similar to: little manipulation on data frame

I have a dataframe D.F1 >dim (D.F1) 14351 9 This dataframe has values and for some 1000 rows it holds NULL values.I hace found the missing values for about 500 and have those in another dataframe D.F.sub.2 >dim(D.F.sub.2) 500 9 as dataframe is a subset of D.F1 the coulmn 1 in D.F.sub.2 is a subset of D.F1.I have to insert the values in D.F1 in other fields while the coulmn 1 in both

Hi, here i have a Max and Min values Min <-3 Max <-6 and also a matrix like this, ABC XYZ PQR ------ ------- ------- 2 4 3 5 4 8 7 1 3 In this i need to check each particular column values are between Max and Min value. If the coulmn value not coming between Max and

Dear all, May I do multiple integration using R? I was looking adapt but it is saying it integrates a scalar function over a multidimensional rectangle. I have integrand of several variable and upper, lower limit too variable. I wanted to see the result using adapt (though it is not for this purpose, I suppose) Func<-function(x){(x[1]*x[2])} adapt(2, lo=c(0,1), up=c(1,x[1]), functn=Func) it

Could you please tell me what is the function or method to get count of elements in all the columns in a matrix ? for eg :- ABC XYZ PQR ------ ----- ------ 2 3 4 4 5 5 4 3 2 Result will be like ABC XYZ PQR ------ ----- ------ 2 4 3 Could you please help me

Hi, I have a data matrix with 13 columns and 55 rows. First coulmn is year and other are monthly values for 55 years. Now i want to create a single column of all the values ( i e a single time series column from column 2 to 13 ). Is it the for loop that works here ...how ? or how i can rearrange the data so that it becomes a times series column. hope i made the question clear . thank you,

Hi, I am trying to build a series of rlm models. I have my data frame and the models will be built using various coulmns of the data frame. Thus a series of models would be m1 <- rlm(V1 ~ V2 + V3 + V4, data) m2 <- rlm(V1 ~ V2 + V5 + V7, data) m3 <- rlm(V1 ~ V2 + V8 + V9, data) I would like to automate this. Is it possible to use a string in place of the formula? I tried doing: fmla

Geneset_name #Chromosome #Hit_in_Biomart original_geneset_len Missing.genes [1,] "AGUIRRE_PANCREAS_CHR12" "1" "51" "59" "8" [3,] "AGUIRRE_PANCREAS_CHR9" "1" "24" "24"

Dear Bert, thank you for your response. here it is the piece of R code : given 3 data frames below --- N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"),

Hi Bogdan, Kinda messy, but: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))

> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > Hi Bogdan, > Kinda messy, but: > > N <- data.frame(N=c("n1","n2","n3","n4")) > M <- data.frame(M=c("m1","m2","m3","m4","m5")) > C <-

Hi, how can I create an array of lists of three components? This approach does not work: n1 <- 2 n2 <- 4 n3 <- 5 res <- array(rep(vector("list",3), n1*n2*n3), dim = c(n1,n2,n3)) res[1,1,1] # is not a list with three components... The goal is that res[1,1,1] is a list with three components. Also, appending the components didn't work. For example, I tried: component

A strange problem with sem: I downloaded the sem library and then, I specified my simple measurement model (below). I highlighted it and ran it. It ran, but it did NOT tell me "22 lines read". And nothing works after that - it looks like it runs, but it does not produce anything... Did I make a mistake somewhere in the model? (notice, TIME has only 1 indicator - t1, and I fixed t1's

Dear R-users, Is there an R function to compute the multinomial beta function? That is, the normalizing constant that arises in a Dirichlet distribution. For example, with three parameters the beta function is Beta(n1,n2,n2) = Gamma(n1)*Gamma(n2)*Gamma(n3)/Gamma(n1+n2+n3) Thanks in advance for any assisstance. Regards, Greg [[alternative HTML version deleted]]

Thank you David. Using xtabs operation simplifies the code very much, many thanks ;) On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > > Hi Bogdan, > > Kinda messy, but: > > > > N <-

foo <- rnorm(30*34*12) dim(foo) <- c(30, 34, 12) I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame. I know how this can be done in a very slow way using for loops, like this: x <- rep(seq(from = 1, to = 30), 34) y <- as.vector(sapply(1:34, function(x) {rep(x, 30)})) month <- as.vector(sapply(1:12,

Simple matrix indexing suffices without any fancier functionality. ## First convert M and N to character vectors -- which they should have been in the first place! M <- sort(as.character(M[,1])) N <- sort(as.character(N[,1])) ## This could be a one-liner, but I'll split it up for clarity. res <-matrix(NA, length(M),length(N),dimnames = list(M,N)) res[as.matrix(C[,2:1])] <-

Dear Contributors, thanks for collaboration. I am trying to reorganize data frame, that looks like this: n1.Index Date PX_LAST n2.Index Date.1 PX_LAST.1 n3.Index Date.2 PX_LAST.2 1 NA 04/02/07 1.34 NA 04/02/07 1.36 NA 04/02/07 1.33 2 NA 04/09/07 1.34 NA 04/09/07

Oh, well, there's no coulmns I can find to show various speed, trackers used, remaining time, ot able to sort on name, speed etc. Basically it's the gui I don't like. It's fine otherwise and does its job excellent. -- /Sorin ________________________________________ From: centos-bounces at centos.org [centos-bounces at centos.org] on behalf of Jeff Allison [jeff.allison at

Hi, I run this code to get the power of the test for modified bartlett's test..but I'm not really sure that my coding is right.. #normal distribution unequal variance asim<-5000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) n1<-20 n2<-20 n3<-20 mu<-0 sd1<-sqrt(25) sd2<-sqrt(50) sd3<-sqrt(100) g1<-rnorm(n1,mu,sd1) g2<-rnorm(n2,mu,sd2)

Here's another approach: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400)) # Rebuild the factors using M and N C$m <-