Re: [R] nth root of matrix
On 7/2/2013 9:24 PM, David Winsemius wrote:
On Jul 2, 2013, at 8:11 PM, Sachinthaka Abeywardana wrote:
Hi all,
I want to do the following:
a=matrix(c(-1,-2,-3))
a^(1/3) #get 3rd root of numbers[,1]
[1,] NaN
[2,] NaN
[3,] NaN
All I get is NaNs, what is the proper way of doing this? Would like to
retain the fact that it is a matrix if possible (not a requirement
though).
?complex
a=matrix(c(-1+0i,-2+0i,-3+0i))
I tried that. The problem is that there are 3 different cube
roots in the complex plane, and a^(1/3) only gives one of them. See
Wikipedia, roots of unity or the examples in the help file for
newton_raphson {elliptic}.
I assume that Sachinthaka wants the real roots. Try the following:
n - 3 # n must be an odd integer for this to work
a=matrix(c(-1,-2,-3))
as - sign(a)
ab - abs(a)
cr - as*(ab^(1/n))
cr
[,1]
[1,] -1.00
[2,] -1.259921
[3,] -1.442250
cr^n
Hope this helps. Spencer Graves
David Winsemius
Alameda, CA, USA
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Re: [R] Recoding variables based on reference values in data frame
Hi,
May be this helps:
Kgeno- read.table(text=
SNP_ID SNP1 SNP2 SNP3 SNP4
Maj_Allele C G C A
Min_Allele T A T G
ID1 CC GG CT AA
ID2 CC GG CC AA
ID3 CC GG nc AA
ID4 _ _ _ _
ID5 CC GG CC AA
ID6 CC GG CC AA
ID7 CC GG CT AA
ID8 _ _ _ _
ID9 CT GG CC AG
ID10 CC GG CC AA
ID11 CC GG CT AA
ID12 _ _ _ _
ID13 CC GG CC AA
,sep=,header=TRUE,stringsAsFactors=FALSE)
library(stringr)
library(car)
fun1- function(x){
MajMin- paste0(x[1],x[2])
MajMaj-str_dup(x[1],2)
MinMin-str_dup(x[2],2)
recode(x,'nc'=NA;'_'=NA;MajMaj=0;MajMin=1;MinMin=2)}
sapply(Kgeno[,-1],fun1)
#or
mat1-sapply(Kgeno[1:2,-1],function(x)
{c(str_dup(x,2),paste(x,collapse=))})[c(1,3,2),]
sapply(seq_len(ncol(Kgeno[,-1])),function(i)
{x-Kgeno[-c(1:2),-1][,i];as.numeric(factor(x,levels=mat1[,i]))-1})
#Speed comparison
KgenoNew- rbind(Kgeno[c(1:2),-1],sapply(Kgeno[-c(1:2),-1],rep,1e4))
system.time(res1- sapply(KgenoNew,fun1))
# user system elapsed
#0.672 0.000 0.674
system.time({
mat1-sapply(Kgeno[1:2,-1],function(x)
{c(str_dup(x,2),paste(x,collapse=))})[c(1,3,2),]
res2- sapply(seq_len(ncol(KgenoNew)),function(i){ x-
KgenoNew[-c(1:2),][,i];as.numeric(factor(x,levels=mat1[,i]))-1})
})
#user system elapsed
# 0.212 0.000 0.214
res1New- res1[-c(1:2),]
res1New1- as.numeric(res1New)
dim(res1New1)- dim(res1New)
identical(res1New1,res2)
#[1] TRUE
A.K.
- Original Message -
From: kathleen askland k.askl...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Tuesday, July 2, 2013 4:46 PM
Subject: [R] Recoding variables based on reference values in data frame
I'm new to R (previously used SAS primarily) and I have a genetics data
frame consisting of genotypes for each of 300+ subjects (ID1, ID2, ID3,
...) at 3000+ genetic locations (SNP1, SNP2, SNP3...). A small subset of
the data is shown below:
SNP_ID SNP1 SNP2 SNP3 SNP4 Maj_Allele C G C A Min_Allele T A T G ID1
CC GG CT AA ID2 CC GG CC AA ID3 CC GG
nc
AA ID4 _ _ _ _ ID5 CC GG CC AA ID6 CC GG CC
AA ID7 CC GG CT AA ID8 _ _ _ _ ID9 CT GG
CC AG ID10 CC GG CC AA ID11 CC GG CT AA
ID12 _ _ _ _ ID13 CC GG CC AA
The name of the data file is Kgeno.
What I would like to do is recode all of the genotype values to standard
integer notation, based on their values relative to the reference rows
(Maj_Allele and Min_Allele). Standard notation sums the total of minor
alleles in the genotype, so values can be 0, 1 or 2.
Here are the changes I want to make:
1. If the genotype= nc or '_ then set equal to NA.
2. If genotype value = a character string comprised of two consecutive
major allele values -- c(Maj_Allele, Maj_Allele) -- then set equal to 0.
3. If genotype value= c(Maj_Allele, Min_Allele) then set equal to 1.
4. If genotype value = c(Min_Allele, Min_Allele) then set equal to 2.
I've tried the following ifelse processing but get error (Warning: Executed
script did not end with R session at the top-level prompt. Top-level state
will be restored) and can't seem to fix the code properly. I've counted the
parentheses. Also, not sure if it would execute properly if I could fix it.
# change 'nc' and '_' to NA, else leave as is:
Kgeno[,2] - ifelse(Kgeno[,2] == nc, NA, Kgeno[,2])
Kgeno[,2] - ifelse(Kgeno[,2] == _, NA, Kgeno[,2])
#convert genotype strings in the first data column to numeric values #(two
major alleles=0, 1 minor and 1 major=1, 2 minor alleles=2), else #leave as
is (to preserve NA values).
Kgeno[,2] -
ifelse(Kgeno[,2] == noquote(paste(as.character(Kgeno[1,2]), as.character(
Kgeno[1,2]), sep=), 0,
ifelse(Kgeno[,2] == noquote(paste(as.character(Kgeno[1,2]), as.character(
Kgeno[2,2]), sep=), 1,
ifelse(Kgeno[,2] == noquote(paste(as.character(Kgeno[2,2]), as.character(
Kgeno[2,2]), sep=), 2,
Kgeno[,2])))
Finally, if above code were corrected, this would only change the first
column of data, but I would like to change all 3000+ columns in the same
way.
I would greatly appreciate some suggestions on how to proceed.
