Re: [R] Issue with using geocode
a) Please post in plain text.
b) Although you mention Excel, it does not seem to be relevant. (This is
a good thing, but mentioning it is a red herring.)
c) Your definition of X is not executable R code. If you have already
imported data from elsewhere, you can use dput to make it easy for us to
read it in (essentially) exactly as you have it loaded in your memory. I
have assumed that you have a vector of strings that you failed to give us.
d) Quoting your variable X is not right. Rather you should index it. See
below.
On Sun, 25 Nov 2012, ioanna ioannou wrote:
Hello,
A very simple question but I am stuck. I have an excel file each row is an
address. However, I cannot make geocode read each line and come up with the
latitude longitude. Could you please correct my code?
library(ggmap)
X-c (2 Afxentiou Ampelokipi Thessaloniki Greece, 2 Afxentiou Ampelokipi
Thessaloniki Greece, 4 Afxentiou Ampelokipi Thessaloniki Greece, 55
Agathonos Ampelokipi Thessaloniki Greece)
For (i in 1:4){
Y-geocode('X')
Y[i] - geocode(X[i])
print Y[i]
}
Or better yet, eliminate the entire for loop and just give geocode the
whole vector:
Y - geocode( X )
Then see what is in Y:
Y
Best wishes,
Ioanna
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Re: [R] compatibility of load() in R 2.15.2
On Nov 25, 2012, at 10:50 PM, Jack Tanner wrote: I have some large-ish files that are the output of save() from R 2.15.1, which that version can load() just fine. After upgrading to 2.15.2, load() no longer works on these files. Is this a known issue? No longer works? -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] bbmle Warning: optimization did not converge
Thank you Uwe Ligges-3 -- View this message in context: http://r.789695.n4.nabble.com/bbmle-Warning-optimization-did-not-converge-tp4650730p4650812.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] R jags clear deviance
Hi all, I'd like to fit different models using a loop together with the jags package. to do this, I load the following packages runjags, R2jags and coda and use the function jags to fit my models. The problem is that the following a windowx with the following message appear between each model fitting inside my loop : the following objects in .GlobalEnv will mask objects in the attached database: deviance. Remove this objests from .GlobalEnv ? I need to click each time on yes or no and that's bad. Do you have an idea to overcome thos issue ? I tried different stuffs using remove function but it did not work Thanking you in advance Guillaume -- View this message in context: http://r.789695.n4.nabble.com/R-jags-clear-deviance-tp4650814.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] R strange behaviour when building huge concatenation
Hello
I'm using some R code in order to use the model BG-NBD implementation
I'm using Java and I call R by using RCaller.
I must admit that i'm really really new to R so maybe I'm doing something
wrong.
I use the code on this URL: http://code.google.com/p/clv-master-thesis/
http://code.google.com/p/clv-master-thesis/
This is my working Java code (a simple Junit test):
@Test
public void testRCaller(){
try {
RCaller caller = new RCaller();
caller.setRscriptExecutable(/usr/bin/Rscript);
caller.cleanRCode();
RCode code = new RCode();
code.clear();
String helper = /dati/helper.R;
String modelNbd = /dati/model-nbd.R;
String modelParetoNbd = /dati/model-pareto-nbd.R;
String modelBgNbd = /dati/model-bg-nbd.R;
String modelCbgCnbd = /dati/model-cbg-cnbd-k.R;
code.R_source(helper);
code.R_source(modelNbd);
code.R_source(modelParetoNbd);
code.R_source(modelBgNbd);
code.R_source(modelCbgCnbd);
code.addRCode(cdData - read.table(\/dati/cdnow.csv\, head=T));
code.addRCode(names(cdData)[2] - \x\;);
code.addRCode(bgMleFit - bgEstimateParameters(cdData, list(r=1, alpha=2,
a=1, b=2)););
code.addRCode(summary(bgMleFit););
code.addRCode(cdBgParams - as.list(coef(bgMleFit)););
code.addRCode(t - 39;);
code.addRCode(cdBgCe - bgConditionalForecast(cdData, cdBgParams, t););
code.addRCode((cdBgSumEstimate - sum(cdBgCe)););
code.addRCode((cdBgMsle - mean((log(cdData$p2x+1)-log(cdBgCe+1))^2)););
code.addRCode((corr - cor(cdData$p2x, cdBgCe)););
caller.setRCode(code);
caller.runAndReturnResult(cdBgCe);
ROutputParser parser = caller.getParser();
ArrayListString nomi = parser.getNames();
for (String nome : nomi) {
double[] previsioni = parser.getAsDoubleArray(nome);
logger.info(Nome +nome+ lunghezza valori +previsioni.length);
for (int i = 0; i previsioni.length; i++) {
logger.info(Valore + previsioni[i]);
}
}
} catch (Exception e) {
logger.error(e.getMessage(), e);
}
}
By running this code all works pretty good; now i didn't want to use the csv
file as data source; I wanted to query DB and pass data to R
As far as I know the R read.table function builds a data.frame from reading
the provided csv file. So what I did is:
I used Java in order to read the csv and I wrote this code (after I read the
csv file and for semplicity I don't write the code related to the csv
reading):
@Test
public void testRCaller(){
try {
RCaller caller = new RCaller();
caller.setRscriptExecutable(/usr/bin/Rscript);
caller.cleanRCode();
RCode code = new RCode();
code.clear();
String helper = /dati/helper.R;
String modelNbd = /dati/model-nbd.R;
String modelParetoNbd = /dati/model-pareto-nbd.R;
String modelBgNbd = /dati/model-bg-nbd.R;
String modelCbgCnbd = /dati/model-cbg-cnbd-k.R;
code.R_source(helper);
code.R_source(modelNbd);
code.R_source(modelParetoNbd);
code.R_source(modelBgNbd);
code.R_source(modelCbgCnbd);
MapString, Object data = this.readCsvData();
StringBuilder userIds = new StringBuilder(ID - c();
long[] utenti = (long[])data.get(userIds);
int[] ordini = (int[]) data.get(ordini);
double[] tx = (double[]) data.get(tx);
double[] t = (double[])data.get(t);
int[] p2x = (int[]) data.get(p2tx);
for(int i = 0; i utenti.length; i++){
userIds.append(utenti[i]+, );
}
//here i check if the stringbuilder ends with, and i clean it...checkSb is
simply an utility method
userIds = checkSb(userIds).append(););
code.addIntArray(p1x, ordini);
code.addDoubleArray(tx, tx);
code.addDoubleArray(t, t);
code.addIntArray(p2x, p2x);
code.addRCode(cdData-data.frame(ID , p1x, tx, t, p2x););
code.addRCode(names(cdData)[2] - \x\;);
code.addRCode(bgMleFit - bgEstimateParameters(cdData, list(r=1, alpha=2,
a=1, b=2)););
code.addRCode(summary(bgMleFit););
code.addRCode(cdBgParams - as.list(coef(bgMleFit)););
code.addRCode(t - 39;);
code.addRCode(cdBgCe - bgConditionalForecast(cdData, cdBgParams, t););
code.addRCode((cdBgSumEstimate - sum(cdBgCe)););
code.addRCode((cdBgMsle - mean((log(cdData$p2x+1)-log(cdBgCe+1))^2)););
code.addRCode((corr - cor(cdData$p2x, cdBgCe)););
caller.setRCode(code);
caller.runAndReturnResult(cdBgCe);
ROutputParser parser = caller.getParser();
ArrayListString nomi = parser.getNames();
for (String nome : nomi) {
double[] previsioni = parser.getAsDoubleArray(nome);
logger.info(Nome +nome+ lunghezza valori +previsioni.length);
for (int i = 0; i previsioni.length; i++) {
logger.info(Valore + previsioni[i]);
}
}
} catch (Exception e) {
logger.error(e.getMessage(), e);
}
}
As you can see the code is totally similar to the previous one except that I
don't use read.table function.
Well by executing this code I have an error. At first I thought in some
error in the Java code but I checked and no error was present
Then I tried the code in the R console. Well I have something really
strange.
Let's start from this instruction (the int array is recovered from the csv
file):
int[] ordini = (int[]) data.get(ordini);
code.addIntArray(p1x, ordini);
This Java-RCaller instruction generates this R code: p1x-c(..).
More exactly the generated R code is the
Re: [R] compatibility of load() in R 2.15.2
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of David Winsemius Sent: Monday, November 26, 2012 9:11 AM To: Jack Tanner Cc: r-h...@stat.math.ethz.ch Subject: Re: [R] compatibility of load() in R 2.15.2 On Nov 25, 2012, at 10:50 PM, Jack Tanner wrote: I have some large-ish files that are the output of save() from R 2.15.1, which that version can load() just fine. After upgrading to 2.15.2, load() no longer works on these files. Is this a known issue? What does mean no longer works? Any error message? I can only guess, but maybe objects in those saved files were created with some packages which are not installed in your new R version. If this is the case just install those packages and all shall be OK. Regards Petr No longer works? -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Help with predict function in glm
Hello, Why mail a question just to me? Post to the list and the odds of getting more answers (and better) are bigger. As for your question, the problem is in the call to glm, you don't need the prefix 'train$' in the formula, the argument 'data' solves that and when predicting R will look for the columns with names in the formula and is unable to find columns called train$Outcome and train$Weight in the new data.frame 'test'. Corrected: mylogit - glm(Outcome ~ Weight, data=train, family = binomial(logit)) predictions - predict(mylogit, newdata = test, type= response) Hope this helps, Rui Barradas Em 26-11-2012 01:42, somnath bandyopadhyay escreveu: Hi, I am trying some basic logistic regression analysis using glm. I just have one dependent variable (Outcome) which is binary in nature and one independent variable (Weight). I fit a model using a training data set (train) which has 85 observations and try to apply it on an independent dataset (test) which has 55 observations. When I apply the predict function on the fitted model for the new dataset, I get the following warning Warning message: 'newdata' had 55 rows but variable(s) found have 85 rows and the predict works on the training observations and not on the test observations. Following is he session info, code and the training and test datasets I am using. What am I doing wrong? Any help would be greatly appreciated. Thanks, S. train - read.table(train_data.txt, header=T, row.names=1, sep=\t) test- read.table(test_data.txt, header=T, row.names=1, sep=\t) mylogit - glm(train$Outcome ~ train$Weight, data=train, family = binomial(logit)) predictions - predict(mylogit, newdata = test, type= response) Warning message: 'newdata' had 55 rows but variable(s) found have 85 rows sessionInfo() R version 2.15.0 (2012-03-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base train Outcome Weight AB256939_21 0 0.331 AB257076_21 0 0.308 AB257079_21 0 0.453 AB415508_21 0 0.303 AB700497_21 0 0.354 AB904508_21 0 0.336 AC048719_21 0 0.420 AC185939_21 0 0.249 AC185940_21 0 1.525 AC445840_21 0 0.261 E7490523_21 0 0.269 E7490524_21 0 0.213 E7659579_21 0 0.360 E7661528_21 0 0.271 E7781094_21 0 0.156 E7781095_21 0 0.221 E7781096_21 0 0.098 E7969081_21 0 0.430 E8117594_21 0 0.321 E8133295_21 0 0.166 E8161578_22 0 0.269 E8483037_21 0 0.162 E8559720_21 0 0.226 L1065550_18 0 0.396 L1065607_17 0 0.541 L1065944_24 0 0.131 L1066017_20 0 0.421 L1069261_12 0 0.357 L1069262_14 0 0.309 L1069263_27 0 0.283 L1069297_24 0 0.620 L1081528_21 0 0.561 L1084066_21 0 0.564 L1086090_21 0 0.649 L1104280_17 0 0.181 L362_22 0 0.199 L1118063_15 0 0.369 L1133550_21 0 0.302 L1144201_14 0 0.249 L1155023_7 0 0.257 L1158386_21 0 0.470 L1163051_4 0 0.446 ... ... ... test Weight AB256870_21 0.364 AB256873_21 0.329 AB415518_21 0.219 AB460669_21 0.481 AB609036_21 0.313 AB609038_21 0.196 AB700495_21 0.402 AB700498_21 0.343 AC112834_21 0.372 AC185937_21 0.270 AC269527_21 0.285 E7352023_21 0.358 E7661554_21 0.471 E7750502_21 0.437 E7845183_21 0.232 E7854155_21 0.474 E7854156_21 0.121 E7924877_21 0.312 E7969079_21 0.423 E8139256_21 0.329 E8161577_22 1.060 E8161580_21 0.157 E8364473_21 0.227 E8364474_21 0.069 L1065940_14 0.256 L1065946_10 0.184 L1066018_25 0.282 L1069260_15 1.094 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] zoo timeseries plot; minor tic mark level control
I cannot figure out how to tune the minor tic marks on the date axis of a zoo plot. I read hundreds of CSV files from a zip archive transparently. The time/date strings I convert to POSIXct format, order them and then make a zoo object as there may be cases which have unequal time stamping. As follows: ## #Transform timestamps and reorder them dat - transform(dat, Timestamp = as.POSIXct(Timestamp, format = %m/%d/%Y %H:%M:%S, tz =Europe/Paris)) dat-dat[order(dat$Timestamp),] #Make a zoo object zdt - zoo(dat[,2:4], dat$Timestamp) #Plot plot.zoo(zdt,main=mttl,xlab=NULL,panel=panel.yaxis, yaxt='n',type = 'l',cex=0.2) # The plot is easy. For this data the timestamping is every 5 seconds over many weeks (2 to 3 months). The standard tic mark labeling I get with the plot.zoo() is just the abbreviated name of the month (Oct and Nov). How can I add minor tic marks for the individual days? All and any help or pointers welcome. Alex van der Spek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] zoo timeseries plot; minor tic mark level control
On Mon, Nov 26, 2012 at 6:36 AM, Alex van der Spek do...@xs4all.nl wrote: I cannot figure out how to tune the minor tic marks on the date axis of a zoo plot. I read hundreds of CSV files from a zip archive transparently. The time/date strings I convert to POSIXct format, order them and then make a zoo object as there may be cases which have unequal time stamping. As follows: ## #Transform timestamps and reorder them dat - transform(dat, Timestamp = as.POSIXct(Timestamp, format = %m/%d/%Y %H:%M:%S, tz =Europe/Paris)) dat-dat[order(dat$Timestamp),] #Make a zoo object zdt - zoo(dat[,2:4], dat$Timestamp) #Plot plot.zoo(zdt,main=mttl,xlab=NULL,panel=panel.yaxis, yaxt='n',type = 'l',cex=0.2) # The plot is easy. For this data the timestamping is every 5 seconds over many weeks (2 to 3 months). The standard tic mark labeling I get with the plot.zoo() is just the abbreviated name of the month (Oct and Nov). How can I add minor tic marks for the individual days? All and any help or pointers welcome. Alex van der Spek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. See the examples in ?plot.zoo . Several of those have major and minor tick marks. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at 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] Problem with glm, gaussian family with log-link
Dear all, I am using the book Generalized Linera Models and Extension by Hardin and Hilbe (second edition, 2007) at the moment. The authors suggest that instead of OLS models, the log link is generally used for response data that take only positive values on the continuous scale. Of course they also suggest residual plots to check whether a normal linera model using an identity link can still be used. I am trying to replicate in R what they do in the book in STATA. Indeed, I have no problems in STATA with the log link. However, when calling the same model using R's glm-function, but specifying *family=gaussian(link=log) *I am asked to provide starting values. When I set them all equal to zero, I always get the message that the algorithm did not converge. Picking other values the message is sometimes the same, but more often I get: * * *Error in glm.fit(x = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, : * * NA/NaN/Inf in 'x' * * * As I said, in STATA I can run these models without setting starting values and without errors. I tried many different models, and different datasets, but the problem is always the same (unless I only include one single independent variable). Could anyone tell me why this is the case, or what I do wrong, or why the suggested models from the book might not be appropriate? I'd appreciate any help! Best, Florian [[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] Webinar signup: Advances in Gradient Boosting: the Power of Post-Processing. December 14, 10-11 a.m., PST
Webinar signup: Advances in Gradient Boosting: the Power of Post-Processing December 14, 10-11 a.m., PST Webinar Registration: http://2.salford-systems.com/gradientboosting-and-post-processing/ Course Outline: *Gradient Boosting and Post-Processing: o What is missing from Gradient Boosting? o Why post-processing techniques are used? *Applications Benefiting from Post-Processing: Examples from a variety of industries. o Financial Services o Biomedical o Environmental o Manufacturing o Adserving *Typical Post-Processing Steps *Techniques: o Generalized Path Seeker (GPS): modern high-speed LASSO-style regularized regression. o Importance Sampled Learning Ensembles (ISLE): identify and reweight the most influential trees. o Rulefit: ISLE on steroids. Identify the most influential nodes and rules. *Case Study Example: o Output/Results without Post-Processing o Output/Results with Post-Processing o Demo [[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] Partial dependence plot in randomForest package (all flat responses)
Not unless we have more information. Please read the Posting Guide to see how
to make it easier for people to answer your question.
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Oritteropus
Sent: Thursday, November 22, 2012 2:02 PM
To: r-help@r-project.org
Subject: [R] Partial dependence plot in randomForest package (all flat
responses)
Hi,
I'm trying to make a partial plot with package randomForest in R. After I
perform my random forest object I type
partialPlot(data.rforest, pred.data=act2, x.var=centroid, C)
where data.rforest is my randomforest object, act2 is the original dataset,
centroid is one of the predictor and C is one of the classes in my response
variable.
