Re: [R] compare strings
Bernd Jagla wrote:
> It helps writing down these question, you are then getting much closer to an
> answer...
>
> summary(as.integer(t3[,2]) == as.integer(t3[,4]) & as.integer(t3[,3]) ==
> as.integer(t3[,5]))
>
> will compare two pairs of column pairs and give a count of flase and true
> rows...
>
>
If they really are factors with different level sets, I think you might
prefer as.character() there.
> x <- factor(c("a","b"))
> y <- factor(c("b","c"))
> x==y
Error in Ops.factor(x, y) : level sets of factors are different
> as.integer(x)==as.integer(y)
[1] TRUE TRUE
> as.character(x)==as.character(y)
[1] FALSE FALSE
Also, extending the above slightly:
> d <- as.character(x)==as.character(y)
> table(d)
d
FALSE TRUE
2 1
> which(!d)
[1] 1 2
> -B
>
> |-Original Message-
> |From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
> |Behalf Of Bernd Jagla
> |Sent: Friday, November 30, 2007 10:25 PM
> |To: r-help@r-project.org
> |Subject: [R] compare strings
> |
> |Sorry for the question, but I really cannot find the right search terms to
> |find an answer..
> |
> |
> |
> |I have a data frame with strings in some of the columns.
> |
> |I want to know all the rows where the strings in both columns are equal.
> |
> |
> |
> |How do I do this?
> |
> |
> |
> |Thanks,
> |
> |
> |
> |Bernd
> |
> |
> | [[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.
>
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] compare strings
Peter Dalgaard wrote:
> Also, extending the above slightly:
>
Oups. Two lines fell out. I meant:
> y <- factor(c("b","c","b"))
> x <- factor(c("a","b","b"))
> d <- as.character(x)==as.character(y)
> table(d)
d
FALSE TRUE
2 1
> which(!d)
[1] 1 2
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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[R] modeling time series with ARIMA
Good afternoon! I'm trying to model a time series on the following data, which represent a monthly consumption of juices: >x<-scan() 1: 2859 3613 3930 5193 4523 3226 4280 3436 3235 3379 3517 6022 13: 4465 4604 5441 6575 6092 6607 6390 6150 6488 5912 6228 10196 25: 7612 7270 8617 9535 8449 8520 9148 8077 7824 7991 7660 12130 37: 9135 9512 9631 12642 11369 12140 13953 12421 11081 46: Read 45 items > arima(x,order=c(2,1,2), seasonal=list(order=c(0,1,0), period=12))->l > acf(l$resid) > sd(l$resid) > Box.test(l$resid) Now, my problem: 1. All the analysis that i have seen regarding ARIMA modeling, had the residuals acf, within the confidence interval, while my residual acf at first lag is very close to one (and going out of the confidence interval), even if the Box.test can not reject the null hypothesis of a significant acf for all my residuals. I imagine that i am doing something wrong with my model. Is the acf at lag 1 a sign that my residuals are not white noise, or what is wrong here? 2. What would be the impact of an inappropriate model on the confidence interval for a future prediction? (disregarding the fact that an inappropriate model would give a bad forecast on future value, could it have also an impact on enlarging the interval?) 3. As a rule of thumb, do you chose your model by selecting the lowest AIC, or by the lowest standard deviation of the residuals ? Thank you and have a great day! - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R function for percentrank
Hi, Does anyone know if R has a built-in function that is equvalent to Excel's percentrank, i.e., returns the rank of a value in a data set as a percentage of the data set? Thanks, -- Tom [[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] Rolling Correlations
Hi R, I want to do some rolling correlations. But before, I searched for "?rollingCorrelation" and tried the example in it. But I was not successful. What could be the problem? Here is the code I tried: > library(zoo) > library(PerformanceAnalytics) > rollingCorrelation([EMAIL PROTECTED],1],[EMAIL PROTECTED],n=12) Error in inherits(object, "zoo") : object "manager.ts" not found BR, Shubha [[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] Correlation among random effects in lme
Dear R-helper: I measured 40 different mouse strains. In 10 of them, I measured 2 males and 4 females for each strain. In another 10 of them, I measured 4 males and 2 females for each strain. In the remaining 20, I measured 3 males and 3 females for each strain. Totally, I have 240 data for 40 strains each of them have 6 data. The model for the data can be written as, pheno=mu+sex+strain+e, where mu (mean) and sex (two levels) are fixed effects, and strain (40 levels) and e are random effects: result1 <- lme(pheno~sex, random=list(strain=~1),data=dat) (model_1) This works and I obtained variances for both strain and residual. However, these 40 strains are highly correlated. I don't know to how to incorporate dependence among these strains into the above model. I already obtained a correlation matrix for the 40 strains. The below R does not work: result2 <- lme(pheno~sex, random=list(strain=~1), correlation=corSymm(value=c(0.8,0.7,...,0.4),form =~ Strain), data=dat) (model_2) I will greatly appreciate if you can give some suggestions Pengyuan Liu Dept of Surgery Washington Univ in St Louis - [[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 with tables
Hello, I'm new using R and developing tables. I have a problem in developing a table. In a questionaire I made I ask this question "Please tell me the first three sympthoms caused by Respiratory tract infection you've caught this year", then the people answer three sympthoms, the first mention (Top of mind) is saved in a variable called "P2_A", the second mention in a variable called "P2_B" and the third mention in "P2_C". Each answer is coded with numbers like this: 1 = Flu 2 = Cough 3 = Asthma 13 = Fever I've already done a K-cluster analysis and segmented my data base. So my first task is to develop tables to develop tables of frequencies crossing Cluster vs. "P2_A" in order to know which are the top of mind products and their frequencies by cluster, then the second mention and third mention. I've used this instruction which worked well: > table(infec1,aglomera) aglomera infec1 1 2 3 4 1 117 88 76 83 2 10 10 9 7 3 15 11 14 14 42 0 1 1 52 3 1 0 61 0 1 0 83 3 0 1 93 1 1 0 11 0 0 1 1 Where "infec1" is a factor of "P2_A" and "aglomera" is a factor of the variable "Cluster" I made. It worked well when I study them separately...however I would like to know the TOTAL mentions of each sympthom by cluster. I've done this exercise in SPSS using the "Multiple Response" instruction first grouping my three variables (i.e. "P2_A", "P2_B" and "P2_C") into a variable called "sick" and cross tabulating it vs. QCL_1 (my cluster variable) and it gave me the table I need in this way (showed at the bottom of this mail): How can I made a table like this in R???. I've tried combining my variables in a matrix and using xtabs, ftable, table, structable and a lot of combination of them but I haven't had succed with any of them. Please help me with this issue, I don't want to keep using SPSS any more. Thanx in advance. P.D. Result from SPSS is shown below. * * * C R O S S T A B U L A T I O N * * * $SICK (group) mr sick by QCL_1 Cluster Number of Case QCL_1 Count I I Row I Total I 1 I 2 I 3 I 4 I $SICK +++++ 1 I 130 I97 I83 I89 I 399 Gripe, influenza, ca IIIII 83.1 +++++ 2 I53 I55 I42 I46 I 196 Tos de cualquier tip IIIII 40.8 +++++ 3 I33 I36 I36 I39 I 144 Dolor irritación IIIII 30.0 +++++ 4 I 5 I 1 I 2 I 3 I11 Bronquitis IIIII 2.3 +++++ 5 I 5 I 4 I 1 I 0 I10 SinusitisIIIII 2.1 +++++ 6 I 1 I 1 I 1 I 3 I 6 Rinitis IIIII 1.3 +++++ 8 I 8 I 6 I 4 I 4 I22 Amigdalitis IIIII 4.6 +++++ 9 I 6 I 4 I 1 I 2 I13 Faringitis IIIII 2.7 +++++ 10 I 1 I 2 I 2 I 3 I 8 Laringitis IIIII 1.7 +++++ 11 I 1 I 1 I 1 I 1 I 4 Neumonia IIIII.8 +++++ 13 I 0 I 0 I 1 I 0 I 1 Asma IIIII.2 +++++ Column 153 116 104 107 480 Total 31.9 24.2 21.7 22.3100.0 Percents and totals based on respondents 480 valid cases; 0 missing cases Act. Calef Alejandro Rodríguez Cuevas Analista de mercado Laboratorios Farmasa S.A. de C.V. Schwabe Mexico, S.A. de C.V. Bufalo Nr. 27 Col. del Valle 03100 Mexico, D.F. Mexico Tel. 52 00 26
[R] How to cbind DF:s with differing number of rows?
#Hi R-users,
#Suppose that I have a data.frame like this:
y1 <- rnorm(10) + 6.8
y2 <- rnorm(10) + (1:10*1.7 + 1)
y3 <- rnorm(10) + (1:10*6.7 + 3.7)
y <- c(y1,y2,y3)
x <- rep(1:3,10)
f <- gl(2,15, labels=paste("lev", 1:2, sep=""))
g <- seq(as.Date("2000/1/1"), by="day", length=30)
DF <- data.frame(x=x,y=y, f=f, g=g)
DF$g[DF$x == 1] <- NA
DF$x[3:6] <- NA
DF$wdays <- weekdays(DF$g)
DF
#For EDA purposes, I would like to calculate frequences in each variable
g <- lapply(DF, function(x) as.data.frame(table(x)))
#After this, I would like to cbind these data.frames (in g) into a
single data.frame (which to export to MS Excel)
#do.call(cbind, g) does not seem to work because of the different
number of rows in each data.frame.
#The resulting data.frame shoul look like this (only two variables
printed here):
Rowid;x;Freq.x;y;Freq.y; # etc...
