Re: [R] Unable to Load package Rcmdr after installation

[Ulrik Stervbo]

Hi Paul,

what fails and how? What did you do from the time the package worked until
is didn't? Did you update packages? Which packages are you trying to load?

Best
Ulrik

On Thu, 9 Mar 2017 at 18:40 paulberna...@gmail.com 
wrote:

> Thanks Ulrik, but the thing is that I tried installing adn loading tve
> Hmisc package but wasn't able to do that either.
>
>
>  Mensaje original 
> Asunto: Re: [R] Unable to Load package Rcmdr after installation
> De: Ulrik Stervbo
> Para: Paul Bernal ,r-help@r-project.org
> CC:
>
>
> Hi Paul,
>
> The error tells you, that the 'Hmisc' does not exist on your system. If
> you install it, everything should work.
>
> Use install.packages with dependencies = TRUE to avoid the problem of
> missing packages.
>
> HTH
>
> Ulrik
>
> On Thu, 9 Mar 2017 at 16:51 Paul Bernal  wrote:
>
> Hello friends,
>
> Has anyone experienced trouble when trying to load package Rcmdr? It was
> working perfectly a couple of days ago, I don´t know why it isn´t working.
>
> > library("Rcmdr")
> Loading required package: splines
> Loading required package: RcmdrMisc
> Loading required package: car
> Loading required package: sandwich
> Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck
> = vI[[j]]) :
>   there is no package called ‘Hmisc’
> Error: package ‘RcmdrMisc’ could not be loaded
>
> [[alternative HTML version deleted]]
>
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> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>

[[alternative HTML version deleted]]

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[R] restructuring data frame

[PIKAL Petr]

Dear all

I have some data with following structure in data frame.

dput(evid[1:2,c(2:4)])

evid <- structure(list(V2 = c("test vodivosti kalcinátu", "impregnace anatasové 
pasty rozprašovací sušárna"
), V3 = c("03.03.2017", "17.03.2017"), V4 = c("EICHLER Věra;#125",
"HOŠŤÁLKOVÁ Jarmila;#119;#BERNÁT Miroslav;#122;#OSTRČIL Marek;#60"
)), .Names = c("V2", "V3", "V4"), row.names = 9:10, class = "data.frame")

Each row in V4 column contain names followed by ;#xxx. I would like to separate 
them like that

mena <- liche(unlist(strsplit(evid[2,4], ";#")))

here is function liche

liche <- function (x)
{
indices <- seq(along = x)
x[indices%%2 == 1]
}

and repeat each respective row of data frame for separated names like following 
(which is only for one row)

temp<-evid[2,][rep(seq_len(nrow(evid[2,])), length(mena)),-4]
cbind(temp, mena)
   V1  V2 V3
10   NEPRAVDA impregnace anatasové pasty rozprašovací sušárna 17.03.2017
10.1 NEPRAVDA impregnace anatasové pasty rozprašovací sušárna 17.03.2017
10.2 NEPRAVDA impregnace anatasové pasty rozprašovací sušárna 17.03.2017
V5V6 V7   mena
10   OSTRČIL Marek OSTRČIL Marek 1kalcinace HOŠŤÁLKOVÁ Jarmila
10.1 OSTRČIL Marek OSTRČIL Marek 1kalcinaceBERNÁT Miroslav
10.2 OSTRČIL Marek OSTRČIL Marek 1kalcinace  OSTRČIL Marek

I probably could do it for each row in cycle (the data frame is not big) but I 
wonder if somebody knows any more elegant/easy/effective solution for such task.

Best regards

Petr Pikal

"Kdo vždy myslí, že se učí,
bude vlasti chlouba.
Kdo si myslí, že dost umí,
začíná být trouba."
Karel Havlíček Borovský



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Re: [R] concatenating range of columns in dataframe

[Ulrik Stervbo]

Hi Evan,

the unite function of the tidyr package achieves the same as Jim suggested,
but in perhaps a slightly more readable manner.

Ulrik

On Fri, 10 Mar 2017 at 07:50 Jim Lemon  wrote:

> Hi Evan,
> How about this:
>
> df2<-data.frame(Trt=df[,1],Conc=apply(df[,2:5],1,paste,sep="",collapse=""))
>
> Jim
>
> On Fri, Mar 10, 2017 at 3:16 PM, Evan Cooch  wrote:
> > Suppose I have the following data frame (call it df):
> >
> > Trt   y1  y2  y3  y4
> > A1A   1001
> > A1B  1100
> > A1 C   0   10   1
> > A1D   111   1
> >
> > What I want to do is concatenate columns y1  -> y4 into a contiguous
> string
> > (which I'll call df$conc), so that the final df looks like
> >
> > Trt  Conc
> > A1A   1001
> > A1B   1100
> > A1C  0101
> > A1D   
> >
> >
> > Now, if my initial dataframe was simply
> >
> >  1   0  0  1
> >  1   1  0  0
> >   0  1  0  1
> >   1  1  1  1
> >
> > then apply(df,1,paste,collapse="") does the trick, more or less.
> >
> > But once I have a Trt column, this approach yields
> >
> > A1A1001
> > A1B1100
> > A1C0101
> > A1D
> >
> > I need to maintain the space between Trt, and the other columns. So, I'm
> > trying to concatenate a subset of columns in the data frame, but I don't
> > want to have to do something like create a cahracter vector of the column
> > names to do it (e.g., c("y1","y2","y3","y4"). Doing a few by hand that
> way
> > is easy, but not if you  have dozens to hundreds of columns to work with.
> >
> >  Ideally, I'd like to be able to say
> >
> > "concatenate df[,2:4], get rid of the spaces, pipe the concatenated
> columns
> > to a new named column, and drop the original columns from the final df.
> >
> > Heuristically,
> >
> > df$conc <- concatenate df[,2:4] # making a new, 5th column in df
> > df[,2:4] <- NULL   # to drop original columns 2 -> 4
> >
> > Suggestions/pointers to the obvious appreciated.
> >
> > Thanks in advance!
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/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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> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

[[alternative HTML version deleted]]

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Re: [R] concatenating range of columns in dataframe

[Bert Gunter]

I think you need to spend some time with an R tutorial or two,
especially with regard to indexing.

Unless I have misunderstood (apologies if I have),

df$Conc <- apply(df[,-1],1,paste,collapse="")

does it.

-- Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Mar 9, 2017 at 8:16 PM, Evan Cooch  wrote:
> Suppose I have the following data frame (call it df):
>
> Trt   y1  y2  y3  y4
> A1A   1001
> A1B  1100
> A1 C   0   10   1
> A1D   111   1
>
> What I want to do is concatenate columns y1  -> y4 into a contiguous string
> (which I'll call df$conc), so that the final df looks like
>
> Trt  Conc
> A1A   1001
> A1B   1100
> A1C  0101
> A1D   
>
>
> Now, if my initial dataframe was simply
>
>  1   0  0  1
>  1   1  0  0
>   0  1  0  1
>   1  1  1  1
>
> then apply(df,1,paste,collapse="") does the trick, more or less.
>
> But once I have a Trt column, this approach yields
>
> A1A1001
> A1B1100
> A1C0101
> A1D
>
> I need to maintain the space between Trt, and the other columns. So, I'm
> trying to concatenate a subset of columns in the data frame, but I don't
> want to have to do something like create a cahracter vector of the column
> names to do it (e.g., c("y1","y2","y3","y4"). Doing a few by hand that way
> is easy, but not if you  have dozens to hundreds of columns to work with.
>
>  Ideally, I'd like to be able to say
>
> "concatenate df[,2:4], get rid of the spaces, pipe the concatenated columns
> to a new named column, and drop the original columns from the final df.
>
> Heuristically,
>
> df$conc <- concatenate df[,2:4] # making a new, 5th column in df
> df[,2:4] <- NULL   # to drop original columns 2 -> 4
>
> Suggestions/pointers to the obvious appreciated.
>
> Thanks in advance!
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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] Negative binomial GAMM using 'by' in factor interactions

[Bert Gunter]

Your queries appear to concern statistical issues. This list is about
R programming and related; statistical issues are typically OT here.
stats.stackexchange.com or a local statistical expert are probably
better places to seek statistical advice.

Cheers,
Bert
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Mar 9, 2017 at 8:14 PM, Eva Maria Leunissen
 wrote:
> I am using a GAMM to model my data (this is as far as I know the only way I
> can use the negative binomial distribution AND a correlation structure
> within the model).
>
> I measured animal detections (including zero detections) per hour at 3
> different locations in an area. location is a factor in my model and the
> other possible explanatory variables are environmental variables and level
> of disturbance.
>
> I'm expecting the response to be different at the 3 different locations for
> each variable so have been modelling the terms as interactions with each of
> the 3 factor levels of location using the 'by' argument in the 'ti'
> smoothing term, as well as 'location' as a variable by itself. Does it make
> sense to include the main effect as well as the interaction term? for
> example for including the variable 'windspeed': if including the term
> ti(windspeed, by=location), is it necessary to also include s(windspeed)?
> or would it only make sense to inlcude them in separate models only and
> compare the models?
>
> As far as I understand the 'by' argument calculates a separate smooth for
> each of the factor levels, so if the effect was the same at each location
> it wouldn't hurt to use the 'ti' smooth with the 'by' argument if the
> effect of the variable was the same at each location.
>
> The issue I'm having is that by including both terms and then doing model
> selection gives me many very similar models within a deltaAIC of 6 of the
> best model, where the differences lie in the inclusion of main effects when
> the 'interaction' is also there. The inclusion of the interaction term
> gives bigger changes in AIC compared to the inclusion of the main effect.
>
> This brings me to my other question. Is it possible to compare GAMMs with a
> negative binomial family using AIC? e.g. using AIC(mod$lme). If not, what
> is the best way to compare them?
>
> Thank you very much for your time
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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] concatenating range of columns in dataframe

[Jim Lemon]

Hi Evan,
How about this:

df2<-data.frame(Trt=df[,1],Conc=apply(df[,2:5],1,paste,sep="",collapse=""))

Jim

On Fri, Mar 10, 2017 at 3:16 PM, Evan Cooch  wrote:
> Suppose I have the following data frame (call it df):
>
> Trt   y1  y2  y3  y4
> A1A   1001
> A1B  1100
> A1 C   0   10   1
> A1D   111   1
>
> What I want to do is concatenate columns y1  -> y4 into a contiguous string
> (which I'll call df$conc), so that the final df looks like
>
> Trt  Conc
> A1A   1001
> A1B   1100
> A1C  0101
> A1D   
>
>
> Now, if my initial dataframe was simply
>
>  1   0  0  1
>  1   1  0  0
>   0  1  0  1
>   1  1  1  1
>
> then apply(df,1,paste,collapse="") does the trick, more or less.
>
> But once I have a Trt column, this approach yields
>
> A1A1001
> A1B1100
> A1C0101
> A1D
>
> I need to maintain the space between Trt, and the other columns. So, I'm
> trying to concatenate a subset of columns in the data frame, but I don't
> want to have to do something like create a cahracter vector of the column
> names to do it (e.g., c("y1","y2","y3","y4"). Doing a few by hand that way
> is easy, but not if you  have dozens to hundreds of columns to work with.
>
>  Ideally, I'd like to be able to say
>
> "concatenate df[,2:4], get rid of the spaces, pipe the concatenated columns
> to a new named column, and drop the original columns from the final df.
>
> Heuristically,
>
> df$conc <- concatenate df[,2:4] # making a new, 5th column in df
> df[,2:4] <- NULL   # to drop original columns 2 -> 4
>
> Suggestions/pointers to the obvious appreciated.
>
> Thanks in advance!
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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] matrix merge, or something else?

[Jeff Newmiller]

test2[ , colnames( test1 ) ] <- test1
-- 
Sent from my phone. Please excuse my brevity.

On March 9, 2017 6:56:13 PM PST, Evan Cooch  wrote:
>Suppose I have the following two matrices, both with same number of
>rows 
>(3), but different number of columns (3 in test1, 4 in test2).
>
>test1 <- matrix(c(1,1,0,1,0,-1,-1,-1,0),3,3,byrow=T);
>test2 <- matrix( rep( 0, len=12), nrow = 3)
>
>I label the rows and columns of the two matrices as follows:
>
>rownames(test1) <- c("row1","row2","row3")
>rownames(test2) <- c("row1","row2","row3")
>
>colnames(test1) <- c("a","b","d")
>colnames(test2) <- c("a","b","c","d")
>
>So, if we look at the matrices, we see
>
>test1
>
>  a  b  d
>row1   1  1  0
>row2   1  0 -1
>row3  -1 -1  0
>
>
>test2
>
> a b c d
>row1  0 0 0 0
>row2  0 0 0 0
>row3  0 0 0 0
>
>So, we see that while both matrices have the same rows, the matrix
>test1 
>has a subset of the columns of test2. In test1, there is no column for 
>'c' -- have columns for 'a', 'b', 'd'.
>
>Now, what I want to do is this -- take the information from each column
>
>in test1, and substitute it into the same row/column in test2. The end 
>result should be a matrix that looks like:
>
> a  b  c  d
>row1  1  1  0  0
>row2  1  0  0 -1
>row3 -1 -1 0   0
>
>My initial though  was some sort of merge by row and column, with some 
>funky sort of intersection, but I couldn't figure out how to get that
>to 
>work.
>
>Any suggestions/pointers to the obvious most appreciated.
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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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[R] concatenating range of columns in dataframe

[Evan Cooch]

Suppose I have the following data frame (call it df):

Trt   y1  y2  y3  y4
A1A   1001
A1B  1100
A1 C   0   10   1
A1D   111   1

What I want to do is concatenate columns y1  -> y4 into a contiguous 
string (which I'll call df$conc), so that the final df looks like


Trt  Conc
A1A   1001
A1B   1100
A1C  0101
A1D   


Now, if my initial dataframe was simply

 1   0  0  1
 1   1  0  0
  0  1  0  1
  1  1  1  1

then apply(df,1,paste,collapse="") does the trick, more or less.

