Stock
and Yogo (2005).
dm.
Daniel Malter wrote
>
> You find code to compute the Cragg-Donald Wald F test statistic to test
> for weak instruments (e.g., Stock and Yogo, 2005) below with a working
> example. The code reproduces the results from Stata 12 using the same data
> (see
It's not quite clear what you want to do. Is it just this?
a<-c(1,2,3,4)
dim(a)<-c(2,2)
a/a
This gives the element by element ratio.
HTH,
Daniel
shukor wrote
>
> Hi,
>
> I have a matrix as below:
>
> mat=
> [,1] [,2] [,3]
> [1,]147
> [2,]258
> [3,]36
This is a matrix algebra problem, not an R problem. ej fails because
betaligne and ZIcolone are not conformable matrices.
dim(betaligne)
1 2
dim(ZIcolone)
76 1
Let a matrix be m x n, then you can only matrix multiply it with a matrix
that is n x k (where m and k are arbitrary positive integers,
.
The Cragg-Donald Wald F statistic is the smallest of the eigenvalues of G
(see the output of "eigen(G)" below). Critical values are found in Stock and
Yogo (2005).
HTH,
Daniel Malter
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As df2 has only one column and is thus effectively a variable in this case,
you don't even need to merge.
df1[df1$gene_name%in%df2$gene_name , ]
should do.
HTH,
Daniel
wong, honkit (Stephen) wrote
>
> Dear Experienced R users,
>
> I have a looks-like simple but complicated problem urgently
Hi,
You can use rbind(). E.g.,
id<-rep(1:10,each=10)
x<-rnorm(100)
y<-rnorm(100)
d<-data.frame(id,x,y)
rm(id,x,y)
newdata<-data.frame()
for(i in 1:10){
newdata<-rbind(newdata,split(d,d$id)[[i]])
print(newdata)
}
The resulting data.frame "newdata" has rbind-ed all split elements of d (in
this ca
The easiest way may be to use lmList in the nlme library:
#simulate data
d<-rep(1:10,each=10)
x<-rnorm(100)
e<-rnorm(100)
y<-2*x+e
require(nlme) #or install and load package
lmList(y~x|d)
#predicted values are obtained with:
predict(lmList(y~x|d)
HTH,
Daniel
jeff6868 wrote
>
> Hi everyb
The function you programed expects you to provide 7 arguments. In the first
case, you explicitly specify each of the seven arguments, i.e., you tell the
function: this is yu, this is yf, and so forth. In the second case, you only
specify 2 arguments, t and par[1,1:16]. So the function thinks that
p
Hi,
In the example below, I am modeling the dependent variable Y as a function
of X when the errors are autoregressive at the first lag. There is a number
of ways/functions with which to model this. arima (tseries), gls (nlme), and
gamm should produce similar results in the simulated example belo
Hi,
I have a data frame whose first row (not the header) contains the true
column names. The same column name can occur multiple times in the dataset.
Columns with equal names are not adjacent, and for each observation only one
of the equally named columns contains the actual data (see the exampl
This just saved me a lot of time.
Thank you!
Daniel
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Assume your year value is
x<-007/A
You want to replace all non-numeric characters (i.e. letters and
punctuation) and all zeros with nothing.
gsub('[[:alpha:]]|[[:punct:]]|0','',x)
Let's say you have a vector with both month and year values (you can
separate them). Now we need to identify the c
#The code of rank 1 in the previous post should have read
#rank1<-apply(iterator1,1,function(x) x+base1)
#corrected code below
siegel.tukey=function(x,y,id.col=TRUE,adjust.median=F,rnd=-1,alternative="two.sided",mu=0,paired=FALSE,exact=FALSE,correct=TRUE,conf.int=FALSE,conf.level=0.95){
if(id.col=
The previously posted code contains bugs. The code below should work:
x: a vector of data
y: Group indicator (if id.col=TRUE); data of the second group (if
id.col=FALSE). If y is the group indicator it MUST take 0 or 1 to indicate
the groups, and x must contain the data for both groups.
id.col:
Hi,
I am running a MacBook Pro with OS X 10.6.8. When I try to load library(cem)
and run any function, R crashes. Any suggestions as to why this is? Thanks in
advance. I think the problem is that I don't have all the "infrastructure" I
need to run CEM, but I have no clue what to do either...