Thank you,
Kathleen
---
Kathleen Askland, MD
Assistant Professor
Department of Psychiatry Human Behavior
The Warren Alpert School of Medicine
Brown University/Butler Hospital
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Re: [R] matching similar character strings
Dear Mario, Not sure if this is what you wanted: F1_ex- read.table(text= Nome.azienda;Indirizzo 17;Alterego;Via Edmondo De Amicis, 18 18;Alterego;Via Edmondo De Amicis, 65 ,sep=;,header=TRUE,stringsAsFactors=FALSE) F2_ex- read.table(text= CODICE;STRADA;AREADICIRCOLAZIONE;NUMBER1;BARRATO1;NUMBER2;BARRATO2;SECTION 1;15620;VIA;DE AMICIS EDMONDO;1;;5;;1288 2;15620;VIA;DE AMICIS EDMONDO;2;;34;;1261 3;15620;VIA;DE AMICIS EDMONDO;7;;17;;1287 4;15620;VIA;DE AMICIS EDMONDO;36;;62;;1264 5;15620;VIA;DE AMICIS EDMONDO;37;;37;;1287 6;15620;VIA;DE AMICIS EDMONDO;64;;84;;1262 ,sep=;,header=TRUE,stringsAsFactors=FALSE) library(stringr) vec1-sapply(lapply(toupper(str_trim(gsub([0-9,],,F1_ex[,2]))),word,c(1,3,4,2)),paste,collapse= ) vec2- as.numeric(gsub(\\D+,,F1_ex[,2])) F1_ex[,1]-F2_ex[sapply(vec2,function(x) which((xF2_ex[,4] x F2_ex[,6]) paste(F2_ex[,2],F2_ex[,3])%in%vec1)),SECTION] F1_ex # Nome.azienda Indirizzo #17 1261 Via Edmondo De Amicis, 18 #18 1262 Via Edmondo De Amicis, 65 A.K. - Original Message - From: A M Lavezzi mario.lave...@unipa.it To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 2, 2013 10:22 AM Subject: Re: [R] matching similar character strings Dear Arun, please excuse me for this late reply, we had to stop working on this temporaririly. Let me reproduce here two examples of rows from F1 and F2 (sorry, but with dput() I am not able to produce a clear example) F1_ex Nome.azienda Indirizzo 17 Alterego Via Edmondo De Amicis, 18 On row 17 of F1 we have a firm named (Nome.azienda) 'Alterego' whose address (indirizzo) is 'Via Edmondo de Amicis, 18' Below I reproduce the portion of F2 with information on the street mentioned in F1_ex$Indirizzo. F2_ex CODICE STRADA AREADICIRCOLAZIONE NUMBER1 BARRATO1 NUMBER2 BARRATO2 SECTION 1 15620 VIA DE AMICIS EDMONDO 1 5 1288 2 15620 VIA DE AMICIS EDMONDO 2 34 1261 3 15620 VIA DE AMICIS EDMONDO 7 17 1287 4 15620 VIA DE AMICIS EDMONDO 36 62 1264 5 15620 VIA DE AMICIS EDMONDO 37 37 1287 6 15620 VIA DE AMICIS EDMONDO 64 84 1262 Line 1 says that the portion of VIA DE AMICIS EDMONDO (STRADA+AREADICIRCOLAZIONE), with street numbers between 1 and 5 belongs to SECTION 1288 (these are census sections). (BARRATO1 and BARRATO2 refer to the letter in street numbers such as 12/A, 28/D, etc. In the present example they are empty) Line 2 says that the portion of VIA DE AMICIS EDMONDO, with street numbers between 2 and 34 belongs to SECTION 1261, etc. Our problem is to assign SECTION 1261 to 'Alterego', exploting the information on its address. The problem is that the syntax of the street address in F1 is different from the syntax in F2. Hope I have clarified the issue thanks a lot Mario On Fri, Jun 21, 2013 at 5:25 PM, arun smartpink...@yahoo.com wrote: Dear Mario, I didn't find any difference between 1st and 2nd row of F2, except for the last three columns. Question is that why should F1 1st row should be merged to 2nd row of F2 instead of 1st row of F2. In your previous example, you mentioned about A1, A2, ... and B1, B2, etc. Here, it is not provided. As I mentioned before, it is better to provide the output of ?dput() from a subset of dataset. dput(head(F1,20)) dput(head(F2,20)) #so that there would be atleast some matching pairs within the example dataset. Also, please post it to r-help as I will be able to check only after a couple of hours Tx. Arun - Original Message - From: Mario Lavezzi mario.lave...@unipa.it To: arun smartpink...@yahoo.com Cc: Sent: Friday, June 21, 2013 11:08 AM Subject: Re: [R] matching similar character strings dear Arun thank you very much. Let me explain the problem: Imagine that a portion of the row in F1 is: F1 1) Street | J.F. Kennedy | 30 it means that our unit of interest (a firm) has address: J.F. Kennedy Street, 30 The F2 database contains the list of all the streets of the city, with additional variables characterizing that street (Census data). The database contains sometimes street divided in some parts, according to the street number. For example: Example of three rows of F2 concerning Kennedy street and Kennedy Road: F2 1) Street | Kennedy John Fitzgerald | 1 | 20 | A12 2) Street | Kennedy John
Re: [R] nth root of matrix
sign(a)*abs(a)^(1/3) # [,1] #[1,] -1.00 #[2,] -1.259921 #[3,] -1.442250 A.K. - Original Message - From: Sachinthaka Abeywardana sachin.abeyward...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Tuesday, July 2, 2013 11:11 PM Subject: [R] nth root of matrix Hi all, I want to do the following: a=matrix(c(-1,-2,-3)) a^(1/3) #get 3rd root of numbers[,1] [1,] NaN [2,] NaN [3,] NaN All I get is NaNs, what is the proper way of doing this? Would like to retain the fact that it is a matrix if possible (not a requirement though). Thanks, Sachin [[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] Multinomial model and p-values
On Jul 2, 2013, at 21:21 , Luciano La Sala wrote: Hello everyone, I have a dataset which consists of Pathology scores (Absent, Mild, Severe) as outcome variable, and two main effects: Age (two factors: twenty / thirty days old) and Treatment Group (four factors: infected without ATB; infected + ATB1; infected + ATB2; infected + ATB3). First I tried to fit an ordinal regression model, which seems more appropriate given the characteristics of my dependent variable (ordinal). However, the assumption of odds proportionality was severely violated (graphically), which prompted me to use a multinomial model instead, using the nnet package. First I chose the outcome level that I need to use as baseline category: Data$Path - relevel(Data$Path, ref = Absent) Then, I needed to set baseline categories for the independent variables: Data$Age - relevel(Data$Age, ref = Twenty) Data$Treat - relevel(Data$Treat, ref = infected without ATB) The model: test - multinom(Path ~ Treat + Age, data = Data) # weights: 18 (10 variable) initial value 128.537638 iter 10 value 80.623608 final value 80.619911 converged summary1 - summary(test1) summary1 Call: multinom(formula = Jej_fact ~ Treat + Age, data = Data) Coefficients: (Intercept) infected+ATB1 infected+ATB2 infected+ATB3 AgeThirty Moderate -2.238106 -1.1738540 -1.709608 -1.599301 2.684677 Severe -1.544361 -0.8696531 -2.991307 -1.506709 1.810771 Std. Errors: (Intercept)infected+ATB1 infected+ATB2 infected+ATB3 AgeThirty Moderate 0.78800460.8430368 0.7731359 0.7718480 0.8150993 Severe 0.61109030.7574311 1.1486203 0.7504781 0.6607360 Residual Deviance: 161.2398 AIC: 181.2398 For a while, I could not find a way to get the p-values for the model and estimates when using nnet:multinom. Yesterday I came across a post where the author put forward a similar issue regarding estimation of p-values for coefficients (http://stats.stackexchange.com/questions/9715/how-to-set-up-and-estimate-a- multinomial-logit-model-in-r). There, one blogger suggested that getting p-values from the summary() result of multinom() is pretty easy, by first getting the t values as follows: pt(abs(summary1$coefficients / summary1$standard.errors), df=nrow(Data)-10, lower=FALSE) (Intercept) infected+ATB1 infected+ATB2 infected+ATB3 AgeThirty Moderate 0.002670340 0.08325396 0.014506395 0.02025858 0.0006587898 Severe 0.006433581 0.12665278 0.005216581 0.02352202 0.0035612114 I AM NOT a statistician, so don't be baffled by a silly question! I this procedure correct? There's at least a factor of 2 missing for a two-tailed p value. It is usually a mistake to use the t-distribution for what is really a z-statistic; for aggregated data, it can be a very bad mistake. However, it's not really an R question, and you obviously know where to find stackexchange... (local, professional advice would be even better, though.) -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] MuMIn package: plotting averaged model estimates with CI
Hiya, I'm fairly new to MuMIn (and plotting predicted values in R, for that matter) and I have quite a basic question: I want to plot model-averaged estimates with their 95% CI's for the supported interaction (Infection x ecs). I used the following data: head(summ12b) ld sum12 Infection ecs 1 13 78.0 Infected Enlarged 2 15 134.5 Uninfected Enlarged 3 14 91.5 Infected Control 4 16 62.0 Infected Reduced 5 17 93.6 Infected Control 6 15 104.6 Uninfected Control s12b-lm(sum12~(Infection+ecs)^2+ld,data=summ12b) ss12b-dredge(s12b,trace = TRUE, rank = AICc) top.ss12b- get.models(ss12b, subset = delta2) ad-summary(model.avg(top.ss12b)) and here's what I've got: ad Call: model.avg.default(object = top.ss12b) Component models: df logLik AICc Delta Weight 1234 8 -112.95 249.10 0.00 0.47 124 7 -115.31 249.95 0.86 0.31 1 4 -120.44 250.55 1.45 0.23 Term codes: ecs Infectionld ecs:Infection 1 2 3 4 Model-averaged coefficients: Estimate Std. Error Adjusted SE z value Pr(|z|) (Intercept) 112.737 30.847 31.505 3.578 0.000346 *** ecsEnlarged23.817 13.376 13.798 1.726 0.084324 . ecsReduced-10.597 8.352 8.740 1.212 0.225364 InfectionUninfected 7.493 9.381 9.893 0.757 0.448793 ld -3.477 1.763 1.865 1.864 0.062252 . ecsEnlarged :InfectionUninfected 23.308 14.740 15.510 1.503 0.132892 ecsReduced :InfectionUninfected -22.112 12.589 13.307 1.662 0.096572 . --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Full model-averaged coefficients (with shrinkage): (Intercept) ecsEnlarged ecsReduced InfectionUninfected ld ecsEnlarged :InfectionUninfected 112.7373 23.8169-10.5966 5.7952 -1.6285 18.0261 ecsReduced :InfectionUninfected -17.1008 Relative variable importance: ecs Infection ecs:Infectionld 1.00 0.77 0.77 0.47 I know that I can obtain 95% Ci's by using ad$avg.model, however, how to derive the estimates and CI's for ALL the groups and their factor levels included in the interaction? Thanks for your help! cheers, Kasia [[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] Package PLM
Hi Giovanni and Yves, My is Eliano Marques, i'm a master degree econometrician from ISEG in Portugal. I was trying to use your R package with a specific output i cannot get the same results that i'm getting in Stata. Could you please let me know if the below is correct or if i'm missing anything? On the below model I want to run a model for 62 companies for a 5 year period: log(revenue+2)it = log(revenue+2)(-1)it*b1 + log(Cost+2)(-1)it*b2 + log(Cost+2)it*b3 + vit modeloutput- pgmm(log(Revenue+2) ~ lag(log(Revenue+2), 1)+ lag(log(Cost+2), 0:1) | lag(log(Cost+2), 2:99) + lag(log(Cost+2), 2:99) , data = BD, effect = twoways, model = onestep, transformation = ld) summary(modeloutput, robust = TRUE) I want to include time.dummies in both equation in level and differences. I get this output: Coefficients Estimate Std. Error z-value Pr(|z|) lag(log(Revenue + 2), 1)0.486593 0.146151 3.3294 0.0008703 *** lag(log(Cost+ 2), 0:1)0 1.165293 0.069329 16.8081 2.2e-16 *** lag(log(Cost + 2), 0:1)1 -0.565378 0.199199 -2.8383 0.0045360 ** From Stata, running the following code I get this: xtabond2 lnrevenue l.lnrevenue lncost l.lncost i.year , iv(i.year ) gmm(l.lnrevenue lncost ) robust small lnrevenue | Coef. Std. Err. tP|t| [95% Conf. Interval] -+ lnrevenue | L1. |.358987 .1176716 3.05 0.003 .123688 .594286 | lncost | --. | 1.185943 .036389132.59 0.000 1.113179 1.258708 L1. | -.4332851 .1551862-2.79 0.007-.7435992 -.1229711 | year | 2009 | -.2100918 .0466175-4.51 0.000-.3033094 -.1168743 2010 | -.0936157 .0365742-2.56 0.013-.1667503 -.020481 2011 | -.0351653.036502-0.96 0.339-.1081556 .037825 | _cons | .3372286.194194 1.74 0.088-.0510863 .7255435 -- Instruments for first differences equation Standard D.(2008b.year 2009.year 2010.year 2011.year 2012.year) GMM-type (missing=0, separate instruments for each period unless collapsed) L(1/4).(L.lnrevenue lncost) Instruments for levels equation Standard 2008b.year 2009.year 2010.year 2011.year 2012.year _cons GMM-type (missing=0, separate instruments for each period unless collapsed) D.(L.lnrevenue lncost) Do you what I be doing wrong to correct this values? Many thanks in advance, Eliano [[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] spped up a function
first thing to do when trying to speed up a function is to see where it is
spending its time. take a subset of the data and use Rprof to profile the
code. my guess is that a lot of time is taken up in the use of dataframes.