Whatever predictor or response class I try I always get a plot with a
straight line (a completely flat response). Similarly, If I set a
categorical variable as predictor, I get a barplot with all the bar with the
same height. I suppose I'm doing something wrong here because all other
analysis on the same rforest object seem correct (e.g. varImp or MDSplot).
Is it possible it is related to some option set in random forest object? Can
somebody see the problem here?
Thanks for your time
--
View this message in context:
http://r.789695.n4.nabble.com/Partial-dependence-plot-in-randomForest-package-all-flat-responses-tp4650470.html
Sent from the R help mailing list archive at Nabble.com.
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[R] puzzling RODBC error
Dear all, I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase). I keep getting an error about the channel when using sqlQuery(). However, sqlTables() does not complain about the channel and lists all tables in the database. If I try sqlFetch(), then R crashes. I'm happy to hear suggestions on how to solve this. Best regards, Thierry MDB - odbcConnectAccess(//inbodata/indata/Projects/PRJ_Watervogels/Geoloket/Telgebieden watervogeltellingen/Watervogellocaties_copy.mdb) sqlQuery(channnel = MDB, SELECT gebiedscode FROM ganzengebieden;) Error in sqlQuery(channnel = MDB, SELECT gebiedscode FROM ganzengebieden;) : first argument is not an open RODBC channel sqlTables(channel = MDB) #truncated output! TABLE_SCHEMTABLE_NAME TABLE_TYPE REMARKS 13NAganzengebiedenTABLENA sqlFetch(channel = MDB, 'ganzengebieden') #makes R crash sessionInfo() R version 2.15.2 (2012-10-26) Platform: i386-w64-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Dutch_Belgium.1252 LC_CTYPE=Dutch_Belgium.1252 [3] LC_MONETARY=Dutch_Belgium.1252 LC_NUMERIC=C [5] LC_TIME=Dutch_Belgium.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] RODBC_1.3-6 loaded via a namespace (and not attached): [1] tools_2.15.2 ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] RExcel, ROOo and LibreOffice inquiry
John Kane Kingston ON Canada -Original Message- From: landronim...@gmail.com Sent: Sun, 25 Nov 2012 15:02:16 +0100 To: jrkrid...@inbox.com Subject: Re: [R] RExcel, ROOo and LibreOffice inquiry On Sun, Nov 25, 2012 at 2:38 PM, John Kane jrkrid...@inbox.com wrote: Can you supply a link for ROOo ? I don't see it anywhere. Oh, sorry. I thought it was obvious: http://rcom.univie.ac.at/download.html#ROOo It should have been. I must need new glasses. Also what do you mean OpenOffice is deprecated? Do you mean in terms of using it with ROOo? Otherwide OOo, now under new management at Apache is under active development under the name Apache Open Office. Sun/Oracle discontinued developing OOo and donated it to Apache, which to my surprise made a recent release. However many Linux distros switched to LibreOffice, originally forked from OOo in 2010. To my understanding LibO is the project that should be followed by end-users. I am not a close follow of the AOO/LibreOffice saga but as far as I know there are good points to each and certainly from my reading of the OOo forums there is no particular reason in general to go with one or the other. Each one has its own peculiarities and minor bugs so which to use seems a matter of personal pereference and specific need at any givin point in time. I suspect the move to LibreOffice with many distros was more a pollcy decision when OOo 's fate was in question. Since the Apache move, it looks like AOO is going to be quite viable. I think that LibreOffice offers a bit better compatibility with MS products. I use both on Ubuntu 12.10 quite happily. Since ROOo is beta and rather outdated, I was wondering if there were an effort to bring an RExcel-like functionality to recent LibO or Gnumeric releases. Not the slightest idea about that. Sorry I cannot help. You might want to post a question on one of the OOo forums about it. I did not see ROOo in the AOO extensions bank so I'd be worried about it. Regards Liviu FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing to find out more! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting an adjusted survival curve
First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an adjusted survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized incidence ratio concept found in epidemiology. To answer your technical question: fit - coxph(Surv(. mysurv - survfit(fit, newdata= mydata) This will give a set of predicted curves, one for each observation in mydata. If we assume 2 treatments and 4 ethnicities, this means that there are 8 possible predicted curves. You can certainly take the curves for trt=1, white and trt=2, white, plot them together on one graph, and call this your adjusted survival curves; the mydata data set would have two observations. This is not a correct label but is certainly common. Terry Therneau On 11/26/2012 05:00 AM, r-help-requ...@r-project.org wrote: Dear R-users I am trying to make an adjusted Kaplan-Meier curve (using the Survival package) but I am having difficulty with plotting it so that the plot only shows the curves for the adjusted results. My data come from a randomised controlled trial, and I would like the adjusted Kaplan-Meier curve to only show two curves for the adjusted survival: one for those on treatment (Treatment==1) and another curve for those on placebo (Treatment==0). My problem is that when I plot the survfit of my coxph, I think it displays a curve for every single individual factor in my coxph, whereas I would like it to only display the adjusted curves for when Treatment==1 and Treatment==0. How can I do this? A simplified example of my code with only one effect-modifier is: simple.cox.ethnicity- coxph(Surv(whenfailed,failed) ~ factor(Treatment) + factor(ethnicity)) #I've my data are attached already survfit.simple.cox.ethnicity- survfit(simple.cox.ethnicity,survmat) #survmat is a data.frame that contains Treatment and ethnicity plot(survfit.simple.cox.ethnicity, col=c(red,black), main=survfit.simple.cox, xlab=survival time, ylab=propotion surviving) Thank you so much for your help. Yours gratefully, Brent Caldwell __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] printing difftime summary
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs, 6.98
days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a part
of the data frame summary?
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51 -0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$delay)
r1
string secs
Min.492.00 ms 0.5
1st Qu. 18.08 min 1085.0
Median 1.77 hrs 6370.0
Mean 8.20 hrs 29530.0
3rd Qu. 8.12 hrs 29250.0
Max.6.98 days 602900.0
str(r1)
Classes 'summary.difftime' and 'data.frame': 6 obs. of 2 variables:
$ string: chr 492.00 ms 18.08 min 1.77 hrs 8.20 hrs ...
$ secs :Classes 'summaryDefault', 'table' num [1:6] 4.92e-01
1.08e+03 6.37e+03 2.95e+04 2.92e+04 ...
--8---cut here---end---8---
but not as a part of data frame:
--8---cut here---start-8---
a - summary(infl)
Error in summary.difftime(X[[22L]], ...) :
unused argument(s) (maxsum = 7, digits = 12)
--8---cut here---end---8---
I guess I should somehow accept a list of options in
summary.difftime()
and pass them on to the inner call to summary() (or should it be
explicitly summary.numeric()?)
In the usual way. If you know that the function will be called with
arguments from the summary.data.frame function then you should allow the
argument list to accept them. You can ignore them or provide provisions
for them. You just can't define your function to have only one argument
if you expect (as you should since you passes summary a dataframe
object) that it might be called within summary.data.frame.
This is the argument list for summary.data.frame:
summary.data.frame
function (object, maxsum = 7, digits = max(3, getOption(digits) -
3), ...)
how do I do that?
summary.difftime - function (v, ... ) {
There are many asked and answered questions on rhelp about how to deal
with the dots arguments.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://www.memritv.org http://memri.org
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People with a good taste are especially appreciated by cannibals.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] puzzling RODBC error
On Nov 26, 2012, at 7:37 AM, ONKELINX, Thierry thierry.onkel...@inbo.be wrote: Dear all, I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase). I keep getting an error about the channel when using sqlQuery(). However, sqlTables() does not complain about the channel and lists all tables in the database. If I try sqlFetch(), then R crashes. I'm happy to hear suggestions on how to solve this. Best regards, Thierry MDB - odbcConnectAccess(//inbodata/indata/Projects/PRJ_Watervogels/Geoloket/Telgebieden watervogeltellingen/Watervogellocaties_copy.mdb) sqlQuery(channnel = MDB, SELECT gebiedscode FROM ganzengebieden;) Error in sqlQuery(channnel = MDB, SELECT gebiedscode FROM ganzengebieden;) : first argument is not an open RODBC channel sqlTables(channel = MDB) #truncated output! TABLE_SCHEMTABLE_NAME TABLE_TYPE REMARKS 13NAganzengebiedenTABLENA sqlFetch(channel = MDB, 'ganzengebieden') #makes R crash sessionInfo() R version 2.15.2 (2012-10-26) Platform: i386-w64-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Dutch_Belgium.1252 LC_CTYPE=Dutch_Belgium.1252 [3] LC_MONETARY=Dutch_Belgium.1252 LC_NUMERIC=C [5] LC_TIME=Dutch_Belgium.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] RODBC_1.3-6 loaded via a namespace (and not attached): [1] tools_2.15.2 Some comments: 1. DB related posts should go to R-SIG-DB: https://stat.ethz.ch/mailman/listinfo/r-sig-db 2. You might try to use odbcConnectAccess2007() to see if that provides a resolution. 3. You might also be sure that you don't have a 32/64 bit mis-match between the ODBC drivers, the DSN configuration and Access. I note that you are running 32 bit R on Windows, so perhaps you have already looked into this. There is some additional info on this in the RODBC vignette and some posts in the archives suggesting that you might get other errors, but worth considering if you have not. The crash suggests that something is amiss in the configuration. 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.
[R] creation of an high frequency series
(from R-help Digest, Vol 117, Issue 26) Date: Sun, 25 Nov 2012 07:49:05 -0800 (PST) From: billycorgcandi...@gmail.com To:r-help@r-project.org Subject: [R] creation of an high frequency series Message-ID:1353858545067-4650744.p...@n4.nabble.com Content-Type: text/plain; charset=us-ascii Hi R Users! I would like to create an high frequency series but I am experiencing some difficulties. My series should start at 09.30 a.m. each day and end at 16.00 for, let's say, 2 years. I don't care on how many observations are for each day. It's ok also one observation each minute. In this case, I would have 390 observations each day. I have tried the following: start - ISOdatetime(year=2001, month=1, day=1, hour=9, min=0, sec=0, tz=GMT) end - ISOdatetime(year=2001, month=1, day=1, hour=16, min=0, sec=0, tz=GMT) z - zooreg(NA, start = start, end = end, frequency=390) For this problem z - zooreg(rep(NA,390), start = start, end = end, frequency=390) works, however, if your real problem is that you have time stamps, possibly not equally spaced, then you should consider zoo() with the order.by argument, for example: z - zoo(rep(NA,390), order.by = start + 30 * 0:389) If your data source actually provides the time stamps then this should be very easy. Note that neither of these do consistency checking between the length of the series and the specified time frame, that is up to you to do. You might get warning messages later indicating that something is not right if you made a mistake. Your zooreg() above seems to just ignore the inconsistency whereas I think zoo() with order.by recycles the data. Paul but the result is a poor, single NA. And I am not considering multiple days.. How could I solve this problem?? Thank you so much!! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Connect R with SQLSERVER
Hi, Here im not able to connect with MS-SQLSERVER database with *.R-File. Previously i was able to do in R in different machine and configuration was R Ver-2.11.1 Package Installed : RODBC_1.3-2 And now where im doinig now there the same code im trying to execute but its not connecting. configuration is, R Ver-2.12 Package Installed : RODBC_1.3-2 - here iam trying to install (RODBC_1.3-6) which will support for R 2.12. But getting error while do installation. ERROR I GETTING== C:\Documents and Settings\mduserR CMD INSTALL C:\Program Files\R\R-2.12.0\bin\ RODBC_1.3-6.tar.gz * installing to library 'C:/PROGRA~1/R/R-212~1.0/library' * installing *source* package 'RODBC' ... ** libs gcc -IC:/PROGRA~1/R/R-212~1.0/include -I.-O3 -Wall -std=gnu99 -c RODB C.c -o RODBC.o gcc: not found make: *** [RODBC.o] Error 127 ERROR: compilation failed for package 'RODBC' * removing 'C:/PROGRA~1/R/R-212~1.0/library/RODBC' == - Could anyone please help me ? Thanks Antony -- View this message in context: http://r.789695.n4.nabble.com/Connect-R-with-SQLSERVER-tp4650822.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] Import/Export excel files to/from R, without changing the file type
You should read your error messages more carefully, especially: could not find function! take a look at loadworkbook and loadWorkbook. Bart -- View this message in context: http://r.789695.n4.nabble.com/Import-Export-excel-files-to-from-R-without-changing-the-file-type-tp4650717p4650823.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to count the number of different elements in a column
Hello, Suppose that i have a dataframe a - read.dta(banca_impresa.dta) i have a column with 17900 obs like 1 2 3 1 6 7 8 3 4 4 and i want to know the number of the different values so in this case it would be 7 How can i do? Thank you -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825.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] Class Not found execption
I am trying to call a r file named es.r which have lotes of R functions. These R functions are internally calling java functions by using .jnew() and .jcall(). I have added necessary jar's to the Classpath and I am able to run es.r from command prompt . But when I tried to call it from anoter r file using source() getting Exception Error in .jnew(com/algoTree/ClientElasticSearch/ElasticSearchLoader) : java.lang.ClassNotFoundException How can I solve this ? How can i execute the file from r-console. ? -- View this message in context: http://r.789695.n4.nabble.com/Class-Not-found-execption-tp4650824.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
Goodmorning, I'moneafazrtrbhoquhasa variablefactorcomtwoNivesCandH.Queoaatravésdlinear regressionrelationshipetrehaSaerifthe variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat showsthe two curvesHandCLIRand Selecionar tudo Thank you Ana C. Rocha Rua __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] merging two specific rows in a DF
Hello members, I have this data frame with 3 columns, C1 C2 TYPE 10 20 A 33 44 B 66 80 A 111 140 B 144 220 B 300 340 A 380 449 A 455 500 B 510 520 A 540 580 B Here, the rows 4 , 5 has type B and similarly 6,7 has A . I need to merge these rows in a way to get the output with unique type, something like below, where the lowest value from DF$C1 and highest value from DF$C2 corresponding to rows 4,5 are picked. C1 C2 TYPE 10 20 A 33 44 B 66 80 A 111 220 B 300 449 A 455 500 B 510 520 A 540 580 B I Request your kind help.. Regards, karthick -- View this message in context: http://r.789695.n4.nabble.com/merging-two-specific-rows-in-a-DF-tp4650826.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 in splitting the records
Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-splitting-the-records-tp4650827.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] some help
Hey again! I finally, after some work done before, had time to apply the code. The sorting of the table did not work well or maybe something was misunderstood. I have a table with 973 rows and 1329 col (ascii/text file). I want to sort the table that all columns are one under each other so that at the end I have 973*1329 rows and 1 col. The col should be sorted in a way that col 2 is under col 1, col 3 under col 2, col 4 under col 3 etc. I applied this code: dat - read.table(filename, sep=separator, header=TRUE) stacked - do.call(rbind, dat) unlist(dat) ..but putting dim(dat), the number of rows and col was still 973 and 1329. So seemingly it did not work as i wanted. Thanks very much for more help. Stefan -- View this message in context: http://r.789695.n4.nabble.com/some-help-tp4648316p4650828.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] How to count the number of different elements in a column
this is wrong because with the command unique it counts the only values that are unique .. in my column, for instance, 1 and 4 are not unique so the formula doesn't work in my case -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825p4650843.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] How to count the number of different elements in a column
length(unique(a)) Hard Core wrote Hello, Suppose that i have a dataframe a - read.dta(banca_impresa.dta) i have a column with 17900 obs like 1 2 3 1 6 7 8 3 4 4 and i want to know the number of the different values so in this case it would be 7 How can i do? Thank you -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825p4650833.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] Performing operations only on selected data
HI Marcel, Sorry, that was a mistake. I guess this should be the one: df$condition2-ifelse(df$condition1=1,runif(length(df$condition1),0,1),df$condition1) df$condition2 #[1] 0.5207776 0.5227388 0.5196520 0.6552820 2.000 2.000 2.000 #[8] 2.000 3.000 3.000 3.000 3.000 4.000 4.000 #[15] 4.000 4.000 A.K. - Original Message - From: Jeff Newmiller jdnew...@dcn.davis.ca.us To: Marcel Curlin cemar...@u.washington.edu; r-help@r-project.org Cc: Sent: Monday, November 26, 2012 1:23 AM Subject: Re: [R] Performing operations only on selected data a) Please read the posting guide. This mailing list is not Nabble, and you are requested to post in plain text and include context from previous messages in the thread. b) arun's solution is wrong in two respects: it fails to add condition1 to the random numbers, and it feeds differently-sized vectors as values to the ifelse function. The length of each argument to ifelse should be as long as the desired result. BTW: df is a function in the base package, always present. It is not a good idea to use that name for your own purposes, as eventually you may want to use that base function or at least not confuse others. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Marcel Curlin cemar...@u.washington.edu wrote: Thank you, this works very well. My only remaining question about this is about how ifelse is working; I understand the basic syntax (df$condition2 gets assigned the value *runif(nrow(df1[df1$condition1=1,]),0,1)* or the value *df$condition1* depending on whether or not df$condition1 meets the criterion =1. As I understand it, runif(nrow(df1[df1$condition1=1,]),0,1) is a vector of random values with vector length equal to the number of rows meeting df$condition1=1 and df$condition1 is just my column of condition1 values. So the command seems to be going down row by row and assigning condition2 values from one of two vectors in an interleaved way. So my question is, how does R keep track of which item in each of the vectors to assign to condition2? For example, if the first 4 entries of condition1 are 1, 3, 4, 1, how does R know to use the *first* entry of vector runif(nrow(df1[df1$condition1=1,]),0,1) then the *second* and *third* values of vector df$condition1, then the *second* value of vector runif(nrow(df1[df1$condition1=1,]),0,1)? -- View this message in context: http://r.789695.n4.nabble.com/Performing-operations-only-on-selected-data-tp4650646p4650803.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 with predict function in glm
