similar to: how to append element at last position in array dynamically

i want hash like A : 1.2, 3.4, 4.5 B : 9.7, 5.6, 4.8 C : 3.4 ,5.7, 4.6 where A key contain all three value at right sight how to append single keys & values if i have predefined hash with known length -- View this message in context: http://r.789695.n4.nabble.com/how-to-make-hash-append-element-if-i-want-following-condition-tp4409761p4409761.html Sent from the R help mailing list

i have two variable > pyar [1] "M" "12" "34" "13" > gita [1] "U" "22" "33" "44" want to append gita object below pyar object as & want to take in anther variable "M" "12" "34" "13" "U" "22" "33" "44" & write it in

is have vector from table stored in s variable > s [1] R P Y A C P V E S C D R R F S R S A D L T R H I R I Levels: A C D E F G H I K L M N P Q R S T V Y -- View this message in context: http://r.789695.n4.nabble.com/want-to-drop-levels-of-table-vector-join-two-vector-appending-tp4356851p4356851.html Sent from the R help mailing list archive at Nabble.com.

Hi, a[length(a)] gives the value of last element. Is there an alternative without using functions ? [[alternative HTML version deleted]]

I wonder what's the right way in R to do the following -- placing objects of the same kind together in subarrays of varying length. Here's what I mean: > word <- c("a","b","c","d","e","f","g","h","i","j") > kind <- c(1,1,1,2,3,4,5,5,7,7) > d <-

I want to create a new array which selects values from an original array based on a function of the indices. That is: I want to create a new matrix Vnew[i,j,k]=Vold[i,j,ks] where ks is a function of the index elements i,j,k. I want to do this WITHOUT a loop. Call the function "ksfunction", and the array dimensions nis,njs,nks. I can do this using a loop as follows: # Loop version:

how to avoid writing index of variable in write.table command, e.g. output---> index,character,state "1" "M" "2" "2" "K" "3" "3" "R" "1" "4" "E" "1" "5" "S" "1" "6" "H" "1" "7" "K" "1"

Hi, There is an object, ".Last.value" to which the result of the most recent evaluation is assigned. This is similar to "ans" in Matlab. In Matlab "ans" can be very useful and time-saving, but typing the larger R version is somewhat clunky and takes away from the usefulness. Is it possible to reassign '.Last.value' to something simpler, like

i m searching on net,material related to R, whenever i get ppt presentation of some topic ,i found it in speical format white backgroud & blue bars,having links for further topics see link 1) https://www.rmetrics.org/files/Meielisalp2007/Presentations/Pfaff.pdf 2)http://grapevine.com.au/~yanchang/docs/Time-Series-Mining-slides.pdf If this is format,how can i made it,if i want to give my R

i have following numbers 0.889046409368551 1.22726162946495 1.22726162946495 1.35785109728356 1.35785109728356 1.10704609982913 1.4424189950435 1.2277843378837 1.35785109728356 0.970883941918588 0.822170913920467 1.35785109728356 0.358815782262543 0.774234247460432 0.822170913920467 0.822170913920467 0.72599976881814 0.671583894425946

Hello I'm fairly new to R. I have a script which produces a 1-column vector (numeric). I can write it to a text file using the write command. What I want to do is to run the script a number of times on different input files, producing a different 1-column vector each time and save all the vectors to a text file so that each vector is one separate column in the text file. I had thought of

days=Sys.Date()-1:70 price=abs(rnorm(70)) regular=rep(c(0,0,0,0,1,0,1,0,0,1),c(7,7,7,7,7,7,7,7,7,7)) y=data.frame(cbind(days,price,regular)) y is like days price regular 1 14990 0.16149463 0 2 14989 1.69519358 0 3 14988 1.57821998 0 4 14987 0.47614311 0 5 14986 0.87016180 0 6 14985 2.55679229 0 7 14984 0.89753533 0 the output I want:

i can see source code of function > cor function (x, y = NULL, use = "everything", method = c("pearson", "kendall", "spearman")) { na.method <- pmatch(use, c("all.obs", "complete.obs", "pairwise.complete.obs", "everything", "na.or.complete"))

checkout the 'officer' package Thanks Jim Holtman *Data Munger Guru* *What is the problem that you are trying to solve?Tell me what you want to do, not how you want to do it.* On Fri, Dec 29, 2023 at 10:14?AM Andy <phaedrusv at gmail.com> wrote: > Hello > > I am trying to work through a problem, but feel like I've gone down a > rabbit hole. I'd very much

On 10/29/07, Luke Kanies <luke@madstop.com> wrote: > On Oct 29, 2007, at 2:35 PM, Brian Finney wrote: > > Although iteration would be really cool if we could also append to an > > array in a global fashion, but thats a completely different dream. > > What do you mean? Thinking some thing like: # define for simple firewall control define port($port, $status){ push

Hi all R users, I am trying to extract elements from an array (x as shown below) where the difference between the consecutive elements does not equal 1. The desired output is: j = 6,17,27,38,58,69,79,90 However, the code below only returns: j = 1,69,79,90 I am assuming that "cbind" function is not extracting all the elements meeting the condition. In

i have > x [1] 1 2 3 4 5 6 > y [1] 34 5 6 > z<-cbind(x,y) > z x y [1,] 1 34 [2,] 2 5 [3,] 3 6 [4,] 4 34 [5,] 5 5 [6,] 6 6 i don't want recycling, instead can we put NA /0 like below> z x y [1,] 1 34 [2,] 2 5 [3,] 3 6 [4,] 4 NA [5,] 5 NA [6,] 6 NA & want distance matrix -- View this message in context:

i want "http://scop.berkeley.edu/astral/pdbstyle/?id=d1fjgc2&output=html",showing information in webpage to be written in .txt file as it is(i don't want any html tag) i am using "RCurl" package >marathi<-htmlTreeParse("http://scop.berkeley.edu/astral/pdbstyle/?id=d1fjgc2&output=html") >marathi

i have uploaded file,but when i am opening it in R,using > u<-file(file.choose(),"r") >k<-readLines(u) >k > k[1:120] is has all /t (tabs)& newlines, how to avoid it, can i take first 3 columns only in table form (lines starts with # not important for me) uploaded file:- http://r.789695.n4.nabble.com/file/n4577757/rabata.txt rabata.txt -- View this message in

Hi, I'm trying to write a small microsimulation in R: that is, I have a dataframe with info on N individuals for the base-year and I have to grow it dynamically for T periods: df = data.frame( id = 1:N, x =.... ) The most straightforward way to solve the problem that came to my mind is to create for every period a new dataframe: for(t in 1:T){ for(i in 1:N){ row = data.frame( id =