see if you can use matrices instead.
Sent from my iPad
On Jul 2, 2013, at 13:47, Santiago Guallar sgual...@yahoo.com wrote:
Hi,
I have written a function to assign the values of a certain variable 'wd'
from a dataset to another dataset. Both contain data from the same time
period but differ in the length of their time intervals: 'GPS' has regular
10-minute intervals whereas 'xact' has irregular intervals. I attached
simplified text versions from write.table. You can also get a dput of 'xact'
in this address: http://www.megafileupload.com/en/file/431569/xact-dput.html).
The original objects are large and the function takes almost one hour to
finish.
Here's the function:
fxG= function(xact, GPS){
l - rep( 'A', nrow(GPS) )
v - unique(GPS$Ring) # the process is carried out for several individuals
identified by 'Ring'
for(k in 1:length(v) ){
I = v[k]
df - xact[xact$Ring == I,]
for(i in 1:nrow(GPS)){
if(GPS[i,]$Ring== I){# the code runs along the whole data.frame for each i;
it'd save time to make it stop with the last record of each i instead
u - df$timepos = GPS[i,]$timepos
# fill vector l for each interval t from xact = each interval from GPS (take
the max if there's 1 interval)
l[i] - df[max( which(u == TRUE) ),]$wd
}
}
}
return(l)}
vwd - fxG(xact, GPS)
My question is: how can I speed up (optimize) this function?
Thank you for your help
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] nlrq with normal copula
Dear all, I want to run a nomal copula based quantile regression with XLS data. I've worked out how to load xls data in r. the regression fumula is simply: y= a+bx+e and the quantile curve is http://r.789695.n4.nabble.com/file/n4670779/%E6%97%A0%E6%A0%87%E9%A2%98.png I use: excel.file-file.path(D:/dailyvix.xls) dailyvix-readWorksheetFromFile(excel.file, sheet=1) names(dailyvix) [1] y x fit1-nlrq(y~qnorm(pnorm(mu*qnorm(pnorm(x))+sqrt(1-mu*mu)*qnorm(tau))), tau=0.5, data=dailyvix) However, I get Error in getInitial.default(func, data, mCall = as.list(match.call(func, : no 'getInitial' method found for function objects I am a greenhand. I don't know how to deal with such error. I really need help Looking forward to any reply Yours, Tong -- View this message in context: http://r.789695.n4.nabble.com/nlrq-with-normal-copula-tp4670779.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] merge dataframes
Hello,
I have two dataframes:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2-data.frame(x=seq(0,6,by=0.1),y=rep(0,60)))
I want to replace the corresponding rows of dat2 with the ones of dat1.
I tried:
for(i in 1:nrow(dat1))
{
dat2[dat2$x==dat1[i,1],2]-dat1[i,2]
}
But I discovered that not every 5.1 is equal:
dat2[52,1][1] 5.1 dat1[5,1][1] 5.1 dat2[52,1]==dat1[5,1][1] FALSE
How to solve this?
Regards,
André
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] merge dataframes
FAQ 7.31
On Wed, Jul 3, 2013 at 7:55 AM, André de Boer rnie...@gmail.com wrote:
Hello,
I have two dataframes:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2-data.frame(x=seq(0,6,by=0.1),y=rep(0,60)))
I want to replace the corresponding rows of dat2 with the ones of dat1.
I tried:
for(i in 1:nrow(dat1))
{
dat2[dat2$x==dat1[i,1],2]-dat1[i,2]
}
But I discovered that not every 5.1 is equal:
dat2[52,1][1] 5.1 dat1[5,1][1] 5.1 dat2[52,1]==dat1[5,1][1] FALSE
How to solve this?
Regards,
André
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
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.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Non-linear modelling with several variables including a categorical variable
If preytype is an independent variable, then models based on it should be OK. If preytype comes into the parameters you are trying to estimate, then the easiest way is often to generate all the possible combinations (integers -- fairly modest number of these) and run all the least squares minimizations. Crude but effective. nlxb from nlmrt or nlsLM from minpack.lm may be more robust in doing this, but less efficient if nls works OK. JN On 13-07-03 06:00 AM, r-help-requ...@r-project.org wrote: Message: 10 Date: Tue, 2 Jul 2013 19:01:55 +0700 From: Robbie Weteringsrobbie.weteri...@gmail.com To:r-help@r-project.org Subject: [R] Non-linear modelling with several variables including a categorical variable Message-ID: cafe5dhzrm+bpg1v77ezhun+tacv64j_9pnsfgh_xne5csz9...@mail.gmail.com Content-Type: text/plain Hello everyone, I am trying to model some data regarding a predator prey interaction experiment (n=26). Predation rate is my response variable and I have 4 explanatory variables: predator density (1,2,3,4 5), predator size, prey density (5,10,15,20,25,30) and prey type (3 categories). I started with several linear models (glm) and found (as expected) that prey and predator density were non-linear related to predation rates. If I use a log transformation on these variables I get really nice curves and an adjusted R2 of 0.82, but it is not really the right approach for modelling non-linear relationships. Therefore I switched to non-linear least square regression (nls). I have several predator-prey models based on existing ecological literature e.g.: model1 - nls(rates ~ (a * prey)/(1 + b * prey), start = list(a = 0.27,b = 0.13), trace = TRUE) ### Holling's type II functional response model2 - nls(rates ~ (a*prey)/(1+ (b * prey) + c * (pred -1 )), start = list(a=0.22451, b=-0.18938, c=1.06941), trace=TRUE, subset=I1) ### Beddington-**DeAngelis functional response These models work perfectly, but now I want to add prey type as well. In the linear models prey type was the most important variable so I don't want to leave it out. I understand that you can't add categorical variables in nls, so I thought I try a generalized additive model (gam). The problem with the gam models is that the smoothers (both spline and loess) don't work on both variables because there are only a very restricted number of values for prey density and predator density. I can manage to get a model with a single variable smoothed using loess. But for two variables it is simply not working. The spline function does not work at all because I have so few values (5) for my variables (see model 4). model3 - gam(rates~ lo(pred, span=0.9)+prey) ## this one is actually working but does not include a smoother for prey. model4 - gam(rates~ s(pred)+prey) ## this one gives problems: *A term has fewer unique covariate combinations than specified maximum degrees of freedom* My question is: are there any other possibilities to model data with 2 non-linear related variables in which I can also include a categorical variable. I would prefer to use nls (model2) with for example different intercepts for each category but I'm not sure how to get this sorted, if it is possible at all. The dataset is too small to split it up into the three categories, moreover, one of the categories only contains 5 data points. Any help would be really appreciated. With kind regards, -- Robbie Weterings *Project Manager Cat Drop Thailand ** Tel: +66(0)890176087 * 65/13 Mooban Chakangrao, Naimuang Muang Kamphaeng Phet 62000, Thailand àÅ¢·Õè 65/13 Á.ªÒ¡Ñ§ÃÒÇ ¶¹¹ ÃÒª´íÒà¹Ô¹2 ã¹àÁ×ͧ ÍíÒàÀÍ/ ࢵ àÁ×ͧ¡íÒàྦྷྪà ¨Ñ§ËÇÑ´ ¡íÒàྦྷྪà 62000 http://www.catdropfoundation.org http://www.catdropfoundation.org/facebook/Facebook.html *www.catdropfoundation.org* http://www.catdropfoundation.org/ *www.facebook.com/catdropfoundation*http://www.facebook.com/catdropfoundation *Boorn 45, 9204 AZ, Drachten, The Netherlands* [[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] One-parameter fitting
