similar to: Conditional Row Sum

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 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, 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 Sachin, llvm-dev, I've just committed a new PBQP solver which, among other things, should take care of this bug. Please let me know how it works out for you. Cheers, Lang. On Tue, Dec 15, 2009 at 5:54 PM, Lang Hames <lhames at gmail.com> wrote: > 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

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%

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

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 Lang, Thanks for your inputs on the problem. I was just curious to know if you got any opportunity to work on the solution for this. Regards Sachin > -----Original Message----- > From: llvmdev-bounces at cs.uiuc.edu [mailto:llvmdev-bounces at cs.uiuc.edu] On > Behalf Of Sachin.Punyani at microchip.com > Sent: Tuesday, November 17, 2009 12:00 PM > Subject: Re: [LLVMdev] Crash

Hi Sachin, Confirmed: This is being caused by a subtle issue in the heuristic PBQP solver. Specifically: R1/R2 reductions as currently implemented can, on rare occasions, lead to the heuristic solver failing to find a finite cost solution, even though one exists. The infinite cost solution will always be in violation of some rule of register allocation (failing to handle an interference, or

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 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, 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 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

Thanks Lang! I think we can use linear scan as work around for short term. Thanks for your help. Regards Sachin > -----Original Message----- > From: Lang Hames [mailto:lhames at gmail.com] > Sent: Sunday, November 15, 2009 10:08 AM > To: Sachin Punyani - I00202 > Cc: llvmdev at cs.uiuc.edu > Subject: Re: [LLVMdev] Crash in PBQP register allocator > > Hi Sachin, >

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, 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 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, 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

This looks like a bug in the PBQP solver. I'm currently investigating. Cheers, Lang. On Thu, Nov 12, 2009 at 12:46 AM, <Sachin.Punyani at microchip.com> wrote: > Hi, > > > > Please see the two “.ll’ files attached. > > > > Command line used > > llc –march=pic16  –pre-RA-sched=list-burr –regalloc=pbqp new.obc > > > > The above test case