search for: 5,6

When I was studying the function cut I found this example: > x <- rep(0:8, tx0) > x [1] 0 0 0 0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5 5 5 5 5 5 5 5 6 [39] 6 6 6 6 7 7 7 8 8 8 8 8 > cut(x, b = 8) [1] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] [6] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (0.994,2] [11] (0.994,2] (0.994,2] (0.994,2] (2,3] (2,...

Hi, Following expression returns only the first element ifelse(T, c(1,2,3), c(5,6)) However I am looking for some one-liner expression like above which will return the entire vector. Is there any way to achieve this?

...rom either 'yes' or 'no' depending on whether the element of 'test' is 'TRUE' or 'FALSE'. This is actually rather startling, because elsewhere in the S (R) language, operands are normally replicated to the length of the longer. Thus c(1,2,3)*10 + c(5,6) first (notionally) replicates 10 to c(10,10,10) and then c(5,6) to c(5,6,5), yielding c(15,26,35). And this *does* happen, sort of. > ifelse(c(F,T,F), c(1,2,3), c(5,6)) => 5 2 5. But it *doesn't* apply to the test. There's another surprise. Years ago I expected that all three arg...

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried frac=1 too.) I then add a month to the date, substract one day from the resultant date. But this wouldn...

Hi, It might be a trivial question but how do you store matrices of different dimensions read from a file or in a loop together? The best solution might be a list but I don't store the first matrix  correctly:  m = rbind(c(1,2),c(3,4),c(5,6)) t=rbind(c(1,2),c(5,6))  l = list(m) > l = list(l,t)    #I assumed that at the begining I don't have m and t at the same time to do list(m,t), but I list them, one after the other > l [[1]] [[1]][[1]]      [,1] [,2] [1,]    1    2 [2,]    3    4 [3,]    5    6 [[2]]   [,1] [,2] t  ...

.... Estoy haciendo calculos por grupos con data,table. Tengo un archivo (zp.res) con tres columnas que clasifican los datos (sol, con, dia) y una columna de datos numericos (media), de la siguiente forma: sol con dia media 1: con 0 1 -22.6 2: con 0 1 -36.6 3: con 0 1 -35.6 y quiero calcular el promedio de "media" (la col 4) agrupando con las variables sol,con,dia. Lo hago de la siguiente forma: med <- zp.res[, mean(media), by="sol,dia,con"] cuando reviso "med" esta todo bien, se han calculado las medias y el objeto tiene solo l...

Dear list, I have somewhat a small problem with a plot in R. I want to produce a plot with the two axis intersection being the exact values of ( 5,6). The code i am using is" BN<-c(14.95,9.03,10.42,9.3,9.52,7.73,6.35,6.73,5.54,5.42,6.24,5.98,6.21,5.83,7.14,5.79,5.57,7.43,9.19,10.17,9.79,9.56,9.87,9.33,10.56,10.08,9.49,8.49,7.34,6.14,5.98,7.27,5.56, 7.3,6.14,6.37,5.14,5.53,5.64,6.55,7.04,9.41,9.59,9.92,10.77,9.77,10.26,8.71,8.91,9.7...

...umber of days since 1/1/1970. How do I either get the first block to recognize the desired format, or the second block to format the x-axis as quarters? Thank you in advance! #block 1 time.val <- as.factor(c ("2004Q1","2004Q2","2004Q3","2004Q4","2005Q1","2005Q2")) inc <- c(0.9903797, 1.3741467, 0.9938702, 0.4252443, 0.7700158, 1.1770313) # returns nice x-axis, but black points plot(time.val, inc, type="b", col="red") #block 2 time2 <- ifelse (substr(time.val,5,6) == "Q1", as.Date(paste(...

Hello !! I have this problem: A matrix on True/False and as many numerical vectors as columns, but of different length. What I 'd like to get is this: set.seed(12) > dat <- as.data.frame(matrix(as.logical(sample(T:F, 30, T)),5,6)) > colnames(dat) <- letters[1:6] > rownames(dat) <- paste(letters[1:5],1:5, sep="") > dat a b c d e f a1 TRUE TRUE TRUE TRUE TRUE TRUE b2 FALSE TRUE FALSE TRUE FALSE FALSE c3 FALSE FALSE TRUE FALSE TRUE TRUE d4 TRUE TRUE TRUE FALS...

...at are factors ( such as sites "A", "B", and "C") that I would like to include in the new data frame. I have done this in a clunky way using match() and a loop, but am wondering if there is a more elegant approach. Here is an example data set. #Example a<-c(rep(1:5,6)); b<-sort(b) b<-c(rep("A",10),rep("B",10),rep("C",10)) a<-c(rep(1:5,6)); b<-sort(b) d<-c(2008,2008,2009,2009,2010,2010);d<-rep(d,5) e<-rnorm(30,2,1) df<-data.frame(a,b,d,e) ; names(df)<-c("ind","site","year",&...