Greetings to every body. we know that for linear regression: two-parameter fitting in R: (for a and b as slope and intercept) [a] for y=ax+b type models we use: lm(y~x) [b] for y=axtype models we use: lm(y~0+x) = forced to pass through origin Now I have question: what about y=b fitting? is there any model to force or impose the ax to be zero Let say x - c(1,2,3,4,5,6,7,8,9) y - c( 0.853,0.852, 0.854, 0.858, 0.862, 0.856, 0.858, 0.857, 0.863) plot(y~x, xlim=c(0,10), ylim=c(0,1)) abline(lm(y~x),col=blue) # this doesn't give exact horizontal but slightly inclined. or is it as simple as, just find the intercept using the lm(y~x) , then abline(h=intercept,col=red) ? any comment or advice is greatly appreciated ;) [[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] One-parameter fitting
Hello, Try ?mean abline(h = mean(y), col = blue) Hope this helps, Rui Barradas Em 03-07-2013 15:07, Birdada Simret escreveu: Greetings to every body. we know that for linear regression: two-parameter fitting in R: (for a and b as slope and intercept) [a] for y=ax+b type models we use: lm(y~x) [b] for y=axtype models we use: lm(y~0+x) = forced to pass through origin Now I have question: what about y=b fitting? is there any model to force or impose the ax to be zero Let say x - c(1,2,3,4,5,6,7,8,9) y - c( 0.853,0.852, 0.854, 0.858, 0.862, 0.856, 0.858, 0.857, 0.863) plot(y~x, xlim=c(0,10), ylim=c(0,1)) abline(lm(y~x),col=blue) # this doesn't give exact horizontal but slightly inclined. or is it as simple as, just find the intercept using the lm(y~x) , then abline(h=intercept,col=red) ? any comment or advice is greatly appreciated ;) [[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] merge dataframes
In addition to reading Frequently Asked Questions Section 7.31, you
should always run your own code before posting. Yours does not work.
Spend some time reading one or more of the R tutorials and learn
about vectorization. It will save you time and typing. There are
many to choose from at
http://cran.r-project.org/other-docs.html
Do not try to match decimal values. Instead create an index value
that is a character string. Then you can match the character
strings:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2 - data.frame(x=seq(0.1,6,by=0.1),y=rep(0,60)) # Errors in
your code corrected here
dat1$xc - sprintf(%1.1f, dat1$x, 1) # Create dat1 index value
as character
dat2$xc - sprintf(%1.1f, dat2$x, 1) # Create dat2 index value
as character
dat2[match(dat1$xc, dat2$xc),] - dat1 # update dat2 with dat1
dat2[dat2$y0,] # Check the results
x y xc
10 1.0 23 1.0
12 1.2 17 1.2
32 3.2 12 3.2
40 4.0 27 4.0
51 5.1 8 5.1
dat1
x y xc
1 1.0 23 1.0
2 1.2 17 1.2
3 3.2 12 3.2
4 4.0 27 4.0
5 5.1 8 5.1
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of jim holtman
Sent: Wednesday, July 3, 2013 7:22 AM
To: André de Boer
Cc: R mailing list
Subject: Re: [R] merge dataframes
FAQ 7.31
On Wed, Jul 3, 2013 at 7:55 AM, Andri de Boer rnie...@gmail.com
wrote:
Hello,
I have two dataframes:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2-data.frame(x=seq(0,6,by=0.1),y=rep(0,60)))
I want to replace the corresponding rows of dat2 with the ones of
dat1.
I tried:
for(i in 1:nrow(dat1))
{
dat2[dat2$x==dat1[i,1],2]-dat1[i,2]
}
But I discovered that not every 5.1 is equal:
dat2[52,1][1] 5.1 dat1[5,1][1] 5.1 dat2[52,1]==dat1[5,1][1]
FALSE
How to solve this?
Regards,
Andri
[[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.
--
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.
[[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] merge dataframes
Hello,
Or use ?all.equal.
all.equal(dat2[52, 1], dat1[5, 1]) # TRUE
Hope this helps,
Rui Barradas
Em 03-07-2013 15:18, David Carlson escreveu:
In addition to reading Frequently Asked Questions Section 7.31, you
should always run your own code before posting. Yours does not work.
Spend some time reading one or more of the R tutorials and learn
about vectorization. It will save you time and typing. There are
many to choose from at
http://cran.r-project.org/other-docs.html
Do not try to match decimal values. Instead create an index value
that is a character string. Then you can match the character
strings:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2 - data.frame(x=seq(0.1,6,by=0.1),y=rep(0,60)) # Errors in
your code corrected here
dat1$xc - sprintf(%1.1f, dat1$x, 1) # Create dat1 index value
as character
dat2$xc - sprintf(%1.1f, dat2$x, 1) # Create dat2 index value
as character
dat2[match(dat1$xc, dat2$xc),] - dat1 # update dat2 with dat1
dat2[dat2$y0,] # Check the results
x y xc
10 1.0 23 1.0
12 1.2 17 1.2
32 3.2 12 3.2
40 4.0 27 4.0
51 5.1 8 5.1
dat1
x y xc
1 1.0 23 1.0
2 1.2 17 1.2
3 3.2 12 3.2
4 4.0 27 4.0
5 5.1 8 5.1
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of jim holtman
Sent: Wednesday, July 3, 2013 7:22 AM
To: André de Boer
Cc: R mailing list
Subject: Re: [R] merge dataframes
FAQ 7.31
On Wed, Jul 3, 2013 at 7:55 AM, Andri de Boer rnie...@gmail.com
wrote:
Hello,
I have two dataframes:
dat1-data.frame(x=c(1.0,1.2,3.2,4.0,5.1),y=c(23,17,12,27,8))
dat2-data.frame(x=seq(0,6,by=0.1),y=rep(0,60)))
I want to replace the corresponding rows of dat2 with the ones of
dat1.
I tried:
for(i in 1:nrow(dat1))
{
dat2[dat2$x==dat1[i,1],2]-dat1[i,2]
}
But I discovered that not every 5.1 is equal:
dat2[52,1][1] 5.1 dat1[5,1][1] 5.1 dat2[52,1]==dat1[5,1][1]
FALSE
How to solve this?
Regards,
Andri
[[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
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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] One-parameter fitting
Now I have question: what about y=b fitting? is there any model to
force or impose the ax to be zero
Rui Barradas has answered correctly for a simple lm case.
You can also do
lm(y~1) to obtain an estimated mean.
That formulation happens to be quite useful if you are interested in a purely
random-effects model* fitted using, say, lme, though you won't generally get a
simple mean for unbalanced data in such a case.
S Ellison
***
This email and any attachments are confidential. Any use...{{dropped:8}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] One-parameter fitting
Thank you all ;)
On Wed, Jul 3, 2013 at 4:28 PM, S Ellison s.elli...@lgcgroup.com wrote:
Now I have question: what about y=b fitting? is there any model to
force or impose the ax to be zero
Rui Barradas has answered correctly for a simple lm case.
You can also do
lm(y~1) to obtain an estimated mean.
That formulation happens to be quite useful if you are interested in a
purely random-effects model* fitted using, say, lme, though you won't
generally get a simple mean for unbalanced data in such a case.
