similar to: transforming year.weeknumber into dates

Hello! If I have in my data frame MyFrame a variable saved as a Date and want to translate it into years, I currently do it like this using "zoo": library(zoo) as.year <- function(x) as.numeric(floor(as.yearmon(x))) myFrame$year<-as.year(myFrame$date) Is there a function that would do it directly - like "as.yearmon" - but for years? Thank you! -- Dimitri

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Dear R-ers! I have a badly organized data base in Excel. Once I read it into R it looks like this (all variables become factors because of many spaces and other characters in Excel):

Hello! I'd like to take Dates and extract from them months and years - but so that it sorts correctly. For example: x1<-seq(as.Date("2009-01-01"), length = 14, by = "month") (x1) order(x1) # produces correct order based on full dates # Of course, I could do "format" - but this way I am losing the "Date" quality of the data:

?hanks for the advice, Jeff. Will keep it in mind. But I am anal - I shy away from using letters and words that "look familiar" to me in R (such as mean, sd, T, etc.) But still, it's a good advice. On Thu, Jul 27, 2017 at 11:53 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > I think you should be more suspicious of yourself, Dimitri. A letter T > variable can

I think you should be more suspicious of yourself, Dimitri. A letter T variable can easily arise in the problem domain when you are not thinking of logical values at all, at which point your cavalier use of T as a synonym for TRUE can suddenly become a bug. -- Sent from my phone. Please excuse my brevity. On July 27, 2017 8:18:03 AM PDT, Dimitri Liakhovitski <dimitri.liakhovitski at

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

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

Just a thought: Did you try na.rm = TRUE in case you have an object named "T" in scope? -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Thu, Jul 27, 2017 at 7:49 AM, Dimitri Liakhovitski <dimitri.liakhovitski at

Thank you, Bert! I do NOT have an object named "T" in scope (I checked - and besides, it would never occur to me to use this name). TRUE or T results in the same unexpected behavior: ggplot(data = md, mapping = aes(x = a)) + geom_bar(na.rm = TRUE) On Thu, Jul 27, 2017 at 10:57 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Just a thought: > > Did you try

To clarify: my question is not about "who could I exclude NAs from being counted" - I know how to do that. My question is: Why na.rm = T is not working for geom_bar in this case? On Thu, Jul 27, 2017 at 8:24 AM, Dimitri Liakhovitski < dimitri.liakhovitski at gmail.com> wrote: > Hello! > > I am trying to understand how ggplot2's geom_bar treats NAs. > The help file

I suspect this is by design. Questions about "why" should probably cc the contributed package maintainer(s). -- Sent from my phone. Please excuse my brevity. On July 27, 2017 7:49:47 AM PDT, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote: >To clarify: my question is not about "who could I exclude NAs from >being >counted" - I know how to do

Hello! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

Hello, I have a matrix that is a product of tapply on a larger data set. Let's assume it looks like this: X<-matrix(c(10,20,30,40,50,60),2,3) dimnames(X)<-list(c("1","2"),c("1","2","3")) (X) 1 2 3 1 10 30 50 2 20 40 60 Is there an efficient way of transforming this matrix into the following matrix: rows columns entries 1

Hello! I am using function "curve" to create a line graph. I was wondering, if it's possible to "turn off" the default tick marks and introduce those tick marks in specific locations. For example, currently in my X axis tick marks are (automatically) at 10, 11, 12, 13 but I want them to be in 5 specific locations, like 9.89, 10.34, etc. Any hint would be greatly

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g., x<-c(1,2,3,4). How can I get a TRUE or FALSE for whether x contains a 2? -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello everyone! I have a list X with 3 elements, each of which is a data frame, for example: a<-data.frame(a=1,b=2,c=3) b<-data.frame(a=c(4,7),b=c(5,8),c=c(6,9)) c<-data.frame(a=c(10,13,16),b=c(11,14,17),c=c(12,15,18)) X<-list() X[[1]]<-a X[[2]]<-b X[[3]]<-c (X) How can I most effectively transform X into a data frame with columns a, b, and c? I would love to find a generic

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello. Until today I've been using R2.9 and since today R2.10 (on a PC). In both of them it takes about 20 sec for the prompt to appear IN R console after I start R. And every time it says: "Previous saved work space restored" - even if I have not saved any workspace or, in case of R2.10 - even though I have not used it once. In the older versions - R would start within 2-3 sec. Is