similar to: question about setdiff()

Hi everybody. The question: I get two vectors 'iFalseFalse' and 'i2'. I think they should be the same but they are not. Is it because R does not handle complicated logical expressions in such cases or I do something wrong? > z1 = c(NA, "", 3, NA, "", 3) > z2 = c("", "", 3, NA, 3, NA) > cV = (as.character(z1)==as.character(z2))

I have two factors l1, l2, and I'd like to merge them. (Remark: The factors can not be converted to charaters) Function c() does not give me the result I want: > l1 = factor(c('aaaa', 'bbbb')) > l2 = factor(c('ccc', 'dd')) > lMerge = factor(c(l1, l2)) > lMerge [1] 1 2 1 2 Levels: 1 2 > I'd like to merge l1 and l2 and to get lMerge

Today is a good day for asking question, I guess. > c() NULL > > length(c())==0 [1] TRUE > > r = ifelse(length(c())!=0, c(), c(1,2)) ### OK > r = c() ### OK > r = ifelse(length(c())==0, c(), c(1,2)) ### why this is not OK (given > the previous two)? Error in "[<-"(`*tmp*`, test, value = rep(yes, length =

Hi, everybody. This was an interesting discussion last time and it helped me a lot. Could you please have a look at some feature and tell me why it was designed this way (my questions are under #########) > x = c(1, 10) > y = c(99, 55) > d <- data.frame(x = x, y = y) > d x y 1 1 99 2 10 55 > add <- data.frame(x = 14, y = 99) > add x y 1 14 99 > d <-

The funciton c() works differently for strings and for factors: For strings: > l = c('a', 'b') > l [1] "a" "b" For factors: > l = c(factor('a'), factor('b')) > l [1] 1 1 What should be the right technique for merging factors? -- Svetlana Eden Biostatistician II School of Medicine

Dear R-devel, Please see the recent thread on R-help, "Odd Behavior Out of setdiff(...) - addition of duplicate entries is not identified" posted by Jason Rupert. I gave an answer, then read David Winsemius' answer, and then did some follow-up investigation. I would like to change my answer. My current version of setdiff() is acting in a way that I do not understand, and a way

Dear R-devel, Please see the recent thread on R-help, "Odd Behavior Out of setdiff(...) - addition of duplicate entries is not identified" posted by Jason Rupert. I gave an answer, then read David Winsemius' answer, and then did some follow-up investigation. I would like to change my answer. My current version of setdiff() is acting in a way that I do not understand, and a way

The Details. In the following version. > version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 1 minor 8.1 year 2003 month 11 day 21 language R > lookup.xport ignores some datasets in sas export file. File "emptySasData3.xpt" (available at http://biostat.mc.vanderbilt.edu/tmp/emptySasData3.xpt) is a

I think I am using the improved version of setdiff(...) that handles data.frames, so I think some odd behavior was expected but this one is escaping me. It appears that the the addition of duplicate entries is not caught by the setdiff(...). Is this expected behavior? If so, is there another method or approach that should be used to identify duplicate row entries between two different data

Hello, I have been interested in setdiff() for data frames that operates row-wise. I looked in the documentation, mailing lists, etc., and didn't find exactly the right thing. Given data frames A, B with the same columns, the goal is to extract the rows that are in A, but not in B. Of course, one can usually do setdiff(rownames(A), rownames(B)) but that is cheating. :-) I played around a

> version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 1 minor 8.1 year 2003 month 11 day 21 language R > aa <- as.POSIXct(c('1/1/1980','1/2/1980','')) > ifelse(is.na(aa),as.POSIXct('1/3/1980'),aa) Error in rep.int(unclass(x), times) : Argument "times" is missing,

Dear List, I have a data structure like: > aa [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 > bb [[1]] [1] 3 [[2]] [1] 5 I would like to set differences between "aa" and "bb" and get as result another list of lists like: > res [[1]] [1] 1 2 [[2]] [1] 4 6 I am trying: lapply(aa, setdiff, bb) but I simply get "aa" back as result. Could you please point me in the

hola, tengo dos vectores de 1134 y 385 elementos que se corresponden con números de accesión de genes. Necesito saber que números son comunes o intersección de vectores. He utilizado intersect(x,y) y me da todo el rato un único valor: V1 V1.1 V1.2 V1.3 V1.4 V1.5 V1.6 V1.7 V1.8 V1.9 V1.10 V1.11 V1.12 1 AJ558305 AJ558305 AJ558305 AJ558305 AJ558305

Hola Marcos. Yo probaría algo del estilo a esto.... Intersecc<-v1[v1%in%v2] Un saludo, _____________________________ Miguel Ángel Rodríguez Muíños Dirección Xeral de Innovación e Xestión da Saúde Pública Consellería de Sanidade Xunta de Galicia http://dxsp.sergas.es -----Mensaje original----- De: r-help-es-bounces en r-project.org [mailto:r-help-es-bounces en r-project.org] En nombre

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key1.=c(1, 2, 3) key2.=c(2) if (identical(key1.,key2.) == "TRUE") { cat("No Errors found") } if (length(setdiff(key1., key2.)) !=0) {

Hi Everyone, To complete the outer() and kronecker() functions in the base, may I suggest the following tensor() function, which allows the multiplication of arrays through sets of conformable dimensions. I am happy to write a help page if required. The code also needs a setdiff() function which prompts me to ask: what about simple set functions? I expect many of us have written our own

hello, I am hoping for some advice regarding comparing variables from 3 versions of a spreadsheet which have been combined into a single dataframe. The aim is to identify which rows have been changed. The dataframe contains 177 rows of data (each cell contains text). 'intersect' produced a file with 35 rows, 'union' a file with 303 rows and 'setdiff' a file with 130

Hi, I am struggling, I have 2 lists with shared elements, one ~600, one ~1000, and I need to determine the difference between them. They are character strings, and to use setdiff(), or unique() I need vectors. I don't know how to force these character strings into a form where you can use functions like setdiff(). Any help would be greatly appreciated. > head(R1) [1]

hi, I have 2 files example 1 and example 2 and would like to know what is in example2 and not in example1 (attached) V1 contain data which could be in duplicated which I am using as identifiers I used setdiff(example2$V1,example1$V1) to find the identifiers which are specific to example2: [1] "rs2276598" "rs17253672" I am looking for a way to get an output with all