That did it! Thanks so much as always. I emailed the question to you because I think you are an R expert based on all the suggestions, feedback and codes I have received from you in the past. Yes, I do look for answers in the open forum but when it comes to a question for which the answers out there are not very clear I prefer to ask somebody whom I can trust. I hope I donât bother you with my questions. Thanks once again. Cheers! -Som. Date: Mon, 26 Nov 2012 03:33:43 -0800 From: ml-node+s789695n4650829...@n4.nabble.com To: genome1...@hotmail.com Subject: Re: Help with predict function in glm Hello, Why mail a question just to me? Post to the list and the odds of getting more answers (and better) are bigger. As for your question, the problem is in the call to glm, you don't need the prefix 'train$' in the formula, the argument 'data' solves that and when predicting R will look for the columns with names in the formula and is unable to find columns called train$Outcome and train$Weight in the new data.frame 'test'. Corrected: mylogit - glm(Outcome ~ Weight, data=train, family = binomial(logit)) predictions - predict(mylogit, newdata = test, type= response) Hope this helps, Rui Barradas Em 26-11-2012 01:42, somnath bandyopadhyay escreveu: Hi, I am trying some basic logistic regression analysis using glm. I just have one dependent variable (Outcome) which is binary in nature and one independent variable (Weight). I fit a model using a training data set (train) which has 85 observations and try to apply it on an independent dataset (test) which has 55 observations. When I apply the predict function on the fitted model for the new dataset, I get the following warning Warning message: 'newdata' had 55 rows but variable(s) found have 85 rows and the predict works on the training observations and not on the test observations. Following is he session info, code and the training and test datasets I am using. What am I doing wrong? Any help would be greatly appreciated. Thanks, S. train - read.table(train_data.txt, header=T, row.names=1, sep=\t) test- read.table(test_data.txt, header=T, row.names=1, sep=\t) mylogit - glm(train$Outcome ~ train$Weight, data=train, family = binomial(logit)) predictions - predict(mylogit, newdata = test, type= response) Warning message: 'newdata' had 55 rows but variable(s) found have 85 rows sessionInfo() R version 2.15.0 (2012-03-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base train Outcome Weight AB256939_21 0 0.331 AB257076_21 0 0.308 AB257079_21 0 0.453 AB415508_21 0 0.303 AB700497_21 0 0.354 AB904508_21 0 0.336 AC048719_21 0 0.420 AC185939_21 0 0.249 AC185940_21 0 1.525 AC445840_21 0 0.261 E7490523_21 0 0.269 E7490524_21 0 0.213 E7659579_21 0 0.360 E7661528_21 0 0.271 E7781094_21 0 0.156 E7781095_21 0 0.221 E7781096_21 0 0.098 E7969081_21 0 0.430 E8117594_21 0 0.321 E8133295_21 0 0.166 E8161578_22 0 0.269 E8483037_21 0 0.162 E8559720_21 0 0.226 L1065550_18 0 0.396 L1065607_17 0 0.541 L1065944_24 0 0.131 L1066017_20 0 0.421 L1069261_12 0 0.357 L1069262_14 0 0.309 L1069263_27 0 0.283 L1069297_24 0 0.620 L1081528_21 0 0.561 L1084066_21 0 0.564 L1086090_21 0 0.649 L1104280_17 0 0.181 L362_22 0 0.199 L1118063_15 0 0.369 L1133550_21 0 0.302 L1144201_14 0 0.249 L1155023_7 0 0.257 L1158386_21 0 0.470 L1163051_4 0 0.446 ... ... ... test Weight AB256870_21 0.364 AB256873_21 0.329 AB415518_21 0.219 AB460669_21 0.481 AB609036_21 0.313 AB609038_21 0.196 AB700495_21 0.402 AB700498_21 0.343 AC112834_21 0.372 AC185937_21 0.270 AC269527_21 0.285 E7352023_21 0.358 E7661554_21 0.471 E7750502_21 0.437 E7845183_21 0.232 E7854155_21 0.474 E7854156_21 0.121 E7924877_21 0.312 E7969079_21 0.423 E8139256_21 0.329 E8161577_22 1.060 E8161580_21 0.157 E8364473_21 0.227 E8364474_21 0.069 L1065940_14 0.256 L1065946_10 0.184 L1066018_25 0.282 L1069260_15 1.094 __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to the discussion below:
Re: [R] puzzling RODBC error
It seems that your sqlTables also give no results. So there must be something wrong with the ODBC connect. I didn't use odbcConnectAccess, but made an ODBC connection, and then used ODBCconnect to connect to our database. Maybe you can define a user DSN and try it this way? Bart -- View this message in context: http://r.789695.n4.nabble.com/puzzling-RODBC-error-tp4650837p4650840.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] error in plot(table(c('a','a')))
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ). With more than one value I don't get
errors, neither with integers (even if only 1 value):
tbl.char1 - table(c('a','a'))
tbl.char2 - table(c('a','a','b'))
tbl.int1 - table(c(1,1))
# error:
plot(tbl.char1)
# no errors:
plot(tbl.char2)
plot(tbl.int1)
sessionInfo()
R version 2.14.0 (2011-10-31)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
Thanks, Ludo
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[R] cluster analysis error - mclust package
I am following instructions online for cluster analysis using the mclust package, and keep getting errors. http://www.statmethods.net/advstats/cluster.html These are the instructions (there is no sample dataset unfortunately): # Model Based Clustering library(mclust) fit - Mclust(mydata) plot(fit, mydata) # plot results print(fit) # display the best model This is what I did and the error I get: library(mclust) fit - Mclust(mydat) plot(fit, mydat) #plot results Error in match.arg(what, c(BIC, classification, uncertainty, density), : 'arg' must be NULL or a character vector My data is arranged so I have each row representing one individual with 9 values for morphological data. I want to see if they will group into 2 clusters, representing gender. I have tried using the instructions from the cran-r website, but they didn't work either Any help would be great, thank you -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-error-mclust-package-tp4650842.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] some help
Hey The code need some corrections and I would kindly ask for help. Say: I have a table with two columns: col1=LST and col2=NDVI i would like to sort all data by NDVI. in reality the NDVI ranges between 0 and 1 (although some values might be minus also). I want to sort by NDVI values and then make 100 intervals of 0.01 ndvi. therefore the first inerval is 0.01-0.02, the second 0.02-0.03 and so on.. for each interval I would like to get the highest (max) and lowest (min) value of LST. It would be very helpful if this values can be written in a seperate table/file. Thanks you very much for help! Stefan -- View this message in context: http://r.789695.n4.nabble.com/some-help-tp4648316p4650844.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] How to count the number of different elements in a column
My apologies. I still do not understand the difference; good luck. Hard Core wrote this is wrong because with the command unique it counts the only values that are unique .. in my column, for instance, 1 and 4 are not unique so the formula doesn't work in my case -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825p4650848.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] wrong data interpretation in R
Hi, maybe somebody would be kind enough to help a bloody and unprofessional beginner like me (and I hope I did not overlook the answer to my question on the website). I've imported a csv data frame into R, but I can't run a regression because R interprets 4 out of 5 variables as factors (rather than numeric vectors). I tried the as.numeric() command, but R says it is invalid to change the storage mode of a factor. Is there any way to change the mode of a factor into numeric. I am very grateful for help. Thank you very much, Robert. -- View this message in context: http://r.789695.n4.nabble.com/wrong-data-interpretation-in-R-tp4650849.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 in splitting the records
You state that you want $$$ as the separator, but your example has $$ and $ so I'm assuming an indeterminate number of consecutive $. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) x - do.call(rbind, x) x - data.frame(x, stringsAsFactors = FALSE) x$X2 - as.numeric(x$X2) I've also made some assumptions about what you want the output to be like, since you didn't specify, and broken it into as many steps as possible so that you can look at what functions I used and read the help files. Sarah On Mon, Nov 26, 2012 at 5:51 AM, arunkumar akpbond...@gmail.com wrote: Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me -- 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.
Re: [R] How to count the number of different elements in a column
x - c(1, 2, 3, 1, 6, 7, 8, 3, 4, 4) length(unique(x)) [1] 7 On Mon, Nov 26, 2012 at 5:20 AM, Hard Core gi...@hotmail.it wrote: Hello, Suppose that i have a dataframe a - read.dta(banca_impresa.dta) i have a column with 17900 obs like 1 2 3 1 6 7 8 3 4 4 and i want to know the number of the different values so in this case it would be 7 How can i do? Thank you -- 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.
Re: [R] wrong data interpretation in R
Most likely there's something amiss in your csv file: R won't convert numeric data to factors unless there are non-numeric characters included. First check your csv file for errors. If that doesn't solve your problem, please provide a reproducible example. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Sarah On Mon, Nov 26, 2012 at 10:59 AM, EcoFranc rob...@gmx.de wrote: Hi, maybe somebody would be kind enough to help a bloody and unprofessional beginner like me (and I hope I did not overlook the answer to my question on the website). I've imported a csv data frame into R, but I can't run a regression because R interprets 4 out of 5 variables as factors (rather than numeric vectors). I tried the as.numeric() command, but R says it is invalid to change the storage mode of a factor. Is there any way to change the mode of a factor into numeric. I am very grateful for help. Thank you very much, Robert. -- 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.
Re: [R] How to count the number of different elements in a column
Hi, Imagine the column is named XX. Type: nrow(table(banca_impresa$XX)) and you'll get how many different categories there are in that column. (If you type table(banca_impresa$XX) you'll get the frequencies). José José Iparraguirre Chief Economist Age UK T 020 303 31482 E jose.iparragui...@ageuk.org.uk -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Hard Core Sent: 26 November 2012 10:20 To: r-help@r-project.org Subject: [R] How to count the number of different elements in a column Hello, Suppose that i have a dataframe a - read.dta(banca_impresa.dta) i have a column with 17900 obs like 1 2 3 1 6 7 8 3 4 4 and i want to know the number of the different values so in this case it would be 7 How can i do? Thank you -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825.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. A Star for Christmas Kick start the festive season by attending one of Age UK’s Carol Concerts, A Star for Christmas. Taking place at Manchester Cathedral on Saturday 1 December and London’s St Pancras Church (opposite Euston Station) on Thursday 6 December, they will feature special musical performances, readings by your favourite celebrities and carols, followed by mince pies and wine. Tickets are priced at £20 full price/ £10 concessions. For more information, please visit http://www.ageuk.org.uk/astarforchristmas or contact the Fundraising Events Team on 020 303 31725. Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] printing difftime summary
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13
hrs, 6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
--
David.
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51
-0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$delay)
r1
string secs
Min.492.00 ms 0.5
1st Qu. 18.08 min 1085.0
Median 1.77 hrs 6370.0
Mean 8.20 hrs 29530.0
3rd Qu. 8.12 hrs 29250.0
Max.6.98 days 602900.0
str(r1)
Classes 'summary.difftime' and 'data.frame':6 obs. of 2 variables:
$ string: chr 492.00 ms 18.08 min 1.77 hrs 8.20 hrs ...
$ secs :Classes 'summaryDefault', 'table' num [1:6] 4.92e-01
1.08e+03 6.37e+03 2.95e+04 2.92e+04 ...
--8---cut here---end---8---
but not as a part of data frame:
--8---cut here---start-8---
a - summary(infl)
Error in summary.difftime(X[[22L]], ...) :
unused argument(s) (maxsum = 7, digits = 12)
--8---cut here---end---8---
I guess I should somehow accept a list of options in
summary.difftime()
and pass them on to the inner call to summary() (or should it be
explicitly summary.numeric()?)
In the usual way. If you know that the function will be called with
arguments from the summary.data.frame function then you should
allow the
argument list to accept them. You can ignore them or provide
provisions
for them. You just can't define your function to have only one
argument
if you expect (as you should since you passes summary a dataframe
object) that it might be called within summary.data.frame.
This is the argument list for summary.data.frame:
summary.data.frame
function (object, maxsum = 7, digits = max(3, getOption(digits) -
3), ...)
how do I do that?
summary.difftime - function (v, ... ) {
There are many asked and answered questions on rhelp about how to
deal
with the dots arguments.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
http://www.childpsy.net/ http://www.memritv.org http://memri.org
http://honestreporting.com http://dhimmi.com http://openvotingconsortium.org
People with a good taste are especially appreciated by cannibals.
David Winsemius, MD
Alameda, CA, USA
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Connect R with SQLSERVER
On Nov 26, 2012, at 3:53 AM, R_Antony antony.akk...@ge.com wrote: Hi, Here im not able to connect with MS-SQLSERVER database with *.R-File. Previously i was able to do in R in different machine and configuration was R Ver-2.11.1 Package Installed : RODBC_1.3-2 And now where im doinig now there the same code im trying to execute but its not connecting. configuration is, R Ver-2.12 Package Installed : RODBC_1.3-2 - here iam trying to install (RODBC_1.3-6) which will support for R 2.12. But getting error while do installation. ERROR I GETTING== C:\Documents and Settings\mduserR CMD INSTALL C:\Program Files\R\R-2.12.0\bin\ RODBC_1.3-6.tar.gz * installing to library 'C:/PROGRA~1/R/R-212~1.0/library' * installing *source* package 'RODBC' ... ** libs gcc -IC:/PROGRA~1/R/R-212~1.0/include -I.-O3 -Wall -std=gnu99 -c RODB C.c -o RODBC.o gcc: not found make: *** [RODBC.o] Error 127 ERROR: compilation failed for package 'RODBC' * removing 'C:/PROGRA~1/R/R-212~1.0/library/RODBC' == - Could anyone please help me ? Thanks Antony First, DB related posts should go to R-SIG-DB: https://stat.ethz.ch/mailman/listinfo/r-sig-db Please use that in the future, not R-Help. Second, why are you trying to install the *source* version of RODBC when there are pre-compiled binary versions of the package available for Windows? Just use: install.packages(RODBC) within an R session. Finally, R version 2.12.0 is now over two years old. 2.15.2 is the current version, so you should also be updating your R installation. 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] How to count the number of different elements in a column
Hello, You're wrong, unique doesn't count unique returns a vector without repetitions. unique(a) # includes 1 and 4 [1] 1 2 3 6 7 8 4 length(unique(a)) [1] 7 Hope this helps, Rui Barradas Em 26-11-2012 14:55, Hard Core escreveu: this is wrong because with the command unique it counts the only values that are unique .. in my column, for instance, 1 and 4 are not unique so the formula doesn't work in my case -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825p4650843.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] wrong data interpretation in R
Robert -- it may be that some of your factor variables cannot be coerced to numeric because they have non-numeric elements. This is why we ask for a reproducible example when posting. I suspect just a few lines from the data frame would tell the story. On 26-Nov-12, at 7:59 AM, EcoFranc wrote: Hi, maybe somebody would be kind enough to help a bloody and unprofessional beginner like me (and I hope I did not overlook the answer to my question on the website). I've imported a csv data frame into R, but I can't run a regression because R interprets 4 out of 5 variables as factors (rather than numeric vectors). I tried the as.numeric() command, but R says it is invalid to change the storage mode of a factor. Is there any way to change the mode of a factor into numeric. I am very grateful for help. Thank you very much, Robert. -- View this message in context: http://r.789695.n4.nabble.com/wrong- data-interpretation-in-R-tp4650849.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. Don McKenzie, Research Ecologist Pacific Wildland Fire Sciences Lab US Forest Service phone: 206-732-7824 Affiliate Professor School of Environmental and Forest Sciences University of Washington __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Fine Tuning Country Map
Lorenzo,
I'd suggest posting such questions to the R-sig-geo list, which seems
more suitable.
The book Applied Spatial Data Analysis with R written by Roger
Bivand, Edzer Pebesma and Virgilio Gómez-Rubio has exactly what you want.
http://www.amazon.com/Applied-Spatial-Data-Analysis-Use/dp/0387781706
Package classInt can help with colouring;
and about the legend: The location may also be specified by setting x
to a single keyword from the list bottomright, bottom, bottomleft,
left, topleft, top, topright, right and center. This places
the legend on the inside of the plot frame at the given location.
Partial argument matching is used. The optional inset argument specifies
how far the legend is inset from the plot margins. If a single value is
given, it is used for both margins; if two values are given, the first
is used for x- distance, the second for y-distance.
Good luck!
Marcio
www.dsr.inpe.br/~mello
On 8/13/12 11:33 PM, Lorenzo Isella wrote:
Dear All,
Please see the short script at the end of the email, which I assembled
looking for bits and pieces on the web.
It essentially does what I need: it plots several countries as a
color-coded map.
I just would like to fine-tune a bit the final image, in particular
(1) Select my own colors for painting the countries (i.e. associate
manually a color to every level)
(2) Be able to control the position of the legend and the size of the
character used in the legend itself.
Any suggestion is welcome.