Your post is unlikely to solicit a helpful response. As you do not adhere to
the posting guide, it is quite impossible for us to figure out from your
description what is going wrong. Please provide a self-contained example
that reproduces the problem (i.e., code with simulated or actual data that
w
The likelihood function is a product. Thus, the log likelihood function is a
sum. Your log.lik statement, however, fails to compute the sum, which it
should minimize. Hence your optim statement does not know what to optimize
because log.lik is a vector of the length of the number of observations in
;-factor(column)
contrasts(row)<-contr.treatment(levels(row))
contrasts(column)<-contr.treatment(levels(column))
# Works for Terps
fit.terp<-glm(counts ~ row + column + row*column,
family=poisson(link="log"))
summary(fit.terp)
HTH,
Daniel Malter
University of Maryland, College Park
There is a friedman.test() function. Any reason you want to do it by hand?
If so, you can do:
#Simulated data matrix
x<-matrix(rnorm(9),3,3,byrow=T)
x
#Rank matrix
r<-matrix(rank(x),dim(x))
HTH,
Daniel
JohnnyJames wrote:
>
> My data looks like this:
> (treatmen
I find your question impossible to answer as you do not provide any
description of what the matrix columns actually capture and how the variable
in your proposed function relate to the columns of this matrix.
Best,
Daniel
alaios wrote:
>
> Dear all
> I have the following time stamps (in the fol
You access columns of a data.frame by column indices as in: X[ ,1], X[ ,2],
etc. The index before the comma would stand for the row if you wanted to
restrict those. The index after the comma captures the column.
That said, you typically would not "extract" rows from the data frame but
draw directl
I am not an expert on this, but there is a way to check this. You can predict
from a gam using predict(ozonea, newdata=...). In the "newdata" argument you
can specify the X-values of interest to you. Thus, you can compare if your
predictions are the same when predicted directly from the gam or when
I think there must be an easier solution, but this works:
y <- c(0,1,1,3,3,3,5,5,6)
x<-matrix(0:6,ncol=1)
apply(x,1,function(x){length(y[y==x])})
HTH,
Daniel
Kathie wrote:
>
> Dear R users,
>
> I'd like to count the number of integers in a vector y.
>
> Here is an example.
>
> y <- c(0,1,
What a pleasant post to respond to - with self-contained code. :)
heat<-matrix(0,nrow=dim(xa)[1],ncol=dim(xa)[2])
heat[lower.tri(heat)]<-xa[lower.tri(xa)]
heat[upper.tri(heat)]<-xb[upper.tri(xb)]
diag(heat)<-1
heat
HTH,
Daniel
1Rnwb wrote:
>
> Hello Gurus
> I have two correlation matrices 'x
You can get an estimate for the omitted baseline category by not estimating
the intercept. To do that, type "-1" on the right-hand side of the
regression statement. If that was your actual question, the simple question,
"How can I omit the baseline in a regression?," would have sufficed.
Moreover,
I need to quote David Winsemius on this one again: "The advancement of
science would be safer if you knew what you were doing."
Note that the whole model screams at you that it is wrongly modeled. You are
running a fully interacted model with factor variables. Thus, you have 19
regressors plus the
You are modeling Condition * Stimulus * Group as fully interacted fixed
effects. I do not actually think that you would need a complex random effect
structure for this. A simple random effect for the individual might suffice.
You could try to model this with lmer (in the lme4 library) and inspect
w
You probably should not do panel data analysis but multiple time series
analysis as your T is much larger than N. You only have seven units of
observation but some 150 observations on each unit.