But once I have a Trt column, this approach yields

A1A1001
A1B1100
A1C0101
A1D

I need to maintain the space between Trt, and the other columns. So, I'm 
trying to concatenate a subset of columns in the data frame, but I don't 
want to have to do something like create a cahracter vector of the 
column names to do it (e.g., c("y1","y2","y3","y4"). Doing a few by hand 
that way is easy, but not if you  have dozens to hundreds of columns to 
work with.


 Ideally, I'd like to be able to say

"concatenate df[,2:4], get rid of the spaces, pipe the concatenated 
columns to a new named column, and drop the original columns from the 
final df.


Heuristically,

df$conc <- concatenate df[,2:4] # making a new, 5th column in df
df[,2:4] <- NULL   # to drop original columns 2 -> 4

Suggestions/pointers to the obvious appreciated.

Thanks in advance!

__
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[R] Negative binomial GAMM using 'by' in factor interactions

[Eva Maria Leunissen]

I am using a GAMM to model my data (this is as far as I know the only way I
can use the negative binomial distribution AND a correlation structure
within the model).

I measured animal detections (including zero detections) per hour at 3
different locations in an area. location is a factor in my model and the
other possible explanatory variables are environmental variables and level
of disturbance.

I'm expecting the response to be different at the 3 different locations for
each variable so have been modelling the terms as interactions with each of
the 3 factor levels of location using the 'by' argument in the 'ti'
smoothing term, as well as 'location' as a variable by itself. Does it make
sense to include the main effect as well as the interaction term? for
example for including the variable 'windspeed': if including the term
ti(windspeed, by=location), is it necessary to also include s(windspeed)?
or would it only make sense to inlcude them in separate models only and
compare the models?

As far as I understand the 'by' argument calculates a separate smooth for
each of the factor levels, so if the effect was the same at each location
it wouldn't hurt to use the 'ti' smooth with the 'by' argument if the
effect of the variable was the same at each location.

The issue I'm having is that by including both terms and then doing model
selection gives me many very similar models within a deltaAIC of 6 of the
best model, where the differences lie in the inclusion of main effects when
the 'interaction' is also there. The inclusion of the interaction term
gives bigger changes in AIC compared to the inclusion of the main effect.

This brings me to my other question. Is it possible to compare GAMMs with a
negative binomial family using AIC? e.g. using AIC(mod$lme). If not, what
is the best way to compare them?

Thank you very much for your time

[[alternative HTML version deleted]]

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[R] R2WinBUGS Error

[Liu, J.]

Hi,

I'm trying to run R2WinBUGS using the R code below (Thinkpad Yoga 260,
Win8, system x86_64, mingw32, R version 3.3.1). It worked fine for several
times but then one error began to pop up in every run: command #Bugs:set
cannot be executed (is greyed out). I've been trying for more than one week
but still can't figure out where is the problem. It would be great if
someone could help me with this. Thanks in advance!

Kind regards,
JLiu

Here's the code:

sink("mod1.txt")
cat("
model {

for( k in 1 : n ) {
for( i in 1 : n - 1 ) {
for( j in i + 1 : n ) {
y[k , i , j] ~ dbern(p[k , i , j])
y[k , j , i] ~ dbern(p[k , j , i])
logit(p[k , i , j]) <- mu + a[i] + b[j] + g[k] + ab[i , j] + ag[i , k]
+ ag[j , k]
logit(p[k , j , i]) <- mu + a[j] + b[i] + g[k] + ab[j , i] + ag[j , k]
+ ag[i , k]
}
}
}

# Compute difference ab[1,2] - ab[2,1] for Figure 3 of the paper
dif12 <- ab[1 , 2] - ab[2 , 1]

# Prior for the overall mean effect mu
mu ~ dnorm(0, 1)

# Tri-normal prior for actor, partner, rater effects (a, b, g)
# with zero-sum constraints
for( i in 1 : n ) {
a[i] <- a1[i , 1] - mean(a1[ , 1])
b[i] <- a1[i , 2] - mean(a1[ , 2])
g[i] <- a1[i , 3] - mean(a1[ , 3])

# without zero-sum constraints will make the code run faster
# for( i in 1 : n ) {
# a[i] <- a1[i,1]
# b[i] <- a1[i,2]
# g[i] <- a1[i,3]

a1[i , 1:3] ~ dmnorm(zero[1:3], S1[ , ])
}

# Wishart prior for precision matrix S1 = Sigma_1-inverse
# degrees of freedom nu = 3
# Om1 = nu*Identity Matrix is provided in the data list
S1[1:3 , 1:3] ~ dwish(Om1[1:3 , 1:3], 3)
Sig1[1:3 , 1:3] <- inverse(S1[1:3 , 1:3])

# Compute the correlations and the variances for the main effects
rho1 <- Sig1[1 , 2] / sqrt(Sig1[1 , 1] * Sig1[2 , 2])
rho2 <- Sig1[1 , 3] / sqrt(Sig1[1 , 1] * Sig1[3 , 3])
rho3 <- Sig1[2 , 3] / sqrt(Sig1[3 , 3] * Sig1[2 , 2])
sig.a <- Sig1[1 , 1]
sig.b <- Sig1[2 , 2]
sig.g <- Sig1[3 , 3]

# Standard deviation values are reported in the paper (Table 2)
sd.a <- sqrt(sig.a)
sd.b <- sqrt(sig.b)
sd.g <- sqrt(sig.g)
# A vector of zero's is needed for the mean values of some vectors
for( i in 1 : 4 ) {
zero[i] <- 0
}

# Priors for interaction effects ab_ij and ag_ij in equation (3) of the
paper
for( i in 1 : n ) {
# alpha_beta[i,i] is not defined in the model, just put  =0
ab[i , i] <- 0

# del[i] = personal bias of subject i in reporting his tendency to
establish friendship ties
# The posterior distribution of del[i]'s is shown as boxplots in Figure
3 of the paper.
# Side-by-side boxplots are created using Inference -> Compare menu in
WinBUGS

del[i] <- ag[i , i] - ag.mean[i]
ag.mean[i] <- (sum(ag[i , ]) - ag[i , i]) / (n - 1)
ag[i , i] ~ dnorm(0, tau.ag)
}