1;1;9;1.69151845313816;1;
2;2;9;5.03748767699799;1;
3;3;8;5.37387749444247;1;
4;Empty;Empty;6.83926626214299;1;
5;Empty;Empty;6.97484558968873;1;
6;Empty;Empty;7.11023821708323;1;
7;Empty;Empty;7.1348316549091;1;
8;Empty;Empty;7.16727166992407;1;
9;Empty;Empty;7.35983428577469;1;
10;Empty;Empty;7.7596470136235;1;
11;Empty;Empty;7.86369414967578;1;
12;Empty;Empty;7.97164674771006;1;
13;Empty;Empty;8.0787295301318;1;
14;Empty;Empty;8.14161030348166;1;
15;Empty;Empty;8.20134832959661;1;
16;Empty;Empty;10.1469115339016;1
17;Empty;Empty;12.7442067301746;1
18;Empty;Empty;14.0865167751202;1
19;Empty;Empty;15.8280312307450;1
20;Empty;Empty;16.0484499360756;1
21;Empty;Empty;17.079522214;1
22;Empty;Empty;18.1254057823357;1
23;Empty;Empty;22.7169729331525;1
24;Empty;Empty;30.7237748005358;1
25;Empty;Empty;37.2141271786934;1
26;Empty;Empty;44.4954633229803;1
27;Empty;Empty;50.2302409305761;1
28;Empty;Empty;57.8913405112114;1
29;Empty;Empty;64.849897477945;1
30;Empty;Empty;71.4205263353053;1
#Anyone have an idea how to do this?
#Thanks,
#Lauri
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Re: [R] Rating R Helpers
On Dec 1, 2007 2:21 AM, <[EMAIL PROTECTED]> wrote: > To me a much more urgent initiative is some kind of user online review > system for packages, even something as simple as that used by Amazon.com > has for customer review of books. > > I think the need for this is rather urgent, in fact. Most packages are > very good, but I regret to say some are pretty inefficient and others > downright dangerous. You don't want to discourage people from > submitting their work to CRAN, but at the same time you do want some > mechanism that allows users to relate their experience with it, good or > bad. You can get a very rough idea of this automatically by making a list of which packages are dependents of other packages. library(pkgDepTools) # from BioC AP <- available.packages() dep <- AP[, "Depends"] deps <- unlist(sapply(dep, pkgDepTools:::cleanPkgField)) sort(table(deps)) Of course some packages are more naturally end-user oriented and so would never make such a list and others may be good but just no one knows about them so they have never been used. Some packages might get on the list because an author has two packages and one uses the other so an enhancement could be to eliminate dependencies with a common author. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 cbind DF:s with differing number of rows?
This should do it for you by padding out the rows so they are the same length:
> # use your 'g' and pad out the rows so they are the same length
> str(g)
List of 5
$ x:'data.frame': 3 obs. of 2 variables:
..$ x : Factor w/ 3 levels "1","2","3": 1 2 3
..$ Freq: int [1:3] 9 9 8
$ y:'data.frame': 30 obs. of 2 variables:
..$ x : Factor w/ 30 levels "4.21178116845085",..: 1 2 3 4 5 6 7 8 9 10 ...
..$ Freq: int [1:30] 1 1 1 1 1 1 1 1 1 1 ...
$ f:'data.frame': 2 obs. of 2 variables:
..$ x : Factor w/ 2 levels "lev1","lev2": 1 2
..$ Freq: int [1:2] 15 15
$ g:'data.frame': 20 obs. of 2 variables:
..$ x : Factor w/ 20 levels "2000-01-02","2000-01-03",..: 1 2 3 4
5 6 7 8 9 10 ...
..$ Freq: int [1:20] 1 1 1 1 1 1 1 1 1 1 ...
$ wdays:'data.frame': 7 obs. of 2 variables:
..$ x : Factor w/ 7 levels "Friday","Monday",..: 1 2 3 4 5 6 7
..$ Freq: int [1:7] 2 3 3 4 3 2 3
> # determine max nrows
> max.rows <- max(sapply(g, nrow))
> g.new <- lapply(g, function(.x){
+ if (nrow(.x) < max.rows) .x <- rbind(.x, matrix(NA, ncol=2,
nrow=max.rows - nrow(.x),
+ dimnames=list(NULL, c('x', 'Freq'
+ .x
+ })
> do.call('cbind', g.new)
x.x x.Freq y.x y.Freq f.x f.Freqg.x g.Freq
wdays.x wdays.Freq
1 1 9 4.21178116845085 1 lev1 15 2000-01-02 1
Friday 2
2 2 9 4.78984323641143 1 lev2 15 2000-01-03 1
Monday 3
3 3 8 5.4787594194582 1 NA 2000-01-05 1
Saturday 3
4 NA 5.5853001128225 1 NA 2000-01-06 1
Sunday 4
5 NA 5.96437138758995 1 NA 2000-01-08 1
Thursday 3
6 NA 5.97953161588198 1 NA 2000-01-09 1
Tuesday 2
7 NA 6.17354618925767 1 NA 2000-01-11 1
Wednesday 3
8 NA 6.49461161284364 1 NA 2000-01-12 1
NA
9 NA 6.98364332422208 1 NA 2000-01-14 1
NA
10 NA 7.12950777181536 1 NA 2000-01-15 1
NA
11 NA 7.28742905242849 1 NA 2000-01-17 1
NA
12 NA 7.3757813516535 1 NA 2000-01-18 1
NA
13 NA 7.53832470512922 1 NA 2000-01-20 1
NA
14 NA 8.39528080213779 1 NA 2000-01-21 1
NA
15 NA 10.6249309181431 1 NA 2000-01-23 1
NA
16 NA 11.1550663909848 1 NA 2000-01-24 1
NA
17 NA 11.3189773716082 1 NA 2000-01-26 1
NA
18 NA 12.8838097369011 1 NA 2000-01-27 1
NA
19 NA 15.5438362106853 1 NA 2000-01-29 1
NA
20 NA 17.1212211950981 1 NA 2000-01-30 1
NA
21 NA 17.8821363007311 1 NANA
NA
22 NA 18.5939013212175 1 NANA
NA
On Dec 1, 2007 8:05 AM, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> #Hi R-users,
> #Suppose that I have a data.frame like this:
>
> y1 <- rnorm(10) + 6.8
> y2 <- rnorm(10) + (1:10*1.7 + 1)
> y3 <- rnorm(10) + (1:10*6.7 + 3.7)
> y <- c(y1,y2,y3)
> x <- rep(1:3,10)
> f <- gl(2,15, labels=paste("lev", 1:2, sep=""))
> g <- seq(as.Date("2000/1/1"), by="day", length=30)
> DF <- data.frame(x=x,y=y, f=f, g=g)
> DF$g[DF$x == 1] <- NA
> DF$x[3:6] <- NA
> DF$wdays <- weekdays(DF$g)
>
> DF
>
> #For EDA purposes, I would like to calculate frequences in each variable
> g <- lapply(DF, function(x) as.data.frame(table(x)))
>
> #After this, I would like to cbind these data.frames (in g) into a
> single data.frame (which to export to MS Excel)
> #do.call(cbind, g) does not seem to work because of the different
> number of rows in each data.frame.
> #The resulting data.frame shoul look like this (only two variables
> printed here):
>
> Rowid;x;Freq.x;y;Freq.y; # etc...
> 1;1;9;1.69151845313816;1;
> 2;2;9;5.03748767699799;1;
> 3;3;8;5.37387749444247;1;
> 4;Empty;Empty;6.83926626214299;1;
> 5;Empty;Empty;6.97484558968873;1;
> 6;Empty;Empty;7.11023821708323;1;
> 7;Empty;Empty;7.1348316549091;1;
> 8;Empty;Empty;7.16727166992407;1;
> 9;Empty;Empty;7.35983428577469;1;
> 10;Empty;Empty;7.7596470136235;1;
> 11;Empty;Empty;7.86369414967578;1;
> 12;Empty;Empty;7.97164674771006;1;
> 13;Empty;Empty;8.0787295301318;1;
> 14;Empty;Empty;8.14161030348166;1;
> 15;Empty;Empty;8.20134832959661;1;
> 16;Empty;Empty;10.1469115339016;1
> 17;Empty;Empty;12.7442067301746;1
> 18;Empty;Empty;14.0865167751202;1
> 19;Empty;Empty;15.8280312307450;1
> 20;Empty;Empty;16.0484499360756;1
> 21;Empty;Empty;17.079522214;1
> 22;Empty;Empty;18.1254057823357;1
> 23;Empty;Empty;22.7169729331525;1
> 24;Empty;Empty;30.7237748005358;1
> 25;Empty;Empty;37.2141271786934;1
> 26;Empty;Empty;44.4954633229803;1
> 27;Empty;Empty;50.2302409305761;1
> 28;Empty;Empty;57.8913405112114;1
> 29;Empty;Empty;64.849897477945;1
> 30;
Re: [R] lmer and method call
On Nov 29, 2007 8:09 PM, M-J Milloy <[EMAIL PROTECTED]> wrote: > > Hello all, > > I'm attempting to fit a generalized linear mixed-effects model using lmer > (R v 2.6.0, lmer 0.99875-9, Mac OS X 10.4.10) using the call: > > vidusLMER1 <- lmer(jail ~ visit + gender + house + cokefreq + cracfreq + > herofreq + borcur + comc + (1 | code), data = vidusGD, family = binomial, > correlation = corCompSymm(form = 1 | ID), method = "ML") > > Although the model fits, the summary indicates the model is a "Generalized > linear mixed model fit using Laplace". I've tried any number of > permutations; is only Laplace supported in lmer, despite the text of the > help file? The help file does say that for a generalized linear mixed model (GLMM), which is what family = binomial implies, the estimation criterion is always "ML" (maximum likelihood) as opposed to "REML" (restricted, or residual, maximum likelihood). So stating method = "ML" is redundant. For a GLMM, however, the log-likelihood cannot not be evaluated directly and must be approximated. Here the help file is misleading because it implies that there are three possible approximations, "PQL" (penalized quasi-likelihood), "Laplace" and "AGQ" (adaptive Gaussian quadrature). AGQ has not yet been implemented so the only effective choices are PQL and Laplace. The default is PQL, to refine the starting estimates, followed by optimization of the Laplace approximation. In some cases it is an advantage to suppress the PQL iterations which can be done with one of the settings for the control argument. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lmer and method call
On Dec 1, 2007 9:26 AM, Douglas Bates <[EMAIL PROTECTED]> wrote: > On Nov 29, 2007 8:09 PM, M-J Milloy <[EMAIL PROTECTED]> wrote: > > > > Hello all, > > > > I'm attempting to fit a generalized linear mixed-effects model using lmer > > (R v 2.6.0, lmer 0.99875-9, Mac OS X 10.4.10) using the call: > > > > vidusLMER1 <- lmer(jail ~ visit + gender + house + cokefreq + cracfreq + > > herofreq + borcur + comc + (1 | code), data = vidusGD, family = binomial, > > correlation = corCompSymm(form = 1 | ID), method = "ML") > > > > Although the model fits, the summary indicates the model is a "Generalized > > linear mixed model fit using Laplace". I've tried any number of > > permutations; is only Laplace supported in lmer, despite the text of the > > help file? > > The help file does say that for a generalized linear mixed model > (GLMM), which is what family = binomial implies, the estimation > criterion is always "ML" (maximum likelihood) as opposed to "REML" > (restricted, or residual, maximum likelihood). So stating method = > "ML" is redundant. > > For a GLMM, however, the log-likelihood cannot not be evaluated > directly and must be approximated. Here the help file is misleading > because it implies that there are three possible approximations, "PQL" > (penalized quasi-likelihood), "Laplace" and "AGQ" (adaptive Gaussian > quadrature). AGQ has not yet been implemented so the only effective > choices are PQL and Laplace. The default is PQL, to refine the > starting estimates, followed by optimization of the Laplace > approximation. In some cases it is an advantage to suppress the PQL > iterations which can be done with one of the settings for the control > argument. I forgot to mention that the correlation argument has no effect in this call. That argument is for the lme function in the nlme package. In lmer it is ignored. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 cbind DF:s with differing number of rows?
