similar to: Help with indexing

Dear R Helpers, I have trouble applying reShape and reshape although I read the documentation and several posts, so I would very much appreciate your help on the two points below. I have a dataframe df = data.frame(Name=c("a", "a", "a", "b", "b", "c"), X1=c("12", "13", "14", "20", "25",

Hi, there, I have a simple data manipulation question for you. Thank you for your help! Suppose that I have this data about people appearing in a class Mary Mary Mary Sam Sam John John John John Then I want to find out what exact time(s) the student appears at the moment such as Mary 1 Mary 2 Mary 3 Sam 1 Sam 2 John 1 John 2 John 3 John 4 the fifth row shows tha Sam show the second times

Dear R Team, I am a new R user and I am currently trying to subset my data under a special condition. I have went through several pages of the subsetting section here on the forum, but I was not able to find an answer. My data is as follows: ID NAME MS Pol. Party 1 John x F 2 Mary s S

I have read a frequency table in to R called "temp." I now want to create a histogram table from it, but I obviously first have to expand the data - to the sample size of 100. I want to use the command rep(), but I'm not sure how to go about it..I tried using the code: temp1<-rep(temp$Chest,100) hist(temp1) But this creates a v. odd histogram so I know it must be wrong!

I realize that this is terribly basic, but I just don't seem to see it at this moment, so I would very much appreciate your help. How shall I transform this dataframe: > df1 ? Name Index Value 1??? a???? 1?? 0.1 2??? a???? 2?? 0.2 3??? a???? 3?? 0.3 4??? a???? 4?? 0.4 5??? b???? 1?? 2.1 6??? b???? 2?? 2.2 7??? b???? 3?? 2.3 8??? b???? 4?? 2.4 into this dataframe: > df2 ??? Index?

Hello, I am having trouble getting the output from the tapply function formatted so that it can be made into a nice table. Below is my question written in R code. Does anyone have any suggestions? Thank you. Geoff #Input the data; name <- c('Tom', 'Tom', 'Jane', 'Jane', 'Enzo', 'Enzo', 'Mary', 'Mary'); year <- c(2008, 2009,

I would like to select rows if a row contains any one of several values. I can do the selection as follows: result[,"Subject"]=="JEFF" | result[,"Subject"]=="BG" But this is very unwieldily if one wishes to select many, many rows as one has to continuously repeat the source: result[,"Subject"]=="JEFF" |

In R-3.6.0 autoprinting was changed so that print methods for the storage modes are not called when there is no explicit class attribute. E.g., % R-3.6.0 --vanilla --quiet > print.function <- function(x, ...) { cat("Function with argument list "); cat(sep="\n ", head(deparse(args(x)), -1)); invisible(x) } > f <- function(x, ...) { sum( x * seq_along(x) ) }

It also is a problem with storage.modes "integer" and "complex": 3.6.0> print.integer <- function(x,...) "integer vector" 3.6.0> 1:10 [1] 1 2 3 4 5 6 7 8 9 10 3.6.0> print(1:10) [1] "integer vector" 3.6.0> 3.6.0> print.complex <- function(x, ...) "complex vector" 3.6.0> 1+2i [1] 1+2i 3.6.0>

Hi all If I have a string, say "ABCDA", and I want to convert this to the sum of the letter values, e.g. A -> 1 B -> 2 etc, so "ABCDA" = 1+2+3+4+1 = 11 Is there an elegant way to do this? Trying something like which(LETTERS %in% unlist(strsplit("ABCDA", ""))) is not quite correct, as it does not count repeated characters. I guess what I need is

Dear R community, is there an easy way to convert an adjacency list (or a data-frame) to a non-symmetric matrix? The adjacency list has the following form: person group 1 Sam a 2 Sam b 3 Sam c 4 Greg a 5 Tom b 6 Tom c 7 Tom d 8 Mary b 9 Mary d I need the data in a matrix with persons as rows and groups as columns: a b c d Sam 1 1 1 0 Greg 1 0 0 0 Tom 0 1 1 1 Mary 0 1 0 1 I know that there

Hmm. Chuck: I don't see how this example represents incomplete/incommensurate recycling. Doesn't TRUE replicate from length-1 to length-3 in this case (mat[c(TRUE,FALSE),2] would be an example of incomplete recycling)? William: clever, but maybe too clever unless you really need the speed? (The clever way is 8 times faster in the following case ...) x <- rep(1,1e6)

Letting a user supply the autoprint function would be nice also. In a way you can already do that, using addTaskCallback(), but that doesn't let you suppress the standard autoprinting. Having the default autoprinting do its own style of method dispatch doesn't seem right. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, May 21, 2019 at 10:50 AM Lionel Henry <lionel at

Hi, I am trying to create a Cumulative Frequency graph and I am using the following example: http://www.r-tutor.com/elementary-statistics/quantitative-data/cumulative-frequency-graph When I plot this data, how do I put in "real values" on the x-axis, rather than the values that are used for the breaks Thanks -- Shane [[alternative HTML version deleted]]

I'm trying to understand the behavior of seq_along in the following example: x <- 1:5; sum(x) y <- 6:10; sum(y) data <- c(x,y) S <- sum( data[seq_along(x)] ) S T <- sum( data[seq_along(y)] ) T Why is T != sum(y) ?

What is the 'idiomatic' way of writing a vectorized switch statement? That is, I would like to write, e.g., vswitch( c('a','x','b','a'), a= 1:4, b=11:14, 100 ) => c(1, 100, 13, 4 ) equivalent to ifelse( c('a','x','b','a') ==

I am moving this from r-help to r-devel. Based on offline communications with Jim, suppose dat is defined as follows: set.seed(123) dat <- data.frame(ID= c(rep(1,2),rep(2,3), rep(3,3), rep(4,4), rep(5,5)), var1 =rnorm(17, 35,2), var2=runif(17,0,1)) # Then this ave call works as expected: ave(dat$ID, dat$ID, FUN = function(x) seq_along(x)) # but this apparently identical calculation

Hi all, I have a problem with passing multiple option(-flag=aaa/bbb) from clang to llvm, I don't know how to parse it, should I parse it in clang and pass it to llvm or should I parse it in llvm pass where I use it, could you help me. Thanks in advance, Sevak.

Using the IRanges package from Bioconductor and somewhat recent R-2.9.1. ov = IRanges(1:3, 4:6) length(ov) # 3 seq(along = ov) # 1 2 3 as wanted seq_along(ov) # 1! I had expected that the last line would yield 1:3. My guess is that somehow seq_along don't utilize that ov is an S4 class with a length method. The last line of the *Details* section of ?seq has a typeo. Currently it is

Hallo, I have got an existing data frame and want to add a new column. The existing data frame was created like this: > df <- rbind( c("Fred", "Mary", 4), c("Fred", "Mary", 7), + c("Fred", "Mary", 9), c("Barney", "Liz", 3), + c("Barney", "Liz", 5) ) > df