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

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, 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, 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 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, 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)

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, 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

...xtx <- xtx + crossprod(cbind(1, x[, -1])) xty <- xty + crossprod(cbind(1, x[, -1]), x[, 1]) } close(f) solve(xtx, xty) coef(lm.fit(cbind(1, dat[,-1]), dat[,1])) ## check result unlink("c:/temp/big.txt") ## clean up. Andy -----Original Message----- From: Sachin J [mailto:sachinj.2006@yahoo.com] Sent: Monday, April 24, 2006 5:09 PM To: Liaw, Andy; R-help@stat.math.ethz.ch Subject: RE: [R] Handling large dataset & dataframe [Broadcast] Hi Andy: I searched through R-archive to find out how to handle large data set using readLines and other related R functions. I coul...

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, 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, 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 accomplish this in R. I have a Dataframe with 3 columns. Column B and C have same elements. But column A has more elements than B and C. I want to compare Column A with B and do the following: If A is not in B then insert a new row in B and C and fill these new rows with B = A and C = 0. So finally I will have balanced dataframe with equal no of rows (entries) in

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

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, 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, I am trying to use rm.outlier method but encountering following error: > y <- rnorm(100) > rm.outlier(y) Error: Error in if (nrow(x) != ncol(x)) stop("x must be a square matrix") : argument is of length zero Whats wrong here? TIA Sachin __________________________________________________ [[alternative HTML version

Hi, I am using parnor function of lmomco package. I believe it provides mean and std. dev for the set of data. But the std. dev provided does not match with the actual std. dev of the data which is 247.9193 (using sd function). Am I missing something here? > lmr <- lmom.ub(c(123,34,4,654,37,78)) > parnor(lmr) $type [1] "nor" $para [1] 155.0000 210.2130 >