similar to: Keep date, strip time?

I've passed a data.frame as an input into a function which does some plotting on a 4x4 matrix of certain parameters within in the data.frame. I'd like to add a small header on top of each plot with the name of the data.frame so that it's clear as I compare these 16 things where each on came from. So far I haven't found the right way to get the name of the data.frame as a string

Dear R guru: If I got a variable aaa<- "up.6.11(16)" how can I extract 16 out of the bracket? I could use substr, e.g. substr(aaa, start=1, stop=2) [1] "up" But it needs start and stop, what if my start or stop is not fixed, I just want the number inside the bracket, how can I achieve this? Many thanks yan

Hi All, I've been struggling on that one for a couple of days. I've tried several things, but nothing is worth publishing. I'm working with a bird nests data frame for which I have a laying date (LD) and a fledgling date (FD) (in calendar Julian date) by nest id. For the period spanning between LD and SD, the nest is termed "active". Each nest id occur once in the data

Hi there I have a string like this i want to extract 9831019 from this string i used a regular expresion \d+ by which i can only make it to see 7 and returns. This type of number(9831019) appears in any part of the string and is definitely more than 5 digits all the time and i want to give that as a condition UV7C11-F9-E1 MCS#9831019 MCS Lot #9512516" how do i go abt it Ramya -- View

Hi, I'm trying to get better at things like lapply but it still stumps me. I have a function I've written, tested and debugged using individual calls to the function, ala: ResultList5 = DoAvgCalcs(IndexData, Lookback=5, SampleSize=TestSamples , Iterations=TestIterations ) ResultList8 = DoAvgCalcs(IndexData, Lookback=8, SampleSize=TestSamples , Iterations=TestIterations ) ResultList13

Hello, today I've updated on the newest R-Version. But sadly a function I needed didnt want to work: The input is e.g. days(as.Date("21-07-2005","%d-%m-%y")) the error is: Fehler in Math.Date(dts): floor nicht definiert f??r Date Objekte (Error in Math.Date(dts): floor not defined for date objects) Same for year. Only months gives me the correct output. In Version

Hi, I'm trying to read some trajectory files (text files) which have the form: # trial n # t X Y 0 1 2 0.2 1 3 0.4 1.2 4 ... # trial n+1 # t X Y 0 1 2 0.2 1.3 3.3 0.4 1.5 5 ... ... where the symbol # means that the line is not a numeric value, but I still need to read the number of the trial (n=1, 2, etc.). I

I've always found the chron library to be useful for tasks like this: > x <- round(runif(10)*100000, digits=0) > y <- as.Date(x, origin="1970-01-01") > library(chron) > days(y) [1] 7 25 26 25 10 24 1 31 12 8 31 Levels: 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10 < 11 < 12 < 13 < ... < 31 Notice that it returns the days as

Hi, guys I don't understand why I can apply as.Date to a single item in the list: > as.Date(alldays[4]) [1] "29-03-20" but when I try to lapply as.Date to all the items, i got a sequence of neg numbers: > sapply(alldays[1:4], FUN=as.Date) 03-04-2012 02-04-2012 30-03-2012 29-03-2012 -718323 -718688 -708492 -708857 does anyone know what's wrong here? i am

Hello All: I am trying to generate a sequence of dates using library(chron). Using the following code, I got most of what I want, dts <- seq.dates("01/02/1998", "01/10/2001", by="day") dts <- dts[!is.weekend(dts)] a sequence of weekdays only (no weekend dates), with first observation appearing as 01/02/98 and the last 01/10/01. But I would

Dear all I use aggregate with variables of type numeric and dates. For type numeric functions, such as sum() are very fast, but similar simple functions, such as min() are much slower for the variables of type 'dates'. The difference gets bigger the larger the 'id' var is - but see this sample code: dts <- dates(c("02/27/92", "02/27/92",

Hi, I wonder if someone can suggest how to create a new data.frame Y from X where X$PL_Pos is summed by each unique X$MyDate. Y should end up with two (or more) columns Y$MyDate and Y$PL_Sum with its value being the cumsum of all the values in X for that date. - a 'daily cumsum'. Thanks, Mark TStoDate = function (TSDate) { X = strptime(TSDate + 19e6L, "%Y%m%d")

R community: Hello, I would to like to convert a character date variable from %m/%d/%Y to %m/%d/%y. Any advice would be greatly appreciated. I have tried functions for changing the formatting and removing the unnecessary digits without success. Mike

Hi, Can strptime (or some other function) help me turn the following column of a data.frame into two new columns, one as date and the other as time, preserving the AM/PM value? Thanks, Mark > B ENTRY DATE 1 3/23/2009 6:30:00 AM 2 3/23/2009 6:30:00 AM 3 3/23/2009 6:39:00 AM 4 3/23/2009 6:39:00 AM 5 3/23/2009 6:48:00 AM 6 3/23/2009 6:48:00 AM 7 3/23/2009 7:00:00 AM

Hi list, Could someone explain to me why the following result is not a integer? > difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004", >format="%d%b%Y"), units="days") Time difference of 195.0417 days I'm using R2.12.0 on WinXP. Thanks! ...Tao

Hi Can anyone explain why a date becomes numeric when you loop over a series of dates? > dt <- Sys.Date() > dt [1] "2017-11-14" > class(dt) [1] "Date" > dts <- dt - 1:0 > class(dts) [1] "Date" > > for (i in dts) { + print(i) + print(class(i)) + print(as.Date(i, "1970-01-01")) + print(class(as.Date(i,

I would like to use POSIX classes to store dates and extract components of dates. Following the example in Spector ("Data Manipulation in R"), I create a date > mydate = as. POSIXlt('2005-4-19 7:01:00') I then successfully extract the day with the command > mydate$day [1] 19 But when I try to extract the month > mydate$mon [1] 3 it returns the wrong month. And

https://bugzilla.netfilter.org/show_bug.cgi?id=926 Summary: icmp: ICMPv6 types are not supported Product: nftables Version: unspecified Platform: x86_64 OS/Version: Debian GNU/Linux Status: NEW Severity: normal Priority: P5 Component: nft AssignedTo: pablo at netfilter.org ReportedBy:

Hello, If I have a character string like d <- c("1990m3", "1992m8") #March 1990 and Aug 1992 what is the easiest way to convert it into any standard date form; for example, d <- c("01/03/1990", "01/08/1992") I looked at as.Date but it doesn't seem to address my problem as I have an "m" stuck in the middle of my character string

Hi, I am trying to find the min of day for each student in each year. Here is the dataset: date studentid year 1/1/05 6:07 AM 236 20082009 3/27/09 9:45 AM 236 20082009 4/29/09 8:44 AM 236 20082009 3/27/09 11:36 AM 310 20082009 4/1/09 10:43 AM 310 20082009 10/15/09 8:54 AM 310