Also, if the values on each unit of observation are very close to each other
between t and t-1, then yo
I have not read the manual, but I drew 1 random normal vectors and 1
random Poisson vectors of length 1 and was unable to reproduce this
behavior. Can you provide an example (self-contained code) that reproduces
this problem?
Thanks,
Daniel
Jeanne M. Spicer wrote:
>
> The summary fu
?plot
?points
You will probably need to get some R basics down as to how to index certain
subsets of your data. This you find in any introductory R manual.
HTH,
Daniel
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And there I caught myself with the next blooper: it wasn't Ben Bolker, it was
Bert Gunter who pointed that out. :)
Daniel Malter wrote:
>
> Ben Bolker sent me a private email rightfully correcting me that was
> factually wrong when I wrote that ML /is/ a numerical method (I
.
Further, only numerical optimization will show the behavior discussed in
this post for the given reasons. (I hope this post isn't yet another blooper
of mine at 5 a.m. in the morning).
Best,
Daniel
Daniel Malter wrote:
>
> With respect, your statement that R's optim does not giv
With respect, your statement that R's optim does not give you a reliable
estimator is bogus. As pointed out before, this would depend on when optim
believes it's good enough and stops optimizing. In particular if you stretch
out x, then it is plausible that the likelihood function will become flat
df2<-melt(df1)
df3<-cast(df2,Index~Name)
df3
HTH,
Daniel
Dana Sevak wrote:
>
> I realize that this is terribly basic, but I just don't seem to see it at
> this moment, so I would very much appreciate your help.
>
>
> How shall I transform this dataframe:
>
>> df1
> Name Index Value
> 1
On a methodological level, if the choices do not correspond on a cardinal or
at least ordinal scale, you don't want to use correlations. Instead you
should probably use Cramer's V, in particular if the choices are
multinomial. Whether the wide format is necessary will depend on the format
the funct
Your questions is pretty opaque. Please adhere to the posting guide. Provide
a self-contained (!) example (i.e., code) that reproduces your problem.
Generally, you would predict like this:
x<-rnorm(100)
e<-rnorm(100)
y<-x+x^2+e
reg<-gam(y~s(x))
plot(reg)
predict(reg,newdata=data.frame(x=2))
wher
Please adhere to the posting guide (i.e., provide a sample of self contained
code and provide the error message). And what does "but it is not working"
mean? Is there an error code?
rueu wrote:
>
> hello:
> my data looks like:
> time1 time2 event catagoria
>
> 2004 2006 1
Why would that be preferable to dropping the variables after importing the
whole dataset?
Daniel
sassorauk wrote:
>
> Is it possible to import only certain variables from a SPSS file.
>
> I know that read.spss in the foreign library will bring the data into R
> but can I choose to important on
x<-rnorm(25,1.5,1)
dim(x)=c(5,5)
x
y<-ifelse(c(x)>1.5,1,0)
dim(y)<-dim(x)
HTH,
Daniel
Spartina wrote:
>
> Hello all,
>
> I'm having trouble creating an adjacency matrix.
>
> Basically, I need to turn the following distance matrix into an adjacency
> matrix based on whether values are >1.5 o
Not satisfactory in which sense?
The survreg(Surv(Value,Censoring)~indvars+strata(id)) should/may work. For a
discussion of Tobit fixed effects, see also Greene's website:
http://pages.stern.nyu.edu/~wgreene/publications.htm under "Fixed Efects and
Bias Due to the Incidental Parameters Problem in
This suggests that this is a dangerous office to be in because this is a
basic question. I am sure somebody in your office knows this. Anyway, the
baseline gives you the average value of the group that constitutes the
baseline when all other covariates are zero. Let's say you measure whether
men or
You can do this using ifelse(). See example below.
x<-rpois(100,100)
NA.x<-sample(1:100,40)
x[NA.x]=NA
y<-rpois(100,100)
NA.y<-sample(1:100,40)
y[NA.y]=NA
z<-ifelse(!is.na(y),y,ifelse(!is.na(x),x,NA))
HTH,
Daniel
holly shakya wrote:
>
> I have 2 columns for weight. There are NAs in each col
I have read it three times and still no concrete idea what you are actually
trying to do, mainly because there is no information as to which
level/variable you are aggregating on. It'd help if you provided the
aggregated data (or sample rows thereof) so that we know what you want the
result to be.