Thanks Jim! That helped me a lot!
Cheers,
Lauri
2007/12/1, jim holtman <[EMAIL PROTECTED]>:
> This should do it for you by padding out the rows so they are the same length:
>
> > # use your 'g' and pad out the rows so they are the same length
> > str(g)
> List of 5
> $ x:'data.frame': 3 obs. of 2 variables:
> ..$ x : Factor w/ 3 levels "1","2","3": 1 2 3
> ..$ Freq: int [1:3] 9 9 8
> $ y:'data.frame': 30 obs. of 2 variables:
> ..$ x : Factor w/ 30 levels "4.21178116845085",..: 1 2 3 4 5 6 7 8 9 10 ...
> ..$ Freq: int [1:30] 1 1 1 1 1 1 1 1 1 1 ...
> $ f:'data.frame': 2 obs. of 2 variables:
> ..$ x : Factor w/ 2 levels "lev1","lev2": 1 2
> ..$ Freq: int [1:2] 15 15
> $ g:'data.frame': 20 obs. of 2 variables:
> ..$ x : Factor w/ 20 levels "2000-01-02","2000-01-03",..: 1 2 3 4
> 5 6 7 8 9 10 ...
> ..$ Freq: int [1:20] 1 1 1 1 1 1 1 1 1 1 ...
> $ wdays:'data.frame': 7 obs. of 2 variables:
> ..$ x : Factor w/ 7 levels "Friday","Monday",..: 1 2 3 4 5 6 7
> ..$ Freq: int [1:7] 2 3 3 4 3 2 3
> > # determine max nrows
> > max.rows <- max(sapply(g, nrow))
> > g.new <- lapply(g, function(.x){
> + if (nrow(.x) < max.rows) .x <- rbind(.x, matrix(NA, ncol=2,
> nrow=max.rows - nrow(.x),
> + dimnames=list(NULL, c('x', 'Freq'
> + .x
> + })
> > do.call('cbind', g.new)
>x.x x.Freq y.x y.Freq f.x f.Freqg.x g.Freq
> wdays.x wdays.Freq
> 1 1 9 4.21178116845085 1 lev1 15 2000-01-02 1
> Friday 2
> 2 2 9 4.78984323641143 1 lev2 15 2000-01-03 1
> Monday 3
> 3 3 8 5.4787594194582 1 NA 2000-01-05 1
> Saturday 3
> 4 NA 5.5853001128225 1 NA 2000-01-06 1
> Sunday 4
> 5 NA 5.96437138758995 1 NA 2000-01-08 1
> Thursday 3
> 6 NA 5.97953161588198 1 NA 2000-01-09 1
> Tuesday 2
> 7 NA 6.17354618925767 1 NA 2000-01-11 1
> Wednesday 3
> 8 NA 6.49461161284364 1 NA 2000-01-12 1
>NA
> 9 NA 6.98364332422208 1 NA 2000-01-14 1
>NA
> 10 NA 7.12950777181536 1 NA 2000-01-15 1
>NA
> 11 NA 7.28742905242849 1 NA 2000-01-17 1
>NA
> 12 NA 7.3757813516535 1 NA 2000-01-18 1
>NA
> 13 NA 7.53832470512922 1 NA 2000-01-20 1
>NA
> 14 NA 8.39528080213779 1 NA 2000-01-21 1
>NA
> 15 NA 10.6249309181431 1 NA 2000-01-23 1
>NA
> 16 NA 11.1550663909848 1 NA 2000-01-24 1
>NA
> 17 NA 11.3189773716082 1 NA 2000-01-26 1
>NA
> 18 NA 12.8838097369011 1 NA 2000-01-27 1
>NA
> 19 NA 15.5438362106853 1 NA 2000-01-29 1
>NA
> 20 NA 17.1212211950981 1 NA 2000-01-30 1
>NA
> 21 NA 17.8821363007311 1 NANA
>NA
> 22 NA 18.5939013212175 1 NANA
>NA
>
>
> On Dec 1, 2007 8:05 AM, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> > #Hi R-users,
> > #Suppose that I have a data.frame like this:
> >
> > y1 <- rnorm(10) + 6.8
> > y2 <- rnorm(10) + (1:10*1.7 + 1)
> > y3 <- rnorm(10) + (1:10*6.7 + 3.7)
> > y <- c(y1,y2,y3)
> > x <- rep(1:3,10)
> > f <- gl(2,15, labels=paste("lev", 1:2, sep=""))
> > g <- seq(as.Date("2000/1/1"), by="day", length=30)
> > DF <- data.frame(x=x,y=y, f=f, g=g)
> > DF$g[DF$x == 1] <- NA
> > DF$x[3:6] <- NA
> > DF$wdays <- weekdays(DF$g)
> >
> > DF
> >
> > #For EDA purposes, I would like to calculate frequences in each variable
> > g <- lapply(DF, function(x) as.data.frame(table(x)))
> >
> > #After this, I would like to cbind these data.frames (in g) into a
> > single data.frame (which to export to MS Excel)
> > #do.call(cbind, g) does not seem to work because of the different
> > number of rows in each data.frame.
> > #The resulting data.frame shoul look like this (only two variables
> > printed here):
> >
> > Rowid;x;Freq.x;y;Freq.y; # etc...
> > 1;1;9;1.69151845313816;1;
> > 2;2;9;5.03748767699799;1;
> > 3;3;8;5.37387749444247;1;
> > 4;Empty;Empty;6.83926626214299;1;
> > 5;Empty;Empty;6.97484558968873;1;
> > 6;Empty;Empty;7.11023821708323;1;
> > 7;Empty;Empty;7.1348316549091;1;
> > 8;Empty;Empty;7.16727166992407;1;
> > 9;Empty;Empty;7.35983428577469;1;
> > 10;Empty;Empty;7.7596470136235;1;
> > 11;Empty;Empty;7.86369414967578;1;
> > 12;Empty;Empty;7.97164674771006;1;
> > 13;Empty;Empty;8.0787295301318;1;
> > 14;Empty;Empty;8.14161030348166;1;
> > 15;Empty;Empty;8.20134832959661;1;
> > 16;Empty;Empty;10.1469115339016;1
> > 17;Empty;Empty;12.7442067301746;1
> > 18;Empty;Empty;14.0865167751202;1
> > 19;Empty;Empty;15.8280312307450;1
> > 20;Empty;Empty;16.0484499360756;1
> > 21;Em
[R] Spellchecking Sweave documents
I have been using Aspell on a Linux system, but it doesn't understand the noweb chunks, which I'd rather it not spellcheck. I can run it on the generated .tex files, but then changes I make during the spellcheck will not be propagated back to the original source. Any suggestions on how to spellcheck Sweave documents? I see from a search that some people seem to be trying Flyspell on Emacs. I'd rather have a solution that runs outside of Emacs, but if anyone is using Flyspell successfully, I'd love to know of their experiences. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lmer and method call
Douglas Bates stat.wisc.edu> writes: (lmer) > The default is PQL, to refine the > starting estimates, followed by optimization of the Laplace > approximation. In some cases it is an advantage to suppress the PQL > iterations which can be done with one of the settings for the control > argument. I had found out the hard way that it is often better to let PQL play the game rather loosely. Yet I never dared to tell someone, for fear the approximation could end up in the wrong slot, Any rules (beside trying variants) if I can trust such a result? Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 data editor when R built with --enable-mcfs
Hi, I encountered this problem when I was looking around in R, by default R is built with --enable-mcfs and when I tried to edit any data I was getting: dataentry(datalist, modes) : invalid device unable to create fontset -*-fixed-medium-r-normal--13-*-*-*-*-*-*-* anyway I looked around code, noticed that it comes from fragment when SUPPORT_MCFS is defined, and decided to try my luck without it, rebuild with: --disable-mfcs and it worked like a charm, just doesn't have UTF-8 support now of course what would be nice... am I missing something? I googled and searched and found one similar message, but then it was stated it works with current versions... my fonts in xorg.conf points to right directories in my system and all other apps works. I run current version of Arch linux I would appreciate any help or hints how to solve that - whatever this is bug or misconfiguration :) thanks in advance, Andrzej Giniewicz. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sweave: Variables in code chunk headers
I would like to be able to do something like this: <>= ... @ with mywidth set in a previous code chunk. Is there a way to do this in Sweave? (Sorry for two questions in a row, I have been saving these up.) -- 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] finding roots (Max Like Est)
Why exactly the same question?
