similar to: binary symmetric matrix combination

Displaying 20 results from an estimated 600 matches similar to: "binary symmetric matrix combination"

2011 Sep 21
3
Reading data in lisp format
Hi, I am trying to read the "credit.lisp" file of the Japanese credit database in UCI repository, but it is in lisp format which I do not know how to read. I have not found how to do that in the foreign library http://archive.ics.uci.edu/ml/datasets/Japanese+Credit+Screening <http://archive.ics.uci.edu/ml/datasets/Japanese+Credit+Screening> Could anyone help me? Best
2007 Feb 08
5
remove component from list or data frame
Sorry to ask such a simple question, but I can't find the answer after extensive searching the docs and the web. How do you remove a component from a list? For example say you have: lst<-c(5,6,7,8,9) How do you remove, for example, the third component in the list? lst[[3]]]<-NULL generates an error: "Error: more elements supplied than there are to replace" Also,
2013 Apr 10
6
means in tables
Hi. I have 2 tables, with same dimensions (8000 x 5). Something like: tab1: V1 V2 V3 V4 V5 14.23 1.71 2.43 15.6 127 13.20 1.78 2.14 11.2 100 13.16 2.36 2.67 18.6 101 14.37 1.95 2.50 16.8 113 13.24 2.59 2.87 21.0 118 tab2: V1 V2 V3 V4 V5 1.23 1.1 2.3 1.6 17 1.20 1.8 2.4 1.2 10 1.16 2.6 2.7 1.6 11 1.37 1.5 2.0 1.8 13 1.24 2.9 2.7 2.0 18 I need generate a table of averages, the
2007 Jul 18
1
Neuman-Keuls
hello, I have programmed this function to calculate the Neuman-Keuls test but I have a problem the function return an empty list and I don't know why. summary(fm1) E <- sqrt((summary(fm1)[[1]]["Residuals","Mean Sq"])/length(LR)) lst <- list() lst1 <- list() lst2 <- list() NK <- function (x) { if (length(x) == 2) { Tstudent <- t.test(subset(exple,
2008 Oct 01
3
lapply where each list object has multiple parts
Hi. I have a list where each object in the list has multiple parts. I'd like to take the mean of just one part of each object. Is it possible to do this with lapply? If not, can you recommend another function? Thanks. eric > x1 <- c(0,1,2,3) > x2 <- c(7,8) > x3 <- c(2,6,6,8) > x4 <- c(4,8) > > Lst1 <- list(label1 = x1,label2 = x2) > Lst2 <-
2013 Sep 20
3
search species with all absence in a presence-absence matrix
Dear list I have a matrix composed of islandID as rows and speciesID as columns. IslandID: Island A, B, C….O (15 islands in total) SpeciesID: D0001, D0002, D0003….D0100 (100 species in total) The cell of the matrix describes presence (1) or absence (0) of the species in an island. Now I would like to search the species with absence (0) in all the islands (Island A to Island O.)
2006 Jul 26
1
.Call question
Writing R Ext says to treat R objects that are arguments to .Call as read only (i.e. don't modify). I have a long list of lists that and I want to avoid the overhead of a copy in my C code. I would just like to modify some of the elements of list by replacing them with elements of exactly the same size/type. below is an example of the essence of the problem. This seems to work. Is this
2013 Apr 18
6
count each answer category in each column
Hey, Is it possible that R can calculate each options under each column and return a summary table? Suppose I have a table like this: Gender Age Rate Female 0-10 Good Male 0-10 Good Female 11-20 Bad Male 11-20 Bad Male >20 N/A I want to have a summary table including the information that how many answers in each category, sth like this: X
2013 Aug 25
1
Capturing the whole output using R
Hi, May be this helps: #Creating some dummy data.? set.seed(24) lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE))) set.seed(49) lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE))) ??Find_Max_CCF() #No vignettes or demos or help files found with alias or concept or #title matching ?Find_Max_CCF? using regular expression matching. Found a function with the same
2013 Apr 24
2
Distance matrices Combinations
Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for
2009 Sep 29
1
Comparing vectors from lists
Hi guys, I still did not solve my problem properly! I have to compare the values of two lists of 250 numbers as a result of using the ?by function! List1 of 250  $ 0   : num [1:28] 22 11 31...  $ 1   : num [1:15] 12 14 9 ... .. .. ..  - attr(*, "dim")= int 250  - attr(*, "dimnames")=List of 1 List2 of 250  $ 0   : num [1:24] 20 12 22...  $ 1   : num [1:17] 11 12 19 ... .. ..
2013 Nov 08
2
making chains from pairs
Hello, having a data frame like test with pairs of characters I would like to create chains. For instance from the pairs A/B and B/I you get the vector A B I. It is like jumping from one pair to the next related pair. So for my example test you should get: A B F G H I C F I K D L M N O P > test V1 V2 1 A B 2 A F 3 A G 4 A H 5 B F 6 B I 7 C F 8 C I 9 C K 10 D L
2013 Apr 13
2
Comparison of Date format
Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???
2010 Mar 25
1
Error using Rcpp
Hi, Im not sure if this is the right place to post this. I am using Xubuntu Karmic Koala and am trying to use the Rcpp package. I am testing it using a simple code that takes in a vector and adds 1 to each element: #include <Rcpp.h> // This file takes in a vector and adds one to each entry RcppExport SEXP addone(SEXP vec){ // create a local copy of vec Rcpp::NumericVector
2013 Jun 07
4
matched samples, dataframe, panel data
I R-helpers #I have a data panel of thousands of firms, by year and industry and #one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not #and another variable the represents the firm dimension (total assets in thousand of euros) #I need to create two separated samples with the same number os firms where #one firm in the first have a corresponding
2013 May 02
3
R issue with unequal large data frames with multiple columns
I'm a bit of an amateur R programmer. I can do simple R scenarios but my handle on complex grammatical issues isn't steady. I have 12 CSV files that I've read into dataframes. Each has 8 columns and over 2000000 rows. Each dataframe has data associated by time component and a date component in the format of: X.DATE and then X.TIME X.DATE is in the format of MMDDYYYY and X.TIME is
2013 Aug 30
4
mean
When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com.
2011 Oct 06
3
Wide to long form conversion
I have some data 'myData' in wide form (attached at the end), and would like to convert it to long form. I wish to have five variables in the result: 1) Subj: factor 2) Group: between-subjects factor (2 levels: s / w) 3) Reference: within-subject factor (2 levels: Me / She) 4) F: within-subject factor (2 levels: F1 / F2) 5) J: within-subject factor (2 levels: J1 / J2) As this is the
2013 Mar 10
2
list + lapply insead of matrix + apply
I need to develop a simple list manipulation. Although it seems easier to do it in matrix form, but I need it in list form. I have a matrix x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2) for list form example, the conversion is x.list <- lapply(seq_len(nrow(x)), function(i) x[i,]) ### list version calcnorm=function(a, b){ diff <- mapply("-", a, b)
2008 Sep 09
1
creating table of averages
Dear Colleagues, I have a dataframe with variables: [1] "ID" "category" "a11" "a12" "a13" "a21" [7] "a22" "a23" "a31" "a32" "b11" "b12" [13] "b13" "b21"