# Generate 4 pair-wise interaction parameters as Normal_4 with
precision matrix S2
for( i in 1 : n - 1 ) {
for( j in i + 1 : n ) {
ab[i , j] <- a2[i , j , 1]
ab[j , i] <- a2[i , j , 2]
ag[i , j] <- a2[i , j , 3]
ag[j , i] <- a2[i , j , 4]
a2[i , j , 1:4] ~ dmnorm(zero[1:4], S2[ , ])
}
}

# Get the precision matrix of interaction parameters from the var-cov
matrix Sig_2 in the paper
# Note that we don't use Inv-Wishart prior for this precision matrix
S2[1:4 , 1:4] <- inverse(Sig2[1:4 , 1:4])

# Now build the var-cov matrix Sig2 from variances and correlations
(phi's) in equation (7)

phi1 ~ dunif(-0.99, 0.99)
phi2 ~ dunif(-0.99, 0.99)
phi3 ~ dunif(-0.99, 0.99)
phi4 ~ dunif(-0.99, 0.99)

# Compute two detrminants in equations (9-10) in the paper
det1 <-  1 -  phi1 * phi1 - phi2 * phi2 - phi3 * phi3 + 2 * phi1 * phi2
* phi3
det2 <-  1 -  phi1 * phi1 - 2 * phi2 * phi2 - 2 * phi3 * phi3 - phi4 *
phi4 + 4 * phi1 * phi2 * phi3
+ 4 * phi2 * phi3 * phi4 + phi1 * phi1 * phi4 * phi4 - 2 * phi2 * phi2
* phi3 * phi3
- 2 * phi1 * phi3 * phi3 * phi4 - 2 * phi1 * phi2 * phi2 * phi4 + phi2
* phi2 * phi2 * phi2
+ phi3 * phi3 * phi3 * phi3

# Check if determinants are positive
cons1 <- step(det1)
cons2 <- step(det2)

# Zeros trick
O1 <- 0
O2 <- 0
q1 <- 1 - cons1
q2 <- 1 - cons2
O1 ~ dbern(q1)
O2 ~ dbern(q2)

# Inverse Gamma priors for variance parameters sig.ab, sig.ag
tau.ab ~ dgamma(100, 10)
tau.ag ~ dgamma(100, 10)
sig.ab <- 1 / tau.ab
sig.ag <- 1 / tau.ag

# Standard deviations sd.ab and sd.ag are reported in the paper (Table
2)
sd.ab <- sqrt(sig.ab)
sd.ag <- sqrt(sig.ag)
Sig2[1 , 1] <- sig.ab
Sig2[3 , 3] <- sig.ag
Sig2[2 , 2] <- Sig2[1,1]
Sig2[4 , 4] <- Sig2[3 , 3]

# Create the off-diagonal entries of Sig2
# If the generated set of phi1, phi2, phi3 and phi4 values don't have
both det1 > 0
# and det2 > 0, the off-diagonal elements of Sig2 are all set equal to
zero
Sig2[1,2] <- 

[R] matrix merge, or something else?

[Evan Cooch]

Suppose I have the following two matrices, both with same number of rows 
(3), but different number of columns (3 in test1, 4 in test2).


test1 <- matrix(c(1,1,0,1,0,-1,-1,-1,0),3,3,byrow=T);
test2 <- matrix( rep( 0, len=12), nrow = 3)

I label the rows and columns of the two matrices as follows:

rownames(test1) <- c("row1","row2","row3")
rownames(test2) <- c("row1","row2","row3")

colnames(test1) <- c("a","b","d")
colnames(test2) <- c("a","b","c","d")

So, if we look at the matrices, we see

test1

 a  b  d
row1   1  1  0
row2   1  0 -1
row3  -1 -1  0


test2

a b c d
row1  0 0 0 0
row2  0 0 0 0
row3  0 0 0 0

So, we see that while both matrices have the same rows, the matrix test1 
has a subset of the columns of test2. In test1, there is no column for 
'c' -- have columns for 'a', 'b', 'd'.


Now, what I want to do is this -- take the information from each column 
in test1, and substitute it into the same row/column in test2. The end 
result should be a matrix that looks like:


a  b  c  d
row1  1  1  0  0
row2  1  0  0 -1
row3 -1 -1 0   0

My initial though  was some sort of merge by row and column, with some 
funky sort of intersection, but I couldn't figure out how to get that to 
work.


Any suggestions/pointers to the obvious most appreciated.

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Re: [R] how apply.monthly() in package xts works

[Joshua Ulrich]

On Thu, Mar 9, 2017 at 3:46 PM, Joshua Ulrich  wrote:
> On Thu, Mar 9, 2017 at 3:31 PM, Waichler, Scott R
>  wrote:
>> Hi,
>>
>> I found that apply.monthly() in xts does not work as I expected in the case 
>> of a sparse timeseries:
>>
>> my.dates <- as.Date(c("1992-06-01", "1992-06-24", "1992-06-30", 
>> "1993-06-22", "1994-06-07", "1995-06-08"))
>> my.xts <- xts(1:6, my.dates)
>> start(my.xts)  # "1992-06-24"
>> end(my.xts)  # "1995-06-08"
>> apply.monthly(my.xts, mean)
>> #   [,1]
>> # 1995-06-08 3.5
>>
>> The endpoints it chooses are based on looking at the month (June) alone.  I 
>> was able to get a value for each (month, year) in the timeseries with the 
>> following use of aggregate():
>>
> Thanks for the minimal, reproducible example!  This is clearly a bug.
>
Now formally documented as such:
https://github.com/joshuaulrich/xts/issues/169

>> my.months <- months(my.dates)
>> my.years <- years(my.dates)
>> df1 <- data.frame(x = coredata(my.xts), dates = my.dates, months = 
>> my.months, years = my.years)
>> df2 <- aggregate(df1[-c(3,4)], df1[c("months", "years")], mean)
>> xts(df2$x, df2$dates)
>> #[,1]
>> # 1992-06-182
>> # 1993-06-224
>> # 1994-06-075
>> # 1995-06-086
>>
>> Two questions:
>> 1) Is there a more elegant way to do this?
>
> Create your own endpoints until endpoints() is fixed.  Here's a quick
> hack, off the top of my head:
>
> endpointsMonthHack <- function(x, on = "months", k = 1) {
>   # yearmon index
>   ymIndex <- as.yearmon(index(x))
>   # month changes
>   monthDiff <- c(0, diff(ymIndex))
>   # locations in index
>   locations <- which(monthDiff != 0)
>   ep <- c(0, locations, nrow(x))
>   unique(ep)
> }
>
The function above is wrong.  That's what I get for posting without
actually running the code.  Here's a function that's actually tested
(only on this example though):

endpointsMonthHack <- function(x, on = "months", k = 1) {
  # yearmon index
  ymIndex <- as.yearmon(index(x))
  # month change locations
  locations <- which(diff(ymIndex) != 0)
  # endpoints
  ep <- c(0, locations, nrow(x))
  unique(ep)
}