You were told what to do.
I am new to R. I did what the previous poster said.
I found it. Write your function in terms of vector operations. Avoid
loops if you can.
Sample input for R follows
-
# par is the thing that has to be found
# x are the observations
f <- function(par,x) sum(2*(x-par)/(1+(x-par)^2))
# trial stuff
x <- 1:10
# use uniroot to find a value for par such that f(par,x) == 0
# interval for par is obvious (lower=1 and upper=10)
paropt <- uniroot(f,c(1,10),tol=1e-8,x)
paropt
It works.
Berend Hasselman
On 30 Nov 2007, at 17:59, stathelp wrote:
>
> I have this maximum liklihood estimate problem
>
> i need to find the roots of the following:
>
> [sum (from i=1 to n) ] ((2(x[i]-parameter)/(1+(x[i]-parameter)^2))=0
>
> given to me is the x vector which has length 100
>
> how would I find the roots using R?
>
> I have 2 thoughts.. 1 is using a grid search ... eg. brute
> force, just
> choosing a whole bunch of different values for my parameter
> such as
> parameter=seq(0,100,.1) and this is what I have so far,
>
>
> john=rep(0,length(x))
> for(i in 1:length(x)) {
> john[i]=((x[i]-0)/(1+(x[i]-0)^2))
> }
> sum(john)
>
> then
>
> john=rep(0,length(x))
> for(i in 1:length(x)) {
> john[i]=((x[i]-.1)/(1+(x[i]-.1)^2))
> }
> sum(john)
>
> then
>
> john=rep(0,length(x))
> for(i in 1:length(x)) {
> john[i]=((x[i]-.2)/(1+(x[i]-.2)^2))
> }
> sum(john)
>
> something like this ...
>
> theta=seq(0,100,.1)
> john=rep(0,length(x))
> for(i in 1:length(x)) {
> john[i]=((x[i]-theta[j])/(1+(x[i]-theta[j])^2))
> }
> sum(john)
>
> but something is wrong with my code because its not working. Anyone
> have any
> ideas? (I am very new to R and statistical software in general)
>
> The 2nd thought was to use the Newton Raphson Method, but, I dont
> even know
> where to start with that.
>
> Any thoughts help.
>
> Thanks
>
>
> --
> View this message in context:
> http://www.nabble.com/finding-roots-%28Max-Like-Est%29-tf4901659.html#a14040895
> 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] Rating R Helpers
I'll throw one more idea into the mix. I agree with Bill that a rating system for respondents is probably not that practical and of not the highest importance. It also seems like a recipe for creating inter-personal problems that the list doesn't need. I do like Bill's idea of a review system for packages, which could be incorporated into my idea that follows... What I would find useful would be some sort of tagging system for messages. I can't count the times I've remembered seeing a message that addresses a question I have down the road but, when Googled, I can't find it. It would be so nice, for example, to reliably be able to find all messages related to a certain package or package function posted within the last X days. This could be implemented as simply as asking posters to provide keywords at the end of a message, but it would be great if they could somehow be pulled out of a message and stored in a DB. For instance keywords could be surrounded by a sequence of special characters, which a parser could then extract and store in a DB along with the message. Of course, this would be work to set up, but how many of our "experts" who so kindly give of their time, get exasperated when similar questions keep popping up on the list? Also, if we had a web-accessable DB, the responses, not the responders, could be rated as to how well a reply takes care of an issue. Thus, over time, a sort of auto-wiki could be born. I can think of more uses for this as well. For example a developer could quickly check to see what usability problems or suggestions have cropped up of on individual package. Mark On Dec 1, 2007 2:21 AM, <[EMAIL PROTECTED]> wrote: > This seems a little impractical to me. People respond so much at random > and most only tackle questions with which they feel comfortable. As > it's not a competition in any sense, it's going to be hard to rank > people in any effective way. But suppose you succeed in doing so, then > what? > > To me a much more urgent initiative is some kind of user online review > system for packages, even something as simple as that used by Amazon.com > has for customer review of books. > > I think the need for this is rather urgent, in fact. Most packages are > very good, but I regret to say some are pretty inefficient and others > downright dangerous. You don't want to discourage people from > submitting their work to CRAN, but at the same time you do want some > mechanism that allows users to relate their experience with it, good or > bad. > > > Bill Venables > CSIRO Laboratories > PO Box 120, Cleveland, 4163 > AUSTRALIA > Office Phone (email preferred): +61 7 3826 7251 > Fax (if absolutely necessary): +61 7 3826 7304 > Mobile: +61 4 8819 4402 > Home Phone: +61 7 3286 7700 > mailto:[EMAIL PROTECTED] > http://www.cmis.csiro.au/bill.venables/ > > -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > On Behalf Of Doran, Harold > Sent: Saturday, 1 December 2007 6:13 AM > To: R Help > Subject: [R] Rating R Helpers > > Since R is open source and help may come from varied levels of > experience on R-Help, I wonder if it might be helpful to construct a > method that can be used to "rate" those who provide help on this list. > > This is something that is done on other comp lists, like > http://www.experts-exchange.com/. > > I think some of the reasons for this are pretty transparent, but I > suppose one reason is that one could decide to implement the advise of > those with "superior" or "expert" levels. In other words, you can trust > the advice of someone who is more experienced more than someone who is > not. Currently, there is no way to discern who on this list is really an > R expert and who is not. Of course, there is R core, but most people > don't actually know who these people are (at least I surmise that to be > true). > > If this is potentially useful, maybe one way to begin the development of > such ratings is to allow the original poster to "rate" the level of help > from those who responded. Maybe something like a very simple > questionnaire on a likert-like scale that the original poster would > respond to upon receiving help which would lead to the accumulation of > points for the responders. Higher points would result in higher levels > of expertise (e.g., novice, ..., wizaRd). > > Just a random thought. What do others think? > > Harold > > > > >[[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 p
Re: [R] Rating R Helpers
I don't think r-help is really intended for packages although for some very popular packages questions appear on it anyways sometimes. On Dec 1, 2007 11:28 AM, Mark Kimpel <[EMAIL PROTECTED]> wrote: > I'll throw one more idea into the mix. I agree with Bill that a rating > system for respondents is probably not that practical and of not the highest > importance. It also seems like a recipe for creating inter-personal problems > that the list doesn't need. > > I do like Bill's idea of a review system for packages, which could be > incorporated into my idea that follows... > > What I would find useful would be some sort of tagging system for messages. > I can't count the times I've remembered seeing a message that addresses a > question I have down the road but, when Googled, I can't find it. It would > be so nice, for example, to reliably be able to find all messages related to > a certain package or package function posted within the last X days. This > could be implemented as simply as asking posters to provide keywords at the > end of a message, but it would be great if they could somehow be pulled out > of a message and stored in a DB. For instance keywords could be surrounded > by a sequence of special characters, which a parser could then extract and > store in a DB along with the message. > > Of course, this would be work to set up, but how many of our "experts" who > so kindly give of their time, get exasperated when similar questions keep > popping up on the list? Also, if we had a web-accessable DB, the responses, > not the responders, could be rated as to how well a reply takes care of an > issue. Thus, over time, a sort of auto-wiki could be born. I can think of > more uses for this as well. For example a developer could quickly check to > see what usability problems or suggestions have cropped up of on individual > package. > > Mark > > On Dec 1, 2007 2:21 AM, <[EMAIL PROTECTED]> wrote: > > > > This seems a little impractical to me. People respond so much at random > > and most only tackle questions with which they feel comfortable. As > > it's not a competition in any sense, it's going to be hard to rank > > people in any effective way. But suppose you succeed in doing so, then > > what? > > > > To me a much more urgent initiative is some kind of user online review > > system for packages, even something as simple as that used by Amazon.com > > has for customer review of books. > > > > I think the need for this is rather urgent, in fact. Most packages are > > very good, but I regret to say some are pretty inefficient and others > > downright dangerous. You don't want to discourage people from > > submitting their work to CRAN, but at the same time you do want some > > mechanism that allows users to relate their experience with it, good or > > bad. > > > > > > Bill Venables > > CSIRO Laboratories > > PO Box 120, Cleveland, 4163 > > AUSTRALIA > > Office Phone (email preferred): +61 7 3826 7251 > > Fax (if absolutely necessary): +61 7 3826 7304 > > Mobile: +61 4 8819 4402 > > Home Phone: +61 7 3286 7700 > > mailto:[EMAIL PROTECTED] > > http://www.cmis.csiro.au/bill.venables/ > > > > -Original Message- > > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > > On Behalf Of Doran, Harold > > Sent: Saturday, 1 December 2007 6:13 AM > > To: R Help > > Subject: [R] Rating R Helpers > > > > Since R is open source and help may come from varied levels of > > experience on R-Help, I wonder if it might be helpful to construct a > > method that can be used to "rate" those who provide help on this list. > > > > This is something that is done on other comp lists, like > > http://www.experts-exchange.com/. > > > > I think some of the reasons for this are pretty transparent, but I > > suppose one reason is that one could decide to implement the advise of > > those with "superior" or "expert" levels. In other words, you can trust > > the advice of someone who is more experienced more than someone who is > > not. Currently, there is no way to discern who on this list is really an > > R expert and who is not. Of course, there is R core, but most people > > don't actually know who these people are (at least I surmise that to be > > true). > > > > If this is potentially useful, maybe one way to begin the development of > > such ratings is to allow the original poster to "rate" the level of help > > from those who responded. Maybe something like a very simple > > questionnaire on a likert-like scale that the original poster would > > respond to upon receiving help which would lead to the accumulation of > > points for the responders. Higher points would result in higher levels > > of expertise (e.g., novice, ..., wizaRd). > > > > Just a random thought. What do others think? > > > > Harold > > > > > > > > > >[[alternative HTML version deleted]] > > > > __ > > R-help@r-project.org mailing list > > https://stat.
