com/Average-for-Huge-file-tp4586926p4587482.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] Intercept between two lines
Dear List, I know this is not the first post on this topic, but I need basic help I guess. Assuming the simple case of two lines with one intercept, how can I make R calculate this intercept, NOT using locator(). par (xaxs=i, yaxs=i) plot( 1, bty=n ,xlim=c(0,300) , ylim=c(0,300) , xlab=X, ylab=Y) curve(100-0.5*x, -50,250, add=T, col=blue) curve(150- x , -50,250, add=T, col=red) I want R to come up with the coordinates X=100, Y=50. Thank you for your help! -- View this message in context: http://r.789695.n4.nabble.com/Intercept-between-two-lines-tp4587343p4587343.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] Equate two (or more) functions
Hello, Try (x - matrix(c(5, 12, 3, -2), ncol=2)) (y - c(30, 26)) solve(x, y) solve.qr(qr(x), y) (Use the second.) Also, see section 5.7.5 of the R-intro.pdf that comes with your (any) installation of R. Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Equate-two-or-more-functions-tp4587302p4587523.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] Intercept between two lines
Hi pannigh, The following might get you started: x0 - uniroot(function(x) 100-0.5*x - (150- x), c(0, 150))$root x0 [1] 100 100- 0.5*x0 [1] 50 HTH, Jorge.- On Wed, Apr 25, 2012 at 1:01 PM, pannigh wrote: Dear List, I know this is not the first post on this topic, but I need basic help I guess. Assuming the simple case of two lines with one intercept, how can I make R calculate this intercept, NOT using locator(). par (xaxs=i, yaxs=i) plot( 1, bty=n ,xlim=c(0,300) , ylim=c(0,300) , xlab=X, ylab=Y) curve(100-0.5*x, -50,250, add=T, col=blue) curve(150- x , -50,250, add=T, col=red) I want R to come up with the coordinates X=100, Y=50. Thank you for your help! -- View this message in context: http://r.789695.n4.nabble.com/Intercept-between-two-lines-tp4587343p4587343.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. [[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] On the Design of the R Language
On Apr 25, 2012, at 20:19 , Marc Schwartz wrote: I have not read it yet, but the acknowledgements at the end of the paper note that: 1. It was supported by an NSF grant. 2. At least two members of R Core are recognized amongst the list of names. Also, most of the overall content is hardly a surprise to R Core (not that I have read it properly either). R as a language has many weaknesses and Luke can certainly attest the difficulties in writing compilers for R due to the fact that semantics can change at any moment (I believe that's what the authors refer to as extreme dynamism). Competent critical review by computer scientist should be welcomed, and in my opinion is long overdue. As has been said, the good _and_ the bad thing about R is that it was designed by statisticians... -pd Regards, Marc Schwartz On Apr 25, 2012, at 1:07 PM, Thomas Adams wrote: Hmmm∑ an 'objective' assessment? Maybe. But it looks to me that some commercial enterprise paid for this study as a means to argue against the use of R in favor of a commercial package∑ On Wed, Apr 25, 2012 at 1:48 PM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: http://www.cs.purdue.edu/homes/jv/pubs/ecoop12.pdf A new paper out on R the language -- I'm not all the way through it but it's been an interesting read so far. Thought it might be of interest to the list. Michael Weylandt [[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. -- 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] Equate two (or more) functions
On 25-04-2012, at 18:52, pannigh wrote: Dear List, I was wondering if there is a simple way to solve (equate) two functions in R. For example: (1) 5x + 3y = 30 (2) 12x - 2y = 26 I would like R to finde x = 3 and y = 5 Is there an implemented function in R or would I have to write an own function? And if your functions are nonlinear there is package nleqslv and BB. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] On the Design of the R Language
Thanks Michael: Interesting! Is it legitimate to comment on this in this list? It would only be my opinions, not real R-Help stuff. Where would be a better place to post such UN-expert opinion? -- Bert On Wed, Apr 25, 2012 at 10:48 AM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: http://www.cs.purdue.edu/homes/jv/pubs/ecoop12.pdf A new paper out on R the language -- I'm not all the way through it but it's been an interesting read so far. Thought it might be of interest to the list. Michael Weylandt [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] Rprofile.site on Windows
On 12-04-25 4:04 PM, Trevor Miles wrote:
Thanks Duncan.
