similar to: Seeking help to define s4 method for 'print'

Displaying 20 results from an estimated 200 matches similar to: "Seeking help to define s4 method for 'print'"

2007 Oct 12
1
use 'lapply' to creat 2 new columns based on old ones in a data frame
There is a dataset 'm', which has 3 columns: 'index', 'old1' and 'old2'; I want to create 2 new columns: 'new1' and 'new2' on this condition: if 'index'==i, then 'new1'='old1'+add[i]. 'add' is a vector of numbers to be added to old columns, e.g. add=c(10,20,30 ...) Like this: index old1 old2 new1
2010 Jun 01
1
loop
Can any one help it will be very kind, loop statements I have this table and some more records, I want to reshape it V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 TP53 Dis1 Dis2 Dis3 Dis4 Dis5 Dis6 DCI New1 New2 New3 New4 FDI Hi2 H3 H4 GHD I1 I3 I4 I5 I6 I7 I8 I want my new table or matrix to be some thing like this V1 V2 V3 Tp53 Dis1 Dis2 Tp53 Dis1 Dis3 Tp53 Dis1 Dis4 Tp53 Dis1 Dis5 Tp53 Dis1 Dis6 Tp53 Dis2
2011 Sep 23
4
replace multiple string
Hi I would like to use a bash script that searches files and subdirectories name in a directory /var/ww/html/web for a specific string, and when it finds the search string, replaces the string (old1) with new string (new1), and so on old2 with new2 ....oldn with newn. replace_string.sh #!/bin/bash for db in $(find /var/www/html/web -name * -exec) do sed -e "s/old1/new1/" \ sed
2008 Jun 10
2
convert characters into integers in a matrix
Hello, I will appreciate any suggestion for this simple problem. I have a matrix of characters "A", "P", "M". I want to convert those characters to integers 0,1, 1 respectively. I am using the following R statements: exprs.new<-gsub("P",1,exprs) exprs.new1 <- gsub("A",0,exprs.new) exprs.new2 <-
2001 Apr 09
5
predict problem
Windows 98 R : Copyright 2001, The R Development Core Team Version 1.2.1 (2001-01-15) Dear friends. How comes this works and produce a single prediction: x <- rnorm(15) y <- x + rnorm(15) predict(lm(y ~ x)) new <- data.frame(x = seq(-3, 3, 0.5)) predict(lm(y ~ x), new, se.fit = TRUE) pred.w.plim <- predict(lm(y ~ x), new, interval="confidence") new1 <- data.frame(x=3)
2010 Apr 14
1
Problem trying to plot Vennerable object
Hi, I'm new to the list so apologies if this has been asked before. I couldn't find any refs in google. I recently installed the Vennerable library from Rforge, this required an upgrade of R to 2.10.0 by the 'pylr' dependency. However, a very simple command fails: Vcomb <- Venn(SetNames= c("EOCT", "EOCM"), Weight=c(0, 11841, 24084, 24660)) plot(Vcomb)
2017 Dec 20
1
utils::unzip ignores overwrite argument, effectively
It does give a warning, but then it overwrites the files, anyway. Reproducible example below. This is R 3.4.3, but it does not seem to be fixed in R-devel: https://github.com/wch/r-source/blob/4a9ca3e5ac6b19d7faa7c9290374f7604bf0ef64/src/main/dounzip.c#L171-L174 FYI, G?bor dir.create(tmp <- tempfile()) setwd(tmp) cat("old1\n", file = "file1") cat("old2\n", file
2008 May 15
2
How to remove autocorrelation from a time series?
