similar to: fridays date to date

This is mostly a RFC [but *not* about the many extra packages, please..]: Noticing to my chagrin how my students work in a project, googling for R code and cut'n'pasting stuff together, accumulating this and that package on the way all just for simple daily time series (though with partly missing parts), using chron, zoo, lubridate, ... all for things that are very easy in base R *IF*

Wanted to raise two questions: 1. Is bugs.r-project.org down? I haven't been able to reach it for two or three days: ``` ping bugs.r-project.org PING rbugs.research.att.com (207.140.168.137): 56 data bytes Request timeout for icmp_seq 0 Request timeout for icmp_seq 1 Request timeout for icmp_seq 2 Request timeout for icmp_seq 3 Request timeout for icmp_seq 4 Request timeout for icmp_seq 5

On converting character variables to ordered factors, regression result has strange names. Is it possible to obtain same variable names with and without intercept? Thanks, Naresh mydf <- data.frame(date = seq.Date(as.Date("2024-01-01"), as.Date("2024-03-31"), by = 1)) mydf[, "wday"] <- weekdays(mydf$date, abbreviate = TRUE) mydf.work <- subset(mydf, !(wday

I''m having difficulty figuring out the ruby neccessary to do this so I thought I would try the list. I''m hoping to get a date_select to only display certain days (specifically only fridays) but rails doesn''t seem to have the flexibility so I''ve been going at it in ruby to no avail. I''ve looked on google but couldn''t find it (because it

R-Help Please can you tell me if there is a function within R that will return the day of any given date? Using 19 March 2006 as an example: given 20060319 (or any other date format) will return the day Sunday. Many thanks Fiona.

I have a xts object made of daily closing prices I have acquired using quantmod. Here is my code: library(xts) library(quantmod) library(lubridate) # Gets SPY data getSymbols("SPY") # Subset Prices to just closing price SP500 <- Cl(SPY) # Show day of the week for each date using 2-6 for monday-friday SP500wd <- wday(SP500) # Add Price and days of week together

Running R 2.13.1 on Windows XP. I would like to get week of the year (1-52) for each date. library(chron) dts <- dates(c("02/27/92", "02/27/92", "01/14/92","02/28/92", "02/01/92")) dts dts.chron <- as.chron(dts) dts.chron class(dts.chron) # all of these component extractions work: months(dts.chron) weekdays(dts.chron) years(dts.chron)

Hello, I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to get every quarter period end date between those dates? Does anyone know how to do this? Speed is important... Here is a small sample: Two dates: "2007-01-31" "2012-01-31" And I'd like to get this: [1] "2007-03-31" "2007-06-30"

Hello, I'm using package RpgSQL. Is there a better way to create a multi-line query/character string? I'm looking for less to type and readability. This is not very readable for large queries: s <- 'create table r.BOD("id" int primary key,"name" varchar(12))' I write a lot of code, so I'm looking to type less than this, but it is more readable from

Hi, I wanted to learn how to solve a date and time manipulation where i can do the following two 1. difference of two dates eg (differnce between 5th jan 2013 and 1st jan 2013) 2.Suppose i have week number of the year, i want to know if i can find out the day it refers to eg( say week 2 of 2013 would be 6th jan 2013 and the day is sunday) i need my result to tell me that its

Dear R-users, I have a factor variable within my data frame which I derive week after week from a POSIXct variable using the cut(var,"weeks") command I have found in the chron package. The levels() command gives me: [1] "2009-03-30 00:00:00" "2009-04-06 00:00:00" "2009-04-13 00:00:00" "2009-04-20 00:00:00" "2009-04-27 00:00:00"

Hello, I have a nasty loop that I have to do 11877 times. The only thing that slows it down really is this merge: xx1 = merge(dt,ua_rd,by.x=1,by.y= 'rt_date',all.x=T) Any ideas on how to speed it up? The output can't change materially (it works), but I'd like it to go faster. I'm looking at getting around the loop (not shown), but I'm trying to speed up the merge first.

Hello, Thanks in advance for any help! How do I pass an unknown number of objects into the "..." (dot dot dot) parameter? Put another way, is there some standard way to pass multiple objects into "..." to "fool" the function into thinking the objects are passed in separately/explicitly with common separation (like "x,y,z" when x, y and z are objects to be

Have you noticed any problems with big dates (>=1/1/2040) in R? Here is the bit of code that I'm having trouble with: > test.date <- strptime("1/1/2040",format="%m/%d/%Y") > > unlist(test.date) sec min hour mday mon year wday yday isdst 0 0 0 1 0 140 0 0 0 > > date.plus.one <- as.POSIXct(test.date) +

Hello, Does anyone know how to get the rounding method used for the axis tick numbers/values in plot()? I'm using mtext() to plot the values used to plot vertical and horizontal lines (using abline()) and I'd like these vertical and horizontal line values to be rounded like the axis tick values are rounded. In other words, I want numbers plotted with mtext() to be rounded in the same

Hello, I'd like to understand 'what' is predicting the response for library(mgcv) gam? For example: library(mgcv) fit <- gam(y~s(x),data=as.data.frame(l_yx),family=binomial) xx <- seq(min(l_yx[,2]),max(l_yx[,2]),len=101) plot(xx,predict(fit,data.frame(x=xx),type="response"),type="l") I want to see the generalized function(s) used to predict the response

I''m using the date_select and datetime_select helpers in my view, and they return Date classes from the params hash. But how do I work with Date classes, they don''t print human readable dates or times, Time classes work well I can use strftime("%H:%M") to print to the screen. Is it possible to convert a Date to a Time, Ive been tinkering in irb but have got

Dear R-Gurus I have a data frame (from CSV file) which has its first column called Date. The Date is in the format mm/dd/yyyy. I was trying to get the weekday for these dates and I tried using wday() and day.of.week() functions and both of them gave me precisely the wrong answers. I think the issue lies in the proper formatting of dates. The class of this column is a factor class and hence I

Hello, Is there a faster way to do this? Basically, I'd like to NA all values in all_data if there are no 1's in the same column of the other matrix, iu. Put another way, I want to replace values in the all_data columns if values in the same column in iu are all 0. This is pretty slow for me, but works: all_data = matrix(c(1:9),3,3) colnames(all_data) =

Hello, What is the best way to get ranks for a vector of values, limit the range of rank values and create equal count in each group? I call this uniform ranking...uniform count/number in each group. Here is an example using three groups: Say I have values: x = c(3, 2, -3, 1, 0, 5, 10, 30, -1, 4) names(x) = letters[1:10] > x a b c d e f g h i j 3 2 -3 1 0 5 10 30 -1 4 I