Re: [R] R function for percentrank
I don't see one but that means nothing. I think you
can write such a function in a few minutes
Will something like this work or am I
missunderstanding what Excel's percentrank does ?
aa <- rnorm(25); aa # data vector
percentrank <- function(x) {
var <- sort(x)
p.rank <- 1:length(var)/length(var)*100
dd <- cbind(var,p.rank)
}
pr <- percentrank(aa); pr
--- tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Does anyone know if R has a built-in function that
> is equvalent to Excel's
> percentrank, i.e., returns the rank of a value in a
> data set as a percentage
> of the data set?
>
> Thanks,
>
> --
> Tom
>
> [[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] Sweave: Variables in code chunk headers
Michael Hoffman sneakemail.com> writes: > > I would like to be able to do something like this: > ><>= >... >@ > > with mywidth set in a previous code chunk. Is there a way to do this in > Sweave? > Not in the <<>>, but you could set a hook for fig: >From Sweave docs: If option "SweaveHooks" is defined as list(fig = foo), and foo is a function, then it would be executed before the code in each figure chunk. This is especially useful to set defaults for the graphical parameters in a series of figure chunks. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R function for percentrank
John,
The Excel's percentrank function works like this: if one has a number, x for
example, and one wants to know the percentile of this number in a given data
set, dataset, one would type =percentrank(dataset,x) in Excel to calculate
the percentile. So for example, if the data set is c(1:10), and one wants to
know the percentile of 2.5 in the data set, then using the percentrank
function one would get 0.166, i.e., 2.5 is in the 16.6th percentile.
I am not sure how to program this function in R. I couldn't find it as a
built-in function in R either. It seems to be an obvious choice for a
built-in function. I am very surprised, but maybe we both missed it.
On 12/1/07, John Kane <[EMAIL PROTECTED]> wrote:
>
> I don't see one but that means nothing. I think you
> can write such a function in a few minutes
>
> Will something like this work or am I
> missunderstanding what Excel's percentrank does ?
>
> aa <- rnorm(25); aa # data vector
> percentrank <- function(x) {
> var <- sort(x)
> p.rank <- 1:length(var)/length(var)*100
> dd <- cbind(var,p.rank)
> }
> pr <- percentrank(aa); pr
>
>
> --- tom soyer <[EMAIL PROTECTED]> wrote:
>
> > Hi,
> >
> > Does anyone know if R has a built-in function that
> > is equvalent to Excel's
> > percentrank, i.e., returns the rank of a value in a
> > data set as a percentage
> > of the data set?
> >
> > Thanks,
> >
> > --
> > Tom
> >
> > [[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.
> >
>
>
>
> Ask a question on any topic and get answers from real people. Go to
> Yahoo! Answers and share what you know at http://ca.answers.yahoo.com
>
--
Tom
[[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] lmer and method call
On Dec 1, 2007 10:08 AM, Dieter Menne <[EMAIL PROTECTED]> wrote: > Douglas Bates stat.wisc.edu> writes: > > (lmer) > > > The default is PQL, to refine the > > starting estimates, followed by optimization of the Laplace > > approximation. In some cases it is an advantage to suppress the PQL > > iterations which can be done with one of the settings for the control > > argument. > > I had found out the hard way that it is often better to let PQL > play the game rather loosely. Yet I never dared to tell someone, for fear > the approximation could end up in the wrong slot, > Any rules (beside trying variants) if I can trust such a result? I'm not sure I understand the sense of your first statement. Do you mean that you have found that you should use PQL or you should not use PQL? I would advise using the Laplace approximation for the final estimates. At one time I thought it would be much slower than the PQL iterations but it doesn't seem to be that bad. I also thought that PQL would refine the starting estimates in the sense that it would take comparatively crude starting values and get you much closer to the optimum before you switched to Laplace. However, because PQL is an algorithm that iterates on both the fixed effects and the random effects with fixed weights, then updates the weights, then goes back to the fixed effects and random effects, etc. there is a possibility that the early weights can force poor values of the fixed effects and later iterations do not recover. I tend to prefer the Laplace approximation directly without any PQL iterations. That is method = "Laplace", control = list(usePQL = FALSE) I would be interested in learning what experiences you or others have had with the different approaches. I am cc:ing this to the R-SIG-mixed-models list and suggest we switch to that list only for further discussion. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Need help on changing a table
Hi all: Im kind of new on R and I need help changing a table. The thing is, i read a file on R using the read.table command and the table looks like this: Item 3d Plot XY plot 001 1 0 001 0 1 001 0 1 002 1 0 002 1 0 002 0 1 ..... .. And what I want to do is generate a new table by item with the sum of the numbres, the next one is an example: Item 3d Plot XY plot 001 1 2 002 2 1 003 ...... Does anyone know how to do this? Thanks in advance, help is greatly appreciated -- View this message in context: http://www.nabble.com/Need-help-on-changing-a-table-tf4929014.html#a14107837 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] Need help on changing a table
Note that read.table() reads a table and returns a _data frame_. So it is
the _data frame_ that you want to change (a table is something else in R).
This is a simple application of aggregate(), e.g.
> aggregate(z[-1], list(Item=z$Item), sum)
Item X3d_Plot XY_plot
111 2
222 1
where I got z from read.table("clipboard", header=TRUE).
On Sat, 1 Dec 2007, gsmkb86 wrote:
>
> Hi all:
> Im kind of new on R and I need help changing a table. The thing is, i read a
> file on R using the read.table command and the table looks like this:
> Item 3d Plot XY plot
> 001 1 0
> 001 0 1
> 001 0 1
> 002 1 0
> 002 1 0
> 002 0 1
> ..... ..
> And what I want to do is generate a new table by item with the sum of the
> numbres, the next one is an example:
>
> Item 3d Plot XY plot
> 001 1 2
> 002 2 1
> 003 ......
>
> Does anyone know how to do this? Thanks in advance, help is greatly
> appreciated
>
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] R function for percentrank
"tom soyer" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> John,
>
> The Excel's percentrank function works like this: if one has a number,
> x for example, and one wants to know the percentile of this number in
> a given data set, dataset, one would type =percentrank(dataset,x) in
> Excel to calculate the percentile. So for example, if the data set is
> c(1:10), and one wants to know the percentile of 2.5 in the data set,
> then using the percentrank function one would get 0.166, i.e., 2.5 is
> in the 16.6th percentile.
>
> I am not sure how to program this function in R. I couldn't find it as
> a built-in function in R either. It seems to be an obvious choice for
> a built-in function. I am very surprised, but maybe we both missed it.
My nomination for a function with a similar result would be ecdf(), the
empirical cumulative distribution function. It is of class "function" so
efforts to index ecdf(.)[.] failed for me.
> df4$V2
[1] 1 1 1 1 1 5 6 7 9
> ecdf.V2<-ecdf(df4$V2)
> ecdf.V2(df4$V2)
[1] 0.2 0.2 0.4 0.4 0.5 0.6 0.7 0.8 1.0 0.9
Don't have Excel, but the OpenOffice.org Calc program has the same
function. It produces:
xpercentrank(x)
1 0.000
1 0.000
3 0.222
3 0.222
4 0.444
5 0.556
6 0.667
7 0.778
10 1.000
9 0.889
(Not that I am saying that the OO.o/Excel function is what one _should_
want. Its behavior seems pathological to me.)
--
David Winsemius
>
> On 12/1/07, John Kane <[EMAIL PROTECTED]> wrote:
>>
>> I don't see one but that means nothing. I think you
>> can write such a function in a few minutes
>>
>> Will something like this work or am I
>> missunderstanding what Excel's percentrank does ?
>>
>> aa <- rnorm(25); aa # data vector
>> percentrank <- function(x) {
>> var <- sort(x)
>> p.rank <- 1:length(var)/length(var)*100
>> dd <- cbind(var,p.rank)
>> }
>> pr <- percentrank(aa); pr
>>
>>
>> --- tom soyer <[EMAIL PROTECTED]> wrote:
>>
>> > Hi,
>> >
>> > Does anyone know if R has a built-in function that
>> > is equvalent to Excel's
>> > percentrank, i.e., returns the rank of a value in a
>> > data set as a percentage
>> > of the data set?