S Ellison
***
This email and any attachments are confidential. Any u...{{dropped:13}}
__
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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] promise already under evaluation
Hi, I asked this question on SO but got no answers:
http://stackoverflow.com/questions/17310825/r-promise-already-under-evaluation
I understand that you are probably sick and tired of answering the same
question again,
but I am still getting the error discussed in several other questions:
promise already under evaluation: recursive default argument reference or
earlier problems?
even though I did follow the cumbersome advice of prepending .:
--8---cut here---start-8---
show.large.objects.threshold - 10
show.large.objects.exclude - c(closure)
show.large.objects - function (.envir = sys.frame(),
threshold = show.large.objects.threshold,
exclude = show.large.objects.exclude) {
for (n in print(ls(.envir, all.names = TRUE))) tryCatch({
o - get(n,envir = .envir)
s - object.size(o)
if (s threshold !(typeof(o) %in% exclude)) {
cat(n,: )
print(s,units=auto)
}
}, error = function(e) { cat(n=,n,\n); print(e) })
}
show.large.objects.stack - function (threshold = show.large.objects.threshold,
skip.levels = 1,# do not examine the last
level - this function
exclude = show.large.objects.exclude) {
for (level in 1:(sys.nframe()-skip.levels)) {
cat(*** show.large.objects.stack(,level,) )
print(sys.call(level))
show.large.objects(.envir = sys.frame(level))
}
}
--8---cut here---end---8---
but I still get errors:
--8---cut here---start-8---
f - function () { c - 1:1e7; d - 1:1e6;
print(system.time(show.large.objects.stack())) }
f()
*** show.large.objects.stack( 1 ) f()
[1] c d
c : 38.1 Mb
d : 3.8 Mb
*** show.large.objects.stack( 2 ) print(system.time(show.large.objects.stack()))
[1] ... x
n= ...
simpleError in get(n, envir = .envir): argument ... is missing, with no
default
n= x
simpleError in get(n, envir = .envir): promise already under evaluation:
recursive default argument reference or earlier problems?
*** show.large.objects.stack( 3 ) system.time(show.large.objects.stack())
[1] exprgcFirst ppt time
n= expr
simpleError in get(n, envir = .envir): promise already under evaluation:
recursive default argument reference or earlier problems?
user systemelapsed
0 (0.00ms) 0 (0.00ms) 0.002 (2.00ms)
--8---cut here---end---8---
So, what am I still doing wrong?
Do I really need the . in .envir?
Why do I get the [[argument ... is missing, with no default]] error?
Why do I get the [[promise already under evaluation]] error?
What is the right way to pass threshold and exclude from
show.large.objects.stack to show.large.objects?
Thanks!
PS. I would prefer an answer on SO, but please feel free to reply using any
venue you like and I will copy your explanation to the other venues.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 13.04 (raring) X 11.0.11303000
http://www.childpsy.net/ http://iris.org.il http://mideasttruth.com
http://honestreporting.com http://openvotingconsortium.org
Linux - find out what you've been missing while you've been rebooting Windows.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] promise already under evaluation
* Sam Steingold f...@tah.bet [2013-07-03 11:33:47 -0400]:
Hi, I asked this question on SO but got no answers:
http://stackoverflow.com/questions/17310825/r-promise-already-under-evaluation
Backlin explained on SO that the errors are to be expected: ... is a
formal argument which was not supplied and expr and x were actually
being evaluated at the time of get() call.
The bottom line is that I must catch and ignore errors.
The remaining problem is: how do I pass the same arguments down?
e.g.,
--8---cut here---start-8---
f - function (... verbose=FALSE ...) { ... }
g - function (... verbose=FALSE ...) { ... f(... verbose=verbose ...) ... }
--8---cut here---end---8---
results in promise already under evaluation (and, yes, I do understand
why).
is there anything better than
--8---cut here---start-8---
f - function ( ... f.verbose=FALSE ... ) { ... }
g - function ( ... g.verbose=FALSE ... ) { ... f(... f.verbose=g.verbose ...)
... }
--8---cut here---end---8---
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 13.04 (raring) X 11.0.11303000
http://www.childpsy.net/ http://www.memritv.org http://mideasttruth.com
http://honestreporting.com http://think-israel.org http://jihadwatch.org
Incorrect time synchronization.
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[R] Splitting a string expression into components
I have a vector of strings that contain mathematical expressions. E.g., x - c(5 = 7, z = 1+2) and I would like to decompose each expression into its left- and right-hand-side components etc., output something like 5 = 7 z = 1 + 2 Is there something built into the R language to accomplish this? Thanks for advance. Dan [[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] Splitting a string expression into components
If your syntax is like R's syntax then parse() can help. R p - parse(text=z = 1 + 2) R p[[1]] # the first ( only) expression in z = 1 + 2 z = 1 + 2 R as.list(p[[1]]) # the 'function' called (`=`) and its 'arguments' [[1]] `=` [[2]] z [[3]] 1 + 2 R as.list(p[[1]][[3]]) # ditto for the 2nd argument above [[1]] `+` [[2]] [1] 1 [[3]] [1] 2 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 Dan Murphy Sent: Wednesday, July 03, 2013 10:38 AM To: R-help@r-project.org Subject: [R] Splitting a string expression into components I have a vector of strings that contain mathematical expressions. E.g., x - c(5 = 7, z = 1+2) and I would like to decompose each expression into its left- and right-hand-side components etc., output something like 5 = 7 z = 1 + 2 Is there something built into the R language to accomplish this? Thanks for advance. Dan [[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] Splitting a string expression into components
Alternatively, perhaps strsplit() with an appropriate regex argument. See ?strsplit ?regexp e.g. strsplit (z=5,split = [=]+) [[1]] [1] z 5 This of course will only work if you know what the characters are that define left and right, how complex the expressions might be, etc. -- i.e. the simple case. -- Bert On Wed, Jul 3, 2013 at 10:55 AM, William Dunlap wdun...@tibco.com wrote: If your syntax is like R's syntax then parse() can help. R p - parse(text=z = 1 + 2) R p[[1]] # the first ( only) expression in z = 1 + 2 z = 1 + 2 R as.list(p[[1]]) # the 'function' called (`=`) and its 'arguments' [[1]] `=` [[2]] z [[3]] 1 + 2 R as.list(p[[1]][[3]]) # ditto for the 2nd argument above [[1]] `+` [[2]] [1] 1 [[3]] [1] 2 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 Dan Murphy Sent: Wednesday, July 03, 2013 10:38 AM To: R-help@r-project.org Subject: [R] Splitting a string expression into components I have a vector of strings that contain mathematical expressions. E.g., x - c(5 = 7, z = 1+2) and I would like to decompose each expression into its left- and right-hand-side components etc., output something like 5 = 7 z = 1 + 2 Is there something built into the R language to accomplish this? Thanks for advance. Dan [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Academic study of R users and developers
Dear all, A group of researchers from Eindhoven Univ of Technology (The Netherlands) and UC Davis (USA) is conducting a study of support activities in R. In this study, our goal is to understand how R user support activities have been evolving over time, in the presence of different information exchange media, such as mailing lists, forums, blogs or questionanswer websites. Your participation is voluntary and confidential. If you agree to participate, you will be asked to complete self-report measures that tell us a bit about your information seeking and information providing experiences related to R. Participation in this study is expected to take about 10 minutes of your time. We thank you in advance for your participation in this study. We plan to include the results of this survey in a scientific publication. Individual responses cannot be traced back to an individual respondent. https://docs.google.com/forms/d/14dXaSdy2NJ94Fq50fAbmEr3_GK8XfbRdys9qPLMMZ-c/viewform https://docs.google.com/forms/d/14dXaSdy2NJ94Fq50fAbmEr3_GK8XfbRdys9qPLMMZ-c/viewform Best regards, Alexander Serebrenik Eindhoven Univ Technology The Netherlands -- View this message in context: http://r.789695.n4.nabble.com/Academic-study-of-R-users-and-developers-tp4670797.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] The R Journal, Volume 5, Issue 1
