similar to: Create blocks or observations

Hi all, Trying to figure out a quick way to do something. I have a yahoo groups list, and it has listings that look like this: </td></tr> <tr><td colspan="2" class="ygrp-g"> <span class="ygrp-nowrap"> <em><a href="/group/freecycledc/">freecycledc</a></em> </span> <a

I have windows version of R v1.6.2 I have downloaded the qtoolbox.zip from http://www.cmis.csiro.au/S-PLUS/qtoolbox/ My R command line is > install.packages("C:/Program Files/R/qtoolbox.zip", .libPaths()[1], CRAN = NULL) and I get the following error message: updating HTML package descriptions Warning message: error -1 in extracting from zip file > Is there another way to

Hi. I have the following code in a loop. It splits a vector into subvectors of equal size. But if the size of the original vector is not an exact multiple of the desired subvector size, then the first few subvectors have one more element than the last few. I know that the cut function could be used to determine where to break up the vector, but it doesn't seem to provide control over

Hi everyone! May I request again for your help? I need to make some codes using if else statements... Can I do an "if-else statement" inside an "if-else statement"? Is this the correct form of writing it? Thank you.=) Example: for (v in 1:6) { for (i in 2:200) { if (v==1) (if max(x*v-y*v)>1 break()) if (v==2) (if max(x*v-y*v)>1.8 break()) if (v==3) (if

Is the following developed in my console output a recognized bug or am I using the cat function incorrectly? Thanks, Stan > ifelse(class(data[[n]])!="factor",{print("yes")},{print("no")}) [1] "yes" [1] "yes" > ifelse(class(data[[n]])!="factor",{cat("yes")},{cat("no")}) yesError in ans[test & !nas] <-

Hi, How can I create an NA? for example, i have this program but it has errors > x=scan() 1: 1 2 3 4 5 6: Read 5 items > "na"*x Error in "na" * x : non-numeric argument to binary operator what should I multiply x so that I will get an NA? And I really need to multiply that with my x vector. I need this output: (na,na,na,na,na) Thank you for your time and help!=) --

Hi all, May I request for your help if you have time and if you have an idea on how to do this. I want to add three vectors... And my goal is to obtain the sum of the vectors, ignoring the vector of "na"... Here is what i did in R.. I'm adding the three vectors, e,z,k, and my objective is to get an answer = -23. I tried putting the na.omit but it did not work. Thanks. > z [1]

Help! model1 <- siarmcmcdirichletv4(data, source, tef, concdep=0, 500000, 50000) Error in matrix(1, ncol = (numsources + numiso) * numgroups, nrow = (siardata$iterations - : invalid 'ncol' value (too large or NA) In addition: Warning message: In Ops.ordered((numsources + numiso), numgroups) : '*' is not meaningful for ordered factors Previous message, but more problems:

Hi, may I ask for your help... What is the R code if I want to count the number of times that "2" appeared in the following R output > k [1] 1 1 1 1 1 1 3 1 2 1 1 1 1 1 1 1 1 3 2 1 1 4 1 1 1 1 1 5 1 1 Thank you for your help. :-) -- View this message in context: http://n4.nabble.com/How-to-count-the-number-of-occurence-tp1661733p1661733.html Sent from the R help mailing list

#is there a way to get NA in the table of descriptive statistics instead of the function stopping Thank you in advance #data x.f <- structure(list(Site = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("BC", "HC", "RM119", "RM148", "RM179", "RM185",

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

Hi, I'm searching for a function to subistute Values in a Matrix to new Values. For example: old value new value 1.1 6 1.2 7 . . . . . . 1.9 14 2.0 15 and 2.1 15.5 2.2 16 . . . . 2.9 19.5 3.0 20 There is a difference

Dear R users, I'd like to change the default order of colors & line types. Especially I am using ggplot2 and using color Set1. In Set1, the default color order is red, blue, green, violet,.. ect. However, I want to put red in fourth (not first). Likewise, I want to change the order of default linetype. I want to put "solid" line in fourth. How can I do thses? R code to draw the

Dear arun and all R users, I will first of all try to simply define my issue.. I have data in the following format Year Discharge dd/mm/yyyy x .. … … … There are some NA values in the discharge which I would like to predict by using “predict command”. I cant figure out the way to write the coding for that. Could you please help me on that??? I have also ,written

Dear All, I would like to use the simprof function (clustsig package) but the available distances do not include Jaccard distance, which is the most appropriate for pres/abs community data. Here is the core of the function: > simprof function (data, num.expected = 1000, num.simulated = 999, method.cluster = "average", method.distance = "euclidean", method.transform =

Hi I am trying to find a parametric bootstrap confidence interval and when I used the boot function I get zero bias and zero st.error? What could be my mistake? Thank you and take care. Laila [[alternative HTML version deleted]]

Hi, my name is Julio, I'm trying to run a script, specifically with the siar package on R (I have attached the script) but the program collapses on during the run. The error message says: *** caught segfault *** address 0xb31941cc, cause 'memory not mapped' Traceback: 1: .C("siarmcmcv4", as.integer(numdata), as.integer(numsources), as.integer(numiso),

Hola, ¿qué tal?: Tengo la siguiente duda: Para el caso del Cosinor Simple: Y = M + A * cos(2*pi*t/T + phi) + error Cuando T conocido, podemos asumir: Y = M + beta * X1 + gamma * X2 + error, siendo X1 = cos(2*pi*t/T) y X2 = sin(2*pi*t/T) En este caso (Cosinor simple) el modelado lo llevo a cabo del siguiente modo: lm( y ~ X1 + X2) Para el caso de componentes múltiples: Y = M + ?Aj *

Given the string > mystr <- "(Preconsuntivo del giorno gas 10 maggio 2007)Tj" I'm trying to detect and eliminate the string ")Tj" at the very end of mystr by means of gsub(rx2,"",mystr) BUT preparing the matching regexp string a warning pops up > rx2 <- "\)Tj$" Warning messages: 1: '\)' is an unrecognized escape in a

Hi, I was wondering if it is possible to get the rowindices without using the function "which" because I don't have a restriction criteria. Here's an example of what I mean: # take 10 randomly selected instances iris[sample(1:nrow(iris), 10),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 76 6.6 3.0 4.4 1.4