>> >
>> > Thanks,
>> >
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Re: [R] R function for percentrank
David Winsemius <[EMAIL PROTECTED]> wrote in news:[EMAIL PROTECTED]: > "tom soyer" <[EMAIL PROTECTED]> wrote in > news:[EMAIL PROTECTED]: > >> John, >> >> The Excel's percentrank function works like this: if one has a number, >> x for example, and one wants to know the percentile of this number in >> a given data set, dataset, one would type =percentrank(dataset,x) in >> Excel to calculate the percentile. So for example, if the data set is >> c(1:10), and one wants to know the percentile of 2.5 in the data set, >> then using the percentrank function one would get 0.166, i.e., 2.5 is >> in the 16.6th percentile. >> >> I am not sure how to program this function in R. I couldn't find it as >> a built-in function in R either. It seems to be an obvious choice for >> a built-in function. I am very surprised, but maybe we both missed it. > > My nomination for a function with a similar result would be ecdf(), the > empirical cumulative distribution function. It is of class "function" so > efforts to index ecdf(.)[.] failed for me. > >> df4$V2 > [1] 1 1 1 1 1 5 6 7 9 #copied wrong line in R session Make that; df4$V2<-c(1,1,3,3,4,5,6,7,10,9) [1] 1 1 3 3 5 6 7 9 >> ecdf.V2<-ecdf(df4$V2) >> ecdf.V2(df4$V2) > [1] 0.2 0.2 0.4 0.4 0.5 0.6 0.7 0.8 1.0 0.9 > > Don't have Excel, but the OpenOffice.org Calc program has the same > function. It produces: > xpercentrank(x) > 1 0.000 > 1 0.000 > 3 0.222 > 3 0.222 > 4 0.444 > 5 0.556 > 6 0.667 > 7 0.778 > 10 1.000 > 9 0.889 > > (Not that I am saying that the OO.o/Excel function is what one _should_ > want. Its behavior seems pathological to me.) > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Spellchecking Sweave documents
On 01/12/2007 11:11 AM, Michael Hoffman wrote: > I have been using Aspell on a Linux system, but it doesn't > understand the noweb chunks, which I'd rather it not spellcheck. I > can run it on the generated .tex files, but then changes I make > during the spellcheck will not be propagated back to the original > source. Any suggestions on how to spellcheck Sweave documents? If you want the concordance between the lines in the original file and those in the .tex file, you can get it (with the concordance=TRUE Sweave option). Then it would be theoretically possible to convert reports about errors in the .tex into reports about the .Rnw. I don't know if Aspell provides information in a form where you could actually make use of this, but if so, it might be a nice contribution to do so. (For sample code decoding the concordance, see my patchDVI package: it converts source links in .dvi files to point to the .Rnw instead of the .tex file.) Duncan Murdoch > > I see from a search that some people seem to be trying Flyspell on > Emacs. I'd rather have a solution that runs outside of Emacs, but if > anyone is using Flyspell successfully, I'd love to know of their > experiences. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/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] R function for percentrank
On Sat, 2007-12-01 at 18:40 +, David Winsemius wrote:
> David Winsemius <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
> > "tom soyer" <[EMAIL PROTECTED]> wrote in
> > news:[EMAIL PROTECTED]:
> >
> >> John,
> >>
> >> The Excel's percentrank function works like this: if one has a number,
> >> x for example, and one wants to know the percentile of this number in
> >> a given data set, dataset, one would type =percentrank(dataset,x) in
> >> Excel to calculate the percentile. So for example, if the data set is
> >> c(1:10), and one wants to know the percentile of 2.5 in the data set,
> >> then using the percentrank function one would get 0.166, i.e., 2.5 is
> >> in the 16.6th percentile.
> >>
> >> I am not sure how to program this function in R. I couldn't find it as
> >> a built-in function in R either. It seems to be an obvious choice for
> >> a built-in function. I am very surprised, but maybe we both missed it.
> >
> > My nomination for a function with a similar result would be ecdf(), the
> > empirical cumulative distribution function. It is of class "function"
> so
> > efforts to index ecdf(.)[.] failed for me.
You can use ls.str() to look into the function environment:
> ls.str(environment(ecdf(x)))
f : num 0
method : int 2
n : int 25
x : num [1:25] -2.215 -1.989 -0.836 -0.820 -0.626 ...
y : num [1:25] 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 ...
yleft : num 0
yright : num 1
You can then use get() or mget() within the function environment to
return the requisite values. Something along the lines of the following
within the function percentrank():
percentrank <- function(x, val)
{
env.x <- environment(ecdf(x))
res <- mget(c("x", "y"), env.x)
Ind <- which(sapply(seq(length(res$x)),
function(i) isTRUE(all.equal(res$x[i], val
res$y[Ind]
}
Thus:
set.seed(1)
x <- rnorm(25)
> x
[1] -0.62645381 0.18364332 -0.83562861 1.59528080 0.32950777
[6] -0.82046838 0.48742905 0.73832471 0.57578135 -0.30538839
[11] 1.51178117 0.38984324 -0.62124058 -2.21469989 1.12493092
[16] -0.04493361 -0.01619026 0.94383621 0.82122120 0.59390132
[21] 0.91897737 0.78213630 0.07456498 -1.98935170 0.61982575
> percentrank(x, 0.48742905)
[1] 0.56
One other approach, which returns the values and their respective rank
percentiles is:
> cumsum(prop.table(table(x)))
-2.2146998871775 -1.98935169586337 -0.835628612410047
0.040.080.12
-0.820468384118015 -0.626453810742333 -0.621240580541804
0.160.200.24
-0.305388387156356 -0.0449336090152308 -0.0161902630989461
0.280.320.36
0.0745649833651906 0.183643324222082 0.329507771815361
0.400.440.48
0.389843236411431 0.487429052428485 0.575781351653492
0.520.560.60
0.5939013212175090.61982574789471 0.738324705129217
0.640.680.72
0.782136300731067 0.821221195098089 0.918977371608218
0.760.800.84
0.94383621068531.124930918143111.51178116845085
0.880.920.96
1.59528080213779
1.00
HTH,
Marc Schwartz
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Re: [R] R function for percentrank
--- David Winsemius <[EMAIL PROTECTED]> wrote:
> "tom soyer" <[EMAIL PROTECTED]> wrote in
>
news:[EMAIL PROTECTED]:
>
>
> > John,
> >
> > The Excel's percentrank function works like this:
> if one has a number,
> > x for example, and one wants to know the
> percentile of this number in
> > a given data set, dataset, one would type
> =percentrank(dataset,x) in
> > Excel to calculate the percentile. So for example,
> if the data set is
> > c(1:10), and one wants to know the percentile of
> 2.5 in the data set,
> > then using the percentrank function one would get
> 0.166, i.e., 2.5 is
> > in the 16.6th percentile.
> >
> > I am not sure how to program this function in R. I
> couldn't find it as
> > a built-in function in R either. It seems to be an
> obvious choice for
> > a built-in function. I am very surprised, but
> maybe we both missed it.
>
> My nomination for a function with a similar result
> would be ecdf(), the
> empirical cumulative distribution function. It is of
> class "function" so
> efforts to index ecdf(.)[.] failed for me.
>
> > df4$V2
> [1] 1 1 1 1 1 5 6 7 9
> > ecdf.V2<-ecdf(df4$V2)
> > ecdf.V2(df4$V2)
> [1] 0.2 0.2 0.4 0.4 0.5 0.6 0.7 0.8 1.0 0.9
>
> Don't have Excel, but the OpenOffice.org Calc
> program has the same
> function. It produces:
> xpercentrank(x)
> 1 0.000
> 1 0.000
> 3 0.222
> 3 0.222
> 4 0.444
> 5 0.556
> 6 0.667
> 7 0.778
> 101.000
> 9 0.889
>
> (Not that I am saying that the OO.o/Excel function
> is what one _should_
> want. Its behavior seems pathological to me.)
>
Excel
x percentrank(x)
1 0
1 0
3 0.222
3 0.222
4 0.444
5 0.555
6 0.666
7 0.777
10 1
9 0.888
It seems that OOo is following Excel's distinguished
footsteps.
How can one have a 0 percentile ranking?
> --
> David Winsemius
>
> >
> > On 12/1/07, John Kane <[EMAIL PROTECTED]> wrote:
> >>
> >> I don't see one but that means nothing. I think
> you
> >> can write such a function in a few minutes
> >>
> >> Will something like this work or am I
> >> missunderstanding what Excel's percentrank does ?
> >>
> >> aa <- rnorm(25); aa # data vector
> >> percentrank <- function(x) {
> >> var <- sort(x)
> >> p.rank <- 1:length(var)/length(var)*100
> >> dd <- cbind(var,p.rank)
> >> }
> >> pr <- percentrank(aa); pr
> >>
> >>
> >> --- tom soyer <[EMAIL PROTECTED]> wrote:
> >>
> >> > Hi,
> >> >
> >> > Does anyone know if R has a built-in function
> that
> >> > is equvalent to Excel's
> >> > percentrank, i.e., returns the rank of a value
> in a
> >> > data set as a percentage
> >> > of the data set?
> >> >
> >> > Thanks,
> >> >
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained,
> reproducible code.
>
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Re: [R] Sweave: Variables in code chunk headers
Dieter Menne wrote: > Michael Hoffman sneakemail.com> writes: > >> I would like to be able to do something like this: >> >><>= >>... >>@ >> >> with mywidth set in a previous code chunk. Is there a way to do this in >> Sweave? >> > > Not in the <<>>, but you could set a hook for fig: > > >>From Sweave docs: > > If option "SweaveHooks" is defined as list(fig = foo), and foo is a function, > then it would be executed before the code in each figure chunk. This is > especially useful to set defaults for the graphical parameters in a series of > figure chunks. Thanks. I guess what I really want to do is switch between one of two settings. Only one value can be in the defaults, and I would like some way of setting the other value. This might not be easily possible, but I thought I would ask. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R function for percentrank
Its a bit tricky if you want to get it to work exactly the same as
Excel even in the presence of runs but in terms of the R approx function
I think percentrank corresponds to ties = "min" if the value is among those
in the table and ties = "ordered" otherwise so:
percentrank <- function(table, x = table) {
table <- sort(table)
ties <- ifelse(match(x, table, nomatch = 0), "min", "ordered")
len <- length(table)
f <- function(x, ties)
(approx(table, seq(0, len = len), x, ties = ties)$y) / (len - 1)
mapply(f, x, ties)
}
# test
tab <- c(1, 1, 2, 2, 3, 3)
percentrank(tab, 2:6/2) # c(0, .3, .4, .7, .8)
which is the same result as Excel 2007 gives.