Cheers
Lorenzo
à
## you will need the sp-package
library('sp')
## load a file from GADM (you just have to specify the countries
special part of the file name, like ARG for Argentina. Optionally
you can specify which level you want to have
loadGADM - function (fileName, level = 0, ...) {
load(url(paste(http://gadm.org/data/rda/;, fileName, _adm,
level, .RData, sep = )))
gadm
}
## the maps objects get a prefix (like ARG_ for Argentina)
changeGADMPrefix - function (GADM, prefix) {
GADM - spChFIDs(GADM, paste(prefix, row.names(GADM), sep = _))
GADM
}
## load file and change prefix
loadChangePrefix - function (fileName, level = 0, ...) {
theFile - loadGADM(fileName, level)
theFile - changeGADMPrefix(theFile, fileName)
theFile
}
## this function creates a SpatialPolygonsDataFrame that contains all
maps you specify in fileNames.
## E.g.:
## spdf - getCountries(c(ARG,BOL,CHL))
## plot(spdf) # should draw a map with Brasil, Argentina and Chile on it.
getCountries - function (fileNames, level = 0, ...) {
polygon - sapply(fileNames, loadChangePrefix, level)
polyMap - do.call(rbind, polygon)
polyMap
}
spdf - getCountries(c(ITA,CHE,FRA, DEU,BEL, LUX))
AP - c(SLS,SLS, NWS, NSLS, NSLS, NWS)
spdf$AP - as.factor(AP)
png(many-countries.png)
## print(spplot(spdf, NAME_ENGLI))
print(spplot(spdf, AP))
## plot(spdf)
dev.off()
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] wrong data interpretation in R
Also when reading a CSV file, if you do not want characters columns converted to factors, use 'as.is = TRUE' as one of the parameters. If you have a column that is a factor, to convert it to numeric you have to do the following: as.numeric(as.character(factorColumn)) notice that you have to convert it to character first; otherwise you will get the numeric value of the factor which is probably not what you want. On Mon, Nov 26, 2012 at 12:02 PM, Sarah Goslee sarah.gos...@gmail.com wrote: Most likely there's something amiss in your csv file: R won't convert numeric data to factors unless there are non-numeric characters included. First check your csv file for errors. If that doesn't solve your problem, please provide a reproducible example. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Sarah On Mon, Nov 26, 2012 at 10:59 AM, EcoFranc rob...@gmx.de wrote: Hi, maybe somebody would be kind enough to help a bloody and unprofessional beginner like me (and I hope I did not overlook the answer to my question on the website). I've imported a csv data frame into R, but I can't run a regression because R interprets 4 out of 5 variables as factors (rather than numeric vectors). I tried the as.numeric() command, but R says it is invalid to change the storage mode of a factor. Is there any way to change the mode of a factor into numeric. I am very grateful for help. Thank you very much, Robert. -- 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. -- 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] wrong data interpretation in R
EcoFranc, di you remember to use the argument header=TRUE ? If not, the variable name is interpreted as data and will flip you into factor rather than numeric. On Mon, Nov 26, 2012 at 12:15 PM, jim holtman jholt...@gmail.com wrote: Also when reading a CSV file, if you do not want characters columns converted to factors, use 'as.is = TRUE' as one of the parameters. If you have a column that is a factor, to convert it to numeric you have to do the following: as.numeric(as.character(factorColumn)) notice that you have to convert it to character first; otherwise you will get the numeric value of the factor which is probably not what you want. On Mon, Nov 26, 2012 at 12:02 PM, Sarah Goslee sarah.gos...@gmail.com wrote: Most likely there's something amiss in your csv file: R won't convert numeric data to factors unless there are non-numeric characters included. First check your csv file for errors. If that doesn't solve your problem, please provide a reproducible example. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Sarah On Mon, Nov 26, 2012 at 10:59 AM, EcoFranc rob...@gmx.de wrote: Hi, maybe somebody would be kind enough to help a bloody and unprofessional beginner like me (and I hope I did not overlook the answer to my question on the website). I've imported a csv data frame into R, but I can't run a regression because R interprets 4 out of 5 variables as factors (rather than numeric vectors). I tried the as.numeric() command, but R says it is invalid to change the storage mode of a factor. Is there any way to change the mode of a factor into numeric. I am very grateful for help. Thank you very much, Robert. -- 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. -- 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. [[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] merging two specific rows in a DF
Hello,
Try the following.
x - read.table(text=
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
, header = TRUE)
x
fun - function(x){
mn - which.min(x$C1)
mx - which.max(x$C2)
c(C1 = x$C1[mn], C2 = x$C2[mx], TYPE = x$TYPE[1])
}
idx - seq_len(nrow(x))[-1]
idx2 - cumsum(c(FALSE, x$TYPE[idx - 1] != x$TYPE[idx]))
y - do.call(rbind, lapply(split(x, idx2), fun))
rownames(y) - seq_len(nrow(y))
y
Hope this helps,
Rui Barradas
Em 26-11-2012 10:24, karthicklakshman escreveu:
Hello members,
I have this data frame with 3 columns,
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
Here, the rows 4 , 5 has type B and similarly 6,7 has A . I need to
merge these rows in a way to get the output with unique type, something like
below, where the lowest value from DF$C1 and highest value from DF$C2
corresponding to rows 4,5 are picked.
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 220 B
300 449 A
455 500 B
510 520 A
540 580 B
I Request your kind help..
Regards,
karthick
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging two specific rows in a DF
try this:
x - read.table(text = C1 C2 TYPE
+ 10 20 A
+ 33 44 B
+ 66 80 A
+ 111 140 B
+ 144 220 B
+ 300 340 A
+ 380 449 A
+ 455 500 B
+ 510 520 A
+ 540 580 B, header = TRUE, as.is = TRUE)
# mark successive rows that are different
x$diff - c(TRUE, head(x$TYPE, -1) != tail(x$TYPE, -1))
# create groups where adjacent rows are the same
x$group - cumsum(x$diff)
# now process each group to get min/max
result - lapply(split(x, x$group), function(.same){
+ c1Min - min(.same$C1)
+ c2Max - max(.same$C2)
+ # put result back into first row and just return its value
+ .same$C1[1L] - c1Min
+ .same$C2[1L] - c2Max
+ .same[1,1:3] # return value drop off the extra columns we added
+ })
# combine back together
do.call(rbind, result)
C1 C2 TYPE
1 10 20A
2 33 44B
3 66 80A
4 111 220B
5 300 449A
6 455 500B
7 510 520A
8 540 580B
On Mon, Nov 26, 2012 at 5:24 AM, karthicklakshman
karthick.laksh...@gmail.com wrote:
Hello members,
I have this data frame with 3 columns,
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
Here, the rows 4 , 5 has type B and similarly 6,7 has A . I need to
merge these rows in a way to get the output with unique type, something like
below, where the lowest value from DF$C1 and highest value from DF$C2
corresponding to rows 4,5 are picked.
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 220 B
300 449 A
455 500 B
510 520 A
540 580 B
I Request your kind help..
Regards,
karthick
--
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http://r.789695.n4.nabble.com/merging-two-specific-rows-in-a-DF-tp4650826.html
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__
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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.
--
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.
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and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: prediction problem
-- Forwarded message -- From: Catarina Maia catarinaramosm...@gmail.com Date: 2012/11/22 Subject: prediction problem To: r-help@r-project.org Hello, I am using the mda package and in particular the fda routine to classify/predict in terms of color to a set of 20 samples for which i don´t know the color. I preformed a flexible discriminant analysis (FDA) using a set of 147 samples for which i know all the information. My script and data follow in attachment. A total of 23 predictors were considered. 20 of the predictors are numeric and 3 are discrete/categorical. The resulting FDA rule was applied to the matrix and for the same predictors in order to predict the color. However a consistent error is occuring: Error in mindist[l] - ndist[l] : **NAs are not allowed in subscripted assignments It is possible the problem being related with the 2 predictors that are discrete/categorical variables? There is already available some rotine to perform a discriminant analysis considering continuous and discrete/categorical variables? If someone can help, would be very grateful. cheers, Catarina n Codecolor GROUP_N Licenca_N type1_W type2_W type3_W type4_W type5_W type6_W type7_W type8_W type9_W type1_V type2_V type3_V type4_V type5_V type7_V type8_V type9_V type10_Vtype11_Vtotal 1 PRT0026940473 verde A8 B7 0,558416119 0 0 0 0,152231706 0,243598713 0 0 0,005176997 0,483991231 0 0 0 0,375554726 0,092127884 0 0 0 0,011391047 714,7 2 PRT0026940480 verde A8 B7 0,431011386 0 0 0 0,090087073 0,389819156 0 0 0 0,27662459 0 0 0 0,329209955 0,254047865 0 0 0 0 597,2 3 PRT0026940500 verde A8 B7 0,349034749 0 0,013127413 0 0,362162162 0,197297297 0 0 0 0,166132511 0 0,005179074 0 0,647152728 0,095422917 0 0 0 0 259 4 PRT0026940507 verde A8 B7 0,38562904 0 0 0 0,235952263 0,245400298 0 0 0 0,195929366 0 0 0 0,561989935 0,149391253 0 0 0 0 402,2 6 PRT0042540249 verde A9 B7 0,160142349 0 0 0,0113879 0 0,098932384 0 0 0 0,071878686 0 0 0,017455771 0 0,127177759 0 0 0 0 140,5 7 PRT0042540358 verde A9 B7 0,258287293 0 0 0 0,04558011 0,165745856 0 0 0 0,308268994 0 0 0 0,147252993 0,073850285 0 0 0 0 72,4 9 PRT0042540498 verde A9 B7 0,535232384 0 0 0 0,230384808 0,143428286 0 0 0 0,220279346 0 0 0 0,50482587 0,113254862 0,089373366 0 0 0 200,1 10 PRT0042540507 verde A9 B7 0,631288766 0 0 0,029634735 0,188146106 0 0 0 0 0,187408732 0 0 0,033825063 0,637267173 0 0 0 0 0 145,1 11 PRT0042540527 verde A9 B7 0,775214835 0 0 0,014925373 0,03256445 0,054726368 0 0 0 0,610476668 0 0 0,031623122 0,225502758 0,110096002 0 0 0 0 221,1 12 PRT0112140260 amarelo A7 B5 0 0 0,009701493 0,664925373 0 0,017164179 0,042537313 0,023880597 0 0 0 0,011007463 0,617276119 0 0,004626866 0 0,120279851 0,012649254 0 134 13 PRT0112140305 azulA7 B5 0 0 0 0 0 0 0,095049505 0 0 0 0 0 0 0 0 0,080972426 0,567469139 0 0 101 14 PRT0112140366 azulA7 B5 0 0 0,014857143 0,102857143 0,040,014857143 0,011428571 0,217142857 0 0 0 0,005893035 0,125398294 0,16692363 0,013019495 0 0,024394422 0,152328091 0 87,5 15 PRT0112140373 azulA7 B5 0 0 0,019488429 0,052375152 0 0,017052375 0,052375152 0 0 0 0 0,020974032 0,041306002 0 0,007062684 0 0,2152573 0 0 82,1 16 PRT0112140382 azulA7 B5 0
Re: [R] some help
try this: (provide sample data next time) # create some test data x - data.frame(LST = runif(1000), NDVI = runif(1000, 0, 1)) head(x,10) # show some data LST NDVI 1 0.86542839 0.95129647 2 0.88910058 0.75971649 3 0.44086718 0.86532140 4 0.99879370 0.05511501 5 0.02401490 0.92282834 6 0.56026534 0.80915721 7 0.65051596 0.03606430 8 0.25897388 0.61624609 9 0.07873261 0.85179368 10 0.09829056 0.91198307 # create partition values partition - seq(0, 1, .01) # add column to the data to define the partition x$part - cut(x$NDVI, partition) head(x) LST NDVIpart 1 0.8654284 0.95129647 (0.95,0.96] 2 0.8891006 0.75971649 (0.75,0.76] 3 0.4408672 0.86532140 (0.86,0.87] 4 0.9987937 0.05511501 (0.05,0.06] 5 0.0240149 0.92282834 (0.92,0.93] 6 0.5602653 0.80915721 (0.8,0.81] # now compute range (min/mx) for the data xRange - tapply(x$LST, x$part, range) head(xRange, 10) $`(0,0.01]` [1] 0.01945995 0.83500402 $`(0.01,0.02]` [1] 0.02267906 0.69770971 $`(0.02,0.03]` [1] 0.01287795 0.75275416 $`(0.03,0.04]` [1] 0.1402162 0.9960408 $`(0.04,0.05]` [1] 0.007688249 0.691519845 $`(0.05,0.06]` [1] 0.1047314 0.9987937 $`(0.06,0.07]` [1] 0.002181767 0.990990999 $`(0.07,0.08]` [1] 0.08271319 0.87609409 $`(0.08,0.09]` [1] 0.1174585 0.8931750 $`(0.09,0.1]` [1] 0.2331289 0.8485212 On Mon, Nov 26, 2012 at 9:56 AM, dattel_palme dattel_pa...@yahoo.de wrote: Hey The code need some corrections and I would kindly ask for help. Say: I have a table with two columns: col1=LST and col2=NDVI i would like to sort all data by NDVI. in reality the NDVI ranges between 0 and 1 (although some values might be minus also). I want to sort by NDVI values and then make 100 intervals of 0.01 ndvi. therefore the first inerval is 0.01-0.02, the second 0.02-0.03 and so on.. for each interval I would like to get the highest (max) and lowest (min) value of LST. It would be very helpful if this values can be written in a seperate table/file. Thanks you very much for help! Stefan -- View this message in context: http://r.789695.n4.nabble.com/some-help-tp4648316p4650844.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. -- 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] Help in splitting the records
Hi, You could do this: Lines-qwer$$12$$qwre ewrtr$7789$ewwe res-unlist(strsplit(Lines,split=\\$|\n)) as.data.frame(matrix(res[res!=],nrow=2,byrow=TRUE),stringsAsFactors=FALSE) V1 V2 V3 #1 qwer 12 qwre #2 ewrtr 7789 ewwe A.K. - Original Message - From: arunkumar akpbond...@gmail.com To: r-help@r-project.org Cc: Sent: Monday, November 26, 2012 5:51 AM Subject: [R] Help in splitting the records Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2 V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-splitting-the-records-tp4650827.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] DF grouping
Hello Petr Savicky, hello all, I have a situation similar to the previous one, I need to group a data.frame in a specific way, col1 col2 score 2873 3192 319 4268 4451 183 5389 5534 145 6622 10622 4000 12631 17853 5222 20408 20615 207 21595 21838 243 23121 2313918 the out put should be like, [1] 2873 3192 4268 4451 5389 5534 6622 10622 [2] 10622 12631 [3] 17853 20408 20615 21595 21838 23121 23139 Basically the split should be based on the DF$score 500, and all the col1 and col2 values should be aggregated. but DF[i, col2] != DF[i+1, col1] Your inputs will be very helpful Thank you Karthick -- View this message in context: http://r.789695.n4.nabble.com/DF-grouping-tp4381310p4650865.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in splitting the records
In the example below, I don't see any $$$ separators. Are you sure this is supposed to be the separate or just a single dollar sign? If this is the case, you don't specify what is to happen when multiple separators appear next to each other. From your example, it appears that two separators are to be treated as one? From: arunkumar akpbond...@gmail.com To: r-help@r-project.org Sent: Monday, November 26, 2012 5:51 AM Subject: [R] Help in splitting the records Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2 V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-splitting-the-records-tp4650827.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.
[R] cacheSweave problem
I noticed the following problem with cacheSweave. If I want to print the result of a list object, with cache=TRUE option, if I just use summary(x), the output would not appear in the tex file. If I use print(summary(x)) instead, the output would appear. With cache=FALSE option, however, summary(x) and print(summary(x)) both have the output. Is this a bug for cacheSweave package? %This chunk would not have output eval=T,echo=T,results=verbatim,cache=TRUE= x=NULL x[[1]]=rnorm(100) x[[2]]=rnorm(100) x[[3]]=rnorm(100) summary(x) @ %This chunk would have output eval=T,echo=T,results=verbatim,cache=TRUE= print(summary(x)) @ %This chunk would also have output eval=T,echo=T,results=verbatim,cache=FALSE= summary(x) @ -- View this message in context: http://r.789695.n4.nabble.com/cacheSweave-problem-tp4650868.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] cosine curve fit
does anybody have a suggestion as to how to use R to fit some date to a cosine function and then have some output statistics that will evaluate the fit? -- View this message in context: http://r.789695.n4.nabble.com/cosine-curve-fit-tp4650866.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] wavelet Variance Error
The message is referring to the format of your input data. The package needs a matrix or vector format. I suggest using this: xm- as.matrix(x) return.modwt-modwt(xm, filter=la8, n.levels=5, boundary=periodic, fast=TRUE) -- View this message in context: http://r.789695.n4.nabble.com/wavelet-Variance-Error-tp4301149p4650871.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] ncvreg question
To whom It May Concern, I am working on a dimensional reduction problem using Smoothly Clipped Asolute Deviation (SCAD) Penalty according to Variable Selection via Nonconcave Penalized Likelihood and its Oracle Properties, J.Fan and R. Li, JASA, Dec 2001. I found an R package named *ncvreg* which is capable to perform this variable selection procedure. I noctice, in function*ncvreg(x, y), * an intercept is automatically added to the model fitting step. I wonder, is there anyway to remove the intercept, something similar to *lars()*where you can specify the option and let *intercept=FALSE.* Or, does intercept has to be included for SCAD penalized regression? Could someone give a clue? Thanks, -Jiayi -- Jiayi Hou Ph.D Candidate Department of Biostatistics School of Medicine Virginia Commonwealth University Tel:(804)-828-2879(office) (804)-274-8757(cell) [[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] How to count the number of different elements in a column
Yes, I'm sorry ... i was checking for another column so i made a mistake. Thank you people ;) -- View this message in context: http://r.789695.n4.nabble.com/How-to-count-the-number-of-different-elements-in-a-column-tp4650825p4650875.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] Plot(x,y) help
Dear All:
I would any appreciate any help with this plot I am struggling with.
