similar to: Lag based on Date objects with non-consecutive values

Can any one help it will be very kind, loop statements I have this table and some more records, I want to reshape it V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 TP53 Dis1 Dis2 Dis3 Dis4 Dis5 Dis6 DCI New1 New2 New3 New4 FDI Hi2 H3 H4 GHD I1 I3 I4 I5 I6 I7 I8 I want my new table or matrix to be some thing like this V1 V2 V3 Tp53 Dis1 Dis2 Tp53 Dis1 Dis3 Tp53 Dis1 Dis4 Tp53 Dis1 Dis5 Tp53 Dis1 Dis6 Tp53 Dis2

Dear all, an hopefully quick table question. I have the following data: Two objects that are 2*9 matrix with nine column names (Dis1, ..., Dis9) and the row names (2010,2020). The content are frequencies (numeric). In want to create a table that is along the lines of ftable(UCBAdmissions) and should looks like this: Dis1 | ...| Dis9 2010|2020|....|2010|2020 (first row,first column is the value

Hello, I analyse some free-sorting data so I use hierarchical clustering. I want to compare my proximity matrix with the tree representation to evalute the fitting. (stress, cophenetic correlation (pearson's correlation)...) "The cophenetic similarity of two objects a and b is defined as the similarity level at wich objects a and b become members of the same cluster during the course of

Hello ! I have a problem of syncing browse lists between DMB and LMB. The topology of ours network is as follows: Linux Network comma(Linux, DMB, WINS) vol(Linux) door(Linux, LMB) | | | | | | | | | --------------(195.226.226.16/28)------ | | | | | | | --(local net

Kirsten, this may not be the most elegant approach, but I think I would create dummy variables on the fly for each disease code you have, then use the "with" function to get means. Give this a try: # setup some fake data as an example icd9 <- c(rep("dis1",4), rep("dis2",5), rep("dis3",3)) n <- length(icd9) exposure <- rnorm(n) working <-

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

I'm currently working with some time series data with the xts package, and would like to generate a forecast 12 periods into the future. There are limited observations, so I am unable to use an ARIMA model for the forecast. Here's the regression setup, after converting everything from zoo objects to vectors. hire.total.lag1 <- lag(hire.total, lag=-1, na.pad=TRUE) lm.model <-

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is

Hi R, I have doubt. >x= c(4,5,6,3,2,4,5) >acf(x,plot=F,lag.max=1) Autocorrelations of series 'x', by lag 0 1 1.000 0.182 But if I actually calculate the autocorrelation at lag1 I get, >cor(x[-1],x[-length(x)]) [1] 0.1921538 Even in excel I get 0.1921538 value. So, I want to know what the 'acf' function is calculating here....

Full_Name: Ben Cooper Version: 1.5.0 OS: linux Submission from: (NULL) (134.174.187.90) The help page for glm.nb (in MASS package) says that it takes "Any other arguments for the glm() function except family" One such argument is start "starting values for the parameters in the linear predictor." However, when called with starting values glm.nb returns: Error in

Hello, I have a question on unit root test with urca toolbox. First, to run a unit root test with lags selected by BIC, I type: > CPILD4UR<-ur.df(x1$CPILD4[5:nr1], type ="drift", lags=12, selectlags ="BIC") > summary(CPILD4UR) The results indicate that the optimal lags selected by BIC is 4. Then I run the same unit root test with drift and 4 lags:

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Dear all, I have weekly data by city (variable citycode). I would like to take the average of the previous two, three, four weeks (without the current week) of the variable called value. This is what I have tried to compute the average of the two previous weeks; df = df %>% mutate(value.lag1 = lag(value, n = 1)) %>% mutate(value .2.previous = rollapply(data = value.lag1,

Hi, I was wondering if there is a more efficient way of handling the following method of creating a lagged value in a data frame without using the recursive 'for(i in 1:n)' loop and without using as.ts #Steps to creating a lag variable in a data frame 'my.dat.fr' # with 275 columns, 2400 rows of numbers and factors . The #variable x is a factor of #with five different levels the

Hello guys .. I would like to have some help about as.table . I made a table with the autocorrelations of the returns whit 10 lags and i get this : autocorrelazione2 <- as.table(c((cor(r2[-1151,],lag(r2))),(cor(r2[- c(1151,1150),],lag(r2, k=2))),(cor(r2[- c(1151,1150,1149),],lag(r2, k=3))),(cor(r2[- c(1151,1150,1149,1148),],lag(r2, k=4))),(cor(r2[- c(1151,1150,1149,1148,1147),],lag(r2,

Suppose we observe N individuals, for each of which we have a time-series. How do we correctly create a lagged value of the time-series variable? As an example, suppose I create: A <- data.frame(year=rep(c(1980:1984),3), person= factor(sort(rep(1:3,5))), wage=c(rnorm(15))) > A year person wage 1 1980 1 0.17923212 2 1981

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates &lt;- as.Date(data.df[,1]) selection&lt;-which(dates&gt;=as.Date("1986-1-1") &amp; dates&lt;=as.Date("2007-12-31")) dep &lt;- ts(data.df[selection,c("dep")]) indep.ret1

I am sure that this sort of thing has been asked and answered before, so in case my suggestions don't work for you, just search the archives a bit more. I am also sure that it can be handled directly by numerous functions in numerous packages, e.g. via time series methods or by calculating running means of suitably shifted series. However, as it seems to be a straightforward task, I'll

Dear Bert, Thank you very much.This works. I was wondering if the fact that I want to create new variables (sorry for not stating that fact) makes any difference? Thank you again. Sincerely, Milu On Sun, Mar 25, 2018 at 10:05 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > I am sure that this sort of thing has been asked and answered before, > so in case my suggestions

Dear Rui and UseRs,[a text file has also been attached, in case the format of my email is difficult to get]I am extremely sorry that I am bothering you once again, but I?ll have to get to the bottom of it. The following command sp <- lapply(split(agg, agg$st), function(x) x[order(x$year, x$month), ]) gave me an output with monthly mean of population(as under). i am not able to apply