similar to: Deleting many list elements in one time

Hi guys, here is a list of names that I have: MyList: > myList<-list("A", "B","C","D") > myList [[1]] [1] "A" [[2]] [1] "B" [[3]] [1] "C" [[4]] [1] "D" I want to turn this list into a matrix of 1 row and 4 columns with those four components (A, B, C, D) so here is what I do: myDataFrame <-

Hello guys, I am working on a matrix which looks like this one: > initialMatrix <- > rbind(cbind(rep("A",3),seq(1,3)),cbind(rep("B",4),seq(1,4)),cbind(rep("C",3),seq(1,3))) > initialMatrix [,1] [,2] [1,] "A" "1" [2,] "A" "2" [3,] "A" "3" [4,] "B" "1" [5,]

Hello, I have a date value in excel: 1/4/2010 which in number format gives me 40182. When I read this with read.xls from R I get same 40182 so what I do is that I use the as.Date() function but here is what the as.Date() function returns me: > as.Date(40182) [1] "2080-01-06" Why don't I get the same thing as in excel? Thanks! ----- Anna Lippel -- View this message in context:

Hello everyone, I use the split function splitting with the f function on a 3 columns and more than 100 000 rows data frame. Once it's split I have a list of data frames still with 3 columns and n rows. I manipulate those list elements and get a list of data frames still with 3 columns but less rows. So when I unsplit it, I get an error as I use the same factor function I used to split ( f in

Hello R lovers, I have a vector of dates and signals. I want to filter the signals per day in a way that only the first signal of the day remains like this: Dates Signals Filtered Signal 2006-11-02 0 0 2006-11-02 1 1 2006-11-02 0 0 2006-11-02 1 0 2006-11-02 1

Hello everyone, I have the following matrix [,1] [,2] [,3] [,4] [1,] 0.002809706 0.0063856960 0.0063856960 0.011749681 [2,] 0.004893124 0.0023118418 -0.0005122951 -0.014646465 [3,] 0.003547897 0.0063355297 0.0030410542 0.011403953 [4,] 0.004838299 -0.0040383645 -0.0090406831 -0.011027569 [5,] 0.035648755 0.0334815590 0.0380977404

Hello my friends, here is a code I wrote with no loops on matrix that is taking too long (2 seconds and I call him 720 times --> 12 minutes): mat1 and mat2 are both matrix with 103 columns and 164 rows. sequence <- matrix(seq(1 : ncol(mat1))) returns <- apply(sequence, 1, function, mat1= mat1, mat2 = mat2, day = 1) function<- function(mat1, mat2, colNb, day){

Hello, I have a matrix mat1 of dim [1,8] and mat2 of dim[30,8], I want to replace the first row of mat2 with mat1, this is what I do: mat2[1,]<-mat1 but it transforms mat2 in a list I don't understand, I want it to stay a matrix... ----- Anna Lippel -- View this message in context: http://n4.nabble.com/Simple-question-on-replace-a-matrix-row-tp1427857p1427857.html Sent from the R help

Hello everyone, I am trying to retrieve the list of distinct values within a whole data frame. I tried to use unique() function but it retrieves the distinct values within each column or row, I want it for the entire data frame, any idea? ----- Anna Lippel -- View this message in context: http://n4.nabble.com/Retrieve-distinct-values-within-a-whole-data-frame-tp1460205p1460205.html Sent from

Hi guys, I am using the blp() function from RBloomberg package which returns a matrix of prices with the columns corresponding to the security name and the columns to the date. When I have a look at the matrix I can see the rownames (dates) on the left of the prices but when I call the rownames() function it returns me a NULL value. It worked perfectly until I had to reinstall the RBloomberg

Hi everyone, I have a matrix with many Na's. Some rows contain some Na's and some others are entirely composed of Na's. I want to delete the rows that are entirely composed of Na's but not the other ones so I can't use a simple removeNA. Has any one an idea? ----- Anna Lippel -- View this message in context:

Hello guys, I have a list L1 of matrix. I have another list L2 with the same number of elements representing the row of the L matrix that I want to delete (L1[[i]][-L2[[i]],]) but I can't do this with lapply as it iterates through L1 (first argument) and not L2. Any idea? ----- Anna Lippel -- View this message in context:

Hello everyone, I have a vector P and I want to replace each of its missing values by its next element, for example: P[i] = NA --> P[i] = P[i+1] To do this I am using the replace() and lag() functions like this: P <- replace(as.ts(P),is.na(as.ts(P)),as.ts(lag(P,1))) but here is the error that I get: Warning message: In NextMethod("[<-") : number of items to replace is not a

Hello, I am having trouble by using the write.table function to write a data frame of 4 columns and 7530 rows. I don?t know if I should just use a sep="\n" and change the .xls file into a .csv file. Thanks in advance ----- Anna Lippel new in R so be careful I should be asking a loooooooot of questions!:teeth: -- View this message in context:

Hello R specialists, I have a base vector called vBase with 102 elements. I have another vector vec1 which elements are only part of vBase but is shorter. I transform vec1 so I get a vector with the same size as vBase and with each common element on the same indexed row. If a value is missing in vec1 then I put a Na like this: vec1 vBase Amsterdam Amsterdam Na

Hello, I am trying to plot a graphic with many lines with the following command: plot(datas[1:n,1],datas[1:n,2],type="l",main="SP500 Prices and Moving Averages",xlab="Date",ylab="Prices",col="black") lines(datas[1:n,1],datas[1:n,3],type="l",col="green",lty="solid") But I just see the first curve. I have tried it on

I can define a list containing NULL elements: > myList <- list("aaa",NULL,TRUE) > names(myList) <- c("first","second","third") > myList $first [1] "aaa" $second NULL $third [1] TRUE > length(myList) [1] 3 However, if I assign NULL to any of the list element then such element is deleted from the list: > myList$second <-

Hello everybody, I?ve been looking for a function that would create an excel file in my working directory where I would write my dataframe but I only found the functions to write or read in an existing file that you gave me on my former post or on some websites. I can?t find either functions to manipulate those datas: for example, I would like some lines to be red or green according to their

I have a list that is made of lists of varying length. I wish to create a new vector that contains the last element of each list. So far I have used sapply to determine the length of each list, but I'm stymied at the part where I index the list to make a new vector containing only the last item of each list mylist =

Hi! I have a list of 24 elements, all of the same type (dataframe, for example). I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... + mylist[[24]] to obtain the sum. Anyone can help me? Thanks in advance. Nicola S. [[alternative HTML version deleted]]