The only .site files I have are Reviron.site and Rprofile.site, which
are both in C:\Program Files\R\R-2.14.1\etc.
My advice was about 2.15.0, but I don't remember any of this changing
recently.
Copying these files to C:\Program Files\R\R-2.14.1\etc\x64 and
C:\Program Files\R\R-2.14.1\etc\i386 makes no difference.
Sys.getenv returns R_ENVIRON = and R_HOME =
C:/PROGRA~1/R/R-214~1.1. Should I set R_ENVIRON in the Windows Registry?
R_ENVIRON doesn't affect the search for the Rprofile.site file, other
than defining where to find environment variables. You want to look at
R_PROFILE, and possibly set it in your Renviron.site file.
Duncan Murdoch
Date: Tue, 24 Apr 2012 22:50:21 -0400
From: murdoch.dun...@gmail.com
To: trevor.mi...@live.ca
CC: r-help@r-project.org
Subject: Re: [R] Rprofile.site on Windows
On 12-04-24 10:43 PM, Duncan Murdoch wrote:
On 12-04-24 4:13 PM, Trevor Miles wrote:
Hi All
I am struggling to get R to read the Rprofile.site file from the
R_HOME/etc folder.
I know it isn't working because I change the prompt in the
Rprofile.site file. In addition, when I run
source('R_HOME'/etc/Rprofile.site) from the prompt, the prompt gets
changed and other environmental variables get set.
Any ideas?
I don't think the documentation has completely caught up, but that
would
now normally be put in R_HOME/etc/i386 or R_HOME/etc/x64, depending
which version you're running.
Sorry, that's not quite right. It will look where R_PROFILE says to
look, then in the spots mentioned above, but finally will look in
R_HOME/etc if a file hasn't already been found. So maybe you have
another Rprofile.site file that is being read instead of your new one.
Duncan Murdoch
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Re: [R] Help confidence interval graphics
rafitoariaz rafitoariaz at hotmail.com writes: Hi everybody. I'm making a confidence interval plot using the function plotCI. I would like to decrease the space between the lines. How can I do that? Thanks alot There may be a better answer than this, but here goes: The lines are plotted at 1, 2, 3, 4, etc on the 'categorical' axis, and that axis automatically displays just the range of values necessary. So if you have four confidence intervals plotted, the 'categorical' axis will range from 1 to 4. You can reduce the distance between the lines by changing the aspect ratio of your plot device, ie, make the plot narrower along that categorical axis and the plot will be adjusted accordingly. Alternatively (or additionally), you can alter the range of the categorical axis, using the xlim or ylim argument (depending on whether your lines run vertically or horizontally). By making the range on that axis larger, you create more 'white space' at each end of the scale and the lines cluster together in the centre more. Note that you can also customise your axes with axis(), like other plots. Hope this helps, Michael Bibo Queensland Health __ 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] Re : Design of experiments for Choice-Based Conjoint Analysis (CBC)
Hi, I came across this paper which is essentially a tutorial on how to generate designs in R using the AlgDesign package. Hope it's helpful. I'm not trying to generate designs in R using priors, but haven't found a way to do it yet. Anyone out there have any ideas? Cheers, John Paul -- View this message in context: http://r.789695.n4.nabble.com/Design-of-experiments-for-Choice-Based-Conjoint-Analysis-CBC-tp2254142p4587666.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] On the Design of the R Language
Hi, On Wed, Apr 25, 2012 at 4:06 PM, Bert Gunter gunter.ber...@gene.com wrote: Thanks Michael: Interesting! Is it legitimate to comment on this in this list? It would only be my opinions, not real R-Help stuff. FWIW, I'd be interested in hearing opinions about it from R-folk ... Where would be a better place to post such UN-expert opinion? I didn't realize you were also an expert on foreign affairs? Nice. -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] Scatter plot / LOESS, or LOWESS for more than one parameter