Hi, all I am trying to make a plot with a axis break and I want the whole plot to be line, not points. However, when I execute the following command half of the graph is points and the other lines. gap.plot(Xdata, Ydata,gap=c(5,6),gap.axis="x",type="l") I think it might be a bug in plotrix. I would greatly appreciate your input. If there is another way to do it, I would greatly appreciate it. Best wishes, Art Roberts University of Washington Department of Medicinal Chemistry Seattle, WA 98195

Hi all, I have a data set with time interval and depending on the interval I want to create 5 more variables . Sample data below obs, Start, End 1,2/1/2015, 1/1/2017 2,4/11/2010, 1/1/2011 3,1/4/2006, 5/3/2007 4,10/1/2007, 1/1/2008 5,6/1/2011, 1/1/2012 6,10/15/2004,12/1/2004 First, I want get interval between the start date and end dates (End-start). obs, Start , end, datediff 1...

...ho my password into some commands inside of a bash script. But I think I'm going about it incorrectly. Here's the top part of my script: #!/bin/bash pub="~/.ssh/id_rsa.pub" dps_pass="my_pass" ssh="/usr/bin/ssh" scp="/usr/bin/scp" for i in 10.10.10.2{5,6} do echo "xfring key up" echo $dps_pass | $scp $PUB digitalplatform@$i: And here's how it executes: #bash -x deploy_key.sh + pub='~/.ssh/id_rsa.pub' + dps_pass='nbcuV01P!' + ssh=/usr/bin/ssh + scp=/usr/bin/scp + for i in 10.10.10.2{5.6} + echo 'xfring key u...

Dear all, I have following codes: colnames(data) = c("var", "var", "var") i = c(1,2,3) Now I want construct a "for" loop starting from 1 to 3 to give the new names of columns for dataframe "data" like below colnames(data) > c("var1", "var2", "var3") Definitely I could do this manually, however I want to put

Hi, Forgive a very basic question... I need to take two lists-of-lists, and apply a function to each pair of elements in the lists to return a single list... For example l1 <- list(1:5,6:10,2:15) l2 <- list(1:8,4:12,1:19,4:20) I could easily do an lapply across each of them, but is there a function that does a sort-of pairwise-apply across both together? Does anybody know of a good document that discusses this kind of basic matrix/list/vector manipulation in R? Regards, cris...

...trouble with the array structure of R. What I want to do is dynamically add/remove elements to an array. For example: Let's say I have created an array: > myArray <- array(c(3,8), dim=c(1,2)) > myArray [,1] [,2] [1,] 3 8 And I now want to, for example, push an element (5,6) on to this array so it will read: [,1] [,2] [1,] 3 8 [2,] 5 6 And then pop the first element of the array so the array now reads: [,1] [,2] [1,] 5 6 How would I do this? So far I've only read how to create an array if you know the dimensions beforehand, but...

...anti-dependence of postRAScheduler in LLVM. when I use command line "clang -target arm -mcpu=cortex-a8 -O2 -integrated-as -c test.c -o test.o" and get objdump file as follows: ldrr1, [r0,#16]----(1 str r1, [r0,#32]----(2 ldr r1, [r0,#12]----(3 str r1, [r0,#36]----(4 ldr r1, [r0,#08]----(5 str r1, [r0,#40]----(6 However, I expect that instruction pairs (1,2), (3,4) and (5,6) use different registers after postRAScheduler pass. Then, I find that postRAScheduler only handle anti-dependence mode after debug LLVM source code. More while I use gcc to compile test.c and get expected result...

Several bugs are present in R-2.15.3 and R-alpha due to naive copying of list elements. The bug below is due to naive copying in subset.c (with similar bugs for matrices and arrays): a<-list(c(1,2),c(3,4),c(5,6)) b<-a[2:3] a[[2]][2]<-9 print(b[[1]][2]) Naive copying in mapply.c leads to the following bug: X<-1+1 f&lt...

...am trying to generate a randomized weekday survey schedule that ensures even coverage of weekdays in the sample, where the distribution of variable DOW is random with respect to WEEK. To accomplish this I need to randomly sample without replacement two weekdays per week for each of 27 weeks (only 5 are shown). However, I need to sample from a sequence (3:7) that needs to be completely depleted and replenished until the final selection is made. Here is an example of what I want to do, beginning at WEEK 1. I would prefer to do this without using a loop, if possible. sample frame: [3,4,5,6,...

Hi All I have a data frame in which there are 4 columns . Column 1 : name Column 2-4 : values I would like to calculate mean/Standard error of values in column 2-4 and store them in column 5,6 respectively. I have done the following but doesn't seem to work mean_N_SE <-function(x) { name <- x[1] vals <- c(x[2:4]) temp_mean <- mean(vals) SE <- sqrt(var(x)/length(x)) } apply(d,1,mean_N_SE) where d = data frame. Can someone help me with this. Thanks! -Abhi...