>> 2) Shouldn't the xts documentation discuss the problem of sparse data?
>
> No, because it shouldn't be a problem. :)
>
>>
>> Regards,
>> Scott Waichler
>> Pacific Northwest National Laboratory
>> Richland, WA  USA
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Joshua Ulrich  |  about.me/joshuaulrich
> FOSS Trading  |  www.fosstrading.com
> R/Finance 2017 | www.rinfinance.com



-- 
Joshua Ulrich  |  about.me/joshuaulrich
FOSS Trading  |  www.fosstrading.com
R/Finance 2017 | www.rinfinance.com

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how apply.monthly() in package xts works

[Joshua Ulrich]

On Thu, Mar 9, 2017 at 3:31 PM, Waichler, Scott R
 wrote:
> Hi,
>
> I found that apply.monthly() in xts does not work as I expected in the case 
> of a sparse timeseries:
>
> my.dates <- as.Date(c("1992-06-01", "1992-06-24", "1992-06-30", "1993-06-22", 
> "1994-06-07", "1995-06-08"))
> my.xts <- xts(1:6, my.dates)
> start(my.xts)  # "1992-06-24"
> end(my.xts)  # "1995-06-08"
> apply.monthly(my.xts, mean)
> #   [,1]
> # 1995-06-08 3.5
>
> The endpoints it chooses are based on looking at the month (June) alone.  I 
> was able to get a value for each (month, year) in the timeseries with the 
> following use of aggregate():
>
Thanks for the minimal, reproducible example!  This is clearly a bug.

> my.months <- months(my.dates)
> my.years <- years(my.dates)
> df1 <- data.frame(x = coredata(my.xts), dates = my.dates, months = my.months, 
> years = my.years)
> df2 <- aggregate(df1[-c(3,4)], df1[c("months", "years")], mean)
> xts(df2$x, df2$dates)
> #[,1]
> # 1992-06-182
> # 1993-06-224
> # 1994-06-075
> # 1995-06-086
>
> Two questions:
> 1) Is there a more elegant way to do this?

Create your own endpoints until endpoints() is fixed.  Here's a quick
hack, off the top of my head:

endpointsMonthHack <- function(x, on = "months", k = 1) {
  # yearmon index
  ymIndex <- as.yearmon(index(x))
  # month changes
  monthDiff <- c(0, diff(ymIndex))
  # locations in index
  locations <- which(monthDiff != 0)
  ep <- c(0, locations, nrow(x))
  unique(ep)
}

> 2) Shouldn't the xts documentation discuss the problem of sparse data?

No, because it shouldn't be a problem. :)

>
> Regards,
> Scott Waichler
> Pacific Northwest National Laboratory
> Richland, WA  USA
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



-- 
Joshua Ulrich  |  about.me/joshuaulrich
FOSS Trading  |  www.fosstrading.com
R/Finance 2017 | www.rinfinance.com

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[R] how apply.monthly() in package xts works

[Waichler, Scott R]

Hi,  

I found that apply.monthly() in xts does not work as I expected in the case of 
a sparse timeseries:

my.dates <- as.Date(c("1992-06-01", "1992-06-24", "1992-06-30", "1993-06-22", 
"1994-06-07", "1995-06-08"))
my.xts <- xts(1:6, my.dates)
start(my.xts)  # "1992-06-24"
end(my.xts)  # "1995-06-08"
apply.monthly(my.xts, mean)
#   [,1]
# 1995-06-08 3.5

The endpoints it chooses are based on looking at the month (June) alone.  I was 
able to get a value for each (month, year) in the timeseries with the following 
use of aggregate():

my.months <- months(my.dates)
my.years <- years(my.dates)
df1 <- data.frame(x = coredata(my.xts), dates = my.dates, months = my.months, 
years = my.years)
df2 <- aggregate(df1[-c(3,4)], df1[c("months", "years")], mean)
xts(df2$x, df2$dates)
#[,1]
# 1992-06-182
# 1993-06-224
# 1994-06-075
# 1995-06-086

Two questions:  
1) Is there a more elegant way to do this? 
2) Shouldn't the xts documentation discuss the problem of sparse data?

Regards,
Scott Waichler
Pacific Northwest National Laboratory
Richland, WA  USA

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Re: [R] Problem with greek characters in R

[Nordlund, Dan (DSHS/RDA)]

Dimitrios,

You need to help us help you. You say you have a file that has Greek characters 
in it that you want to "open with the program RGui".
1. You need to provide us with a sample of the problem file.  Since we are 
talking about text here, you can create a file with just a few lines in it that 
contain the problematic text.  Give the file a '.txt' extension and then you 
can attach that to your next post to R-help.
2. What do mean by "open the file with RGui"?  Are you trying to source the 
file, or open it some other way?

If you provide some sample text, and describe in more detail what you are 
trying to do with it, someone may be able to offer a solution.


Dan

Daniel Nordlund, PhD
Research and Data Analysis Division
Services & Enterprise Support Administration
Washington State Department of Social and Health Services


On March 9, 2017 2:34:40 AM PST, Dimitrios Mousenikas rotondor...@hotmail.gr> 
wrote:
>Hello,
>
>My computer functions with the version of English windows 10. The
>problem that I face is that when I open an R file that includes Greek
>characters, with the program RGui, the text that appears does not make
>sense. The only way the problem can be solved is by setting Greek as a
>default language in Windows. Although I prefer as a default language
>English.
>I am looking forward for your response.
>
>Thank you in advance
>Dimitrios Mousenikas
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with greek characters in R

[Ista Zahn]

On Thu, Mar 9, 2017 at 11:48 AM, Jeff Newmiller
 wrote:
> The standard response is that RStudio is not R, and has its own forum or 
> discussion areas on stackexchange.com.

But the OP didn't mention RStudio, but rather RGui.

R has its own mechanisms for dealing with alternate character sets, so
be clear when the problem is in the processing of such code by R
versus the display of input or output files associated with your use
of R. Using R or RGui (that are on topic here) can help you discern
where the "problem" is.

Perhaps you can offer some advice in this regard, I've never been able
to figure it out. Ironically, that's one of the reasons I prefer to
use RStudio on Windows!