I am not a Bayesian. In the non-Bayesian case you would use SUR to model both
equations simultaneously. If both use the exact same matrix of data, X
(i.e., the value are numerically absolutely identical), then SUR will
collapse to OLS. In that sense you get a "combined" estimate using SUR that
resp
It will never hurt your chances to get an answer if you tell what the code in
the different language actually does. Is it doing t-est for all of the
variables and the group indicators are in a separate column? If, imagine you
have the following data:
id<-rep(c(1,2),each=100)
x<-c(rnorm(100,0,1),rn
Would:
tapply(mpg,list(year,manufacturer), function(x) length(unique(x)))
do what you are trying to accomplish? Or do you need very specific subsets?
If so, you can do:
year<-c(rep(1980,4),rep(1981,4))
manufacturer<-rep(c('A','B'),4)
mpg<-sample(c(1:2),8,replace=T)
guc<-function(x,y,z){length(u
fit1, the unrestricted model, includes 1 more regressor than fit2, the
restricted model. Testing the models against each other means that fit2 is
"equal to" fit1, assuming that the coefficient on the additional regressor
that is included in fit1 is restricted to zero.
Your F-test is thus whether t
install.packages("bootstrap")
library(bootstrap)
?
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Please provide the data as self-contained code, as requested in the posting
guide, so that helpers can directly paste it into R. Alternatively, you can
provide the dilomaaethiops.txt.
Best,
Daniel
Watching19 wrote:
>
> hello... I am trying to get this code to work, but as I get to the predict
>
y is the dependent variable, not a predictor or independent variable. since
this is a binomial model, y should be 0/1 or, atypically, a proportion.
HTH,
Daniel
Samuel Okoye wrote:
>
> Dear all,
>
> I am using glm with quasibinomial. What does the following error message
> mean:
>
> Error in e
Hi, I answered this question before in this post:
http://r.789695.n4.nabble.com/Regressions-with-fixed-effect-in-R-td2173314.html
, specifically in my message from May 11, 2010; 4:30pm.
However, I believe the newer version of plm shows an R-squared, which should
be the within R-squared. Why the pr
Take a look here: http://www.jstatsoft.org/v25/i05/paper
HTH,
Da.
andy1234 wrote:
>
> Dear everyone,
>
> I am new to R, and I am looking at doing text classification on a huge
> collection of documents (>500,000) which are distributed among 300 classes
> (so basically, this is my training dat
#Do
apply(y,1,print)
#Note the space that is inserted before the "1." If you insert this space in
your function
apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k' &
!is.na(x[1]) & x[1]==' 1') 1 else 0} )
#you get the result you expect.
#Also, note that your !is.na conditions are
In your first line, you write "ARMA(2,2)." However, what you fit in R is
ARIMA(2,1,2). What you fit in eview, I can't tell. Could that explain the
difference?
HTH,
Daniel
Young Gyu Park wrote:
>
> When I do ARMA(2,2) using one lag of LCPIH data
>
>
>
> This is eview result
>
>>
>> *Dependen
This seems to work, as well:
http://www.mathworks.com/matlabcentral/fx_files/5051/1/content/Rdemo.html
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They seem to have a workaround. I don't know whether anything better is
available by now.
http://www.mathworks.com/matlabcentral/newsreader/view_thread/163726
HTH,
Daniel
sarak wrote:
>
> Is it possible for anyone to upload a youtube video showing how to execute
> R commands in Matlab , it's s
It will greatly help if you provide a self-contained example, as requested in
the posting guide. Most of the times this will in fact lead to you figuring
out your problem yourself, but if not it will greatly enhance your chances
that we can help you in a meaningful, unambiguous way.
