similar to: two minor bugs in rowsum()

> with the entirely different rowSums, but it has been around > for a long time.) A lot longer than rowSums ... > Bill Dunlap > Spotfire, TIBCO Software --- This made me smile. The rowsums function was originally an internal part of the survival package, used for fast computation of certain sums when there is a cluster() statement. It was Statistical

Hi, The result of rowsum() in R doesn't have the dimnames I'd expect, e.g.: > rowsum(matrix(1:12, 3,4), c("Y","X","Y")) [,1] [,2] [,3] [,4] 1 2 5 8 11 2 4 10 16 22 whereas S-Plus gives the more useful result: [,1] [,2] [,3] [,4] X 2 5 8 11 Y 4 10 16 22 This is because R's rowsum() code gives

The result of rowsum() in R doesn't have the dimnames I'd expect, e.g.: > rowsum(matrix(1:12, 3,4), c("Y","X","Y")) [,1] [,2] [,3] [,4] 1 2 5 8 11 2 4 10 16 22 whereas S-Plus gives the more useful result: [,1] [,2] [,3] [,4] X 2 5 8 11 Y 4 10 16 22 This is because R's rowsum() code gives the

Hi-- This is a question with a trivial and obvious answer, I'm sure, but I can't seem to find it in the help files and books that I have handy. I have a dataframe consisting of two columns, "Gene_Name," a list of gene symbols, and "Number," a numeric measure of how frequently a tag representing that gene showed up in a SAGE library. Several of the genes are

If all rows are in the same "group", rowsum() returns a vector instead of a (1xN) matrix, contrary to documentation: R> print(z <- rowsum(matrix(1:12, 3,4), rep("x",3))) [1] 6 15 24 33 R> dim(z) NULL It worked correctly in version 1.4.0 but was broken by version 1.6.1. I'm currently using 1.7.1 under Solaris 2.8. --please do not edit the information

'rowsum()' seems to add row names to the resulting matrix, corresponding to the respective 'group' values. This is very handy, but it is not documented. Should the documentation mention it so it could be relied upon as part of API? Cheers, Ott -- Ott Toomet Visiting Researcher School of Information Mary Gates Hall, Suite 310 University of Washington Seattle, WA 98195

Hello, I am trying to achieve something which I *think* is possible using rowsum, but a little help should be useful: Consider the following dataframe DF0: A B C 89 1 140 89 06 20 89 29 137 89 52 13 89 57 10 89 97 23 89 1 37 89 1 12 89 1 3 52 1 11 52 1 31 52 1 16 52 1 6 52 1 10 52 1 13 52 1 10 52 1 25 52 1 2 52 59 38 52 97 75 57 1 14 57 1 13 57 1 14 57 114 12 57 1 23 57 06 26 I need create a

Dear list, I have some trouble generating a frequency table over a number of vectors. Creating these tables over simple numbers is no problem with table() > table(c(1,1,1,3,4,5)) 1 3 4 5 3 1 1 1 , but how can i for example turn: 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 into 0 0 1 1 1 0 0 2 0 1 0 3 My second problem is, sorting rows and columns of a matrix by the rowSums/colSums. I did it

Dear R-List, I am sure there must be a very simple way to do this - I just do not know how... This is what I want to do: #my dataframe df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA),sum=c(3,9,15,24,15,6))    id    a     b     c   sum 1 x01  1     2    NA   3 2

I'm having a very frustrating problem, trying to find the inverse distance squared weighted interpolants of some weather data. I have a data frame of weights, which sum to 1. I have attached the weights data. I also have a data frame of temperatures at 48 grid points, which I have also attached. Now, all I need to do is multiply all of the rows of the temperature data frame by the weights

Hi, in my packages/functions/code I tend to remove large temporary variables as soon as possible, e.g. large intermediate vectors used in iterations. I sometimes also have the habit of doing this to make it explicit in the source code when a temporary object is no longer needed. However, I did notice that this can add a noticeable overhead when the rest of the iteration step does not take that

I am trying to construct a data set with some sequences for example: a = seq(0,1,0.1) m = matrix(nrow = 1331, ncol = 3) m[,1] = rep(a,121) m[,2] = rep(a,11,each = 11) m[,3] = rep(a,1,each = 121) I realize that there may be better ways of doing this, but this approach demonstrates the problem I'm having. I then want to get the sum of the rows and delete any row with a sum of greater than 1.

any suggestions on the following gratefully welcome, I have a dataframe, which I am subsetting via labels atpi[, creativity] where (for example) atpi = as.data.frame(matrix(1:50, ncol = 5, nrow = 10)) names(atpi) = c("Q1", "Q2", "Q3", "Q4", "Q5") and creativity = c("Q1", "Q3", "Q4") I want to add an extra column

Hi, I have been experimenting with the new Survey package. Specifically, I was trying to use some of the functions on the public-use survey data from NHIS (2000 Sample Adult file). Error 1): The first error I get is when I try to specify the complex survey design. nhis.design<-svydesign(ids=~psu, probs=~probs, strata=~strata, data=nhis.df, check.strata=TRUE) Error in svydesign(ids =

Dear Sir/Madam, I'm getting a problem with a R-code which converts a data frame to a matrix. It first generate a (m^(n-m) * m) matrix A and then regenerate another matrix B having less dimension than A which satisfy some condition. Now I wish to assign each row of B to a vector as individual. My problem is when I set any choice of (n,m) except m=1 it works fine but setting m=1 I got the

I have a matrix 700x2000 which is sampled in each cycle from another matrix 788x2000 with the numbers 0,1 and 9 There is one specific collumn of this matrix, dart[,1977], that usually, after the samplimg procedure has only 1 and 9 (because the zero frequency in this collumn is low). However, when this happens, I want to include an IF conditional in my code. so basically what i wanted to do was to

Hi, I have a matrix, say m=matrix(c( 983,679,134, 383,416,84, 2892,2625,570 ),nrow=3 ) i can find its row/col sum by rowSums(m) colSums(m) How do I divide each row/column by its rowSum/colSums and still return in the matrix form? (i.e. the new rowSums/colSums =1) Thanks. Casper -- View this message in context:

Hi Eik, thank you so much - it works perfectly! Thank you and best wishes Alain   Eik Vettorazzi <E.Vettorazzi@uke.de> hat am 6. Februar 2013 um 17:01 geschrieben: > Hi Alain, > here is a one-liner for a df without the rowSum column > >

--============_-1179735293==_ma============ Content-Type: text/plain; charset="us-ascii" ; format="flowed" To: r-help-request@lists.R-project.org From: "Dr. Chris Wills" <cwills@ucsd.edu> Subject: Questions about sorting and functions Cc: Bcc: X-Attachments: Dear R-Gang - Two questions for you: 1) I cannot figure out how to sort one column in an array, and

Dear R List, I am a new to R, so my question may be easy to answer for you: I have a dataframe, for example: df<-data.frame(loc=c("A","B","A","A","A"), year=as.numeric(c("1970","1970","1970","1976","1980"))) and I want to create the following table without using loops: 1970-74 ;