similar to: Dataframe modification

Hi, How can I achieve this in R. Dataset is as follows: >df x 1 2 2 4 3 1 4 3 5 3 6 2 structure(list(x = c(2, 4, 1, 3, 3, 2)), .Names = "x", row.names = c("1", "2", "3", "4", "5", "6"), class = "data.frame") I want to recreate a new data frame whose rows are sum of (1&2, 3&4, 5&6)

Here's a skeletal example. Embellish as needed: p <- 5 n <- 300 set.seed(1) dat <- cbind(rnorm(n), matrix(runif(n * p), n, p)) write.table(dat, file="c:/temp/big.txt", row=FALSE, col=FALSE) xtx <- matrix(0, p + 1, p + 1) xty <- numeric(p + 1) f <- file("c:/temp/big.txt", open="r") for (i in 1:3) { x <- matrix(scan(f, nlines=100), 100,

Hi How can do this in R. >df 48 1 35 32 80 If df < 30 then replace it with 30 and else if df > 60 replace it with 60. I have a large dataset so I cant afford to identify indexes and then replace. Desired o/p: 48 30 35 32 60 Thanx in advance. Sachin

Hi, I have a dataset consisting of 350,000 rows and 266 columns. Out of 266 columns 250 are dummy variable columns. I am trying to read this data set into R dataframe object but unable to do it due to memory size limitations (object size created is too large to handle in R). Is there a way to handle such a large dataset in R. My PC has 1GB of RAM, and 55 GB harddisk space running

Hi, How can I read data of unequal number of observations (rows) as is (i.e. without introducing NA for columns of less observations than the maximum. Example: A B C D 1 10 1 12 2 10 3 12 3 10 4 12 4 10 5 10 Thanks in advance. Sachin --------------------------------- [[alternative HTML version deleted]]

Hi, 1. How do I construct 95% prediction interval for new x values, for example - x = 30000? 2. How do I construct 95% confidence interval? my dataframe is as follows : >dt structure(list(y = c(26100000, 60500000, 16200000, 30700000, 70100000, 57700000, 46700000, 8600000, 10000000, 61800000, 30200000, 52200000, 71900000, 55000000, 12700000 ), x = c(108000, 136000,

Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin ____________________________________________________________________________________ [[alternative HTML version deleted]]

Hi, How can I accomplish this task in R? V1 10 20 30 10 10 20 Create a new column V2 such that: If V1 = 10 then V2 = 4 If V1 = 20 then V2 = 6 V1 = 30 then V2 = 10 So the O/P looks like this V1 V2 10 4 20 6 30 10 10 4 10 4 20 6 Thanks in advance. Sachin

Hi, How can plot a frequency distribution curve for the following data.    V1      V2 1   1 160.54% 2   1 201.59% 3   1  18.45% 4   1 179.03% 5   1 274.37% 6   1   0.00% 7   1  24.52% 8   1  39.17% 9   3  43.72% 10  1  53.06% 11  1  64.97% 12  1  79.84% 13  1  98.08% 14  1 115.32% 15  1 127.96% 16  1 155.38% 17  1 157.25% 18  1 193.17% 19  1  51.53% 20 15  99.32% 21  1 106.86% 22  1 219.44%

Hi, How can I produce the following output in .csv format using write.table function. for(i in seq(1:2)) { df <- rnorm(4, mean=0, sd=1) write.table(df,"C:/output.csv", append = TRUE, quote = FALSE, sep = ",", row.names = FALSE, col.names = TRUE) } Current O/p: x 0.287816 -0.81803 -0.15231 -0.25849 x 2.26831 0.863174

Hi, How can I write the output to an excel (csv) file without printing row names (i.e without breaks). Here is my code: library( fn <- function() { q <- c(1,2,3) write.csv(q,"C:/Temp/op.xls", append = TRUE, row.names = FALSE,quote = FALSE) } # Function Call for(i in 1:3) { fn() } Present Output : x 1 2 3 x 1 2

I have a Controller named Orders which has a pending_orders method which is expected to fetch some records from the database. If i dont write a route for this method, I get the following error when i call this method. Couldn''t find Order with ID=pending_orders I am using rails 2.3.5, in the previous versions i use to get this I am not getting whether its new version requirement... Help

Hi, Am I interpreting the results properly? Are my conclusions correct? > KPSS.test(df) ---- ---- KPSS test ---- ---- Null hypotheses: Level stationarity and stationarity around a linear trend. Alternative hypothesis: Unit root. ---- Statistic for the null hypothesis of level stationarity: 1.089 Critical values: 0.10 0.05 0.025 0.01 0.347 0.463

Hi, I have following dataframe. Column A indicates months. DF <- structure(list(A = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 7, 8, 11, 12, 1, 2, 3, 4, 5, 8), B = c(0, 0, 0, 8, 0, 19, 5, 19, 0, 0, 0, 11, 0, 8, 5, 11, 19, 8, 11, 10, 0, 8, 36, 10, 16, 10, 22)), .Names = c("A", "B"), class = "data.frame", row.names = c("1",

Hi, How can I accomplish this in R. Example: R1 R2 3 101 4 102 3 102 18 102 11 101 I want to find Sum(101) = 14 - i.e SUM(R1) where R2 = 101 Sum(102) = 25 - SUM(R2) where R2 = 102 TIA Sachin --------------------------------- [[alternative HTML version deleted]]

Hi, I have a data set which is assumed to follow weibull distr'. How can I find of cdf for this data. For example, for normal data I used (package - lmomco) >cdfnor(15,parnor(lmom.ub(c(df$V1)))) Also, lmomco package does not have functions for finding cdf for some of the distributions like lognormal. Is there any other package, which can handle these distributions?

Hi, I am trying to extract subset of data from my original data frame based on some condition. For example : (mydf -original data frame, submydf - subset dada frame) >submydf = subset(mydf, a > 1 & b <= a), here column a contains values ranging from 0.01 to 100000. I want to extract only those matching condition 1 i.e a > . But when i execute this command it is

Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword "remove AND library" but got timed out.(why?)....) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R

Hi, I am using 'best.arima' function from forecast package to obtain point forecast for a time series data set. The documentation says it utilizes AIC value to select best ARIMA model. But in my case the sample size very small - 26 observations (demand data). Is it the right to use AIC value for model selection in this case. Should I use AICc instead of AIC. If so how can I modify

Hi Sachin, Yes. Bernhard Scholz and I have just discussed a fix for this. I hope to commit it in the next few days. I will let you know as soon as it goes in to the mainline. Regards, Lang. On Tue, Dec 15, 2009 at 5:34 PM, <Sachin.Punyani at microchip.com> wrote: > Hi Lang, > > Thanks for your inputs on the problem. I was just curious to know if you > got any opportunity to