I have 4 estimates (95% CIs) I want to plot. I want the CI lines to be
horizontal on each plotted point. I was trying to tweak some old codes (was
for a vertical CI lines) into horizontal but not much dice.
Many thanks in advance for your help.
YA
My working codes:
x=c(1,1,1,1.1,1.1,1.1,2,2,2,2.1,2.1,2.1)
y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17)
ptidx = seq(1,12,by=3)
lciidx = seq(2,12,by=3)
uciidx = seq(3,12,by=3)
plot(x,y, type=n,axes=F, xlab=PR(95% CI),ylab= )
points(x[ptidx],y[ptidx],pch=19,cex=4.5)
points(x[lciidx],y[lciidx],pch=_,cex=4.5)
points(x[uciidx],y[uciidx],pch=_,cex=4.5)
box()
for(i in 1:4)
{
lines(c(x[lciidx[i]],x[uciidx[i]]),c(y[lciidx[i]],y[uciidx[i]],lwd=4,cex=4.5))
}
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] reclassification function
Hi everyone, I have tried to produce a table with reclassification (continuous NRI, p value) using the reclassification() function in the PredictABEL package. My problem is that I can only get the results printed on the screen, rather than stored in an object (as when doing a regular regression), and thus need to do a lot of manual copy-paste. Also, it seems that I end up with 0 when the p-value is more significant than 10^-6. copied from screen: NRI(Continuous) [95% CI]: 0.3487 [ 0.2701 - 0.4274 ] ; p-value: 0 Any piece of advise would be highly appreciated. Thanks! /Joel -- View this message in context: http://r.789695.n4.nabble.com/reclassification-function-tp4650851.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 in splitting the records
Sarah: You may not agree, but the following avoids the IMHO ugly rbind() and separate conversion to numeric by using scan(): ## first, as before.. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) ## Then, instead ... x.convert - data.frame(scan(textConnection(unlist(x)),wh=list(a,0,a))) ## You'll need to replace with suitable column names, of course Cheers, Bert On Mon, Nov 26, 2012 at 9:00 AM, Sarah Goslee sarah.gos...@gmail.comwrote: You state that you want $$$ as the separator, but your example has $$ and $ so I'm assuming an indeterminate number of consecutive $. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) x - do.call(rbind, x) x - data.frame(x, stringsAsFactors = FALSE) x$X2 - as.numeric(x$X2) I've also made some assumptions about what you want the output to be like, since you didn't specify, and broken it into as many steps as possible so that you can look at what functions I used and read the help files. Sarah On Mon, Nov 26, 2012 at 5:51 AM, arunkumar akpbond...@gmail.com wrote: Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me -- 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. -- 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 [[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] Savitzky-Golay filtering with missing data
I have a timeseries with some missing data points that I need smoothed using a Savitzky-Golay. Right now I am using the sgolayfilt function in the signal package. It fails if I have NA values in the timeseries. If I replace NA with NULL sgolayfilt does not fail but it drops the NULL values which is not really a valid solution. Is there anything I can do about this? -- Ryan Hope, M.S. CogWorks Lab Cognitive Science Department Rensselaer Polytechnic Institute __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot(x,y) help
Try this code that uses segments to draw in the bars # put data into a nicer form y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17) ym - matrix(y , ncol = 3 , byrow = TRUE , dimnames = list(NULL, c(Val, Lower, Upper)) ) ym - cbind(ym, x = 1:4) # add the x-coord barWidth - .1 # plot the data points plot(ym[, 'x'] , ym[, 'Val'] , pch = 19 , cex = 2 , ylim = range(ym[, Upper], ym[, Lower]) ) # draw the CI values as segments # three sets of coords - top bar, bottom bar, connecting line segments(c(ym[, 'x'] - barWidth, ym[, 'x'] - barWidth, ym[, 'x']) , c(ym[, 'Upper'], ym[, 'Lower'], ym[, 'Upper']) , c(ym[, 'x'] + barWidth, ym[, 'x'] + barWidth, ym[, 'x']) , c(ym[, 'Upper'], ym[, 'Lower'], ym[, 'Lower']) ) On Mon, Nov 26, 2012 at 1:31 PM, YAddo link...@gmail.com wrote: y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17) -- 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] Plot(x,y) help
Hello,
You had a typo in the lines() instruction, the parenthesis didn't close
after c(y...).
Anyway, I'm not sure I understand but to have horizontal lines, just
reverse the roles of x and y. (And change pch = _ to pch = |).
plot(y, x, type=n,axes=F, xlab=PR(95% CI),ylab= )
points(y[ptidx],x[ptidx], pch=19,cex=4.5)
points(y[lciidx],x[lciidx], pch=|,cex=4.5)
points(y[uciidx],x[uciidx], pch=|,cex=4.5)
box()
for(i in 1:4){
lines(c(y[lciidx[i]], y[uciidx[i]]),
c(x[lciidx[i]], x[uciidx[i]]), lwd=4,cex=4.5)
}
Hope this helps,
Rui Barradas
Em 26-11-2012 18:31, YAddo escreveu:
Dear All:
I would any appreciate any help with this plot I am struggling with.
I have 4 estimates (95% CIs) I want to plot. I want the CI lines to be
horizontal on each plotted point. I was trying to tweak some old codes (was
for a vertical CI lines) into horizontal but not much dice.
Many thanks in advance for your help.
YA
My working codes:
x=c(1,1,1,1.1,1.1,1.1,2,2,2,2.1,2.1,2.1)
y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17)
ptidx = seq(1,12,by=3)
lciidx = seq(2,12,by=3)
uciidx = seq(3,12,by=3)
plot(x,y, type=n,axes=F, xlab=PR(95% CI),ylab= )
points(x[ptidx],y[ptidx],pch=19,cex=4.5)
points(x[lciidx],y[lciidx],pch=_,cex=4.5)
points(x[uciidx],y[uciidx],pch=_,cex=4.5)
box()
for(i in 1:4)
{
lines(c(x[lciidx[i]],x[uciidx[i]]),c(y[lciidx[i]],y[uciidx[i]],lwd=4,cex=4.5))
}
--
View this message in context:
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in splitting the records
Another approach is to use gsub() followed by read.table(): x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - gsub(\\$+, \\$, x) x - read.table(text=x, sep=$, stringsAsFactors=FALSE) x V1 V2 V3 1 qwer 12 qwre 2 ewrtr 7789 ewwe -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Bert Gunter Sent: Monday, November 26, 2012 1:00 PM To: Sarah Goslee Cc: r-help Subject: Re: [R] Help in splitting the records Sarah: You may not agree, but the following avoids the IMHO ugly rbind() and separate conversion to numeric by using scan(): ## first, as before.. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) ## Then, instead ... x.convert - data.frame(scan(textConnection(unlist(x)),wh=list(a,0,a))) ## You'll need to replace with suitable column names, of course Cheers, Bert On Mon, Nov 26, 2012 at 9:00 AM, Sarah Goslee sarah.gos...@gmail.comwrote: You state that you want $$$ as the separator, but your example has $$ and $ so I'm assuming an indeterminate number of consecutive $. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) x - do.call(rbind, x) x - data.frame(x, stringsAsFactors = FALSE) x$X2 - as.numeric(x$X2) I've also made some assumptions about what you want the output to be like, since you didn't specify, and broken it into as many steps as possible so that you can look at what functions I used and read the help files. Sarah On Mon, Nov 26, 2012 at 5:51 AM, arunkumar akpbond...@gmail.com wrote: Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me -- 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. -- 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 [[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] remote connection to an Oracle database - using RODBC - RMySQL..?
Thank you Marc, I will study the material you sent and follow up on this at R-SIG-DB - should I still have problems. I'm using a Mac OS X 10.6.8 Thanks. Raff. Subject: Re: [R] remote connection to an Oracle database - using RODBC - RMySQL..? From: marc_schwa...@me.com Date: Wed, 21 Nov 2012 14:16:15 -0600 CC: r-help@r-project.org To: r_varda...@hotmail.com On Nov 21, 2012, at 1:52 PM, Raffaello Vardavas r_varda...@hotmail.com wrote: Dear users, I can access an database oracle database using sql developer. This was done by importing an xml file that contains the login details - username, password and specifies that it uses the KERBEROS_AUTHENTICATION. I'm trying to connect R - so that it can access this database - do sql queries and convert the resulting tables into dataframes. I am a novice in SQL and database access - but a friend provided me with the following approach: library(DBI) library(RMySQL) drvr-dbDriver(MySQL) #Or another driver, say from the RODBC package? acon-dbConnect(drvr, user=ENTER_USERID, dbname=ENTER_NAME, host=ENTER_HOST, port=1521,password=NULL) #password maybe non-null? cmds-dbSendQuery(acon,statement=YOUR SQL QUERY HERE) yourdata-fetch(cmds, n=-1) #Collects all rows and columns of data requested query. I have provided this info changing the relevant info in the dbConnect command and provided the password. However this doesn't work. I suspect because in this command there is not specification of the encryption of the password (i.e., KERBEROS_AUTHENTICATION) When I look at the details of the connection in SQL developer - what is specified is the follow: connection name, username, password (that I cannot see), hostname, port and the SID. Note that although the password here cannot be seen - I believe it is computed by the longer password displaced the the xml file I use to set up the connection with sql developer using the KERBEROS_AUTHENTICATION. Any ideas on how to proceed. Please help. Thank you. Raff. Several comments: 1. Future posts on this subject should be made to R-SIG-DB, not here. More info: https://stat.ethz.ch/mailman/listinfo/r-sig-db 2. Why would you expect to use an R package and driver for MySQL when attempting to access an Oracle server? 3. There is a good starting point on this subject generally in the R Data Import/Export manual: http://cran.r-project.org/doc/manuals/r-release/R-data.html#Relational-databases 4. I would recommend using RODBC, which is what I use. You will of course need to have an ODBC driver for Oracle installed on your system and properly configured. You may need to get that from Oracle or other parties depending upon your OS which is unstated here. You may also need to get assistance with that process from your SysAdmin or DBAdmin. 5. If you use RODBC, there is additional, quite good information in the package vignette, which is accessible by using: vignette(RODBC) post package installation. 6. I don't have any experience using Kerberos authentication on my Oracle server here, so you may have to follow up on the R-SIG-DB list on that point. A search of the archives did not reveal anything material on that point. 7. Alternatives to RODBC would include ROracle and RJDBC via CRAN. Regards, Marc Schwartz [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in splitting the records
Hi, Just a modification: x1-gsub([$], ,x) read.table(text=x1,sep=,header=FALSE,stringsAsFactors=FALSE) # V1 V2 V3 #1 qwer 12 qwre #2 ewrtr 7789 ewwe A.K. - Original Message - From: David L Carlson dcarl...@tamu.edu To: 'Bert Gunter' gunter.ber...@gene.com; 'Sarah Goslee' sarah.gos...@gmail.com Cc: 'r-help' r-help@r-project.org Sent: Monday, November 26, 2012 2:33 PM Subject: Re: [R] Help in splitting the records Another approach is to use gsub() followed by read.table(): x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - gsub(\\$+, \\$, x) x - read.table(text=x, sep=$, stringsAsFactors=FALSE) x V1 V2 V3 1 qwer 12 qwre 2 ewrtr 7789 ewwe -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Bert Gunter Sent: Monday, November 26, 2012 1:00 PM To: Sarah Goslee Cc: r-help Subject: Re: [R] Help in splitting the records Sarah: You may not agree, but the following avoids the IMHO ugly rbind() and separate conversion to numeric by using scan(): ## first, as before.. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) ## Then, instead ... x.convert - data.frame(scan(textConnection(unlist(x)),wh=list(a,0,a))) ## You'll need to replace with suitable column names, of course Cheers, Bert On Mon, Nov 26, 2012 at 9:00 AM, Sarah Goslee sarah.gos...@gmail.comwrote: You state that you want $$$ as the separator, but your example has $$ and $ so I'm assuming an indeterminate number of consecutive $. x - c(qwer$$12$$qwre, ewrtr$7789$ewwe) x - strsplit(x, \\$+) x - do.call(rbind, x) x - data.frame(x, stringsAsFactors = FALSE) x$X2 - as.numeric(x$X2) I've also made some assumptions about what you want the output to be like, since you didn't specify, and broken it into as many steps as possible so that you can look at what functions I used and read the help files. Sarah On Mon, Nov 26, 2012 at 5:51 AM, arunkumar akpbond...@gmail.com wrote: Hi I have set of records seperated by a separator say $$$ i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2 V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me -- 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. -- 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 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remote connection to an Oracle database - using RODBC - RMySQL..?
On Nov 26, 2012, at 1:47 PM, Raffaello Vardavas r_varda...@hotmail.com wrote: Thank you Marc, I will study the material you sent and follow up on this at R-SIG-DB - should I still have problems. I'm using a Mac OS X 10.6.8 Thanks. Raff. In that case, if you use RODBC, you will want to get the Oracle ODBC driver for OSX from Actual Technologies, which is also what I use on 10.8.2. More info here: http://www.actualtech.com/product_oracle.php Note that it is not free ($34.95 US) and Oracle does not provide OSX ODBC drivers. This is covered in the RODBC vignette. The good news is that it makes it fairly easy to set up the DSN connection to Oracle, as you don't have to worry about a lot of the low level configuration issues. You can download an eval version of the driver for free. The limitation of the free version is that a query will only return the first 3 rows. You would then need to pay for a fully functional license. Regards, Marc Subject: Re: [R] remote connection to an Oracle database - using RODBC - RMySQL..? From: marc_schwa...@me.com Date: Wed, 21 Nov 2012 14:16:15 -0600 CC: r-help@r-project.org To: r_varda...@hotmail.com On Nov 21, 2012, at 1:52 PM, Raffaello Vardavas r_varda...@hotmail.com wrote: Dear users, I can access an database oracle database using sql developer. This was done by importing an xml file that contains the login details - username, password and specifies that it uses the KERBEROS_AUTHENTICATION. I'm trying to connect R - so that it can access this database - do sql queries and convert the resulting tables into dataframes. I am a novice in SQL and database access - but a friend provided me with the following approach: library(DBI) library(RMySQL) drvr-dbDriver(MySQL) #Or another driver, say from the RODBC package? acon-dbConnect(drvr, user=ENTER_USERID, dbname=ENTER_NAME, host=ENTER_HOST, port=1521,password=NULL) #password maybe non-null? cmds-dbSendQuery(acon,statement=YOUR SQL QUERY HERE) yourdata-fetch(cmds, n=-1) #Collects all rows and columns of data requested query. I have provided this info changing the relevant info in the dbConnect command and provided the password. However this doesn't work. I suspect because in this command there is not specification of the encryption of the password (i.e., KERBEROS_AUTHENTICATION) When I look at the details of the connection in SQL developer - what is specified is the follow: connection name, username, password (that I cannot see), hostname, port and the SID. Note that although the password here cannot be seen - I believe it is computed by the longer password displaced the the xml file I use to set up the connection with sql developer using the KERBEROS_AUTHENTICATION. Any ideas on how to proceed. Please help. Thank you. Raff. Several comments: 1. Future posts on this subject should be made to R-SIG-DB, not here. More info: https://stat.ethz.ch/mailman/listinfo/r-sig-db 2. Why would you expect to use an R package and driver for MySQL when attempting to access an Oracle server? 3. There is a good starting point on this subject generally in the R Data Import/Export manual: http://cran.r-project.org/doc/manuals/r-release/R-data.html#Relational-databases 4. I would recommend using RODBC, which is what I use. You will of course need to have an ODBC driver for Oracle installed on your system and properly configured. You may need to get that from Oracle or other parties depending upon your OS which is unstated here. You may also need to get assistance with that process from your SysAdmin or DBAdmin. 5. If you use RODBC, there is additional, quite good information in the package vignette, which is accessible by using: vignette(RODBC) post package installation. 6. I don't have any experience using Kerberos authentication on my Oracle server here, so you may have to follow up on the R-SIG-DB list on that point. A search of the archives did not reveal anything material on that point. 7. Alternatives to RODBC would include ROracle and RJDBC via CRAN. Regards, Marc Schwartz [[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] aggregate() runs out of memory
Hi, * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03 -0800]: For instance, if you want the min and max of `delay` within each group defined by `share.id`, and let's assume `infl` is a data.frame, you can do something like so: R as.data.table(infl) R setkey(infl, share.id) R result - infl[, list(min=min(delay), max=max(delay)), by=share.id] perfect, thanks. alas, the resulting table does not contain the share.id column. do I need to add something like id=unique(share.id) to the list? also, if there is a field in the original table infl which only depends on share.id, how do I add this unique value to the summary? it appears that count=unique(country) in list() does what I need, but it slows down the process. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://openvotingconsortium.org http://jihadwatch.org http://thereligionofpeace.com http://palestinefacts.org http://dhimmi.com Why use Windows, when there are Doors? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 on function please
Dear All,
I could use a bit of help here, this function is hard to figure out (for me at
least) I have the following so far:
PKindex-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,28,25,22,18,14,11))
Dose-200
Tinf -0.5
defun- function(time, y, parms) {
dCpdt - -parms[kel] * y[1]
list(dCpdt)
}
modfun - function(time,kel, Vd) {
out -
lsoda(((Dose/Tinf)*(1/(kel*Vd)))*(1-exp(-kel*time)),c(0,time),defun,parms=c(kel=kel,Vd=Vd),rtol=1e-3,atol=1e-5)
out[-1,2]
}
objfun - function(par) {
out - modfun(PKindex$time, par[1], par[2])
gift - which( PKindex$conc != 0 )
sum((PKindex$conc[gift]-out[gift])^2)
}
gen-genoud(objfun,nvars=2,max=FALSE,pop.size=30,max.generations=100,wait.generations=100,starting.value=c(0.7,60),BFGS=FALSE,print.level=0,boundary.enforcement=2,Domains=matrix(c(0.01,0.01,100,100),2,2),MemoryMatrix=TRUE)
but get the following:
Error in lsoda(((Dose/Tinf) * (1/(kel * Vd))) * (1 - exp(-kel * time)), :
The number of derivatives returned by func() (1must equal the length of the
initial conditions vector (7)
i figured that having the time parameter in the equation screws things up,
but do not now how to fix it bc I do not understand the warning message.
your help is greatly apreciated,
Sincerely,
Andras
[[alternative HTML version deleted]]
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Re: [R] aggregate() runs out of memory
Hi Sam, On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold s...@gnu.org wrote: Hi, * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03 -0800]: For instance, if you want the min and max of `delay` within each group defined by `share.id`, and let's assume `infl` is a data.frame, you can do something like so: R as.data.table(infl) R setkey(infl, share.id) R result - infl[, list(min=min(delay), max=max(delay)), by=share.id] perfect, thanks. alas, the resulting table does not contain the share.id column. do I need to add something like id=unique(share.id) to the list? also, if there is a field in the original table infl which only depends on share.id, how do I add this unique value to the summary? it appears that count=unique(country) in list() does what I need, but it slows down the process. Hmm ... I think it should be there, but I'm having a hard time remember what you want. Could you please copy paste the output of `dput(head(infl, 20))` as well as an approximation of what the result is that you want. It will make it easier for us to talk more concretely about how to get what you want. -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] printing difftime summary
On Mon, Nov 26, 2012 at 4:46 PM, David Winsemius dwinsem...@comcast.net wrote:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
Surely reversed no? summary.difftime inherits from data.frame I would
have assumed.