On Dec 1, 2007 6:37 AM, tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Does anyone know if R has a built-in function that is equvalent to Excel's
> percentrank, i.e., returns the rank of a value in a data set as a percentage
> of the data set?
>
> Thanks,
>
> --
> Tom
>
>[[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.
>
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Re: [R] Need help on changing a table
That's no table that's a data.frame :). For what you want probably the easiest way is aggregate. Type ?aggregate for the help information Based on your example and assuming the data is called "mydata" this shold do what you want. aggregate(mydata[,2:3], by=list(item=mydata$Item), sum) --- gsmkb86 <[EMAIL PROTECTED]> wrote: > > Hi all: > Im kind of new on R and I need help changing a > table. The thing is, i read a > file on R using the read.table command and the table > looks like this: > Item 3d Plot XY plot > 001 1 0 > 001 0 1 > 001 0 1 > 002 1 0 > 002 1 0 > 002 0 1 > ..... .. > And what I want to do is generate a new table by > item with the sum of the > numbres, the next one is an example: > > Item 3d Plot XY plot > 001 1 2 > 002 2 1 > 003 ...... > > Does anyone know how to do this? Thanks in advance, > help is greatly > appreciated __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] creating conditional means
Hi all- I have a dataset (year, month, hour, co2(ppm), num1,num2) [49,] 2006 110 383.3709 28 28 [50,] 2006 111 383.3709 28 28 [51,] 2006 112 383.3709 28 28 [52,] 2006 113 383.3709 28 28 [53,] 2006 114 383.3709 28 28 [54,] 2006 115 383.3709 28 28 [55,] 2006 116 383.3709 28 28 [56,] 2006 117 383.3709 28 28 [57,] 2006 118 383.3709 28 28 [58,] 2006 119 383.3709 27 27 [59,] 2006 11 10 383.3709 28 28 that repeats in this style for each month. I would like to compute the mean for each hour in three month intervals. i.e. average all 2pms for each day for months march, april and may. and then do this for each hour interval. i have been messing around with 'for loops' but can't seem to get the output I want. thanks in advance for any help- s.heck CU, Boulder __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Rating R Helpers
Mark Kimpel <[EMAIL PROTECTED]> [Sat, Dec 01, 2007 at 05:28:28PM CET]: > What I would find useful would be some sort of tagging system for messages. Hrm. I find tags immensely useful for entities which do not contain primarily text, such as photos. I am at doubt how keywords are important when they are not found in the text. There are situations where the first keyword that comes to mind is tiptoed around in the message, but I don't know if this is often the case. > I can't count the times I've remembered seeing a message that addresses a > question I have down the road but, when Googled, I can't find it. Is it a problem of way too many false positives or a problem of false negatives? Tags may help out in the second case, but in my experiencd it is rare. [...] > > Of course, this would be work to set up, but how many of our "experts" who > so kindly give of their time, get exasperated when similar questions keep > popping up on the list? Do you think that people who keep asking similar questions do so because they didn't do their homework first, or that they Googled and failed? -- Johannes H�sing There is something fascinating about science. One gets such wholesale returns of conjecture mailto:[EMAIL PROTECTED] from such a trifling investment of fact. http://derwisch.wikidot.com (Mark Twain, "Life on the Mississippi") __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Specify var-covar matrix in mixed linear model using lme?
Hi,All: I have a question about specifying var-cov matrix in mixed linear model using lme. For example, for single-level mixed linear model: yi = XiB + Ziui + ei ui ~ N(0,D), ei ~ N(0,sigma^2R), var(yi)=sigma^2(ZiAZi^T + R). R is an identify matrix and A is a known var-covar matrix in my data. In my data, there is only one random effect besides ei. But this random effect is dependent among different subjects within group (this dependence is characterized in A which is known). corStruct class in nlme can specify correlation structure for within group errors (that is, specify R for ei). But I don't know how to specify A for a specific random effect. In other words, I want to fix matrix A in the analysis. Thanks a lot for your help. Pengyuan Liu Dept of Surgery Washington Univ in St Louis Be a better sports nut! Let your teams follow you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] creating conditional means
Try aggregate: Lines <- "Year Month Hour co2 num1 num2 2006 110 383.3709 28 28 2006 111 383.3709 28 28 2006 112 383.3709 28 28 2006 113 383.3709 28 28 2006 114 383.3709 28 28 2006 115 383.3709 28 28 2006 116 383.3709 28 28 2006 117 383.3709 28 28 2006 118 383.3709 28 28 2006 119 383.3709 27 27 2006 11 10 383.3709 28 28 " DF <- read.table(textConnection(Lines), header = TRUE) aggregate(DF[4:6], with(DF, data.frame(Year, Qtr = (Month - 1) %/% 3 + 1, Hour)), mean) On Dec 1, 2007 3:57 PM, Sherri Heck <[EMAIL PROTECTED]> wrote: > Hi all- > > I have a dataset (year, month, hour, co2(ppm), num1,num2) > > > [49,] 2006 110 383.3709 28 28 > [50,] 2006 111 383.3709 28 28 > [51,] 2006 112 383.3709 28 28 > [52,] 2006 113 383.3709 28 28 > [53,] 2006 114 383.3709 28 28 > [54,] 2006 115 383.3709 28 28 > [55,] 2006 116 383.3709 28 28 > [56,] 2006 117 383.3709 28 28 > [57,] 2006 118 383.3709 28 28 > [58,] 2006 119 383.3709 27 27 > [59,] 2006 11 10 383.3709 28 28 > > that repeats in this style for each month. I would like to compute the > mean for each hour in three month intervals. > i.e. average all 2pms for each day for months march, april and may. and > then do this for each hour interval. > i have been messing around with 'for loops' but can't seem to get the > output I want. > > thanks in advance for any help- > > s.heck > CU, Boulder > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/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] creating conditional means
Hi Gabor, Thank you for your help. I think I need to clarify a bit more. I am trying to say average all 2pms for months march + april + may (for example). I hope this is clearer. here's a larger subset of my data set: year, month, hour, co2(ppm), num1,num2 2006 1 0 384.2055 14 14 2006 1 1 384.0304 14 14 2006 1 2 383.9672 14 14 2006 1 3 383.8452 14 14 2006 1 4 383.8594 14 14 2006 1 5 383.7318 14 14 2006 1 6 383.6439 14 14 2006 1 7 383.7019 14 14 2006 1 8 383.7487 14 14 2006 1 9 383.8376 14 14 2006 1 10 383.8684 14 14 2006 1 11 383.8301 14 14 2006 1 12 383.8058 14 14 2006 1 13 383.9419 14 14 2006 1 14 383.7876 14 14 2006 1 15 383.7744 14 14 2006 1 16 383.8566 14 14 2006 1 17 384.1014 14 14 2006 1 18 384.1312 14 14 2006 1 19 384.1551 14 14 2006 1 20 384.099 14 14 2006 1 21 384.1408 14 14 2006 1 22 384.3637 14 14 2006 1 23 384.1491 14 14 2006 2 0 384.7082 27 27 2006 2 1 384.6139 27 27 2006 2 2 384.7453 26 26 2006 2 3 384.9224 28 28 2006 2 4 384.8581 28 28 2006 2 5 384.9208 28 28 2006 2 6 384.9086 28 28 2006 2 7 384.837 28 28 2006 2 8 384.6163 27 27 2006 2 9 384.7406 28 28 2006 2 10 384.7468 28 28 2006 2 11 384.6992 28 28 2006 2 12 384.6388 28 28 2006 2 13 384.6346 28 28 2006 2 14 384.6037 28 28 2006 2 15 384.5295 28 28 2006 2 16 384.5654 28 28 2006 2 17 384.6466 28 28 2006 2 18 384.6344 28 28 2006 2 19 384.5911 28 28 2006 2 20 384.6084 28 28 2006 2 21 384.6318 28 28 2006 2 22 384.6181 27 27 2006 2 23 384.6087 27 27 thanks you again for your assistance- s.heck Gabor Grothendieck wrote: > Try aggregate: > > > Lines <- "Year Month Hour co2 num1 num2 > 2006 110 383.3709 28 28 > 2006 111 383.3709 28 28 > 2006 112 383.3709 28 28 > 2006 113 383.3709 28 28 > 2006 114 383.3709 28 28 > 2006 115 383.3709 28 28 > 2006 116 383.3709 28 28 > 2006 117 383.3709 28 28 > 2006 118 383.3709 28 28 > 2006 119 383.3709 27 27 > 2006 11 10 383.3709 28 28 > " > DF <- read.table(textConnection(Lines), header = TRUE) > aggregate(DF[4:6], >with(DF, data.frame(Year, Qtr = (Month - 1) %/% 3 + 1, Hour)), >mean) > > On Dec 1, 2007 3:57 PM, Sherri Heck <[EMAIL PROTECTED]> wrote: > >> Hi all- >> >> I have a dataset (year, month, hour, co2(ppm), num1,num2) >> >> >> [49,] 2006 110 383.3709 28 28 >> [50,] 2006 111 383.3709 28 28 >> [51,] 2006 112 383.3709 28 28 >> [52,] 2006 113 383.3709 28 28 >> [53,] 2006 114 383.3709 28 28 >> [54,] 2006 115 383.3709 28 28 >> [55,] 2006 116 383.3709 28 28 >> [56,] 2006 117 383.3709 28 28 >> [57,] 2006 118 383.3709 28 28 >> [58,] 2006 119 383.3709 27 27 >> [59,] 2006 11 10 383.3709 28 28 >> >> that repeats in this style for each month. I would like to compute the >> mean for each hour in three month intervals. >> i.e. average all 2pms for each day for months march, april and may. and >> then do this for each hour interval. >> i have been messing around with 'for loops' but can't seem to get the >> output I want. >> >> thanks in advance for any help- >> >> s.heck >> CU, Boulder >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/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] creating conditional means
