similar to: Identifying US holidays

John Putz wrote: > Thanks. > > */ > email: /**/johnputz3655@yahoo.com/* <mailto:johnputz3655@yahoo.com> > */home: 206-632-6522 > cell: 206-910-5229/* > > > ----- Original Message ---- > From: Joe W. Byers <ecjbosu@aol.com> > To: John Putz <johnputz3655@yahoo.com> > Cc: r-sig-finance@stat.math.ethz.ch > Sent: Friday, December 7, 2007

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

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! 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 dates for the beginning of each week, e.g.: weekly<-data.frame(week=seq(as.Date("2010-04-01"), as.Date("2011-12-26"),by="week")) week # each week starts on a Monday I also have a vector of dates I am interested in, e.g.: july4<-as.Date(c("2010-07-04","2011-07-04")) I would like to flag the weeks in my weekly$week that

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! 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! 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

Hi One of the most important calculation in applied finance is the number of days between dates. That kind of calculus become annoying when a specific calendar must be used. That is the case for the business days calculus. The package timeDate has a function isBizday to perform that kind of thing. The problem is that using this function to calculate the number of business days between dates has

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"

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

Hello everybody, it looks like the presense of some (do know know which) S4 methods for a given S4 class degrades the performance of xtfrm (used in 'order' in new R-devel) by a factor of millions. This is for classes that ARE derived from numeric directly and thus should be quite trivial to convert to numeric. Consider the following example: setClass("TimeDateBase",

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 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?

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,

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