Best,
Ista

>
> That said, Googling "RStudio character encoding" might lead to help with your 
> immediate concern.
> --
> Sent from my phone. Please excuse my brevity.
>
> On March 9, 2017 2:34:40 AM PST, Dimitrios Mousenikas 
>  wrote:
>>Hello,
>>
>>My computer functions with the version of English windows 10. The
>>problem that I face is that when I open an R file that includes Greek
>>characters, with the program RGui, the text that appears does not make
>>sense. The only way the problem can be solved is by setting Greek as a
>>default language in Windows. Although I prefer as a default language
>>English.
>>I am looking forward for your response.
>>
>>Thank you in advance
>>Dimitrios Mousenikas
>>
>>__
>>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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] Reverse the scoring of some Columns of a Data Set

[Nordlund, Dan (DSHS/RDA)]

Another alternative (which didn't work last night when I was tired and 
obviously doing something wrong) is to use the built-in function, rev():

df[,1:3] <- apply(df[,1:3], 2, rev)


Dan

Daniel Nordlund, PhD
Research and Data Analysis Division
Services & Enterprise Support Administration
Washington State Department of Social and Health Services

> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Daniel
> Nordlund
> Sent: Wednesday, March 08, 2017 11:35 PM
> To: AbouEl-Makarim Aboueissa; r-help@r-project.org
> Subject: Re: [R] Reverse the scoring of some Columns of a Data Set
> 
> On 3/8/2017 6:14 AM, AbouEl-Makarim Aboueissa wrote:
> > Dear All: goods morning
> >
> > Is there is a way to reverse the scoring of the first three columns
> x1, x2,
> > and x3 and keep the original scores for the fourth column x4.
> >
> >
> > *Here is an example of the data set:*
> >
> > x1 x2 x3 x4
> > 2  5   4   4
> > 1  1   1   6
> > 1  2   1   6
> > 2  3   2   4
> > 1  2   1   6
> > 1  3   1   6
> > 2  2   2   5
> > 2  1   1   6
> > 2  2   4   5
> > 5  5   2   1
> >
> > I am expecting the output to be:
> > x1 x2 x3 x4
> > 5  5   2   4
> > 2  2   4   6
> > 2  1   1   6
> > 2  2   2   4
> > 1  3   1   6
> > 1  2   1   6
> > 2  3   2   5
> > 1  2   1   6
> > 1  1   1   5
> > 2  5   4   1
> >
> >
> >
> > thank you very much for your help and support
> > abou
> > __
> > AbouEl-Makarim Aboueissa, PhD
> > Department of Mathematics and Statistics
> > University of Southern Maine
> >
> 
>   If your data is in a data frame called df, you could do something like
> this:
> 
> df[,1:3] <- apply(df[,1:3], 2, function(x) x[length(x):1])
> 
> 
> Hope this helps,
> 
> Dan
> 
> --
> Daniel Nordlund
> Port Townsend, WA  USA
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see
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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] Problem with greek characters in R

[Jeff Newmiller]

The standard response is that RStudio is not R, and has its own forum or 
discussion areas on stackexchange.com. R has its own mechanisms for dealing 
with alternate character sets, so be clear when the problem is in the 
processing of such code by R versus the display of input or output files 
associated with your use of R. Using R or RGui (that are on topic here) can 
help you discern where the "problem" is. 

That said, Googling "RStudio character encoding" might lead to help with your 
immediate concern. 
-- 
Sent from my phone. Please excuse my brevity.

On March 9, 2017 2:34:40 AM PST, Dimitrios Mousenikas  
wrote:
>Hello,
>
>My computer functions with the version of English windows 10. The
>problem that I face is that when I open an R file that includes Greek
>characters, with the program RGui, the text that appears does not make
>sense. The only way the problem can be solved is by setting Greek as a
>default language in Windows. Although I prefer as a default language
>English. 
>I am looking forward for your response. 
>
>Thank you in advance
>Dimitrios Mousenikas
>
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Re: [R] Transport and Earth Mover's Distance

[Schuhmacher, Dominic]

> Am 08.03.2017 um 11:28 schrieb Schuhmacher, Dominic 
> :
> 
> ...
>>> 
>>> If you have no particular need for binning, check out the function
>>> pppdist in the R-package spatstat, which offers a more flexible way
>>> to deal with point patterns of different size.
>> 
>> 
>> Well, this is not clear, but possibly very important for me.
>> My raw data consists of 2 univariate samples of unequal length.
>> 
>> suppose that
>> 
>> x<-rnorm(100)
>> 
>> and
>> 
>> y<-rnorm(90)
>> 
>> Is there a way to define the Wasserstein distance between them without
>> going through the binning procedure?
>> 
> Define, yes: the 1-Wasserstein distance in one-dimension is the area between 
> the empirical cumulative distribution functions. If the samples had the same 
> lengths this could be directly computed by
> 
> mean(abs(sort(x)-sort(y)))
> 
> Otherwise this needs some lines of code. I will include it in the next 
> version of the transport package (soon).
> 
> Best regards,
> Dominic
> 
> 
Following up on this earlier post: transport 0.8-2, which is on CRAN now, 
offers the possibility to compute the Wasserstein distance between univariate 
samples of differing lengths (more precisely their empirical distributions).

library(transport)
x <- rnorm(100)
y <- rnorm(90)
wasserstein1d(x,y) 

Cheers, Dominic



Dominic Schuhmacher
Professor of Stochastics
University of Goettingen
http://www.dominic.schuhmacher.name

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Re: [R] Unable to Load package Rcmdr after installation

[Jeff Newmiller]

To the question of why it used to work but now it doesn't, I have noticed that 
often when I update packages with dependencies but some error occurs during the 
update, some existing packages that used to work are removed and must be 
re-installed manually. I have not tried to make a reproducible example but the 
pattern of breaking depencies has been noticeable for the last year or so. 
-- 
Sent from my phone. Please excuse my brevity.

On March 9, 2017 8:11:05 AM PST, Ulrik Stervbo  wrote:
>Hi Paul,
>
>The error tells you, that the 'Hmisc' does not exist on your system. If
>you
>install it, everything should work.
>
>Use install.packages with dependencies = TRUE to avoid the problem of
>missing packages.
>
>HTH
>
>Ulrik
>
>On Thu, 9 Mar 2017 at 16:51 Paul Bernal  wrote:
>
>Hello friends,
>
>Has anyone experienced trouble when trying to load package Rcmdr? It
>was
>working perfectly a couple of days ago, I don´t know why it isn´t
>working.
>
>> library("Rcmdr")
>Loading required package: splines
>Loading required package: RcmdrMisc
>Loading required package: car
>Loading required package: sandwich
>Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()),
>versionCheck
>= vI[[j]]) :
>  there is no package called ‘Hmisc’
>Error: package ‘RcmdrMisc’ could not be loaded
>
>[[alternative HTML version deleted]]
>
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
>
>   [[alternative HTML version deleted]]
>
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[R] modify additional parameters, glht and summary

[Cristiano Alessandro]

Hi all,

first of all, thanks a lot in advance for your help. I am running a
sequence of post-hoc tests with glht (mutcomp package), but the function
summary warns me that the algorithm ends with an error > abseps.