Just adjust the formula for Qtr appropriately if your quarters are not Jan/Feb/Mar, Apr/May/Jun, Jul/Aug/Sep, Oct/Nov/Dec as I assumed. On Dec 1, 2007 5:21 PM, Sherri Heck <[EMAIL PROTECTED]> wrote: > Hi Gabor, > > Thank you for your help. I think I need to clarify a bit more. I am > trying to say > > average all 2pms for months march + april + may (for example). I hope this is > clearer. > > here's a larger subset of my data set: > > year, month, hour, co2(ppm), num1,num2 > > 2006 1 0 384.2055 14 14 > 2006 1 1 384.0304 14 14 > 2006 1 2 383.9672 14 14 > 2006 1 3 383.8452 14 14 > 2006 1 4 383.8594 14 14 > 2006 1 5 383.7318 14 14 > 2006 1 6 383.6439 14 14 > 2006 1 7 383.7019 14 14 > 2006 1 8 383.7487 14 14 > 2006 1 9 383.8376 14 14 > 2006 1 10 383.8684 14 14 > 2006 1 11 383.8301 14 14 > 2006 1 12 383.8058 14 14 > 2006 1 13 383.9419 14 14 > 2006 1 14 383.7876 14 14 > 2006 1 15 383.7744 14 14 > 2006 1 16 383.8566 14 14 > 2006 1 17 384.1014 14 14 > 2006 1 18 384.1312 14 14 > 2006 1 19 384.1551 14 14 > 2006 1 20 384.099 14 14 > 2006 1 21 384.1408 14 14 > 2006 1 22 384.3637 14 14 > 2006 1 23 384.1491 14 14 > 2006 2 0 384.7082 27 27 > 2006 2 1 384.6139 27 27 > 2006 2 2 384.7453 26 26 > 2006 2 3 384.9224 28 28 > 2006 2 4 384.8581 28 28 > 2006 2 5 384.9208 28 28 > 2006 2 6 384.9086 28 28 > 2006 2 7 384.837 28 28 > 2006 2 8 384.6163 27 27 > 2006 2 9 384.7406 28 28 > 2006 2 10 384.7468 28 28 > 2006 2 11 384.6992 28 28 > 2006 2 12 384.6388 28 28 > 2006 2 13 384.6346 28 28 > 2006 2 14 384.6037 28 28 > 2006 2 15 384.5295 28 28 > 2006 2 16 384.5654 28 28 > 2006 2 17 384.6466 28 28 > 2006 2 18 384.6344 28 28 > 2006 2 19 384.5911 28 28 > 2006 2 20 384.6084 28 28 > 2006 2 21 384.6318 28 28 > 2006 2 22 384.6181 27 27 > 2006 2 23 384.6087 27 27 > > > thanks you again for your assistance- > > s.heck > > > > Gabor Grothendieck wrote: > > Try aggregate: > > > > > > Lines <- "Year Month Hour co2 num1 num2 > > 2006 110 383.3709 28 28 > > 2006 111 383.3709 28 28 > > 2006 112 383.3709 28 28 > > 2006 113 383.3709 28 28 > > 2006 114 383.3709 28 28 > > 2006 115 383.3709 28 28 > > 2006 116 383.3709 28 28 > > 2006 117 383.3709 28 28 > > 2006 118 383.3709 28 28 > > 2006 119 383.3709 27 27 > > 2006 11 10 383.3709 28 28 > > " > > DF <- read.table(textConnection(Lines), header = TRUE) > > aggregate(DF[4:6], > >with(DF, data.frame(Year, Qtr = (Month - 1) %/% 3 + 1, Hour)), > >mean) > > > > On Dec 1, 2007 3:57 PM, Sherri Heck <[EMAIL PROTECTED]> wrote: > > > >> Hi all- > >> > >> I have a dataset (year, month, hour, co2(ppm), num1,num2) > >> > >> > >> [49,] 2006 110 383.3709 28 28 > >> [50,] 2006 111 383.3709 28 28 > >> [51,] 2006 112 383.3709 28 28 > >> [52,] 2006 113 383.3709 28 28 > >> [53,] 2006 114 383.3709 28 28 > >> [54,] 2006 115 383.3709 28 28 > >> [55,] 2006 116 383.3709 28 28 > >> [56,] 2006 117 383.3709 28 28 > >> [57,] 2006 118 383.3709 28 28 > >> [58,] 2006 119 383.3709 27 27 > >> [59,] 2006 11 10 383.3709 28 28 > >> > >> that repeats in this style for each month. I would like to compute the > >> mean for each hour in three month intervals. > >> i.e. average all 2pms for each day for months march, april and may. and > >> then do this for each hour interval. > >> i have been messing around with 'for loops' but can't seem to get the > >> output I want. > >> > >> thanks in advance for any help- > >> > >> s.heck > >> CU, Boulder > >> > >> __ > >> R-help@r-project.org mailing list > >> https://stat.ethz.ch/mailman/listinfo/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 tables
I guess you can get the result by 1) concatenating all the variables (P2_A, P2_B, P2_C) into one variable , 2) replicating segment membership properly, 3) make the table of 1) and 2) For example, the following may do the job. > ## (1) Generate data set > # Set random seed > set.seed(0) > > n.obs <- 15 # Number of patients > disease.lv <- 1:10 ## different types of disease > data1 <- as.data.frame(t(replicate(n.obs, sample(disease.lv, 3 # Create data set > ## Create segment membership > segment <- sample(LETTERS[1:3], n.obs, replace=TRUE) > > cbind(data1, segment) V1 V2 V3 segment 1 9 3 10 B 2 6 9 2 C 3 9 10 6 A 4 7 1 2 B 5 2 7 4 C 6 8 5 6 C 7 10 4 7 B 8 10 2 6 C 9 2 3 4 B 10 1 4 7 A 11 4 5 9 A 12 5 2 7 A 13 7 8 1 A 14 8 4 7 B 15 7 8 5 B > > > ## (2) Table > infec.all <- unlist(data1) # Concatenate all variables into one variable > segment.all <- rep(segment, ncol(data1)) # Replicate the segment membership as necessary > table(infec.all, segment.all) segment.all infec.all A B C 1 2 1 0 2 1 2 3 3 0 2 0 4 2 3 1 5 2 1 1 6 1 0 3 7 3 4 1 8 1 2 1 9 2 1 1 10 1 2 1 On Nov 30, 2007 10:10 AM, Alejandro Rodríguez <[EMAIL PROTECTED]> wrote: > Hello, I'm new using R and developing tables. > > I have a problem in developing a table. In a questionaire I made I ask > this > question "Please tell me the first three sympthoms caused by Respiratory > tract infection you've caught this year", then the people answer three > sympthoms, the first mention (Top of mind) is saved in a variable called > "P2_A", the second mention in a variable called "P2_B" and the third > mention > in "P2_C". Each answer is coded with numbers like this: > > 1 = Flu > 2 = Cough > 3 = Asthma > > > 13 = Fever > > I've already done a K-cluster analysis and segmented my data base. So my > first task is to develop tables to develop tables of frequencies crossing > Cluster vs. "P2_A" in order to know which are the top of mind products and > their frequencies by cluster, then the second mention and third mention. > I've used this instruction which worked well: > > > table(infec1,aglomera) > aglomera > infec1 1 2 3 4 >1 117 88 76 83 >2 10 10 9 7 >3 15 11 14 14 >42 0 1 1 >52 3 1 0 >61 0 1 0 >83 3 0 1 >93 1 1 0 >11 0 0 1 1 > > Where "infec1" is a factor of "P2_A" and "aglomera" is a factor of the > variable "Cluster" I made. It worked well when I study them > separately...however I would like to know the TOTAL mentions of each > sympthom by cluster. I've done this exercise in SPSS using the "Multiple > Response" instruction first grouping my three variables (i.e. "P2_A", > "P2_B" > and "P2_C") into a variable called "sick" and cross tabulating it vs. > QCL_1 > (my cluster variable) and it gave me the table I need in this way (showed > at > the bottom of this mail): > > How can I made a table like this in R???. I've tried combining my > variables > in a matrix and using xtabs, ftable, table, structable and a lot of > combination of them but I haven't had succed with any of them. > > Please help me with this issue, I don't want to keep using SPSS any more. > > Thanx in advance. > > P.D. Result from SPSS is shown below. > > > >* * * C R O S S T A B U L A T I O N * * * > > $SICK (group) mr sick > by QCL_1 Cluster Number of Case > > > QCL_1 > >Count I > I Row > I Total > I 1 I 2 I 3 I 4 I > $SICK +++++ >1 I 130 I97 I83 I89 I 399 > Gripe, influenza, ca IIIII 83.1 > +++++ >2 I53 I55 I42 I46 I 196 > Tos de cualquier tip IIIII 40.8 > +++++ >3 I33 I36 I36 I39 I 144 > Dolor irritación IIIII 30.0 > +++++ >4 I 5 I 1 I 2 I 3 I11 > Bronquitis IIIII 2.3 > +++++ >5 I 5 I 4 I 1 I 0 I10 > SinusitisIIIII 2.1 > +++++ >6 I 1 I 1 I 1 I 3