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
What is this supposed to do exactly? If you have inheritance why have
the subclass method do nothing other than call the parent method?
Michael
--8---cut here---end---8---
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a part
of the data frame summary?
If you like a particular format from an existing print method then why not
look it up and copy the code?
methods(print)
--
David.
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51 -0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$delay)
r1
string secs
Min.492.00 ms 0.5
1st Qu. 18.08 min 1085.0
Median 1.77 hrs 6370.0
Mean 8.20 hrs 29530.0
3rd Qu. 8.12 hrs 29250.0
Max.6.98 days 602900.0
str(r1)
Classes 'summary.difftime' and 'data.frame':6 obs. of 2 variables:
$ string: chr 492.00 ms 18.08 min 1.77 hrs 8.20 hrs ...
$ secs :Classes 'summaryDefault', 'table' num [1:6] 4.92e-01
1.08e+03 6.37e+03 2.95e+04 2.92e+04 ...
--8---cut here---end---8---
but not as a part of data frame:
--8---cut here---start-8---
a - summary(infl)
Error in summary.difftime(X[[22L]], ...) :
unused argument(s) (maxsum = 7, digits = 12)
--8---cut here---end---8---
I guess I should somehow accept a list of options in
summary.difftime()
and pass them on to the inner call to summary() (or should it be
explicitly summary.numeric()?)
In the usual way. If you know that the function will be called with
arguments from the summary.data.frame function then you should allow the
argument list to accept them. You can ignore them or provide provisions
for them. You just can't define your function to have only one argument
if you expect (as you should since you passes summary a dataframe
object) that it might be called within summary.data.frame.
This is the argument list for summary.data.frame:
summary.data.frame
function (object, maxsum = 7, digits = max(3, getOption(digits) -
3), ...)
how do I do that?
summary.difftime - function (v, ... ) {
There are many asked and answered questions on rhelp about how to deal
with the dots arguments.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
http://www.childpsy.net/ http://www.memritv.org http://memri.org
http://honestreporting.com http://dhimmi.com
http://openvotingconsortium.org
People with a good taste are especially appreciated by cannibals.
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Help R
On Mon, Nov 26, 2012 at 10:38 AM, alanaro...@sapo.pt wrote: Goodmorning, I'moneafazrtrbhoquhasa variablefactorcomtwoNivesCandH.Queoaatravésdlinear regressionrelationshipetrehaSaerifthe variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat showsthe two curvesHandCLIRand Selecionar tudo Thank you Ana C. Rocha Rua I think this might be offered as a very good example of why one shouldn't send HTMangLed mail in lieu of plain text. Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error in plot(table(c('a','a')))
On Mon, Nov 26, 2012 at 2:41 PM, Ludo Pagie l.pa...@nki.nl wrote:
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ). With more than one value I don't get
errors, neither with integers (even if only 1 value):
tbl.char1 - table(c('a','a'))
tbl.char2 - table(c('a','a','b'))
tbl.int1 - table(c(1,1))
# error:
plot(tbl.char1)
# no errors:
plot(tbl.char2)
plot(tbl.int1)
Confirmed in current R-devel. It seems to arise from plot.table's use
of seq.int(x) when the dimnames of the table are not integers. I'm not
sure if this shouldn't be seq_along(x) instead, but I'm not sure I
totally follow the internal logic, so perhaps someone can offer second
opinion?
Michael
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Re: [R] aggregate() runs out of memory
hi Steve, * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 16:08:59 -0500]: On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold s...@gnu.org wrote: * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03 -0800]: For instance, if you want the min and max of `delay` within each group defined by `share.id`, and let's assume `infl` is a data.frame, you can do something like so: R as.data.table(infl) R setkey(infl, share.id) R result - infl[, list(min=min(delay), max=max(delay)), by=share.id] perfect, thanks. alas, the resulting table does not contain the share.id column. do I need to add something like id=unique(share.id) to the list? also, if there is a field in the original table infl which only depends on share.id, how do I add this unique value to the summary? it appears that count=unique(country) in list() does what I need, but it slows down the process. Hmm ... I think it should be there, but I'm having a hard time remember what you want. Could you please copy paste the output of `(head(infl, 20))` as well as an approximation of what the result is that you want. this prints all the levels for all the factor columns and takes megabytes. --8---cut here---start-8--- f - data.frame(id=rep(1:3,4),country=rep(6:8,4),delay=1:12) f id country delay 1 1 6 1 2 2 7 2 3 3 8 3 4 1 6 4 5 2 7 5 6 3 8 6 7 1 6 7 8 2 7 8 9 3 8 9 10 1 610 11 2 711 12 3 812 f - as.data.table(f) setkey(f,id) delays - f[,list(min=min(delay),max=max(delay),count=.N,country=unique(country)),by=id] delays id min max count country 1: 1 1 10 4 6 2: 2 2 11 4 7 3: 3 3 12 4 8 --8---cut here---end---8--- this is still too slow, apparently because of unique. how do I speed it up? Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://iris.org.il http://ffii.org http://pmw.org.il http://mideasttruth.com Programming is like sex: one mistake and you have to support it for a lifetime. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which function prints this:
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
I added cat()s to print.summary.difftime and I do not see them, so it
appears that I have no direct control over how a summary.difftime is
printed as a part of a summary of a data.frame.
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(summary.difftime,data.frame)
invisible(r)
}
print.summary.difftime - function (sd, ...) {
cat([[[print.summary.difftime]]]\n)
print(list(...))
print.data.frame(sd, ...)
}
--8---cut here---end---8---
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://palestinefacts.org http://think-israel.org
http://www.memritv.org http://openvotingconsortium.org http://mideasttruth.com
The force of gravity doubles when acting on a body on a couch.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] aggregate() runs out of memory
Hi,
On Mon, Nov 26, 2012 at 4:57 PM, Sam Steingold s...@gnu.org wrote:
[snip]
Could you please copy paste the output of `(head(infl, 20))` as
well as an approximation of what the result is that you want.
Don't know how dput got clipped in your reply from the quoted text I
wrote, but I actually asked for `dput(head(infl, 20))`
The dput makes a world of difference because I can easily copy/paste
the output into R and get a working table.
this prints all the levels for all the factor columns and takes
megabytes.
Try using droplevels, eg:
R dput(droplevels(head(infl, 20)))
--8---cut here---start-8---
f - data.frame(id=rep(1:3,4),country=rep(6:8,4),delay=1:12)
f
id country delay
1 1 6 1
2 2 7 2
3 3 8 3
4 1 6 4
5 2 7 5
6 3 8 6
7 1 6 7
8 2 7 8
9 3 8 9
10 1 610
11 2 711
12 3 812
f - as.data.table(f)
setkey(f,id)
delays -
f[,list(min=min(delay),max=max(delay),count=.N,country=unique(country)),by=id]
delays
id min max count country
1: 1 1 10 4 6
2: 2 2 11 4 7
3: 3 3 12 4 8
--8---cut here---end---8---
this is still too slow, apparently because of unique.
how do I speed it up?
I think I'm missing something.
Your call to `min(delay)` and `max(delay)` will return the minimum and
maximum delays within the particular id you are grouping by. I guess
there must be several values for country within each id group --
do you really want the same min and max values to be replicated as
many times as there are unique countrys?
Do you perhaps want to iterate over a combo of id and country?
Anyway: if you don't use `unique` inside your calculation, I guess it
goes significantly faster, like so:
R result - f[, list(min=min(delay), max=max(delay),
count=.N,country=country[1L]), by=share.id]
If that's bearable, and you really want the way you suggest (or, at
least, what I'm interpreting), I wonder if this two-step would be
faster?
R setkeyv(f, c('share.id', 'country'))
R r1 - f[, list(min=min(delay), max=max(delay), count=.N), by='share.id']
R result - unique(f)[r1] ## I think
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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Re: [R] aggregate() runs out of memory
Hi, * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 17:32:21 -0500]: --8---cut here---start-8--- f - data.frame(id=rep(1:3,4),country=rep(6:8,4),delay=1:12) f id country delay 1 1 6 1 2 2 7 2 3 3 8 3 4 1 6 4 5 2 7 5 6 3 8 6 7 1 6 7 8 2 7 8 9 3 8 9 10 1 610 11 2 711 12 3 812 f - as.data.table(f) setkey(f,id) delays - f[,list(min=min(delay),max=max(delay),count=.N,country=unique(country)),by=id] delays id min max count country 1: 1 1 10 4 6 2: 2 2 11 4 7 3: 3 3 12 4 8 --8---cut here---end---8--- this is still too slow, apparently because of unique. how do I speed it up? I think I'm missing something. Your call to `min(delay)` and `max(delay)` will return the minimum and maximum delays within the particular id you are grouping by. I guess there must be several values for country within each id group -- do you really want the same min and max values to be replicated as many times as there are unique countrys? there is precisely one country for each id. i.e., unique(country) is the same as country[1]. thanks a lot for the suggestion! R result - f[, list(min=min(delay), max=max(delay), count=.N,country=country[1L]), by=share.id] -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://thereligionofpeace.com http://pmw.org.il http://honestreporting.com http://americancensorship.org Why do you never call me back after I scream that I will never talk to you again?! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] printing difftime summary
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime (which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame. Once you've written it, have your
print.summary.difftime
call it too.
E.g., with the following methods
summary.difftime - function(x, ...) {
ret - quantile(x, p=(0:2)/2, na.rm=TRUE)
class(ret) - c(summary.difftime, class(ret))
ret
}
format.summary.difftime - function(x, ...) c(Min.Med.Max =
paste(collapse=..., NextMethod(format)))
print.summary.difftime - function(x, ...){ print(format(x), quote=FALSE) ;
invisible(x) }
I get
d - data.frame(Num=1:5, Date=as.Date(2012-11-26)+(0:4),
Delta=diff(as.Date(2012-11-26)+2^(0:5)))
summary(d)
Num DateDelta
Min. :1 Min. :2012-11-26 Min.Med.Max: 1 days... 4 days...16 days
1st Qu.:2 1st Qu.:2012-11-27
Median :3 Median :2012-11-28
Mean :3 Mean :2012-11-28
3rd Qu.:4 3rd Qu.:2012-11-29
Max. :5 Max. :2012-11-30
summary(d$Delta)
Min.Med.Max
1 days... 4 days...16 days
My summary.difftime inherits from difftime so the format method is not really
needed, as format.difftime does a reasonable job (except that it does not copy
the input names to its output). I put it in to show how it gets called.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which function prints this:
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
I added cat()s to print.summary.difftime and I do not see them, so it
appears that I have no direct control over how a summary.difftime is
printed as a part of a summary of a data.frame.
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(summary.difftime,data.frame)
invisible(r)
}
print.summary.difftime - function (sd, ...) {
cat([[[print.summary.difftime]]]\n)
print(list(...))
print.data.frame(sd, ...)
}
--8---cut here---end---8---
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
http://www.childpsy.net/ http://palestinefacts.org http://think-israel.org
http://www.memritv.org http://openvotingconsortium.org http://mideasttruth.com
The force of gravity doubles when acting on a body on a couch.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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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] error in plot(table(c('a','a')))
Hello,
Also gives an error in R 2.15.2 on Windows 7. I'd change when the
dimnames of the table are not integers to not numeric as that's what
the code for plot.table tests. And it seems to come from seq.int, since
with table value of 2, seq.int produces a vector of length 2 but the
table length is 1:
seq.int(tbl.char1)
[1] 1 2
seq_along(tbl.char1)
[1] 1
So the plot command will have 2 values for the x axis but just one for
the y axis.
seq_along seems to solve the matter.
Hope this helps,
Rui Barradas
Em 26-11-2012 21:39, R. Michael Weylandt escreveu:
On Mon, Nov 26, 2012 at 2:41 PM, Ludo Pagie l.pa...@nki.nl wrote:
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ). With more than one value I don't get
errors, neither with integers (even if only 1 value):
tbl.char1 - table(c('a','a'))
tbl.char2 - table(c('a','a','b'))
tbl.int1 - table(c(1,1))
# error:
plot(tbl.char1)
# no errors:
plot(tbl.char2)
plot(tbl.int1)
Confirmed in current R-devel. It seems to arise from plot.table's use
of seq.int(x) when the dimnames of the table are not integers. I'm not
sure if this shouldn't be seq_along(x) instead, but I'm not sure I
totally follow the internal logic, so perhaps someone can offer second
opinion?
Michael
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Re: [R] printing difftime summary
Thanks a lot - almost there!
--8---cut here---start-8---
format.summary.difftime - function(sd, ...) {
t - matrix(sd$string)
rownames(t) - rownames(sd)
print(t)
format(as.table(t))
}
print.summary.difftime - function (sd, ...) {
print(format(sd), quote=FALSE)
invisible(sd)
}
--8---cut here---end---8---
this almost works:
--8---cut here---start-8---
summary(delays)
share.id min max
12cf12372b87cce9: 1 NULL:492.00 ms NULL:492.00 ms
12cf36060bdb9581: 1 NULL:3.70 minNULL:21.80 min
12d2665c906bb232: 1 NULL:20.32 min NULL:3.26 hrs
12d2802f1435b4cd: 1 NULL:5.52 hrsNULL:13.78 hrs
12d292988f5f8422: 1 NULL:2.81 hrsNULL:16.20 hrs
12d29dd2894e2790: 1 NULL:6.95 days NULL:6.98 days
--8---cut here---end---8---
why do I see NULLs?!
--8---cut here---start-8---
t - matrix(sd$string)
rownames(t) - rownames(sd)
t
[,1]
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
as.table(t)
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
format(as.table(t))
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
--8---cut here---end---8---
* William Dunlap jqha...@gvopb.pbz [2012-11-26 23:02:48 +]:
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime
(which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame. Once you've written it, have your
print.summary.difftime
call it too.
E.g., with the following methods
summary.difftime - function(x, ...) {
ret - quantile(x, p=(0:2)/2, na.rm=TRUE)
class(ret) - c(summary.difftime, class(ret))
ret
}
format.summary.difftime - function(x, ...) c(Min.Med.Max =
paste(collapse=..., NextMethod(format)))
print.summary.difftime - function(x, ...){ print(format(x), quote=FALSE) ;
invisible(x) }
I get
d - data.frame(Num=1:5, Date=as.Date(2012-11-26)+(0:4),
Delta=diff(as.Date(2012-11-26)+2^(0:5)))
summary(d)
Num DateDelta
Min. :1 Min. :2012-11-26 Min.Med.Max: 1 days... 4 days...16 days
1st Qu.:2 1st Qu.:2012-11-27
Median :3 Median :2012-11-28
Mean :3 Mean :2012-11-28
3rd Qu.:4 3rd Qu.:2012-11-29
Max. :5 Max. :2012-11-30
summary(d$Delta)
Min.Med.Max
1 days... 4 days...16 days
My summary.difftime inherits from difftime so the format method is not really
needed, as format.difftime does a reasonable job (except that it does not copy
the input names to its output). I put it in to show how it gets called.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which function prints this:
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
I added cat()s to print.summary.difftime and I do not see them, so it
appears that I have no direct control over how a summary.difftime is
printed as a part of a summary of a data.frame.