You can also use range( MC.pH, MV.pH, na.rm=TRUE). On Tue, Apr 24, 2012 at 1:29 PM, David Doyle kydaviddo...@gmail.com wrote: Hi Greg, Sloved my own problem. I had some missing data NA in the datasets. So I manually entered the ylim=range(4,6) and it worked!!! Thanks!! David On Tue, Apr 24, 2012 at 1:55 PM, David Doyle kydaviddo...@gmail.com wrote: Hi Greg, Thanks, I got the 1st example to work using the following code: data - read.csv(http://doylesdartden.com/Monthly-pH-example.csv;, sep=,) attach(data) par(mfrow=c(2,1)) scatter.smooth( Year, MC.pH ) scatter.smooth( Year, MV.pH ) This is good but what I'm really looking for is to have them on the same graph. I tried your second example using the code below but got: Error in plot.window(...) : need finite 'ylim' values here is the code I used data - read.csv(http://doylesdartden.com/Monthly-pH-example.csv;, sep=,) attach(data) plot( Year, MC.pH, ylim=range(MC.pH,MV.pH) , col='blue') points( Year, MV.pH, col='green' ) lines( loess.smooth(Year,MC.pH), col='blue') lines( loess.smooth(Year,MV.pH), col='green') Thanks again David On Tue, Apr 24, 2012 at 1:45 PM, Greg Snow 538...@gmail.com wrote: Assuming that you want event as the x-axis (horizontal) you can do something like (untested without reproducible data): par(mfrow=c(2,1)) scatter.smooth( event, pH1 ) scatter.smooth( event, pH2 ) or plot( event, pH1, ylim=range(pH1,pH2) , col='blue') points( event, pH2, col='green' ) lines( loess.smooth(event,pH1), col='blue') lines( loess.smooth(event,pH2), col='green') Only do the second one if pH1 and pH2 are measured on the same scale in a way that the comparison and any crossings are meaningful or if there is enough separation (but not too much) that there is no overlap, but still enough detail. On Mon, Apr 23, 2012 at 10:40 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: The scatter plot is easy: plot(pH1 ~ pH2, data = OBJ) When you say a loess for each -- how do you break them up? Are there repeat values for pH1? If so, this might be hard to do in base graphics, but ggplot2 would make it easy: library(ggplot2) ggplot(OBJ, aes(x = pH1, y = pH2)) + geom_point() + stat_smooth() + facet_wrap(~factor(pH1)) or something similar. Michael On Mon, Apr 23, 2012 at 11:26 PM, David Doyle kydaviddo...@gmail.com wrote: Hi folks. If I have the following in my data event pH1 pH2 1 4.0 6.0 2 4.3 5.9 3 4.1 6.1 4 4.0 5.9 and on and on. for about 400 events Is there a way I can get R to plot event vs. pH1 and event vs. pH2 and then do a loess or lowess line for each?? Thanks in advance David [[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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com -- Gregory (Greg) L. Snow Ph.D. 538...@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] How to test if a slope is different than 1?
Doesn't the p-value from using offset work for you? if you really need a p-value. The confint method is a quick and easy way to see if it is significantly different from 1 (see Rolf's response), but does not provide an exact p-value. I guess you could do confidence intervals at different confidence levels until you find the level such that one of the limits is close enough to 1, but that seems like way to much work. You could also compute the p-value by taking the slope minus 1 divided by the standard error and plug that into the pt function with the correct degrees of freedom. You could even write a function to do that for you, but it still seems more work than adding the offset to the formula. On Tue, Apr 24, 2012 at 8:17 AM, Mark Na mtb...@gmail.com wrote: Hi Greg. Thanks for your reply. Do you know if there is a way to use the confint function to get a p-value on this test? Thanks, Mark On Mon, Apr 23, 2012 at 3:10 PM, Greg Snow 538...@gmail.com wrote: One option is to subtract the continuous variable from y before doing the regression (this works with any regression package/function). The probably better way in R is to use the 'offset' function: formula = I(log(data$AB.obs + 1, 10)-log(data$SIZE,10)) ~ log(data$SIZE, 10) + data$Y formula = log(data$AB.obs + 1) ~ offset( log(data$SIZE,10) ) + log(data$SIZE,10) + data$Y Or you can use a function like 'confint' to find the confidence interval for the slope and see if 1 is in the interval. On Mon, Apr 23, 2012 at 12:11 PM, Mark Na mtb...