Dear all, The latest issue of The R Journal is now available at http://journal.r-project.org/archive/2013-1/ Many thanks to all contributors. Hadley -- Editor-in-chief, The R Journal ___ r-annou...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] String based chemical name identification
The problem is the following: I have two big databases one look like this: 2-Methyl-4-trimethylsilyloxyoct-5-yne Benzoic acid, methyl ester Benzoic acid, 2-methyl-, methyl ester Acetic acid, phenylmethyl ester 2,7-Dimethyl-4-trimethylsilyloxyoct-7-en-5-yne etc. The second one looks like this: Name: D-Tagatose 1,6-bisphosphate Name: 1-Phosphatidyl-D-myo-inositol;: 1-Phosphatidyl-1D-myo-inositol;: 1-Phosphatidyl-myo-inositol;: Phosphatidyl-1D-myo-inositol;: (3-Phosphatidyl)-1-D-inositol;: 1,2-Diacyl-sn-glycero-3-phosphoinositol;: Phosphatidylinositol Name: Androstenedione;: Androst-4-ene-3,17-dione;: 4-Androstene-3,17-dione Name: Spermine;: N,N'-Bis(3-aminopropyl)-1,4-butanediamine Name: H+;: Hydron Name: 3-Iodo-L-tyrosine etc. Both of them have more then 3000 lines. Matching their name by hand is not an option because I don't know chemistry. *Possible solution I came up with*: Go through all the names of the first database and then try to match with the other one. I'm using *regexec *and *strsplit *functions for the matching. Basically I split the name into small chunks and try to get some hit in the other database. I can supply code If needed but I did not want to spam in the first mail. Any solution is welcome! It can be in pseudo-cod also or in any type of logical arguing. It does not matter. Laszlo-Andras Zsurzsa Msc. Informatics, Technical University Munchen [[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] Subsetting multiple rows of a data frame at once
Hi, Try this: set.seed(24) df- data.frame(x=sample(seq(0.25,4.25,by=.05),1e5,replace=TRUE),y= sample(seq(0.10,1.05,by=.05),1e5,replace=TRUE),z=rnorm(1e5)) #Used a shorter vector x1- c(1.05,2.85,3.40,4.25,0.25) y1- c(0.25,0.10,0.90,0.25,1.05) res-do.call(rbind,lapply(seq_along(x1),function(i) subset(df,x==x1[i]y==y1[i]))) head(res,2) # x y z #466 1.05 0.25 0.7865224 #4119 1.05 0.25 -1.5679096 tail(res,2) # x y z #98120 0.25 1.05 -2.1239596 #98178 0.25 1.05 0.3321464 A.K. Hi Everyone, First time poster so any posting rules i should know about feel free to advise... I've got a data frame of 250 000 rows in columns of x y and z. i need to extract 20-30 rows from the data frame with specific x and y values, such that i can find the z value that corresponds. There is no repeated data. (its actually 250 000 squares in a 5x5m grid) to find them individually i can use subset successfully result-subset(df,x==1.05 y==c0.25) gives me the row in the dataframe with that x and y value. so if i have x = 1.05 2.85 3.40 4.25 0.25 3.05 3.70 0.20 0.30 0.70 1.05 1.20 1.40 1.90 2.70 3.25 3.55 4.60 2.05 2.15 3.70 4.85 4.90 1.60 2.45 3.20 3.90 4.45 and y= 0.25 0.10 0.90 0.25 1.05 1.70 2.05 2.90 2.35 2.60 2.55 2.15 2.75 2.05 2.70 2.25 2.55 2.05 3.65 3.05 3.00 3.50 3.75 4.85 4.50 4.50 3.35 4.90 then how can i retrieve the rows for all those values at once. if i name x=xt and y=yt and then result-subset(df,x==xt y==yt) then i get result [1] x y Height 0 rows (or 0-length row.names) i dont understand why zero rows are selected. obviously im applying the vectors inappropriately, but i cant seem to find anything on this method of subsetting online. Thanks for any replies! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Meta-analysis on a repeated measures design with multiple trials per subject using metafor
Hi all, I am currently attempting to compile a summary of a series of five psychological experiments, and I am trying to do this using the metafor package. However, I am quite unsure which of the scenarios described in the metafor help pages applies to these data, because it is a repeated measures design, with multiple trials in each condition. Assume that for every participant, I have a basic contingency table such as this one: treatment 1 2 response 1 10 20 2 20 10 (if this ASCII version does not work, I have 30 trials in each treatment, and participants give either response 1 or 2; the exact numbers don't matter) The problem that I am trying to solve is how to convert these numbers to an effect size estimate that I can use with metafor. As far as I understand it, I can only use it to get an effect size for outcomes that are dichotomous; i.e., either 1 or 0 for any subject. However, I have proportion data for every participant. I have considered and tried these strategies: 1. Base the effect size on within-participant proportion differences. That is, in the table above, the treatment effect would be (20/30)-(10/30) = 1/3; and I would take the M and SD of these values to estimate a study-level effect (MN measure in metafor). 2. Use the overall treatment * response contingency table, ignoring the fact that these counts come from different participants (PHI or OR measures in metafor). In a study with 10 participants, I would get cell counts around 150. However, from the research I've done into this topic, I know that 1) is not applicable to (as far as I understand) an odds ratio, and I suspect 2) overestimates the effect. A third method would be to use the regression coefficients, that I can easily obtain since I have all the raw data that I need. However, it is unclear to me whether and if yes, how I can use these in the metafor package. From my understanding of another message about this topic I found on this list (1), I understand that having access to the raw data is an advantage, but I am not sure whether the scenario mentioned applies to my situation. 1: http://r.789695.n4.nabble.com/meta-analysis-with-repeated-measure-designs-td2252644.html I would very much appreciate any suggestions or hints on this topic. Regards, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] nls.profile
Hello Daniel, I have the same problem than you : i use nls.profile to compute confidence intervals of estimated parameters but it fails and stop... I tried to increase alphamax value as Douglas suggested that it should work but it's not... Did you find a method to tell nls.profile to ignore errors and proceed to the next parameter value ? Thank you very much! -- Thomas de Marchin PhD Student Laboratory of Bioenergetics University of Liège, Institute of Botany (B.22-P.70) Bld du Rectorat, 27 Sart Tilman, 4000 Liège, Belgium Tel: +32 (0)4 366 38 08 http://www.biophoto.ulg.ac.be __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] qtlnet mcmc.qtlnet sample genetic architeture and QTL network
Im working with the qtlnet package and want to sample the genetic architecture and QTL network. I aiming to do this with the function qtlnet:::mcmc.qtlnet. So far so good. However I cant do it with my own cross dataset. data(Pscdbp) works fine, but not my own. I mean it is from class cross. However, I receive the error: Error in dimnames(geno.dat)[[2]] - tmp : 'dimnames' must be a list Can anybody help me out? I uploaded my crossdata here: http://www.risclab.com/cross_gpsm.RData Would be awesome if somebody could help me out __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] change cell values
Dear R experts, I have two matrices (mat1 mat2) with the same dimension the cells (row and column) are corresponding to each other. I want to change cell values to NA given values of the corresponding cells in mat1 and mat2 are both 1. E.g. both mat1[2,3] and mat2[2,3] are 1, I will put mat1[2,3]=NA, and mat2[2,3]=NA; if either mat1[2,3]=1 or mat2[2,3]=1, I will save both cells. I tried the code, but not working. Could anyone can help fix the problem? mat1[mat11mat21]=NA mat2[mat11mat21]=NA mat1=matrix(rnorm(12),3) mat2=matrix(rnorm(12),3) mat1 [,1] [,2] [,3] [,4] [1,] -1.3387075 -0.7142333 -0.5614211 0.1846955 [2,] -0.7936087 -0.2215797 -0.3686067 0.7328731 [3,] 0.6505082 0.1826019 1.5577883 -1.5580384 mat2 [,1] [,2] [,3] [,4] [1,] 0.4331573 -1.8086826 -1.7688123 -1.4278934 [2,] -0.1841451 0.1738648 -1.1086942 1.3065109 [3,] -1.0827245 -0.4143808 -0.6889405 0.4046203 [[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] change cell values