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(summary.difftime,data.frame)
invisible(r)
}
Re: [R] Adding a new variable to each element of a list
Thanks, Arun! This pretty much does what I was looking for. Looks like I
should get more familiar with the apply functions, it seems like they can
solve a lot of these problems!
Best,
Steve
On Sun, Nov 25, 2012 at 8:09 PM, arun smartpink...@yahoo.com wrote:
HI Steve,
You could try this:
ylist-lapply(y,function(x) x)
res-lapply(mapply(c,result,Thing=ylist,SIMPLIFY=FALSE),function(x)
do.call(cbind,x))
res
#$`Error: subject`
# Df Sum Sq Mean Sq F value Pr(F) Thing
#Residuals 4 12.4 3.1 NA NA 0.5
#$`Error: subject:myfactor`
# Df Sum Sq Mean Sq F value Pr(F) Thing
#myfactor 2 14.9 7.47 13.57576 0.002682772 0.7724138
#Residuals 8 4.4 0.55 NA NA 0.7724138
#But, it removes the *, which BTW gets removed by individually inserting
as you showed.
A.K.
From: Stephen Politzer-Ahles politzerahl...@gmail.com
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Sunday, November 25, 2012 8:35 PM
Subject: Re: [R] Adding a new variable to each element of a list
Hi Arun,
Thanks a lot for the help. I think I didn't make my request clear, though;
sorry about that. What your example does is applies the values of y to each
element in the list result. However, what I'm actually trying to do is
apply the nth element of y to the nth element of result. In other words,
I'm trying to find a way in one command to do the equivalent of the
following:
result[[1]][[1]]$Thing - y[1]
result[[2]][[1]]$Thing - y[2]
...etc.
Is this even possible? (Granted, for my purpose result is not a very
long list, so it would not be difficult to just do those commands one at a
time or in a for loop. At this point I'm just curious to figure out a more
straightforward way to do it just for the sake of trying to learn more
about how R works!)
If it helps put things into context, in my example the list result is a
summary of an aov object, and each element in the list is one of the ANOVA
effects; y is a vector of the partial eta squareds I calculated
corresponding to each effect, and I'm trying to figure out a way to add
those into the result object so it shows up when my colleague looks at
the summary.
Best,
Steve
On Sat, Nov 24, 2012 at 5:22 PM, arun smartpink...@yahoo.com wrote:
HI,
The example data you gave have only one row for the 1st element of list.
So, it would be better to add it as:
lapply(result,function(x) {x[[2]]-y
return(x)})
#$`Error: subject`
#Component 1 :
# Df Sum Sq Mean Sq F value Pr(F)
#Residuals 4 12.4 3.1
#
#Component 2 :
#[1] 0.500 0.7724138
-
You can also do:
lapply(result,function(x) {x[[1]][6]-y[1]
x[[1]][7]-y[2]
return(x)})
#$`Error: subject`
# Df Sum Sq Mean Sq F value Pr(F) V6 V7
#Residuals 4 12.4 3.10.5 0.7724
#$`Error: subject:myfactor`
# Df Sum Sq Mean Sq F value Pr(F) V6 V7
#myfactor 2 14.93 7.467 13.58 0.002683 0.5 0.7724
#Residuals 8 4.40 0.550 0.5 0.7724#Here it got
repeated
May be there are better methods
A.K.
- Original Message -
From: Stephen Politzer-Ahles politzerahl...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Saturday, November 24, 2012 5:33 PM
Subject: [R] Adding a new variable to each element of a list
Hello,
I have a list of data with multiple elements, and each element in the list
has multiple variables in it. Here's an example:
### Make the fake data
dv - c(1,3,4,2,2,3,2,5,6,3,4,4,3,5,6)
subject - factor(c(s1,s1,s1,s2,s2,s2,s3,s3,s3,
s4,s4,s4,s5,s5,s5))
myfactor - factor(c(f1,f2,f3,f1,f2,f3,f1,f2,f3,
f1,f2,f3,f1,f2,f3))
mydata - data.frame(dv, subject, myfactor)
### Do the anova and store the summary in result
mydata.aov - aov( dv ~ myfactor + Error(subject/myfactor), mydata )
( result - summary( mydata.aov ) ) # see the anova
str(result)
List of 2
$ Error: subject :List of 1
..$ :Classes âanovaâ and 'data.frame':1 obs. of 5 variables:
.. ..$ Df : num 4
.. ..$ Sum Sq : num 12.4
.. ..$ Mean Sq: num 3.1
.. ..$ F value: num NA
.. ..$ Pr(F) : num NA
..- attr(*, class)= chr [1:2] summary.aov listof
$ Error: subject:myfactor:List of 1
..$ :Classes âanovaâ and 'data.frame':2 obs. of 5 variables:
.. ..$ Df : num [1:2] 2 8
.. ..$ Sum Sq : num [1:2] 14.9 4.4
.. ..$ Mean Sq: num [1:2] 7.47 0.55
.. ..$ F value: num [1:2] 13.6 NA
.. ..$ Pr(F) : num [1:2] 0.00268 NA
..- attr(*, class)= chr [1:2] summary.aov listof
As you can see, each element in result has several variables (Df, Sum
Sq,
Mean Sq, F value, Pr(F)):
str( result[[2]][[1]] )
Classes âanovaâ and 'data.frame': 2 obs. of 5 variables:
$ Df : num 2 8
$ Sum Sq : num 14.9 4.4
$ Mean Sq: num 7.47 0.55
$ F value: num 13.6 NA
$
Re: [R] printing difftime summary
why do I see NULLs?!
because
... format.difftime does a reasonable job (except that it does not copy
the input names to its output).
Replace your call of the form
format(difftimeObject)
with
structure(format(difftimeObject), names=names(difftimeObject))
to work around this.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: Sam Steingold [mailto:sam.steing...@gmail.com] On Behalf Of Sam
Steingold
Sent: Monday, November 26, 2012 4:09 PM
To: William Dunlap
Cc: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
Thanks a lot - almost there!
--8---cut here---start-8---
format.summary.difftime - function(sd, ...) {
t - matrix(sd$string)
rownames(t) - rownames(sd)
print(t)
format(as.table(t))
}
print.summary.difftime - function (sd, ...) {
print(format(sd), quote=FALSE)
invisible(sd)
}
--8---cut here---end---8---
this almost works:
--8---cut here---start-8---
summary(delays)
share.id min max
12cf12372b87cce9: 1 NULL:492.00 ms NULL:492.00 ms
12cf36060bdb9581: 1 NULL:3.70 minNULL:21.80 min
12d2665c906bb232: 1 NULL:20.32 min NULL:3.26 hrs
12d2802f1435b4cd: 1 NULL:5.52 hrsNULL:13.78 hrs
12d292988f5f8422: 1 NULL:2.81 hrsNULL:16.20 hrs
12d29dd2894e2790: 1 NULL:6.95 days NULL:6.98 days
--8---cut here---end---8---
why do I see NULLs?!
--8---cut here---start-8---
t - matrix(sd$string)
rownames(t) - rownames(sd)
t
[,1]
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
as.table(t)
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
format(as.table(t))
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
--8---cut here---end---8---
* William Dunlap jqha...@gvopb.pbz [2012-11-26 23:02:48 +]:
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime
(which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame. Once you've written it, have your
print.summary.difftime
call it too.
E.g., with the following methods
summary.difftime - function(x, ...) {
ret - quantile(x, p=(0:2)/2, na.rm=TRUE)
class(ret) - c(summary.difftime, class(ret))
ret
}
format.summary.difftime - function(x, ...) c(Min.Med.Max =
paste(collapse=..., NextMethod(format)))
print.summary.difftime - function(x, ...){ print(format(x), quote=FALSE) ;
invisible(x) }
I get
d - data.frame(Num=1:5, Date=as.Date(2012-11-26)+(0:4),
Delta=diff(as.Date(2012-11-26)+2^(0:5)))
summary(d)
Num DateDelta
Min. :1 Min. :2012-11-26 Min.Med.Max: 1 days... 4 days...16 days
1st Qu.:2 1st Qu.:2012-11-27
Median :3 Median :2012-11-28
Mean :3 Mean :2012-11-28
3rd Qu.:4 3rd Qu.:2012-11-29
Max. :5 Max. :2012-11-30
summary(d$Delta)
Min.Med.Max
1 days... 4 days...16 days
My summary.difftime inherits from difftime so the format method is not
really
needed, as format.difftime does a reasonable job (except that it does not
copy
the input names to its output). I put it in to show how it gets called.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which
Re: [R] aggregate() runs out of memory
On Monday, November 26, 2012, Sam Steingold wrote: [snip] there is precisely one country for each id. i.e., unique(country) is the same as country[1]. thanks a lot for the suggestion! R result - f[, list(min=min(delay), max=max(delay), count=.N,country=country[1L]), by=share.id] And is it performant? It just occurred to me that this is even better: R setkeyv(f, c(share.id, delay)) R result - f[, list(min=delay[1L], max=delay[.N], count=.N, country=country[1L]), by=share.id] -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://thereligionofpeace.com http://pmw.org.il http://honestreporting.com http://americancensorship.org Why do you never call me back after I scream that I will never talk to you again?! -- 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 [[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] Psych package: fa.diagram, how to re-arrange layout so numbers do not over-write each other
Dear R help I have conducted a fa() analysis, and I want to use fa.diagram to assess the extent to which the 11 latent factors predict the 37 items in a psychological battery. However, the display on the screen has very large font size for the coefficients of the relationship between the 11 factors and the 37 items, so the numbers overlap and are therefore illegible. When I output this to a pdf the problem is exacerbated even further (see attachments). I tried to use cex=0.8 and cex=0.4 in the pdf argument but it made no difference - what can I do? The code I used is: fa.diagram(fa.11factors.rawdata) # Start PDF device driver to save output to fa_diagram.pdf pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram.pdf, height=10, width=5) fa.diagram(fa.11factors.rawdata) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram_cex.pdf, height=8, width=5) fa.diagram(fa.11factors.rawdata,cex=0.8) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) Thank you so much for your time Yours sincerely Brent Caldwell fa_diagram_cex.pdf Description: fa_diagram_cex.pdf fa_diagram.pdf Description: fa_diagram.pdf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Psych package: fa.diagram, how to re-arrange layout so numbers do not over-write each other
Dear Bill Wow! Thank you so much for your rapid reply - you are such a kind person, thank you! I'll try fa.rgraph - thanks Thanks Best wishes Brent -Original Message- From: William R Revelle [mailto:reve...@northwestern.edu] Sent: Tuesday, 27 November 2012 5:54 p.m. To: Brent Caldwell Cc: r-help@R-project.org Subject: Re: [R] Psych package: fa.diagram, how to re-arrange layout so numbers do not over-write each other Brent, No, cex doesn't work (as you have discovered). That is a bug. I will work on it. In the meantime, try Rgraphviz called from fa.rgaph or use the output from fa.graph which produces a dot file for processing with any graphic package (including graphviz) which handles the dot language. Bill On Nov 26, 2012, at 10:08 PM, Brent Caldwell brent.caldw...@otago.ac.nz wrote: Dear R help I have conducted a fa() analysis, and I want to use fa.diagram to assess the extent to which the 11 latent factors predict the 37 items in a psychological battery. However, the display on the screen has very large font size for the coefficients of the relationship between the 11 factors and the 37 items, so the numbers overlap and are therefore illegible. When I output this to a pdf the problem is exacerbated even further (see attachments). I tried to use cex=0.8 and cex=0.4 in the pdf argument but it made no difference - what can I do? The code I used is: fa.diagram(fa.11factors.rawdata) # Start PDF device driver to save output to fa_diagram.pdf pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram.pdf, height=10, width=5) fa.diagram(fa.11factors.rawdata) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram_cex.pdf, height=8, width=5) fa.diagram(fa.11factors.rawdata,cex=0.8) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) Thank you so much for your time Yours sincerely Brent Caldwell fa_diagram_cex.pdffa_diagram.pdf__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. William Revellehttp://personality-project.org/revelle.html Professor http://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern Universityhttp://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 5 minutes to midnighthttp://www.thebulletin.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.
Re: [R] Problem with glm, gaussian family with log-link
On Mon, Nov 26, 2012 at 5:33 AM, Florian Weiler fweile...@jhubc.it wrote: Dear all, I am using the book Generalized Linera Models and Extension by Hardin and Hilbe (second edition, 2007) at the moment. The authors suggest that instead of OLS models, the log link is generally used for response data that take only positive values on the continuous scale. snip specifying *family=gaussian(link=log) *I am asked to provide starting values. When I set them all equal to zero, I always get the message that the algorithm did not converge. Picking other values the message is sometimes the same, but more often I get: * * *Error in glm.fit(x = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, : * * NA/NaN/Inf in 'x' * * * As I said, in STATA I can run these models without setting starting values and without errors. I tried many different models, and different datasets, And yet you've failed to provide even one of them together with your code as a reproducible example ... # This works without starting values: set.seed(2341) x - rep(1:10,3) ; y - jitter(rpois(30,5+x)) plot(x,y) (gausslog - glm(y~x,family=gaussian(link='log'))) exp(coef(gausslog)) # This works only with starting values set.seed(2341) x - rep(1:10,3) ; y - jitter(rpois(30,x)) plot(x,y) ; summary(y) # yes,yes, some y 0, just trying to reproduce the error... (gausslog - glm(y~x,family=gaussian(link='log'))) (gausslog - glm(y~x,family=gaussian(link='log'),start=c(0,0))) # also set.seed(2341) x - rep(1:10,3) ; y - rnorm(30,0+0.1*x) plot(x,y) ; summary(y) (gausslog - glm(y~x,family=gaussian(link='log'),start=c(0,0))) So really this is a non issue without the offending data set and code. but the problem is always the same (unless I only include one single independent variable). Oh, more information... way to build up the suspense set.seed(2341) x - rep(1:10,3) ; xx - rep(seq(20,50,l=5),6) ; y - rnorm(30,5+3*x-2*xx) (gausslog - glm(y~x+xx,family=gaussian(link='log'),start=c(0,0,0))) No joy. Still fits. Could anyone tell me why this is the case, or what I do wrong, No or why the suggested models from the book might not be appropriate? I'd appreciate any help! Personally I don't care for reproducing some results from STATA and have no comment on the validity of the above but maybe someone in the list would have a better answer if you repost. Best, Florian Also this: [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factor function for coded numerical values
The question says that there is an experiement to investigate the effect on breathing rate when doing different types of exercise with wearing more clothes or less clothes (factor A, coded 1,2). The duration of exercise was 10min, 20min, 30min (factor B, coded 1,2,3). Could you give m anymore help? I tried using your R code but I couldn't get it to run. -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-for-coded-numerical-values-tp4650483p4650883.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] creation of an high frequency series
Thank you Paul, It works. Now I will study for repeating the same code for different days. Regards, Vincent -- View this message in context: http://r.789695.n4.nabble.com/creation-of-an-high-frequency-series-tp4650850p4650898.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] merging two specific rows in a DF
HI,
Just a modification of Rui's solution:
x1-within(x,TYPE-as.character(TYPE))
group-cumsum(c(TRUE,x1$TYPE[-1]!=x1$TYPE[-length(x1$TYPE)]))
res-as.data.frame(do.call(rbind,lapply(split(x1,group),function(x)
c(C1=min(x[,1]),C2=max(x[,2]),TYPE=x[,3][1]))),stringsAsFactors=FALSE)
res[,-3]-sapply(res[,-3],as.numeric)
res
# C1 C2 TYPE
#1 10 20 A
#2 33 44 B
#3 66 80 A
#4 111 220 B
#5 300 449 A
#6 455 500 B
#7 510 520 A
#8 540 580 B
A.K.
- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: karthicklakshman karthick.laksh...@gmail.com
Cc: r-help@r-project.org
Sent: Monday, November 26, 2012 12:37 PM
Subject: Re: [R] merging two specific rows in a DF
Hello,
Try the following.
x - read.table(text=
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
, header = TRUE)
x
fun - function(x){
mn - which.min(x$C1)
mx - which.max(x$C2)
c(C1 = x$C1[mn], C2 = x$C2[mx], TYPE = x$TYPE[1])
}
idx - seq_len(nrow(x))[-1]
idx2 - cumsum(c(FALSE, x$TYPE[idx - 1] != x$TYPE[idx]))
y - do.call(rbind, lapply(split(x, idx2), fun))
rownames(y) - seq_len(nrow(y))
y
Hope this helps,
Rui Barradas
Em 26-11-2012 10:24, karthicklakshman escreveu:
Hello members,
I have this data frame with 3 columns,
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
Here, the rows 4 , 5 has type B and similarly 6,7 has A . I need to
merge these rows in a way to get the output with unique type, something like
below, where the lowest value from DF$C1 and highest value from DF$C2
corresponding to rows 4,5 are picked.
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 220 B
300 449 A
455 500 B
510 520 A
540 580 B
I Request your kind help..