Best,
Daniel
x<-letters[1:3]
y<-1:3
d<-expand.grid(x,y)
g<-apply(d,1,function(x) paste(x[1],x[2],sep=""))
HTH,
Daniel
Campbell, Desmond-2 wrote:
>
> Dear R-help readers,
>
> I'm sure this problem has been answered but I can't find the solution.
>
> I have two vectors
> v1 <- c("a","b")
> v2 <- c(1,2,3)
#Here is I think an easier way of coding all the components you need.
#The within-group variances, you get with this function:
apply(iris[,1:4],2,function(x) tapply(x,iris$Species,var))
#You can get what you computed by taking the column means.
apply(apply(iris[,1:4],2,function(x) tapply(x,iri
Let's say your fixed vector is x, and y is the list of vectors that you want
to create the squared distance to x with, then:
x<-c(1:5)
y<-list()
y[[1]]<-sample(c(1:5),5)
y[[2]]<-sample(c(1:5),5)
y[[3]]<-sample(c(1:5),5)
y
distances<-lapply(y,function(a,b) crossprod(a-b), b=x)
#lapply goes over
ame(id,obs,d,happy)
Daniel Malter wrote:
>
> Hi all,
>
> I am statistically confused tonight. When the assumptions to a random
> effects estimator are warranted, random effects should be the more
> efficient estimator than the fixed effects estimator because it uses fewer
>
Hi all,
I am statistically confused tonight. When the assumptions to a random
effects estimator are warranted, random effects should be the more efficient
estimator than the fixed effects estimator because it uses fewer degrees of
freedom (estimating just the variance parameter of the normal rathe
B is the specification for time-varying covariates. Otherwise, your model
will think that each row is one independent observation that either had an
event or was censored at "time" or "total_time."
HTH,
Daniel
javier palacios wrote:
>
> Dear R-community,
>
> which of the following two format
The "problem" with your first solution is that it relies on that the each
'year x group' combination is present in both data frames. To avoid this, I
would recommend to use merge()
df3<-merge(df1,df2,by.x=c("Year","Group"),by.y=c("Year","Group"))
df3$ratio<-with(df3,Value.x/Value.y)
df3
HTH,
Dani
It looks like it. However, you provide very little information. Do you have
measurements before and after the intervention and did the intervention
occur at the same point in time for all treated? If so, you could do a
simple difference in differences estimation.
HTH,
Daniel
Troy S wrote:
>
> D
I should have written "the standard errors of the coefficients are the SQUARE
ROOT of the diagonal entries of the variance-covariance matrix," as I
programmed it in the code.
Daniel Malter wrote:
>
> Pick up a book or the like on ordinary least squares regression, which is
&g
Pooling nominal with numeric variables and running pca on them sounds like
conceptual nonsense to me. You use PCA to reduce the dimensionality of the
data if the data are numeric. For categorical data analysis, you should use
latent class analysis or something along those lines.
The fact that your
Pick up a book or the like on ordinary least squares regression, which is
what lm() in its plain vanilla application does. The t-value is the
estimated coefficient divided by the standard error. The standard errors of
the coefficients are the diagonal entries of the variance-covariance matrix.
x<
look into the *apply series of functions. In your case
apply(name.of.your.data.frame,2,min)
or
apply(name.of.your.data.frame,2,max)
will do. You can also put any summary function to your liking instead of
min/max.
Best,
Daniel
Lao Meng wrote:
>
> Hi all:
> If I have a dataframe of N column
if your vector of data is x, use x[x!=-]. Subseting entire data frames
works analogously.
HTH,
Daniel
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General suggestions: avoid cbind() and avoid accessing data frames. Convert
data frames to matrices before accessing them. Also, why do you print? You
don't really want to print(t) for every iteration of the loop, do you? Also
avoid defining elements within the loop that need to be defined only onc
Q1 is very opaque because you are not even saying what kind of plot you want.