$ hr.ph <- glht(hr.lm, linfct = ph_conditional);
$ summary(hr.ph)

Warning messages:
1: In RET$pfunction("adjusted", ...) : Completion with error > abseps
2: In RET$pfunction("adjusted", ...) : Completion with error > abseps
3: In RET$pfunction("adjusted", ...) : Completion with error > abseps
4: In RET$pfunction("adjusted", ...) : Completion with error > abseps

I think it is a matter of modifying the parameters of the algorithm that
summary runs. Looking at the documentation, I could not find a way of
actually doing so. Do you have any suggestion? Also, how does one know the
default values of these algorithms?

Thanks a lot!

Cristiano

[[alternative HTML version deleted]]

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[R] Problem with greek characters in R

[Dimitrios Mousenikas]

Hello,

My computer functions with the version of English windows 10. The problem that 
I face is that when I open an R file that includes Greek characters, with the 
program RGui, the text that appears does not make sense. The only way the 
problem can be solved is by setting Greek as a default language in Windows. 
Although I prefer as a default language English. 
I am looking forward for your response. 

Thank you in advance
Dimitrios Mousenikas

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Re: [R] Unable to Load package Rcmdr after installation

[Ulrik Stervbo]

Hi Paul,

The error tells you, that the 'Hmisc' does not exist on your system. If you
install it, everything should work.

Use install.packages with dependencies = TRUE to avoid the problem of
missing packages.

HTH

Ulrik

On Thu, 9 Mar 2017 at 16:51 Paul Bernal  wrote:

Hello friends,

Has anyone experienced trouble when trying to load package Rcmdr? It was
working perfectly a couple of days ago, I don´t know why it isn´t working.

> library("Rcmdr")
Loading required package: splines
Loading required package: RcmdrMisc
Loading required package: car
Loading required package: sandwich
Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck
= vI[[j]]) :
  there is no package called ‘Hmisc’
Error: package ‘RcmdrMisc’ could not be loaded

[[alternative HTML version deleted]]

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[[alternative HTML version deleted]]

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[R] Unable to Load package Rcmdr after installation

[Paul Bernal]

Hello friends,

Has anyone experienced trouble when trying to load package Rcmdr? It was
working perfectly a couple of days ago, I don´t know why it isn´t working.

> library("Rcmdr")
Loading required package: splines
Loading required package: RcmdrMisc
Loading required package: car
Loading required package: sandwich
Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck
= vI[[j]]) :
  there is no package called ‘Hmisc’
Error: package ‘RcmdrMisc’ could not be loaded

[[alternative HTML version deleted]]

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Re: [R-es] Asignación de objetos

[Fernando Macedo]

Creo que este link te puede servir:

http://stackoverflow.com/questions/1741820/assignment-operators-in-r-and

Fernando Macedo

El 09/03/17 a las 05:57, Álvaro Díez Ansotegui escribió:


Gracias José Antonio,

Entiendo por lo tanto que se puede usar indistintamente y aconsejan utilizar 
“<-“ para no llevar a un posible error. Gracias. Saludos!!

Álvaro D.



El 9 mar 2017, a las 9:49, José Antonio Palazón Ferrando  
escribió:

Hola:

Puedes considerar que:

objeto <- expresión
objeto =  expresión

1. no es una igualdad, en realidad asignas el resultado de la expresión al 
objeto.

2. dado que existe los operadores '==',  '!='  y usamos '=' para
definir los valores de los argumentos en las funciones, pare bueno
evitar otro uso para un símbolo frecuente y que podría llevar a algún error:

función( objeto, argumento = expresión )
función( objeto1, objeto2 = expresión )

¿No sé si te parece suficiente con esto?

Seguimos


El 09/03/17 a las 09:25, Álvaro Díez Ansotegui escribió:

Buenas!

Estoy introduciéndome a R desde el enfoque de las CC Sociales. Estando trabajando los 
manuales introductores (vectores, matrices, importación y exportación de bbdd, 
tratamiento de variables, etc..) me ha quedado una duda que los manuales no 
profundizan en resolver. Pocos textos no distinguen entre la asignación, tratan el “ 
= “ de la misma forma que “ <- “. Sin embargo, la mayoría aconseja utilizar en la 
asignación “<-“, pero no profundizan en el por qué.

¿Podría alguien explicar (sin restaros mucho tiempo) el porqué?

Saludos

Álvaro D.
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--


José Antonio Palazón Ferrando
Profesor Titular. Departamento de Ecología e Hidrología.
Facultad de Biología. Universidad de Murcia.
Campus Universitario de Espinardo
30100 MURCIA-SPAIN
Telf: +34 868 88 49 80
Fax : +34 868 88 39 63
Email: pala...@um.es

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Re: [R] Why is merge sorting even when sort = F?

[DIGHE, NILESH [AG/2362]]

Using the "join" function from the plyr package preserves the data order 
library(plyr)
join(grades2, info, by="grade", type="left", match="all")

Nilesh
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri 
Liakhovitski
Sent: Wednesday, March 08, 2017 12:45 PM
To: Jeff Newmiller 
Cc: r-help 
Subject: Re: [R] Why is merge sorting even when sort = F?

I understood your answer.
The point is that sort = TRUE that doesn't sort is plain confusing.
Instead, the option should have been something like efficient = TRUE or FALSE. 
At least then no one would stupidly expect sort = TRUE to sort and sort = FALSE 
to NOT sort.

On Wed, Mar 8, 2017 at 12:51 PM, Jeff Newmiller  
wrote:
> If you are still wondering, try re-reading my answer. FALSE is more 
> efficient, TRUE is sorted. Lack of sorting has nothing to do with preserving 
> order.
> --
> Sent from my phone. Please excuse my brevity.
>
> On March 8, 2017 8:55:06 AM PST, Dimitri Liakhovitski 
>  wrote:
>>Thank you. I was just curious what sort=FALSE had no impact.
>>Wondering what it is there for then...
>>
>>On Wed, Mar 8, 2017 at 11:43 AM, Jeff Newmiller 
>> wrote:
>>> Merging is not necessarily an order-preserving operation, but 
>>> sorting
>>can make the operation more efficient. The sort=TRUE argument forces 
>>the result to be sorted, but sort=FALSE is in not a promise that order 
>>will be preserved. (I think the imperfect sorting occurs when there 
>>are multiple keys but am not sure.) You can add columns to the input 
>>data that let you restore some semblance of the original ordering 
>>afterward, or you can roll your own possibly-less-efficient merge 
>>using match and
>>indexing:
>>>
>>> info[ match( grades2$grade, info$grade ), ]
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>>
>>> On March 8, 2017 8:07:27 AM PST, Dimitri Liakhovitski
>> wrote:
Hello!
I have a vector 'grades' and a data frame 'info':

grades2 <- data.frame(grade = c(1,2,2,3,1)) info <- data.frame(
  grade = 3:1,
  desc = c("Excellent", "Good", "Poor"),
  fail = c(F, F, T)
)

I want to get the info for all grades I have in info:

This solution resorts everything in the order of column 'grade':
merge(grades2, info, by = "grade", all.x = T, all.y = F)

Could you please explain why this solution also resorts - despite
>>sort
= FALSE?
merge(grades2, info, by = "grade", all.x = T, all.y = F, sort =
>>FALSE)

Thanks a lot!



