similar to: counting entries in vector

Hi, I am sure there is an easy way to do it, but I can't find it. I have a data frame that has 15 columns and 7000 rows. The only values inside the data.frame are "aa", "ab", "bb" as you can see an example bellow. 1 2 3 1 aa ab ab 2 ab ab ab 3 aa aa aa 4 bb bb bb What I would like to do, is to generate a vector (or another data.frame) with 7000 rows, and 3

Dear list, Given a vector of logical values, say >a <- c(TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE) Are there any R functions that can tell whether there are two or more "TRUE" in a row in this vector? Thanks, Wuming

Hello, I have a very long (~50,000) sequence of repeating numbers. The first 100 are: [1] 0 0 0 0 0 0 0 0 0 0 0 429 [13] 429 429 429 429 429 429 429 858 858 858 858 858 [25] 858 1287 1287 1287 1287 1287 1716 2145 2145 2574 2574 3003 [37] 3003 3432 3432 3861 4290 4719 5148 5577 5577 6006 6006

I think I asked a similar question 3 years ago to the Splus list and I think the answer was no or noone answered so noone should spend more than 5 minutes on this because it could definitely be a waste of time. My question is whether the function below can be rewritten without a for loop. apply is fine if it can be done that way but i doubt it. I call it a lot and would prefer to not loop.

Hello I have this problem. I have a large matrix of this sort: > prova [,1] [,2] [,3] [,4] [1,] 3 3 3 3 [2,] 3 3 3 1 [3,] 1 3 3 3 [4,] 1 1 1 3 [5,] 3 1 1 3 [6,] 3 1 1 3 [7,] 1 3 1 3 [8,] 1 3 3 3 What I want to do is to count the number of sequences of ones and stack the results in a

Hello, I have the following problem. I am running simulations on possible states of a set of agents (1=employed, 0=unemployed). I store these simulated time series in a matrix like the following, where rows indicates time periods, columns the number of agents (4 agents and 8 periods in this case): Atr=[ 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0

Hi, I need help to find an efficient way to transform a vector like: a<-c(1,1,0,1,0,0,0,1,1,1,1,0,1,0,1,1) in a vector that counts only di 1 elements, like: b<-c(1,2,0,1,0,0,0,1,2,3,4,0,1,0,1,2) Thank you! -- View this message in context: http://r.789695.n4.nabble.com/counts-of-a-vector-tp2232047p2232047.html Sent from the R help mailing list archive at Nabble.com.

Hi everybody, I have the following problem. I have a vector containing character elements, such as: list = c("aa","bb","cc","dd","ee") I want to create an index which identifies the elements that are different from, e.g. "aa" and "bb". When I do the following: jj = list!="aa" & list!="bb" > jj

Dear R users, Consider the first two columns of a data frame like this: > z[,1:2] x y 1 1 1 2 2 2 3 3 3 4 1 4 Imagine that y represents the times that the value x happens in a population. But z is not exactly a frequency table, because in z we have x=1 twice. So, the x=1 in the first line and the x=1 in the fourth are not the same, differing according to a third variable in

Hallo! I have a vector of ID's like so, id <- c(1,2,2,3,3,3,4,5,5) I would like to create a [start,stop] pair of vectors that index the first and last observation per ID. For the ID list above, it would look like 1 1 2 3 4 6 7 7 8 9 I haven't worked with indexes/data manipulation much in R, so any pointers would be helpful. Many thanks! ~~~~~~~~~~~~~~~~~~~ -Robin Jeffries Dr.P.H.

Hello there R-help, I'm not sure if this should be posted here - so apologies if this is the case. I've found a problem while using rle and am proposing a solution to the issue. Description: I ran into a niggle with rle today when working with vectors with NA values (using R 2.31.0 on Windows 7 x64). It transpires that a run of NA values is not encoded in the same way as a run of other

well! excuse me again but... your.string <- "aaacdf" nc1 <- nchar(your.string)-1 x <- unlist(strsplit(your.string, NULL)) ######## CORRECT x2 <- c() for (i in 1:nc1) x2 <- c(x2, paste(x[i], x[i+1], sep="")) ######## ERRATA 2 cat("ocurrences of <aa> in <your.string>: ", length(grep("aa", x2)), sep="", fill=TRUE) Fran

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I am performing a precipitation analysis. data is in the form of daily precipitation amounts, e.g. x<- c(4,5,3,0,0,0,2,4,6,4,0,0,0,2,2,0,3,4,1,0,...) I would like to find the length of the "storm", length of storm would be defined as the number of days with continues precipitation. in this case the returned vector would be: (3,4,2,3,...) I would also like the amount of

This is not? serious problem but I just wonder if someone can explain what is happening. The same command within a dataframe is giving me a factor and as a plain vector is giving me a character.? It's probably something simple that I have read and forgotten but I thought I'd ask. Thanks #================================================ dat1 <- data.frame(aa = letters[1:10]) str(dat1)

> On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help at r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is happening. > The same command within a dataframe is giving me a factor and as a plain vector is giving me a character. It's probably something simple that I have read and forgotten but I thought I'd ask.

Hallo, suppose I have a vector: x <- c(1,1,1,2,2,3,3,3,3,3,4) How can I generate a vector/sequence in which a fixed number of zeroes (say 3) is inserted between the consecutive values, so I get 1,1,1,0,0,0,2,2,0,0,0,3,3,3,3,3,0,0,0,4 thanks a lot, Serguei [[alternative HTML version deleted]]

Given a vector of numeric of length n, I need to find segments that are >= 0.2, compute the average of individual segments, and replace the original values in each segment by their corresponding averages. For example, there are three segments that are >= 0.2, the average of 1st segment is 0.3, 2nd is 0.5, and the 3rd is 0.5333333 >

Thank you Pikal and Bert. My apology for posting parts of my previous email in HTML. Bert's suggestion will work but i am wondering if there is an alternative especially in the case where the data frames are big; that is the difference in lengths among them is large. Below is a list of sample date frames and desired result. EK

Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame.? I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwartz at me.com> wrote: > On Jul 7, 2017, at 6:03