similar to: How to count the number of occurence?

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

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]

LS, I would really like to know how to count the frequency/occurrence of chachters inside a dataset. I am working with extreemly large datasets of forest inventory data with a large variety of different species inside it. Each row inside the dataframe represents one individual tree and the simplified dataframe looks something like this: num species dbh 1 sp1

LS, I would really like to know how to count the frequency/occurrence of chachters inside a dataset. I am working with extreemly large datasets of forest inventory data with a large variety of different species inside it. Each row inside the dataframe represents one individual tree and the simplified dataframe looks something like this: num species dbh 1 sp1

Dear R experts, I have the data in the following fomat(from some kind of card sorting process) ID Category Card numbers 1 1 1,2,5 1 2 3,4 2 1 1,2 2 2 3 2 3 4,5 I want to transform this data into two co-occurence matrix (one for each ID) -- For ID 1 1 2 3 4 5 1 1 1 0 0 1 2 1 1 0 0 1 3 0 0 1 1 0 4 0 0 1 1 0 5 1

Hi, anyone here who knows how to determine if an integer is "odd" or "even" in R? Thanks. tj -- View this message in context: http://n4.nabble.com/ODD-and-EVEN-numbers-tp1747032p1747032.html Sent from the R help mailing list archive at Nabble.com.

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]]

Sorry for the verbose example. I want to row bind two matrices, and all works except I want the column labelled "row" to be sequential in the new matrix, shown as "mat3" here, i.e. needs to be 1:6 and not 1:3 repeated twice. Any suggestions? Thanks J > colnm1 <- c("row","ti","counti") > colnm2 <-

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

?unique says Value: An object of the same type of 'x'. but if an element is equal to one with a smaller index, it is removed. However, I need to keep the one with the LARGEST index. Can someone please show me the light? I thought about reversing the row order twice, but I couldn't get it to work right (My data frame has 125000 rows and 7 columns, and I'm

Each day the daily balance in the following link http://www. snamretegas.it/italiano/business/gas/bilancio/pdf/bilancio.pdf is updated. I would like to set up an R procedure to be run daily in a server able to read the figures in a couple of lines only ("Industriale" and "Termoelettrico", towards the end of the balance) and put the data in a table. Is that possible? If

How can achieve this in R using seq, or rep function c(-1,0,1,0,-1,0,1,0,-1,0) The range value is between-1 and 1, and I want it such that there could be n number of points between -1 and 1 Anyone? Please help Thanks Rusty [[alternative HTML version deleted]]

Hi, Could you please let me know to use a list in a for loop here geneset is a loop.I am trying to match the names of the list with 1st row of the output. result<- list() for(i in 1:length(output) { result[[i]] <- geneset(which(geneset %n% output[,1])) } Kindly help me out -- View this message in context: http://www.nabble.com/For-loop-tp18163665p18163665.html Sent from the R

Hello, I'm looking for a function that counts the occurences of a given character within a string. f('|','ab|c|d') => 2 More precisely, I need to complete a vector of strings to ensure that all elements have the same count of a "separator": a|b|c a |a|b|c|d => a|b|c|| a|||| |a|b|c|d I guess that scan makes use of an internal function that would do

Hello, I have to code a small application for a car repair shop running under Windows (98 SE but may be my client could upgrade to 2000) and using RoR is my definite choise ! The app will be powered by WebRick (which is enough powerfull for a small app used by only ONE user at a time) My questions are targeted to the best database to use... I had a look to SQLite and Firebird but support

Hello, R-helpers. I have a data frame below called df. I want to add a variable c, based on the following logic: if df$a < 5 then the first two times this condition is met place a 1 in c, the next two times this condition is met place a 2 in c and when the condition is not met let c equal df$b. The result would look like, dfnew, shown below. a <- rep(c(3,5,7), 4) b <- rep(c(NA,1,2),

Sorry - working on a simple tutorial but something is throwing me here. I''m supposed to get a Time.now() print to the page but it''s not working and I can''t figure out why. No errors are thrown my the page shows: Howdy, world! My name is Stuart Felenstein. This was verified at . So no time stamp showing. I went into IRB , did a Time.now() and it''s working

Version 2 of these patches includes optimisations suggested by H Peter Alvin to save bytes: * Use memory-mapped I/O register to read keyboard shift flags * Don't reload DX from memory; move immediate byte to register Tested using mbr.bin. TJ (3): mbr: Ctrl-key press forces load from first hard disk gptmbr: Ctrl-key press forces load from first hard disk isohdpfx: Ctrl-key press