Hello, It's not working because you are changing mat1 before the second condition is evaluated, and so mat1 is full of NA values. Try the following. tmp - mat1 # Make a copy tmp[mat1 1 mat2 1] - NA # And change that copy mat2[mat1 1 mat2 1] - NA mat1 - tmp rm(tmp) Hope this helps, Rui Barradas Em 03-07-2013 22:27, JiangZhengyu escreveu: Dear R experts, I have two matrices (mat1 mat2) with the same dimension the cells (row and column) are corresponding to each other. I want to change cell values to NA given values of the corresponding cells in mat1 and mat2 are both 1. E.g. both mat1[2,3] and mat2[2,3] are 1, I will put mat1[2,3]=NA, and mat2[2,3]=NA; if either mat1[2,3]=1 or mat2[2,3]=1, I will save both cells. I tried the code, but not working. Could anyone can help fix the problem? mat1[mat11mat21]=NA mat2[mat11mat21]=NA mat1=matrix(rnorm(12),3) mat2=matrix(rnorm(12),3) mat1 [,1] [,2] [,3] [,4] [1,] -1.3387075 -0.7142333 -0.5614211 0.1846955 [2,] -0.7936087 -0.2215797 -0.3686067 0.7328731 [3,] 0.6505082 0.1826019 1.5577883 -1.5580384 mat2 [,1] [,2] [,3] [,4] [1,] 0.4331573 -1.8086826 -1.7688123 -1.4278934 [2,] -0.1841451 0.1738648 -1.1086942 1.3065109 [3,] -1.0827245 -0.4143808 -0.6889405 0.4046203 [[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] change cell values
Hi Rui, Yes, I see. You are right. Thanks for your help. Zhengyu Date: Wed, 3 Jul 2013 22:38:37 +0100 From: ruipbarra...@sapo.pt To: zhyjiang2...@hotmail.com CC: r-help@r-project.org Subject: Re: [R] change cell values Hello, It's not working because you are changing mat1 before the second condition is evaluated, and so mat1 is full of NA values. Try the following. tmp - mat1 # Make a copy tmp[mat1 1 mat2 1] - NA # And change that copy mat2[mat1 1 mat2 1] - NA mat1 - tmp rm(tmp) Hope this helps, Rui Barradas Em 03-07-2013 22:27, JiangZhengyu escreveu: Dear R experts, I have two matrices (mat1 mat2) with the same dimension the cells (row and column) are corresponding to each other. I want to change cell values to NA given values of the corresponding cells in mat1 and mat2 are both 1. E.g. both mat1[2,3] and mat2[2,3] are 1, I will put mat1[2,3]=NA, and mat2[2,3]=NA; if either mat1[2,3]=1 or mat2[2,3]=1, I will save both cells. I tried the code, but not working. Could anyone can help fix the problem? mat1[mat11mat21]=NA mat2[mat11mat21]=NA mat1=matrix(rnorm(12),3) mat2=matrix(rnorm(12),3) mat1 [,1] [,2] [,3] [,4] [1,] -1.3387075 -0.7142333 -0.5614211 0.1846955 [2,] -0.7936087 -0.2215797 -0.3686067 0.7328731 [3,] 0.6505082 0.1826019 1.5577883 -1.5580384 mat2 [,1] [,2] [,3] [,4] [1,] 0.4331573 -1.8086826 -1.7688123 -1.4278934 [2,] -0.1841451 0.1738648 -1.1086942 1.3065109 [3,] -1.0827245 -0.4143808 -0.6889405 0.4046203 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] modify timestemp
Hey All, I want to standardize my timestamp which is formatted as hh:mm:ss My data looks like this: Date Time 01/01/2013 00:09:01 01/02/2013 00:10:14 01/03/2013 00:11:27 01/04/2013 00:12:40 01/05/2013 00:13:53 01/06/2013 00:15:06 01/07/2013 00:16:19 01/08/2013 00:17:32 01/09/2013 00:18:45 01/10/2013 00:19:58 Dataset - structure(list(Date = c(01/01/2013, 01/02/2013, 01/03/2013, 01/04/2013, 01/05/2013, 01/06/2013, 01/07/2013, 01/08/2013, 01/09/2013, 01/10/2013), Time = c(00:09:01, 00:10:14, 00:11:27, 00:12:40, 00:13:53, 00:15:06, 00:16:19, 00:17:32, 00:18:45, 00:19:58)), .Names = c(Date, Time), class = data.frame, row.names = c(NA, -10L)) I would like to change all the records in Time column uniformed as hh:mm:00, then the output would be this: Date Time 01/01/2013 00:09:00 01/02/2013 00:10:00 01/03/2013 00:11:00 01/04/2013 00:12:00 01/05/2013 00:13:00 01/06/2013 00:15:00 01/07/2013 00:16:00 01/08/2013 00:17:00 01/09/2013 00:18:00 01/10/2013 00:19:00 Thanks for your help! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] change cell values
On Jul 3, 2013, at 2:27 PM, JiangZhengyu wrote: Dear R experts, I have two matrices (mat1 mat2) with the same dimension the cells (row and column) are corresponding to each other. I want to change cell values to NA given values of the corresponding cells in mat1 and mat2 are both 1. E.g. both mat1[2,3] and mat2[2,3] are 1, I will put mat1[2,3]=NA, and mat2[2,3]=NA; if either mat1[2,3]=1 or mat2[2,3]=1, I will save both cells. I tried the code, but not working. Could anyone can help fix the problem? mat1[mat11mat21]=NA mat2[mat11mat21]=NA I beleive the problem is that after the first NA assignment that mat1 will have changed. You need to record the status of both mat1 and mat2 before the change: both - mat11mat21 both [,1] [,2] [,3] [,4] [1,] FALSE FALSE TRUE TRUE [2,] TRUE FALSE TRUE TRUE [3,] FALSE TRUE FALSE TRUE mat1[both] - NA mat2[both] - NA mat1 [,1] [,2] [,3] [,4] [1,] 1.5599872 2.209537 NA NA [2,] NA -1.144140 NA NA [3,] -0.8516326NA 1.000368 NA mat2 [,1] [,2] [,3] [,4] [1,] 0.8863107 -0.3863741 NA NA [2,]NA 1.3185811 NA NA [3,] 1.4487338 NA 1.051689 NA There is also formalism: is.na(object) - logical.vector so it could have been: is.na(mat1) - both is.na(mat2) - both mat1=matrix(rnorm(12),3) mat2=matrix(rnorm(12),3) mat1 [,1] [,2] [,3] [,4] [1,] -1.3387075 -0.7142333 -0.5614211 0.1846955 [2,] -0.7936087 -0.2215797 -0.3686067 0.7328731 [3,] 0.6505082 0.1826019 1.5577883 -1.5580384 mat2 [,1] [,2] [,3] [,4] [1,] 0.4331573 -1.8086826 -1.7688123 -1.4278934 [2,] -0.1841451 0.1738648 -1.1086942 1.3065109 [3,] -1.0827245 -0.4143808 -0.6889405 0.4046203 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius 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] converting a list of loglin terms to a model formula
I'm developing some functions to create symbolic specifications for
loglinear models of different types.
I don't really know how to 'compute' with model formulas, so I've done
this in the notation
for stats::loglin(), which is a list of high-order terms in the model.
What I'd like is a function to turn the results of these into a model
formula, suitable for
MASS::loglm. That's the reverse of what loglm does.
For example, the simplest versions of models for 3-way tables for joint,
conditional, and marginal independence can be computed as follows.
After each, I indicated
the WANTED model formula I'd like from the result
joint(3)
$term1
[1] 1 2
$term2
[1] 3
WANTED: ~ 1:2 + 3
condit(3)
$term1
[1] 1 3
$term2
[1] 2 3
WANTED: ~ 1:2 + 2:3
mutual(3)
$term1
[1] 1
$term2
[1] 2
$term3
[1] 3
WANTED: ~ 1 + 2 + 3
In case anyone want to play with the code, here are the current, not too
elegant definitions
of the functions, and some further test cases,
# models of joint independence
joint - function(nf, factors=1:nf, with=nf) {
if (nf == 1) return (list(term1=factors[1]))
if (nf == 2) return (list(term1=factors[1], term2=factors[2]))
others - setdiff(1:nf, with)
result - list(term1=factors[others], term2=factors[with])
result
}
# conditional independence
condit - function(nf, factors=1:nf, with=nf) {
if (nf == 1) return (list(term1=factors[1]))
if (nf == 2) return (list(term1=factors[1], term2=factors[2]))
main - setdiff(1:nf, with)
others - matrix(factors[with], length(with), length(main))
result - rbind(factors[main], others)
result - as.list(as.data.frame(result, stringsAsFactors=FALSE))
names(result) - paste('term', 1:length(result), sep='')
result
}
# mutual independence
mutual - function(nf, factors=1:nf) {
result - sapply(factors[1:nf], list)
names(result) - paste('term', 1:length(result), sep='')
result
}
### some comparisons
loglin(HairEyeColor, list(c(1, 2), c(1, 3), c(2, 3)))$lrt
loglm(~1:2 + 1:3 +2:3, HairEyeColor)
# use factor names
joint(3, factors=names(dimnames(HairEyeColor)))
condit(3, factors=names(dimnames(HairEyeColor)))
loglin(HairEyeColor, joint(3))$lrt
loglm(~1:2 + 3, HairEyeColor)
loglin(HairEyeColor, condit(3))$lrt
loglm(~1:3 + 2:3, HairEyeColor)
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept. Chair, Quantitative Methods
York University Voice: 416 736-2100 x66249 Fax: 416 736-5814
4700 Keele StreetWeb: http://www.datavis.ca
Toronto, ONT M3J 1P3 CANADA
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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] String based chemical name identification
