similar to: colname refered by a variable

Hi,   I have a matrix. >a<-cbind(c("a","b","a"),c(4,3,6))      [,1] [,2] [1,] "a"  "4" [2,] "b"  "3" [3,] "a"  "6" I want to remove rows in matrix a whose first column has frequency less than 2. in about example matrix a becomes      [,1] [,2] [1,] "a"  "4" [2,] "a" 

Hello,   I could not find a clear solution for the follow question. please allow me to ask. thanks   mynames=cbind(c('a','b'),c(11,22)) lst=list(a=c(1,2), b=5) now I try to combine mynames and lst: a  1  11   a  2  11 b  5  22   thanks jian   [[alternative HTML version deleted]]

Hello,   I have a dataframe: dd <- data.frame(b = c("chr2", "chr1", "chr15", "chr13"),         x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),         z = c(1, 1, 1, 2))   >dd       b x y z 1  chr2 A 8 1 2  chr1 D 3 1 3 chr15 A 9 1 4 chr13 C 9 2 Now I want to sort them according column "b", but only its

Hi,   how to remove levels that have less than a specific number such as 2. i.e..   > f<-as.factor(c("a","b","a")) > f [1] a b a Levels: a b I want to remove level b because level b has less than 2. > f [1] a a Levels: a   [[alternative HTML version deleted]]

Hello, I have a simple question: I want to list numbers 1:k, but if k <1, I hope nothing listed. how should we do? k=2 for (i in 1:k)  print(i) [1] 1    # <-correct [1] 2 k=0 for (i in 1:k) print(i) [1] 1      #<---- wrong [1] 0   thanks jian [[alternative HTML version deleted]]

Hello R experts, can you tell me a simple way to convert a charactor matrix to numeric matrix? if I use as.numeric, the matrix is converted to a vector. a<-cbind(c("23","54","65"),c("1","2","3")) a      [,1] [,2] [1,] "23" "1" [2,] "54" "2" [3,] "65" "3"  as.numeric(a)

Hi there, how can I test every element in a vector whether appears in another vector? such as a<-c("aa", "bb", "ff", "cc") b <-c("oo","jj","bb","cc") somefunction(a,b) gives [False, True, False, True] thanks yuan jian [[alternative HTML version deleted]]

Hi,   it seems a simple problem, but I can not find a clear way. I have a list: lst=list(m=c('a','b','c'),n=c('c','a'),l=c('a','bc')) > lst $m [1] "a" "b" "c" $n [1] "c" "a" $l [1] "a"  "bc" how can I get list elements that include a given subset? for example, for given

hi, can anyone tell me how to merge a vector and a matrix? > v=c(1,4,2) > names(v)=c("e","r","t") > m=matrix(c("r","t","r","s","e",5,6,7,8,9),nr=5) > colnames(m)=c("c1","c2") I want to do like merge(v, m, by.x="names",by.y="c1") I got error Error in fix.by(by.x, x) :

Hi, if I know the colnames x and y in the following example, I can easily to do lm. tmp <- data.frame(x=c(1,1.2),y=c(1,2)) lm(y ~ x, data=tmp) when the colnames are variable, what should I do? for example colnames(tmp)[1] <- paste("aa",1,sep="_") lm(y ~ paste("aa",1,sep="_"), data = tmp) it gives me error. [[alternative HTML version

Hello, how to merge a data.frame and a matrix by one column in the data.frame and rownames of the matrix? df <- data.frame(col1=c("kk","yy","kk"),col2=c(6,4,3)) > df   col1 col2 1   kk    6 2   yy    4 3   kk    3 m<-matrix(c(3,8,56,9), nrow=2, dimnames = list(c("aa","kk"),c("col1","col2"))) > m    col1 col2 aa   

Hi, is it possible to call R functions from visual studio .net such as c#? --------------------------------- [[alternative HTML version deleted]]

I'm attempting to create an array of treatment comparisons for modelling data generation. This involves comparison of one treatment (c2) with another (c3), relative to a common comparator (c1). Attached code gives me the correct array but need to remove duplicates. Duplicates relate only to c2 and c3 such that I need to remove r3 because c2 and c3 are same as r1 with c2 and c3 swapped r5

Dear All, I am trying to create a empty structure that I want to fill gradually through the code. I want to use something like rbind to create the basic structure first. I am looking for a possibility to do an rbind where the columns names dont match fully (but the missing columns can be defaulted to either zero or n/a) (my actual data has a lot of columns). Please see the data frames below I

Hello, I have a data frame as below ... in cases where I have N.A. I want to use an average of the past date and next date .. any help? 13/10/2010 A 23 13/10/2010 B 12 13/10/2010 C 124 14/10/2010 A 43 14/10/2010 B 54 14/10/2010 C 65 15/10/2010 A 43 15/10/2010 B N.A. 15/10/2010 C 65 ---------------------------------------------------------------------------- -------------------------- Thanks

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,

Dear Group, I have a following dataset: > a A B C D 1 22 3 31 40 2 26 31 36 32 3 3 7 49 16 4 24 40 27 26 5 20 45 47 0 6 34 43 11 18 7 48 48 24 2 8 3 16 39 48 9 20 49 7 21 10 17 36 47 10 > dput(a) structure(list(A = c(22L, 26L, 3L, 24L, 20L, 34L, 48L, 3L, 20L, 17L), B = c(3L, 31L, 7L, 40L, 45L, 43L, 48L, 16L, 49L, 36L), C = c(31L, 36L, 49L, 27L, 47L, 11L, 24L,

I'm unable to find the OR operator like other language .. any suggestions? I want to do If (condition1 OR condition 2){ do something } Thanks for answering this elementary question. [[alternative HTML version deleted]]

Dear R-Group, I am looking for suggestions for the "best" IDE for R. Best is obviously subjective but I need just the basic features that should function well (and I looked through the threads already). - Proper integration with R 2.11.1 - Good key shortcuts ... similar to the R Gui - Manageability of Projects, etc. - Neat formatting features I tried Revolution R but it seems huge in

Hello Group, I need a modification in the data.table example to get my intended result shown below ... is there a more simple way! dt <- data.table(A = rep(1:3, each=4), B = rep(1:4, each=3), C = rep(1:2, 6)) dt[, transform(.SD,D=mean(A)), by="B"] The result I want is below ... which is probably long winded! data.table(unique(as.data.frame(dt[, transform(.SD,D=mean(A)),