Regards,
karthick
--
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Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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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 strange behaviour when building huge concatenation
Any idea on the reason why the instruction p1x-c(2, 1, 0, 0, 0, 7, 1, 0, 2, 0, 5, 0, 0, 0, 0, 0, 10, 1, 3, 0, 2, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 12, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 4, 0, 1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 7, 0, 10, 0, 0, 1, 6, 0, 0, 0, 0, 0, 2, 4, 1, 5, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 4, 3, 0, 1, 2, 0, 1, 2, 0, 2, 1, 1, 0, 5, 2, 7, 2, 0, 4, 13, 0, 4, 4, 0, 0, 1, 0, 1, 29, 0, 3, 0, 0, 1, 0, 10, 0, 0, 13, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 6, 8, 0, 0, 0, 1, 0, 4, 0, 2, 3, 3, 0, 0, 0, 0, 0, 6, 0, 1, 0, 0, 2, 0, 2, 2, 0, 0, 1, 0, 1, 0, 0, 7, 0, 0, 0, 2, 0, 4, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 1, 1, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 5, 0, 0, 2, 0, 1, 2, 0, 0, 1, 0, 1, 0, 5, 0, 0, 1, 2, 0, 1, 0, 0, 1, 2, 1, 0, 1, 0, 1, 0, 3, 1, 13, 0, 0, 0, 0, 0, 3, 3, 1, 0, 0, 3, 0, 5, 0, 2, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 3, 1, 1, 0, 1, 0, 0, 0, 4, 0, 2, 0, 3, 0, 0, 1, 0, 1, 0, 2, 0, 1, 3, 25, 0, 0, 0, 0, 5, 0, 2, 0, 0, 0, 5, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 4, 0, 0, 1, 3, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 11, 0, 0, 4, 0, 0, 1, 2, 0, 0, 0, 0, 1, 1, 0, 3, 1, 0, 0, 2, 0, 8, 1, 0, 2, 1, 0, 0, 1, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 4, 0, 0, 2, 0, 6, 0, 1, 0, 1, 1, 1, 2, 1, 0, 7, 4, 0, 0, 0, 7, 0, 1, 1, 0, 2, 1, 0, 4, 0, 1, 0, 0, 2, 0, 0, 4, 0, 0, 2, 1, 0, 1, 0, 11, 0, 4, 0, 0, 0, 4, 0, 3, 0, 1, 1, 0, 0, 6, 3, 0, 0, 0, 0, 2, 0, 2, 0, 18, 0, 1, 0, 1, 0, 0, 0, 5, 0, 1, 0, 6, 0, 2, 0, 0, 2, 0, 1, 1, 0, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 2, 0, 2, 0, 0, 0, 0, 1, 10, 0, 1, 3, 3, 0, 2, 0, 0, 12, 0, 1, 2, 2, 0, 0, 0, 0, 5, 0, 0, 2, 0, 5, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 1, 7, 0, 0, 2, 0, 0, 2, 0, 4, 0, 0, 3, 0, 1, 0, 2, 2, 0, 1, 1, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 0, 3, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 2, 1, 1, 0, 0, 0, 0, 5, 1, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 8, 1, 0, 6, 1, 2, 0, 3, 6, 0, 1, 0, 2, 5, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 13, 0, 0, 1, 3, 0, 5, 0, 2, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 4, 6, 1, 2, 0, 0, 2, 0, 7, 0, 0, 0, 0, 0, 3, 0, 5, 0, 2, 1, 3, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 1, 10, 0, 0, 0, 0, 3, 3, 0, 0, 2, 5, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 3, 1, 0, 0, 0, 0, 6, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 0, 1, 0, 2, 0, 0, 7, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 1, 13, 7, 0, 0, 3, 0, 1, 0, 1, 1, 2, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 19, 2, 2, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 7, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 0, 1, 1, 0, 0, 4, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 5, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 7, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 4, 0, 1, 1, 0, 1, 0, 1, 2, 0, 1, 0, 0, 0, 2, 0, 4, 0, 1, 7, 1, 0, 1, 0, 0, 4, 0, 1, 0, 0, 0, 0, 1, 0, 7, 1, 0, 6, 0, 5, 0, 2, 0, 1, 0, 6, 0, 2, 0, 0, 0, 2, 2, 0, 0, 6, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 4, 0, 0, 1, 0, 1, 0, 0, 1, 0, 5, 2, 0, 0, 0, 3, 0, 12, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 1, 0, 0, 3, 0, 1, 0, 1, 0, 0, 0, 0, 4, 0, 0, 2, 0, 5, 0, 3, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 3, 0, 0, 0, 0, 11, 0, 0, 2, 0, 1, 0, 7, 0, 0, 1, 0, 3, 0, 2, 0, 1, 0, 4, 2, 2, 1, 0, 0, 0, 0, 0, 1, 3, 0, 0, 1, 1, 6, 0, 0, 4, 0, 0, 1, 0, 2, 0, 1, 3, 7, 2, 0, 5, 0, 0, 0, 0, 5, 0, 12, 1, 0, 1, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 4, 0, 1, 0, 1, 0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 3, 4, 1, 0, 2, 1, 2, 1, 0, 0, 2, 1, 0, 0, 0, 1, 7, 0, 6, 0, 6, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 3, 0, 8, 0, 1, 0, 0, 2, 0, 0, 11, 0, 1, 8, 0, 1, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 3, 1, 0, 1, 6, 0, 2, 1, 1, 0, 1, 1, 0, 3, 2, 0, 0, 0, 2, 1, 1, 0, 0, 0, 0, 3, 2, 0, 3, 0, 0, 1, 4, 0, 6, 0, 1, 1, 4, 0, 2, 0, 4, 0, 0, 0, 0, 2, 0, 1, 0, 1, 2, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 1, 1, 3, 2, 0, 0, 0, 1, 0, 0, 0, 4, 0, 1, 3, 0, 0, 0, 0, 0, 0, 3, 1, 0, 1, 1, 2, 0, 2, 0, 0, 0, 3, 0, 1, 0, 1, 0, 9, 0, 0, 4, 0, 2, 0, 5, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 4, 1, 0, 0, 1, 5, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 3, 0, 0, 2, 0, 14, 4, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 3, 0, 2, 0, 0, 1, 1, 6, 1, 6, 0, 1, 2, 0, 0, 2, 1, 0, 2, 1, 0, 1, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 1, 0, 0, 5, 0, 2, 0, 1, 0, 5, 0, 1, 2, 0, 2, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 3, 0, 4, 0, 1, 0, 0, 3, 1, 0, 2, 0, 2, 1, 0, 0, 0, 4, 0, 0, 0, 26, 1, 6, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 1, 2, 0, 1, 0, 0, 0, 14, 0, 1, 0, 2, 4, 0, 1, 6, 0, 0, 1, 3, 1, 0, 0, 0, 0, 1, 1, 0, 8, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 3, 0, 1, 1, 1, 1, 0, 1, 9, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 2, 0, 4, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 1, 0, 3, 2, 0, 2, 5, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 5, 1, 0, 1, 0, 4, 0, 0, 0, 0, 0, 4, 3, 0,
[R] A problem subsetting a data frame
Hi all, I have this microarray large microarray data set (ALL) from which I would like to subset or extract a set of data based on a factor ($mol.biol).I looked up some example of subsetting in, picked up two commands and tried both but I got error messages as follows testset - subset(ALL, ALL$mol.biol %in% c(BCR/ABL,ALL1/AF4)) Error in c(BCR/ABL, ALL1/AF4) : unused argument(s) (ALL1/AF4) testset - ALL[ALL$mol.biol %in% c(BCR/ABL,NEG), ] Error in ALL[ALL$mol.biol %in% c(BCR/ABL, NEG), ] : error in evaluating the argument 'i' in selecting a method for function '[': Error in c(BCR/ABL, NEG) : unused argument(s) (NEG) At this point, I really appreciate any inputs to move forward. …. str(ALL) Formal class 'ExpressionSet' [package Biobase] with 7 slots ..@ experimentData :Formal class 'MIAME' [package Biobase] with 13 slots .. .. ..@ name : chr Chiaretti et al. .. .. ..@ lab : chr Department of Medical Oncology, Dana-Farber Cancer Institute, Department of Medicine, Brigham and Women's Hospital, Harvard Med| __truncated__ .. .. ..@ contact : chr .. .. ..@ title: chr Gene expression profile of adult T-cell acute lymphocytic leukemia identifies distinct subsets of patients with different respo| __truncated__ .. .. ..@ abstract : chr Gene expression profiles were examined in 33 adult patients with T-cell acute lymphocytic leukemia (T-ALL). Nonspecific filteri| __truncated__ .. .. ..@ url : chr .. .. ..@ pubMedIds: chr [1:2] 14684422 16243790 .. .. ..@ samples : list() .. .. ..@ hybridizations : list() .. .. ..@ normControls : list() .. .. ..@ preprocessing: list() .. .. ..@ other: list() .. .. ..@ .__classVersion__:Formal class 'Versions' [package Biobase] with 1 slots .. .. .. .. ..@ .Data:List of 1 .. .. .. .. .. ..$ : int [1:3] 1 0 0 ..@ assayData:environment: 0x1078636e8 ..@ phenoData:Formal class 'AnnotatedDataFrame' [package Biobase] with 4 slots .. .. ..@ varMetadata :'data.frame': 21 obs. of 1 variable: .. .. .. ..$ labelDescription: chr [1:21] Patient ID Date of diagnosis Gender of the patient Age of the patient at entry ... .. .. ..@ data :'data.frame': 128 obs. of 21 variables: .. .. .. ..$ cod : chr [1:128] 1005 1010 3002 4006 ... .. .. .. ..$ diagnosis : chr [1:128] 5/21/1997 3/29/2000 6/24/1998 7/17/1997 ... .. .. .. ..$ sex : Factor w/ 2 levels F,M: 2 2 1 2 2 2 1 2 2 2 ... .. .. .. ..$ age : int [1:128] 53 19 52 38 57 17 18 16 15 40 ... .. .. .. ..$ BT: Factor w/ 10 levels B,B1,B2,..: 3 3 5 2 3 2 2 2 3 3 ... .. .. .. ..$ remission : Factor w/ 2 levels CR,REF: 1 1 1 1 1 1 1 1 1 1 ... .. .. .. ..$ CR: chr [1:128] CR CR CR CR ... .. .. .. ..$ date.cr : chr [1:128] 8/6/1997 6/27/2000 8/17/1998 9/8/1997 ... .. .. .. ..$ t(4;11) : logi [1:128] FALSE FALSE NA TRUE FALSE FALSE ... .. .. .. ..$ t(9;22) : logi [1:128] TRUE FALSE NA FALSE FALSE FALSE ... .. .. .. ..$ cyto.normal : logi [1:128] FALSE FALSE NA FALSE FALSE FALSE ... .. .. .. ..$ citog : chr [1:128] t(9;22) simple alt. NA t(4;11) ... .. .. .. ..$ mol.biol : Factor w/ 6 levels ALL1/AF4,BCR/ABL,..: 2 4 2 1 4 4 4 4 4 2 ... .. .. .. ..$ fusion protein: Factor w/ 3 levels p190,p190/p210,..: 3 NA 1 NA NA NA NA NA NA 1 ... .. .. .. ..$ mdr : Factor w/ 2 levels NEG,POS: 1 2 1 1 1 1 2 1 1 1 ... .. .. .. ..$ kinet : Factor w/ 2 levels dyploid,hyperd.: 1 1 1 1 1 2 2 1 1 NA ... .. .. .. ..$ ccr : logi [1:128] FALSE FALSE FALSE FALSE FALSE FALSE ... .. .. .. ..$ relapse : logi [1:128] FALSE TRUE TRUE TRUE TRUE TRUE ... .. .. .. ..$ transplant: logi [1:128] TRUE FALSE FALSE FALSE FALSE FALSE ... .. .. .. ..$ f.u : chr [1:128] BMT / DEATH IN CR REL REL REL ... .. .. .. ..$ date last seen: chr [1:128] NA 8/28/2000 10/15/1999 1/23/1998 ... .. .. ..@ dimLabels: chr [1:2] sampleNames sampleColumns .. .. ..@ .__classVersion__:Formal class 'Versions' [package Biobase] with 1 slots .. .. .. .. ..@ .Data:List of 1 .. .. .. .. .. ..$ : int [1:3] 1 1 0 ..@ featureData :Formal class 'AnnotatedDataFrame' [package Biobase] with 4 slots .. .. ..@ varMetadata :'data.frame': 0 obs. of 1 variable: .. .. .. ..$ labelDescription: logi(0) .. .. ..@ data :'data.frame': 12625 obs. of 0 variables .. .. ..@ dimLabels: chr [1:2] featureNames featureColumns .. .. ..@ .__classVersion__:Formal class 'Versions' [package Biobase] with 1 slots .. .. .. .. ..@ .Data:List of 1 .. .. .. .. .. ..$ : int [1:3] 1 1 0 ..@ annotation : chr hgu95av2 ..@ protocolData :Formal class
[R] Help with graphics in gamm4 library
My problem is relatively straight forward, but I cannot seem to find a way to make it work. I have a RCBD with repeated measurements over time. I have created a fit using the gamm4 package. My model is: fit4a - gamm4(Rate ~ s(Time,by=trt,bs=cr)+trt,data=qual.11.dat, random=~(1|block),correlation=corARH1()) What I would like to create is plots with the X-axis being time, the Y-axis being the fitted Rates for each individual treatment with the smoothed curves overlayed on the plot. Every idea I have had to do this has resulted in some errors, and I have reached my wits end. Can anyone steer me in the right direction? -- View this message in context: http://r.789695.n4.nabble.com/Help-with-graphics-in-gamm4-library-tp4650908.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] Plot(x,y) help
Thanks Jim and Rui. -- View this message in context: http://r.789695.n4.nabble.com/Plot-x-y-help-tp4650874p4650906.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] Problem with cbind() and arguments
Yeah, I meant 3 instead of 5. This was just an example, it's not what Im really doing. I am using a 'arules' package for data mining, and I have to pass and 'arg[]' element and use it as the new column name of a data.frame. It's a bit complicated, so I used this example, and I would like to use 'args[1]' as 'c', in a new column name using cbind.. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-cbind-and-arguments-tp4650808p4650912.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] Problem with cbind() and arguments
In this case, instead of naming the column 'c', it names it 'args[1]' as a string, not a variable. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-cbind-and-arguments-tp4650808p4650913.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] glm convergence warning
Hello, When I run the following glm model: modelresult=glm(CID~WS+SS+DV+DS, data=kimu, family=binomial) I get the following warning messages: 1: glm.fit: algorithm did not converge 2: glm.fit: fitted probabilities numerically 0 or 1 occurred What I am trying to do is model my response variable (CID: correct bird identification) as a function of the predictor variables weather state (WS), sea state (SS), distance from the vessel (DV) and duration of the sighting (DS). I defined both sea state and weather state as factors with three levels (0, 1, or 2). Distance of the vessel values are 100, 80, 60, 40, and 20. Duration of the sighting ranges from 0 to 58 seconds. The output R is giving me is: Deviance Residuals: Min 1Q Median 3Q Max -3.562e-05 -2.100e-08 2.100e-08 2.100e-08 3.632e-05 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) -2.000e+02 1.067e+06 0.0001.000 WSf1 7.744e+01 9.086e+04 0.0010.999 WSf2 1.285e+01 6.199e+04 0.0001.000 SSf1-1.042e+02 1.683e+05 -0.0011.000 SSf2-1.859e+02 1.432e+05 -0.0010.999 DV 6.770e-01 9.394e+03 0.0001.000 DS 9.822e+00 1.884e+04 0.0011.000 What do the warning messages mean? Can I still use coefficient estimates and standard error values? 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.
[R] Class Not found execption
I am trying to call a r file named es.r which have lotes of R functions. These R functions are internally calling java functions by using .jnew() and .jcall(). I have added necessary jar's to the Classpath and I am able to run es.r from command prompt . But when I tried to call it from anoter r file using source() getting Exception Error in .jnew(com/algoTree/ClientElasticSearch/ElasticSearchLoader) : java.lang.ClassNotFoundException How can I solve this ? How can i execute the file from r-console. ? Anybody can help me.. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Class-Not-found-execption-tp4650916.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] Psych package: fa.diagram, how to re-arrange layout so numbers do not over-write each other
Brent, No, cex doesn't work (as you have discovered). That is a bug. I will work on it. In the meantime, try Rgraphviz called from fa.rgaph or use the output from fa.graph which produces a dot file for processing with any graphic package (including graphviz) which handles the dot language. Bill On Nov 26, 2012, at 10:08 PM, Brent Caldwell brent.caldw...@otago.ac.nz wrote: Dear R help I have conducted a fa() analysis, and I want to use fa.diagram to assess the extent to which the 11 latent factors predict the 37 items in a psychological battery. However, the display on the screen has very large font size for the coefficients of the relationship between the 11 factors and the 37 items, so the numbers overlap and are therefore illegible. When I output this to a pdf the problem is exacerbated even further (see attachments). I tried to use cex=0.8 and cex=0.4 in the pdf argument but it made no difference - what can I do? The code I used is: fa.diagram(fa.11factors.rawdata) # Start PDF device driver to save output to fa_diagram.pdf pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram.pdf, height=10, width=5) fa.diagram(fa.11factors.rawdata) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) pdf(file=I:\\ZAPi\\Tables-of-R-output\\fa_diagram_cex.pdf, height=8, width=5) fa.diagram(fa.11factors.rawdata,cex=0.8) #box() # Create box around plot dev.off() # Turn off device driver (to flush output to PDF) Thank you so much for your time Yours sincerely Brent Caldwell fa_diagram_cex.pdffa_diagram.pdf__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. William Revellehttp://personality-project.org/revelle.html Professor http://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern Universityhttp://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 5 minutes to midnighthttp://www.thebulletin.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.