@gmail.com wrote: Dear R-helpers, I would like to test if the slope corresponding to a continuous variable in my model (summary below) is different than one. I would appreciate any ideas for how I could do this in R, after having specified and run this model? Many thanks, Mark Na Call: lm(formula = log(data$AB.obs + 1, 10) ~ log(data$SIZE, 10) + data$Y) Residuals: Min 1Q Median 3Q Max -0.94368 -0.13870 0.04398 0.17825 0.63365 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -1.18282 0.09120 -12.970 2e-16 *** log(data$SIZE, 10) 0.56009 0.02564 21.846 2e-16 *** data$Y2008 0.16825 0.04366 3.854 0.000151 *** data$Y2009 0.20310 0.04707 4.315 0.238 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.2793 on 228 degrees of freedom Multiple R-squared: 0.6768, Adjusted R-squared: 0.6726 F-statistic: 159.2 on 3 and 228 DF, p-value: 2.2e-16 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com -- Gregory (Greg) L. Snow Ph.D. 538...@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] Accessing a list
I believe that fortune(312) applies here. As my current version of
fortunes does not show this I am guessing that it is in the
development version and so here is what fortune(312) will eventually
print (unless something changes or I got something wrong):
The problem here is that the $ notation is a magical shortcut and like
any other magic if used
incorrectly is likely to do the programmatic equivalent of turning
yourself into a toad.
—Greg Snow (in response to a user that wanted to access a column whose name is
stored in y via x$y rather than x[[y]])
R-help (February 2012)
On Tue, Apr 24, 2012 at 9:42 PM, Jim Silverton jim.silver...@gmail.com wrote:
Hi,
I have the following problem- I want to access a list whose elements are
imp1, imp2, imp3 etc I tried theusing the paste comand in a for loop see
the last for loop below. But I keep calling it df but df = imp1 (for the
first run). Any ideas on how I can access the elements of the list?
Isaac
require(Amelia)
library(Amelia)
data.use - read.csv(multiplecarol.CSV, header=T)
names(data.use) = c(year, dischargex1, y, pressurex2 , windx3)
ts - c (c(1:12), c(1:12), c(1:12), c(1:12), c(1:12), c(1:12), c(1:12),
c(1:6) )
length(ts)
data.use = cbind(ts, data.use)
#a.out2 - amelia(data.use, m = 1000, idvars = year)
n.times = 100
a.out.time - amelia(data.use, m = n.times, ts=ts, idvars=year,
polytime=2)
constant.col = dischargex1.col = pressurex2.col = windx3.col =
rep(0,n.times)
for (i in 1: n.times)
{
x = c(imp,i)
df = paste(x, collapse = )
data1 = a.out.time[[1]]$df
attach(data1)
y = as.numeric(y)
dischargex1 = as.numeric(dischargex1)
pressurex2 = as.numeric(pressurex2)
windx3 = as.numeric(windx3)
multi.regress = lm(y~ dischargex1 + pressurex2 + windx3)
constant.col[i] = as.numeric(multi.regress[[1]][1])
dischargex1.col[i] = as.numeric(multi.regress[[1]][2])
pressurex2.col[i] = as.numeric(multi.regress[[1]][3])
windx3.col[i] = as.numeric(multi.regress[[1]][4])
}
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
__
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] comparison of bivariate normal distributions
sorry for cross-posting Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about how different these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little code example: library(mvtnorm) c-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1)) c2-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1)) xx=seq(0,20,0.1) yy=seq(0,20,0.1) xmult=cbind(rep(yy,201),rep(xx,each=201)) dens=dmvnorm(xmult,mean(c),cov(c)) dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F) dens2=dmvnorm(xmult,mean(c2),cov(c2)) dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F) contour(xx,yy,dmat,lwd=2) contour(xx,yy,dmat2,lwd=2,add=T) ## Is their an easy way to do this (maybe with dmvnorm()?) and could I interpret the intersect (shared volume) in the sense of a t-statistic? Thanks a lot for your help! _ Fabian Roger, Ph.D. student Dept of Biological and Environmental Sciences University of Gothenburg Box 461 SE-405 30 Göteborg Sweden Tel. +46 31 786 2933 [[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.