For a regular scatterplot, you have multiple options.
a.) select only the data in the given intervals and plot the data
b.) plot the entire data, but restrict the graph region to the intervals you
are interested in, or
I am afraid this is one of these posts where I have to quote David Winsemius:
"The advancement of science would be safer if you knew what you were doing."
Moreover, these are questions best addressed to your local statistician
rather than the R-help list. With exceptions, the R-help list helps to s
I am not sure what purpose the while loop has. However, the main problem
seems to be that you need to put:
i<-sample(1:(n-40),1) #This sample from 1 to n-40
rather than
i<-sample(1:n-40,1) #this samples one 1:n and then subtracts 40
Otherwise, you may get negative index values
Best,
Daniel
The kappa2() function in the irr library takes an n x 2 matrix as input,
where the two columns are the ratings by two raters. Let x and y below be
the ratings of the two raters:
x<-sample(c(0,1,2),100,replace=T)
x
o<-sample(c(0,0,0,1),100,replace=T)
y<-x+o
y
#Then kappa is computed as:
kappa2(
o do that? May you look at my look. It is calculating
> the estimates for one row. How do I incorporate the other data that has
> all other columns?
>
> Thanks
>
> Ed
> Sent from BlackBerry® wireless device
>
> -Original Message-
> From: "Daniel Malte
If A has more columns than in your example, you could always try to only
merge those columns of A with B that are relevant for the merging. You could
then cbind the result of the merging back together with the rest of A as
long as the merged data preserved the same order as in A.
Alternatively, yo
This is much clearer. So here is what I think you want to do. In theory and
practice:
Theory:
Check if AA[i] is in BB
If AA[i] is in BB, then take the row where BB[j] == AA[i] and check whether
A1 and A2 are in B1 to B3. Is that right? Only if both are, you want the
indicator to take 1.
Here i
This is not very confusing. It is the exact same error in the sense that this
time the values of x1 are not only outside the interval (0-1) but within
[0-1] as in your first example, but this time they are also outside [0-1].
The reason is that you did not divide x1 by sum(x1) this time. In other
w
This is a theoretical issue. It is impossible for beta-distributed values to
take the value of 0 or 1. Hence, an attempt to fit a beta distribution to a
vector containing these values fails.
HTH,
Daniel
baxy77 wrote:
>
> Hi,
>
> Well, i need some help, practical and theoretical. I am wonderi
Steven's solution is great, but it will only work if the rows are really
duplicates. If the data frame contains another variable whose values vary,
it will not work because then the rows are obviously unique.
df<-data.frame(df,value=rnorm(11))
unique(df)
You would then have to make a decision, w
For question (a), do:
which(AA%in%BB)
Question (b) is very ambiguous to me. It makes little sense for your example
because all values of BB are in AA. Therefore I am wondering whether you
meant in question (a) that you want to find all values in BB that are in AA.
That's not the same thing. I am
Hi,
I hope this is for private purposes. Otherwise, I may cite David Winsemius:
"The advancement of science would be safer if you knew what you were doing."
First, your regression command is inverted. You ought to regress SoloKills
on range, not vice versa.
abline(lm(graph~range)) #does the tric
b<-c("1","2","3","4","<1")
grep('<',b)
HTH,
Daniel
Ryan Utz-2 wrote:
>
> Hi all,
>
> I'm trying to identify a particular digit or value within a vector of
> factors. Specifically, this is environmental data where in some cases the
> minimum value reported is "<" a particular number (and I wan
My recommendation would be to not "subset out" the data, because you are
introducing a potential source of error when binding the new data back
together with the old data. Preferably, I would work on selecting subsets of
the dataset using indices (as suggested in the previous post) and just do
the
If you just want to apply the function over successive columns of a data
frame use
apply(name.of.data.frame, 2 , llik)
Daniel
EdBo wrote:
>
> Hi
>
> I have a code that calculate maximisation using optimx and it is working
> just fine. I want to extend the code to run several colomns of R_j wh
Hi,
The blunt answer is: by learning R. In particular, you will need pattern
matching techniques as in ?grep and (somewhat advanced, some would call it
basic) knowledge of R. So if you aren't familiar with either, I would
suggest an introductory manual or one of the many websites you find online
a
Check the *apply() series of functions. tapply() will do what you want.
attach(mtcars)
tapply(hp,list(cyl,gear),mean)
HTH,
Daniel
Mark Alen wrote:
>
> I know commands like xtabs and table allow a user to do cross-tabulation
> For example the following command generates a pivot table that sh
Typically not, unless the device has not closed properly. Does the second
plot show up in the pdf?
