similar to: identifying weeks (dates) that certain days (dates) fall into

Hello! I am trying to identify which ones of a vector of dates are US holidays. And, ideally, which is which. And I do not know (a-priori) which dates those should be. I have, for example: x<-seq(as.Date("2011-01-01"),as.Date("2011-12-31"),by="day") (x) I think chron should help me here - but maybe I am not using it properly: library(chron) is.holiday(chron) #

Dear R-ers, I have a list of data frames such that the length of the list is unknown in advance (it could be 1 or 2 or more). Each element of the list contains a data frame. I need to loop through all rows of the list element 1 AND (if applicable) of the list element 2 etc. and do something at each iteration. I am trying to figure out how to write a code that is generic, i.e., loops through the

Hello! I have a data frame with a factor and a numeric variable: x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200)) For each level of "factor" - I would like to divide each value of "values" by the mean of "values" that corresponds to the level of "factor" In other

Hello! I have a list of variable length. One example is: X=vector("list",3) X[[1]]=1:2 X[[2]]=1:2 X[[3]]=1:2 How could I run expand.grid on the elements of X so that the results would be the same as expand.grid(1:2,1:2,1:2)? Thank you! Dimitri -- Dimitri Liakhovitski gfk.com <http://marketfusionanalytics.com/> [[alternative HTML version deleted]]

Hello! What function would allow me to "round" down, rather than up? For example, x<-1.98 I'd like to get 1.9 - rather than 2.0. Reason - I am creating a minimum for an axis for a plot, and I need it to be lower than x (which, in turn, is the lowest number already). Thank you! -- Dimitri Liakhovitski marketfusionanalytics.com

* I am currently reading in a series of files, applying the same functions to them one at a time, and then merging the resulting data frames e.g.: >MyRows <- c("RowA", "RowB", "RowC")>>File1_DF <- read.delim("\\\\DirectoryToFiles\\File1_Folder\\File1.txt", stringsAsFactors=FALSE, check.names=FALSE)>File1_DF <-

Hello! I have a data frame with some NAs. test<-data.frame(a=c(1,2,NA),b=c(10,NA,20)) I need to sum up values in 2 variables. However: test$a+test$b procudes NAs in rows that have NAs. How could I sum up columns while ignoring NAs (the way the function sum(..., na.rm=T) works? Thank you! -- Dimitri Liakhovitski marketfusionanalytics.com

Hello! I have a large data frame x: x<-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has 1000 rows x$item<-as.character(x$item) I also have a small data frame y with just 1 row: y<-data.frame(item="f",a=3,b=10) y$item<-as.character(y$item) I have to decide if y$a is larger than the smallest of all the values in x$a. If it is, I want y to replace the whole

Hello! # I have a list with several data frames: mylist<-list(data.frame(a=1:2,b=2:3), data.frame(a=3:4,b=5:6),data.frame(a=7:8,b=9:10)) (mylist) # I want to grab only one specific column from each list element neededcolumns<-c(1,2,0) # number of the column I need from each element of the list # Below, I am doing it using a loop: newlist<-NULL for(i in 1:length(mylist) ) {

Hello! I have a data set: set.seed(123) y<-data.frame(week=seq(as.Date("2010-01-03"), as.Date("2011-01-31"),by="week")) y$var1<-c(1,2,3,round(rnorm(54),1)) y$var2<-c(10,20,30,round(rnorm(54),1)) # All I need is to create lagged variables for var1 and var2. I looked around a bit and found several ways of doing it. They all seem quite complicated - while in

Hello! It's a shoot in the dark, but I'll try. If one has a total of 100 (e.g., %), and three components of the total, e.g., mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with 3 sectors representing x, y, and z according to their proportions in the total. I am wondering if it's possible to build something very similar, but not on a circle but in a square - such that

Hello! I am trying to create an R optimization routine for a task that's currently being done using Excel (lots of tables, formulas, and Solver). However, otpim seems to be finding a local minimum. Example data, functions, and comparison with the solution found in Excel are below. I am not experienced in optimizations so thanks a lot for your advice! Dimitri ### 2 Inputs:

Hello! I have 2 maps - both created in ssplot and both identical in terms of outline. Is there any way to superimpose Map1 (which has black borders between Canadian provinces) onto Map2 (which is also a map of Canada)? Thanks a lot for your hints! Dimitri ### A. Reading in Canada data at the province and then at the county level: library(raster) getData('ISO3') # Canada's code is

Hello! I have a data frame with several rows, for example: x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) I would like to find y - a data frame that only has the unique rows from x, i.e.: 1,2,3 1,2,2 1,1,1 Thanks a lot for your hints! Dimitri -- Dimitri Liakhovitski gfk.com <http://marketfusionanalytics.com/> [[alternative HTML

Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer! -- Dimitri Liakhovitski marketfusionanalytics.com

Hello! Hope you could help me split the strings. I have a set of strings: x<-c("name_a1_2.5.o","name_a2_2.53.o","name_a3_bla_1.o") I need to extract from each string: 1. Its unique part that comes before the last "_", i.e.: "a1","a2","a3_bla". 2. The part that comes after the last "_" and before ".o"

Dear everybody, I have the following challenge. I have a data set with 2 subgroups, dates (days), and corresponding values (see example code below). Within each subgroup: I need to aggregate (sum) the values by week - for weeks that start on a Monday (for example, 2008-12-29 was a Monday). I find it difficult because I have missing dates in my data - so that sometimes I don't even have the

Dear R-ers, I have 3 vectors - x, y, and z and want to plot a surface (z as the 3rd dimension): x<- c(-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1) y<- c(-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1) z<- c(0.226598762, 0.132395904, 0.14051906, 0.208607098, 0.320840304, 0.429423216, 0.54086732, 0.647792527, 0.256692375, 0.256403273, 0.172881269, 0.121978079,

Hello! I have run a simple regression - lm and created a regression object "myreg". I can see all the coefficients when I print(myreg). Then I tried to run vif(myreg) from the package "car". However, it's giving me an error: in vif.lm(regr.f) : there are aliased coefficients in the model Very sorry for my question: Is there any way to get the vif's for all predictors?

Apologies for a naive question: Can R be installed and run on a server (operating system Windows Server 2008)? Thank you! -- Dimitri Liakhovitski