Dear R users, someone knows how to remove auto-correlation from a frequencies time series? I've tried by differencing (lag 1) the cumulative series (in order to have only positive numbers) , but I can't remove all auto-correlation. If it's useful I can send my db. x <- # autocorrelated series new1<-cumsum(x) new2<-diff(new1,lag=1,differences = 1) acf(new2) #
2012 Feb 09
2
fill an array by rows
I've dug around but not been able to find anything, am probably missing something obvious. How can I fill a three-dimensional (or higher dimension) array by rows instead of columns. eg new1 <- array(c(1:125), c(5,5,5)) works fine for me but fills it by columns and new2 <- array(c(1:125), c(5,5,5), byrow=TRUE) throws an error. Am I missing something obvious? I also tried
2011 Jan 11
1
Writing diagonal matrix in opposite direction
Hi, is there any direct R function to write an diagonal matrix in an opposite way? for example I want to get like:   > diag(rnorm(5))[,5:1]            [,1]       [,2]       [,3]      [,4]       [,5] [1,]  0.0000000  0.0000000  0.0000000  0.000000 -0.1504687 [2,]  0.0000000  0.0000000  0.0000000 -2.139669  0.0000000 [3,]  0.0000000  0.0000000 -0.2102133  0.000000  0.0000000 [4,]  0.0000000
2011 Nov 08
2
Selecting 3 different hours in a day
Hello, I have a csv with 5months of hourly data for 4 years. I would like to get 9am, 12pm and 3pm from each day and create a subset or a new data frame that I can analyze. The time are from hour 0-23 for each day. I am not sure how to create a loop which will take out each of these hours and create a subset. I was thinking of doing a for loop using the row number since: 9am= row 10 12pm= row
2011 Feb 05
1
Seeking help to define method for '+'
Dear all, I am trying to define "+" method for my newly defined s4 class which is as follows: setClass("Me", sealed=F,representation(x1 = "numeric", x2 = "character")) new1 <- new("Me", x1=2, x2="comment1") new2 <- new("Me", x1=3, x2="comment1") setMethod("+", "Me",
2009 Apr 09
2
failed when merging two dataframes, why
Hi, R-listers, Failed, when I tried to merge df1 and df2 by "codetot" in df1 and "codetoto" in df2. I want to know the reason and how to merge them together. Data frames and codes I have used were listed as followed. Thanks a lot in advance. df1: popcode codetot p3need BCPy01-01 BCPy01-01-1 100.0000 BCPy01-01 BCPy01-01-2 100.0000 BCPy01-01 BCPy01-01-3 100.0000 BCPy01-02
2013 Feb 27
2
matrix multiplication
Hi, Try this: #mat1 is the data res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i) {new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j) {x1<-rbind(mat1[i,],mat1[j,]); x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7]-x1[2,7]))+(abs(x1[1,4]-x1[2,4])*abs(x1[1,8]-x1[2,8]))}));new1}))
2003 Jun 06
0
Problems building RELENG_4_8
On a 4.8-PRERELEASE, I have cvsup'ed my sources to the latest RELENG_4_8. While doing buildworld, I get this: cc -c -O -pipe -fexceptions -DIN_GCC -D_PTHREADS -DGTHREAD_USE_WEAK -I/usr/src/gnu/lib/libgcc/../../../contrib/gcc/config -I/usr/src/gnu/lib/libgcc/../../../contrib/gcc -I. -DL_pure -o _pure.o /usr/src/gnu/lib/libgcc/../../../contrib/gcc/libgcc2.c c++ -c -O -pipe -fexceptions
2012 Aug 02
3
text search in r
I am trying to count the number of times that the characters in a string change For example- new=c(AAAABBBBBABBBABB) I want to find the number of times that B changes to A and the number of times that A changes to B. I tried the grep command but I only figured out the positions of when B changes to A when I only need the number of times it occurs. -- View this message in context:
2012 Sep 18
2
Formula in a data-frame
Hello all, I'm new in R, and I have a data-frame like this (dput information below): Specie Fooditem Occurrence Volume 1 Schizodon vegetal 1 0.05 2 Schizodon sediment 1 0.60 3 Schizodon vegetal 1 0.15 4 Schizodon alga 1 0.05 5 Schizodon sediment 1 0.90 6 Schizodon
2009 Jul 01
4
mbox format and UIDVALIDITY
My base concern may be illustrated with the help of that simple telnet session: # telnet 127.0.0.1 imap Trying 127.0.0.1... Connected to localhost. Escape character is '^]'. * OK [CAPABILITY IMAP4rev1 LITERAL+ SASL-IR LOGIN-REFERRALS ID ENABLE AUTH=PLAIN] Dovecot ready. a1 login testuser ****** a1 OK [CAPABILITY IMAP4rev1 LITERAL+ SASL-IR LOGIN-REFERRALS ID ENABLE SORT
2008 Jul 25
0
nlminb--lower bound for parameters are dependent on each others
Hello I'm trying to solve two sets of equations (each set has four equations and all of them share common parameters) with nlminb procedure. I minimize one set and use their parameters as initial values of other set, repeating this until their parameters become very close to each other. I have several parameters (say,param1, param2) and their constraints are given as inequality and depend
2013 Jan 25
1
Recoding variables (without recode() )
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