Daniel
cherron wrote:
>
> Hello,
>
>>I am currently working on a script and I output plots to pdf using
>
> pdf(...)
> plot(...)
> dev.off()
>
>>then later I was trying to plot something and w
For the first part, use the col and pch arguments:
id<-rep(c(0,2),each=50)
e<-rnorm(100)
x<-rnorm(100)
y<-id+x+e
xyplot(y~x,groups=id,col=c(3,4),pch=c(12,13))
For the second part, I do not know what exactly mean by superimpose the mean
level? Should the mean for each group be displayed as a horiz
aniel
From: David Winsemius [dwinsem...@comcast.net]
Sent: Wednesday, July 20, 2011 9:01 AM
To: Daniel Malter
Cc: r-help@r-project.org
Subject: Re: [R] loops and simulation
On Jul 20, 2011, at 1:34 AM, Daniel Malter wrote:
snipped
> requests, except that
Yes, there are in Europe. And there are summer classes in the US, as well.
And no, this list is not so much about helping beginners to learn R. For
that, there is a myriad of online sources. Rather, this list is for people
who have exhausted their ability to (elegantly) solve a problem.
Also, it s
First, it would have helped if you had posted the actual results for us to
see how far they are off (and, more specifically, by which factor).
Second, given your epiphany, you will find that that's exactly what David
(and others before him) said or suggested. It is not about standardizing a
nomina
I dare the conjecture that if you had written the code, you would know how to
do this. This suggests that you are asking us to do your homework, which is
not the purpose of this list. A simple inclusion of the code in a for or
while loop and storing the estimated parameters with the index of the
it
Please read the posting guide (requires a self-contained example of code) and
consult the help pages before posting. If you type ?predict.lm the help page
clearly states that the argument 'newdata' takes "[a]n optional data frame
in which to look for variables with which to predict..."
x1<-rnorm(1
P1-tapply(P1,Experiment,mean)[Experiment]
HTH,
Daniel
ronny wrote:
>
> Hi,
>
> I would like to center P1 and P2 of the following data frame by the factor
> "Experiment", i.e. substruct from each value the average of its
> experiment, and keep the original data structure, i.e. the experiment a
As long as you just want to display it, use print()
GG<- c(1,2,3)
print(summary(GG),str(GG))
Output:
num [1:3] 1 2 3
Min. 1st Qu. MedianMean 3rd Qu.Max.
1.0 1.5 2.0 2.0 2.5 3.0
HTH,
Daniel
andrewH wrote:
>
> Using str() in a function.
>
> I am in th
Try TukeyHSD(anova). TukeyHSD is implemented in the base package, i.e., ready
to use with the base installation of R. Also, the TukeyHSD for the
interaction term should have a colon ":" not a "*". The "which" argument in
TukeyHSD() must be identical to the name of the coefficient in the summary
tab
Probably not the most elegant, but a workable solution. Assume you have a
matrix x of dimensions 10 x 10. Assume further you want to calculate the
mean for each successive block of two columns. One way to do this is to
create a matrix that indicates the column numbers from/to which to apply the
fun
Not that I know of, but the paper says that they are easy to compute. If you
did, you could contribute the code.
Best,
Daniel
David Hugh-Jones-3 wrote:
>
> Hi all,
>
> Is there any code to run fixed effects Tobit models in the style of Honore
> (1992) in R?
> (The original Honore article is he
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