--
Dimitri Liakhovitski

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[R] Antwort: Re: Approach for Storing Result Data

[G . Maubach]

Hi Gunter,
Hi Jeff,
Hi Readers,

many thanks for your reply.

My questions seems to be a little off topic cause it is not about using 
the programming language itself but how to use it in a analytics context. 
It is about processes and approaches of how to do things in R from a 
conception point of view. That is a subject I don't see in the community 
but would help me a lot to enhance my work.

Do you know I place where these things are discussed?

Kind regards

Georg



Von:Jeff Newmiller 
An: r-help@r-project.org, g.maub...@weinwolf.de, 
Datum:  08.03.2017 17:54
Betreff:Re: [R] Approach for Storing Result Data



Seems pretty normal except that your one-by-one lookup process usually 
gets old eventually, and comparing results is much easier if you merge the 
study data with the lookup data all at once and then use aggregate() (or 
any of numerous equivalents from contributed packages) to collect results 
or color/linetype/panel/etc plotted graphical presentations with lattice 
or ggplot2.



Von:Bert Gunter 
An: g.maub...@weinwolf.de, 
Kopie:  R-help 
Datum:  08.03.2017 17:43
Betreff:Re: [R] Approach for Storing Result Data



This does not appear to be a legitimate topic for r-help: it is are
not a consulting service. Please see the posting guide.

Of course, others may disagree and reply. Wouldn't be the first time I'm 
wrong.

-- Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, Mar 8, 2017 at 7:27 AM,   wrote:
> Hi All,
>
> today I have a more general question concerning the approach of storing
> different values from the analysis of multiple variables.
>
> My task is to compare distributions in a universe with distributions 
from
> the respondents using a whole bunch of variables. Comparison shall be 
done
> on relative frequencies (proportions).
>
> I was thinking about the structure I should store the results in and 
came
> up with the following:
>
> -- cut --
>
> library(stringi)
>
> # Result data frame
> # Some sort of tidytidy data set where
> # each value is stored as an identity.
> # This way all values for all variables could be stored in
> # one unique data structure.
> # If an additional variable added for the name of the
> # research one could also build result data set across
> # surveys.
> # Values for measure could be "number" for 'raw' values or
> # "freq" for frequencies/counts.
> # Values for unit could be "n" for 'numbers' and
> # "%" for percentages.
> d_test <- data.frame(
> group = rep(c("Universe", "Respondents"), each = 16),
> variable = rep("State", 32),
> value = rep(c(11.3,
> 12.7,
> 3.3,
> 5,
> 0.6,
> 8.1,
> 6.2,
> 5.8,
> 6.4,
> 14.5,
> 8.3,
> 0.3,
> 3.8,
> 2.5,
> 8.1,
> 3), 2),
> label = rep(c("Baden-Wuerttemberg",
> "Bayern",
> "Berlin",
> "Brandenburg",
> "Bremen",
> "Hamburg",
> "Hessen",
> "Mecklenburg-Vorpommern",
> "Niedersachsen",
> "Nordrhein-Westfalen",
> "Rheinland-Pfalz",
> "Saarland",
> "Sachsen",
> "Sachsen-Anhalt",
> "Schleswig-Holstein",
> "Thueringen"),2),
> measure = rep("freq", 32),
> unit = rep("%", 32),
> stringsAsFactors = FALSE
> )
>
> # This way the variables can be selected using simple
> # value selection from Base R functionality.
> data <- d_test[d_test$variable == "State" ,]
>
> # And plot results for every variable.
> ggplot(
>   data = data,
>   aes(
> x = label,
> y = value,
> fill = group)) +
>   geom_bar(stat = "identity", position = "dodge") +
>   theme(axis.text.x = element_text(angle = 45, hjust = 1)) +
>   scale_fill_discrete(name = 
stringi::stri_trans_totitle(names(data)[1]))
> +
>   scale_x_discrete(name = data$variable[1]) +
>   scale_y_discrete(name = data$unit[1])
>
> -- cut --
>
> The reporting / presentation is done in R Markdown. I would load the
> result data set once at the beginning and running the comparisons as 
plots
> on each variable named in the results data set under "variable".
>
> If I follow this approach for my customer relationship survey, do think 
I
> would face drawbacks or run into serious trouble?
>
> I am interested in your opinion and open for other approaches and
> suggestions.
>
> Kind regards
>
> Georg
>
> __
> 

Re: [R-es] Asignación de objetos

[Álvaro Díez Ansotegui]

Gracias José Antonio, 

Entiendo por lo tanto que se puede usar indistintamente y aconsejan utilizar 
“<-“ para no llevar a un posible error. Gracias. Saludos!!

Álvaro D. 


> El 9 mar 2017, a las 9:49, José Antonio Palazón Ferrando  
> escribió:
> 
> Hola:
> 
> Puedes considerar que:
> 
> objeto <- expresión
> objeto =  expresión
> 
> 1. no es una igualdad, en realidad asignas el resultado de la expresión al 
> objeto.
> 
> 2. dado que existe los operadores '==',  '!='  y usamos '=' para
>definir los valores de los argumentos en las funciones, pare bueno
>evitar otro uso para un símbolo frecuente y que podría llevar a algún 
> error:
> 
>función( objeto, argumento = expresión )
>función( objeto1, objeto2 = expresión )
> 
> ¿No sé si te parece suficiente con esto?
> 
> Seguimos
> 
> 
> El 09/03/17 a las 09:25, Álvaro Díez Ansotegui escribió:
>> Buenas!
>> 
>> Estoy introduciéndome a R desde el enfoque de las CC Sociales. Estando 
>> trabajando los manuales introductores (vectores, matrices, importación y 
>> exportación de bbdd, tratamiento de variables, etc..) me ha quedado una duda 
>> que los manuales no profundizan en resolver. Pocos textos no distinguen 
>> entre la asignación, tratan el “ = “ de la misma forma que “ <- “. Sin 
>> embargo, la mayoría aconseja utilizar en la asignación “<-“, pero no 
>> profundizan en el por qué.
>> 
>> ¿Podría alguien explicar (sin restaros mucho tiempo) el porqué?
>> 
>> Saludos
>> 
>> Álvaro D.
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> 
> -- 
> 
> 
> José Antonio Palazón Ferrando
> Profesor Titular. Departamento de Ecología e Hidrología.
> Facultad de Biología. Universidad de Murcia.
> Campus Universitario de Espinardo
> 30100 MURCIA-SPAIN
> Telf: +34 868 88 49 80
> Fax : +34 868 88 39 63
> Email: pala...@um.es
> 
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