Might be better off using a web service like ChemSpider to do the matching for you http://www.chemspider.com/AboutServices.aspx?. The idea that you can identify the synonyms by name is probably optimistic unless they are exact matches. Here's some python code that seems to make it pretty easy: https://github.com/mcs07/ChemSpiPy. Search the names, extract the InChI for the best match and then you can match them in R via the InChI. Might require some fixing by hand afterwards. HTH, Jason Law -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Zsurzsa Laszlo Sent: Wednesday, July 03, 2013 7:28 AM To: r-help@r-project.org Subject: [R] String based chemical name identification The problem is the following: I have two big databases one look like this: 2-Methyl-4-trimethylsilyloxyoct-5-yne Benzoic acid, methyl ester Benzoic acid, 2-methyl-, methyl ester Acetic acid, phenylmethyl ester 2,7-Dimethyl-4-trimethylsilyloxyoct-7-en-5-yne etc. The second one looks like this: Name: D-Tagatose 1,6-bisphosphate Name: 1-Phosphatidyl-D-myo-inositol;: 1-Phosphatidyl-1D-myo-inositol;: 1-Phosphatidyl-myo-inositol;: Phosphatidyl-1D-myo-inositol;: (3-Phosphatidyl)-1-D-inositol;: 1,2-Diacyl-sn-glycero-3-phosphoinositol;: Phosphatidylinositol Name: Androstenedione;: Androst-4-ene-3,17-dione;: 4-Androstene-3,17-dione Name: Spermine;: N,N'-Bis(3-aminopropyl)-1,4-butanediamine Name: H+;: Hydron Name: 3-Iodo-L-tyrosine etc. Both of them have more then 3000 lines. Matching their name by hand is not an option because I don't know chemistry. *Possible solution I came up with*: Go through all the names of the first database and then try to match with the other one. I'm using *regexec *and *strsplit *functions for the matching. Basically I split the name into small chunks and try to get some hit in the other database. I can supply code If needed but I did not want to spam in the first mail. Any solution is welcome! It can be in pseudo-cod also or in any type of logical arguing. It does not matter. Laszlo-Andras Zsurzsa Msc. Informatics, Technical University Munchen [[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] change cell values
Hi, set.seed(24) mat1=matrix(rnorm(12),3) set.seed(28) mat2=matrix(rnorm(12),3) indx- mat11 mat21 mat1[indx]-NA mat2[indx]-NA mat1 # [,1] [,2] [,3] [,4] #[1,] NA NA NA 0.002311942 #[2,] NA NA NA NA #[3,] NA NA NA 0.598269113 mat2 # [,1] [,2] [,3] [,4] #[1,] NA NA NA 1.841481 #[2,] NA NA NA NA #[3,] NA NA NA 1.520367 A.K. - Original Message - From: JiangZhengyu zhyjiang2...@hotmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Wednesday, July 3, 2013 5:27 PM Subject: [R] change cell values Dear R experts, I have two matrices (mat1 mat2) with the same dimension the cells (row and column) are corresponding to each other. I want to change cell values to NA given values of the corresponding cells in mat1 and mat2 are both 1. E.g. both mat1[2,3] and mat2[2,3] are 1, I will put mat1[2,3]=NA, and mat2[2,3]=NA; if either mat1[2,3]=1 or mat2[2,3]=1, I will save both cells. I tried the code, but not working. Could anyone can help fix the problem? mat1[mat11mat21]=NA mat2[mat11mat21]=NA mat1=matrix(rnorm(12),3) mat2=matrix(rnorm(12),3) mat1 [,1] [,2] [,3] [,4] [1,] -1.3387075 -0.7142333 -0.5614211 0.1846955 [2,] -0.7936087 -0.2215797 -0.3686067 0.7328731 [3,] 0.6505082 0.1826019 1.5577883 -1.5580384 mat2 [,1] [,2] [,3] [,4] [1,] 0.4331573 -1.8086826 -1.7688123 -1.4278934 [2,] -0.1841451 0.1738648 -1.1086942 1.3065109 [3,] -1.0827245 -0.4143808 -0.6889405 0.4046203 [[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] modify timestemp
Hi, May be this helps: dat1# dataset dat1[,2]-gsub(\\d+$,00,dat1[,2]) dat1 # Date Time #1 01/01/2013 00:09:00 #2 01/02/2013 00:10:00 #3 01/03/2013 00:11:00 #4 01/04/2013 00:12:00 #5 01/05/2013 00:13:00 #6 01/06/2013 00:15:00 #7 01/07/2013 00:16:00 #8 01/08/2013 00:17:00 #9 01/09/2013 00:18:00 #10 01/10/2013 00:19:00 A.K. Hey All, I want to standardize my timestamp which is formatted as hh:mm:ss My data looks like this: Date Time 01/01/2013 00:09:01 01/02/2013 00:10:14 01/03/2013 00:11:27 01/04/2013 00:12:40 01/05/2013 00:13:53 01/06/2013 00:15:06 01/07/2013 00:16:19 01/08/2013 00:17:32 01/09/2013 00:18:45 01/10/2013 00:19:58 Dataset - structure(list(Date = c(01/01/2013, 01/02/2013, 01/03/2013, 01/04/2013, 01/05/2013, 01/06/2013, 01/07/2013, 01/08/2013, 01/09/2013, 01/10/2013), Time = c(00:09:01, 00:10:14, 00:11:27, 00:12:40, 00:13:53, 00:15:06, 00:16:19, 00:17:32, 00:18:45, 00:19:58)), .Names = c(Date, Time), class = data.frame, row.names = c(NA, -10L)) I would like to change all the records in Time column uniformed as hh:mm:00, then the output would be this: Date Time 01/01/2013 00:09:00 01/02/2013 00:10:00 01/03/2013 00:11:00 01/04/2013 00:12:00 01/05/2013 00:13:00 01/06/2013 00:15:00 01/07/2013 00:16:00 01/08/2013 00:17:00 01/09/2013 00:18:00 01/10/2013 00:19:00 Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting a list of loglin terms to a model formula
Try this:
as.formula(sprintf( ~ %s, do.call(paste, c(lapply(mutual(3), paste,
collapse = :), sep = +
On Wed, Jul 3, 2013 at 6:55 PM, Michael Friendly frien...@yorku.ca wrote:
I'm developing some functions to create symbolic specifications for
loglinear models of different types.
I don't really know how to 'compute' with model formulas, so I've done
this in the notation
for stats::loglin(), which is a list of high-order terms in the model.
What I'd like is a function to turn the results of these into a model
formula, suitable for
MASS::loglm. That's the reverse of what loglm does.
For example, the simplest versions of models for 3-way tables for joint,
conditional, and marginal independence can be computed as follows. After
each, I indicated
the WANTED model formula I'd like from the result
joint(3)
$term1
[1] 1 2
$term2
[1] 3
WANTED: ~ 1:2 + 3
condit(3)
$term1
[1] 1 3
$term2
[1] 2 3
WANTED: ~ 1:2 + 2:3
mutual(3)
$term1
[1] 1
$term2
[1] 2
$term3
[1] 3
WANTED: ~ 1 + 2 + 3
In case anyone want to play with the code, here are the current, not too
elegant definitions
of the functions, and some further test cases,
# models of joint independence
joint - function(nf, factors=1:nf, with=nf) {
if (nf == 1) return (list(term1=factors[1]))
if (nf == 2) return (list(term1=factors[1], term2=factors[2]))
others - setdiff(1:nf, with)
result - list(term1=factors[others], term2=factors[with])
result
}
# conditional independence
condit - function(nf, factors=1:nf, with=nf) {
if (nf == 1) return (list(term1=factors[1]))
if (nf == 2) return (list(term1=factors[1], term2=factors[2]))
main - setdiff(1:nf, with)
others - matrix(factors[with], length(with), length(main))
result - rbind(factors[main], others)
result - as.list(as.data.frame(result, stringsAsFactors=FALSE))
names(result) - paste('term', 1:length(result), sep='')
result
}
# mutual independence
mutual - function(nf, factors=1:nf) {
result - sapply(factors[1:nf], list)
names(result) - paste('term', 1:length(result), sep='')
result
}
### some comparisons
loglin(HairEyeColor, list(c(1, 2), c(1, 3), c(2, 3)))$lrt
loglm(~1:2 + 1:3 +2:3, HairEyeColor)
# use factor names
joint(3, factors=names(dimnames(**HairEyeColor)))
condit(3, factors=names(dimnames(**HairEyeColor)))
loglin(HairEyeColor, joint(3))$lrt
loglm(~1:2 + 3, HairEyeColor)
loglin(HairEyeColor, condit(3))$lrt
loglm(~1:3 + 2:3, HairEyeColor)
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept. Chair, Quantitative Methods
York University Voice: 416 736-2100 x66249 Fax: 416 736-5814
4700 Keele StreetWeb: http://www.datavis.ca
Toronto, ONT M3J 1P3 CANADA
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and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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[R] setClass confusion
Dear R People:
I am experimenting with S4 classes and methods but am having trouble with
setting up a class.
Here is an example:
buzz - setClass(buzz,slots=c(x=matrix),
+ validity - function(object) {
+ if(is.matrix(object)==FALSE)stop(Input must be a matrix)
+ TRUE
+ })
Error in setClass(buzz, slots = c(x = matrix), validity -
function(object) { :
Argument representation cannot be used if argument slots is supplied
I know that there is something simple that I'm just not seeing.
This is R-3.0.1 for Windows.
Thanks for any help.
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] setClass confusion
On 13-07-03 7:54 PM, Erin Hodgess wrote:
Dear R People:
I am experimenting with S4 classes and methods but am having trouble with
setting up a class.
Here is an example:
buzz - setClass(buzz,slots=c(x=matrix),
+ validity - function(object) {
+ if(is.matrix(object)==FALSE)stop(Input must be a matrix)
+ TRUE
+ })
Error in setClass(buzz, slots = c(x = matrix), validity -
function(object) { :
Argument representation cannot be used if argument slots is supplied
I know that there is something simple that I'm just not seeing.
Since you used -, your third argument uses the positional name
representation, not validity, which is presumably what you intended.
Duncan Murdoch
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