similar to: Creation of a new data frame

Dear all, I currently have a data frame of dimensions 18556 rows by 19 columns. I want to convert this into a grid of dimensions 720 rows by 360 columns. The problem in this case is that not all rows in the initial data frame are complete (there are gaps). Therefore I am perhaps looking for a way of filling a 720 x 360 grid by reading in all values in each row until one is encountered which does

Dear R Users, I have a data frame of 360 rows by 720 columns (259200 values). For each value in this grid I am hoping to apply an equation to, to generate a new grid. One of the parts of the equation (called 'p') relies on reading from a separate reference table. This is Table 4 at: http://www.fao.org/docrep/s2022e/s2022e07.htm#3.1.3%20blaney%20criddle%20method (scroll down a little).

Hi, I need to resample rows in a data frame by subsets L3 <- LETTERS[1:3] d <- data.frame(cbind(x=1, y=1:10), fac=sample(L3, 10, repl=TRUE)) x y fac 1 1 1 A 2 1 2 A 3 1 3 A 4 1 4 A 5 1 5 C 6 1 6 C 7 1 7 B 8 1 8 A 9 1 9 C 10 1 10 A I have seen this used to sample rows with replacement d[sample(nrow(d), replace=T), ] x y fac 7 1 7 B 2

Hi, Excuse me for this simple question. How to convert as.data.frame to as.character? ?data.frame > L3 <- LETTERS[1:3] > L10 <- LETTERS[1:10] > d <- data.frame(cbind(x=c("XYZ"), y=L10), fac=sample(L3, 10, repl=TRUE)) > d x y fac 1 XYZ A A 2 XYZ B A 3 XYZ C A 4 XYZ D A 5 XYZ E B 6 XYZ F C 7 XYZ G A 8 XYZ H C 9 XYZ I B 10 XYZ

Dear Sir or Madam, I have the following data frame (which is just a toy example of my larger dataset) L3 <- LETTERS[1:3] x=c(1,1,2,2,3,3,4,4,5,5) y=1:10 d <- data.frame(cbind(x,y), fac=sample(L3, 10, replace=TRUE)) This data frame produces the following output x y fac 1 1 1 C 2 1 2 C 3 2 3 B 4 2 4 B 5 3 5 C 6 3 6 B 7 4 7 B 8 4 8 C 9 5 9 B 10 5

I have a question, which very easy to solve, but I can't find a solution. I want to convert a data frame to matrix. Here my toy example: > L3 <- c(1:3) > L10 <- c(1:6) > d <- data.frame(cbind(x=c(10,20), y=L10), fac=sample(L3, + 6, repl=TRUE)) > d x y fac 1 10 1 1 2 20 2 1 3 10 3 1 4 20 4 3 5 10 5 2 6 20 6 2 > is.data.frame(d) [1] TRUE > sapply(d,

Dear all, I have a grid (data frame) dataset at 0.5 x 0.5 degrees spatial resolution (720 columns x 360 rows; regular spacing) and wish to coarsen this to a resolution of 2.5 x 2.5 degrees. A simple calculation which takes the mean of a block of points to form the regridded values would do the trick. Values which should be excluded from the calculation are -9999 (unless all points within a block

Dear all, I have 30 arrays, each with dimensions 720,360,12. The naming format for each of these 30 objects is: mrunoff_5221, mrunoff_5222... mrunoff_5250. For example: > str(mrunoff_5221) ? num [1:720, 1:360, 1:12] NA NA NA NA NA NA NA NA NA NA ...? (the initial NA's are nothing to worry about) I am looking for a way by which I can extract each of the third dimension of these grids

Hi, It is easy to understand the types vector and frame. But I am wondering why the type factor is designed in R. What is the advantage of factor compare with other data types in R? Can somebody give an example in which case the type factor is much better than other data types? Regards, Peng

Hi Nick! In r207467 you added code(libunwind: DwarfInstructions.hpp): assert(lastReg <= (int)cieInfo.returnAddressRegister && "register range does not contain return address register"); for (int i = 0; i <= lastReg; ++i) { ..... else if (i == (int)cieInfo.returnAddressRegister) There is misprint here: lastReg should be >=

I'm trying to merge two data frames. One of them is a zero rows data frame. I'm using the merge parameter 'all.x = TRUE' so I'd expect to obtain all the rows of x. In fact the merge help says: all.x: logical; if 'TRUE', then extra rows will be added to the output, one for each row in 'x' that has no matching row in 'y'. These rows

Hi All, I have a doubt regarding join tables I''m having 2 models 1)Fac 2)Cont and both models have " has and belong to many" relationships so there are 3 tables 1)facs 2)conts 3)conts_facs then i''m fetching the data in controller as @conts=Cont.find(:all]) @cfacs=Fac.all(:joins=>:conts, :select=>"facs.name") but i dont know how to get the

Hi, I am trying to plot an interaction.plot with different color for each level of a factor. It has an erratic behavior. For example, it works for the first interaction.plot below, with the example from the ALDA book, but not with the other plots, from the NPK dataset: # from http://www.ats.ucla.edu/stat/r/examples/alda/ch2.htm tolerance <-

I found myself wanting to average a vector [vec] within each level of a factor [Fac], returning a vector of the same length as vec. After a while I realised that lm1 <- lm(vec ~ Fac) fitted(lm1) did what I want. But there must be another way to do this, and it would be good to be able to apply other functions than mean() in this way. Cheers, Murray -- Dr Murray Jorgensen

Hi, I am analyzing a data set with greater than 1000 independent cases (collected in an unrestricted manner), where each case has 3 variables associated with it: one, a factor variable with 0/1 levels (called XX), another factor variable with 8 levels (X) and a third response variable with two levels (Y: 0/1). I am trying to see if X1 has an effect on the relationship between X2 and the

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

Hello, I'm using SPSS at work but really would like to switch to R. Right now I'm trying to learn R in reproducing calculations I did with SPSS but am stuck with something that is quite simple and comprehensible in SPSS-Syntax: IF (variable1.fac = 0 AND variable2.num = 0) variable3=1. IF (variable1.fac = 0 AND variable2.num >= 1) variable3=2. IF (variable1.fac = 1 AND variable2.num =

Hi all, I noticed something strange when I ran aov and anova. vtot=c(7.29917, 7.29917, 7.29917) #identical values fac=as.factor(c(1,1,2)) #group 1 has first two elements, group 2 has the 3rd element When I run: > anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum Sq Mean Sq F value Pr(>F) fac 1 1.6818e-30 1.6818e-30 0.3333 0.6667 Residuals 1

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Hello everyone, Quoting from Samba Team Blog #2 (25 Sept 2009): "Volker showed how to get more than 700MB/sec from Samba using smbclient and a modern Samba server, which shows what you can really do when you understand the protocol thoroughly and don't feel you have to invent a new one (SMB2 :-)." Would it be possible to get a complete accounting of how this was achieved? Thanks,