Displaying 20 results from an estimated 600 matches similar to: "Basic date time arithmetics operations"
2006 May 29
4
Install Other OS under DOMU
Hello,
Anbody has already installed one version fedora 3 or 2 under DomU ?
I have dom0 kernel xen vesion 3 under fedora core 4 and i would like install other version on domU.
Thanks by avance
Eric
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2003 Nov 14
1
What goodness-of-fit measure for robust regression ?
Hi,
i. After estimating some coefficients using robust regression with rlm() or lqs(), I wonder if there exist some measures of the goodness-of-fit as those for standard linear model(R2)... or evenly if it's a statistics non-sense to look for since I do not find any mention of that in differents chapters on robust and resistant regression or in severals R documentation (Fox, Ripley and
2003 Nov 19
5
ISOdate returns incorrect date?
Dear all,
I have found the following (for me) incomprehensible behaviour of
ISOdate (POSIXct):
> ISOdate(1900,6,16)
[1] "1900-06-15 14:00:00 Westeurop?ische Sommerzeit"
> ISOdate(1950,6,16)
[1] "1950-06-16 14:00:00 Westeurop?ische Sommerzeit"
Note that in the first case I get the 15th of June back, not the 16th as
I would have expected!
This happened under R-1.7.1 on
2006 Apr 06
1
Bell Canada Requests $987.14 Rate increase 911 / VOIP Providers
From the bend me over news department.
2 March 2006
Mr. Leonard Katz
Executive Director
Broadcasting and Telecommunications
Canadian Radio-television and
Telecommunications Commission
Ottawa, Ontario
K1A 0N2
Dear Mr. Katz:
Associated with Bell Canada Tariff Notice No. 6929
1. Attached for the Commission's approval are proposed revisions to
Bell Canada's Access Services Tariff Item
2003 Nov 14
5
ISOdate() and strptime()
Dear R-people!
I am using R 1.8.0, under Windows XP.
While using ISOdate() and strptime(), I noticed the following behaviour when
"wrong" arguments (e.g., months>12) are given to these functions:
> ISOdate(year=2003,month=2,day=20) #ok
[1] "2003-02-20 13:00:00 Westeurop?ische Normalzeit"
> ISOdate(year=2003,month=2,day=30) #wrong day, but returns a value
[1]
2007 Mar 22
2
difftime / RBloomberg
hi,
I've troubles with some difftime objects. e.g.
ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) - 2
works, telling me "Time difference of 57 days". But when I'd like to add
days, such as
ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) + 2
the function gives me an error. Function "as.COMDate.chron" of the
Rbloomberg package doesn't work for that reason.
I'm
2007 May 10
3
Getting the last day of the month.
Hi,
Given a date, how do I get the last date of that month? I have
data in the form YYYYMM, that I've read as a date using
> x$Date <-
as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1))
But this gives the first day of the month. To get the last day of the
month, I tried
> as.Date(as.yearmon(x$Date,frac=0))
But I don't get the last day of the month here. (Tried
2002 May 21
1
I() fails on objects of class POSIXct (PR#1587)
Although the documentation is somewhat sketchy, I() can be used to create
objects of class AsIs:
> I("a")
[1] "a"
attr(,"class")
[1] "AsIs" "character"
> I(4)
[1] 4
attr(,"class")
[1] "AsIs" "numeric"
> I(4 + 0i)
[1] 4+0i
attr(,"class")
[1] "AsIs" "complex"
>
This
2006 Apr 10
3
timeAlign
I use POSIXct for datetimes. Is thee a timeAlign function that I can
use where :
align by year
direction -1 ==> start of this year
direction 1 ==> start of next year
align by week
direction -1 ==> date on last sunday
direction 1 ==> date on next sunday
align by day
direction -1 ==> time at past midnight
direction 1 ==> time at this comming
2018 Jan 22
2
Manipulating two large dataset differing by date and time
Dear Members,
Compliments of the Season!!
Below is a part of a code I use for Fourier analysis of signals. The code
handles data with the format 05 01 01 8628 (year, month, day and count)
05 01 02 8589 (year, month, day and count)
The sample data is attached as 2005daily.txt.
I would like to adapt the code to handle data of the form:
05 01 01 00 4009
2002 May 28
2
histogramming dates
I'd like to make a plot showing frequency of an event. The data
is in a data from that includes Year, Month and Day (of month)
fields, so I created a Date with ISOdate(Year, Month, Day,
tz=''). I can plot frequencies for the year 2002 with
> thisyear <- Date[Year==2002]
> hist( thisyear, xaxt='n' )
> axis.POSIXct( 1, at=seq(min(thisyear), max(thisyear),
2003 Sep 12
2
Sorting a vector by date
Hello out there....
Again I have a problem and I stuck...
How can I sort a vector of dates?
For example I have the vector
a<-ISOdate(2001, 1, 1) + 70*86400*runif(10)
How can this vector be sorted chronological?
And what's the function I should work with to handle these entries?
(in sense of: which(a>2001-01-04) or somehting like that)
Thank you for helping
M.Kirschbaum
2004 Nov 03
3
cut POSIX results in NA - bug?
Dear all
I try to make hourly average by cut() function, which almost works
as *I* expected. What puzled me is that if there is only one item at
the end of your data it results in NA.
Example will explain what I mean
datum<-seq(ISOdate(2004,8,31), ISOdate(2004,9,1), "min")
cut(datum[1370:1381],"hour", labels=F)
[1] 1 1 1 1 1 1 1 1 1 1 1 NA
2003 Apr 30
2
ylab in plot.POSIXct
I am using R-1.7.0 and have some data which consist of one vector of
numbers and a second corresponding vector of dates belonging to the
POSIXct class. I would like to plot the numbers against the dates.
What is the best way to do this?
It almost works to just call `plot.' However if I do this while using
the `ylab' parameter I get a warning message:
parameter "ylab"
2016 Apr 18
4
as.Date
Dear All,
I have a data set containing year, month, day and counts as shown below:
data <- read.table("data.txt", col.names = c("year", "month", "day", "counts"))
Using the formula below, I converted the data to as date and plotted.
new.century <- data$year < 70
data$year <- ifelse(new.century, data$year + 2000, data$year + 1900)
2018 Jan 22
0
Manipulating two large dataset differing by date and time
Hi Ogbos,
You can just use ISOdate. If you pass more values, it will process them:
ISOdate(2018,01,22)
[1] "2018-01-22 12:00:00 GMT"
> ISOdate(2018,01,22,18,17)
[1] "2018-01-22 18:17:00 GMT"
Add something like:
if(is.null(data$hour),data$hour<-12
then pass data$hour as it will default to the same value as if you
hadn't passed it.
Jim
On Mon, Jan 22, 2018 at 6:01
2003 Sep 07
4
data manipulation
Hi,
I am new to R, coming from a few years using Stata. I've been twisting my
brain and checking several R and S references over the last few days to
try to solve this data management problem: I have a data set with a unique
patient identifier that is repeated along multiple rows, a variable with
month of patient encounter, and a continous variable for cost of
individual encounters. The data
2008 Jan 02
3
Find missing days
Hi,
I have a data.frame like this:
y <- rnorm(60)
lev <- gl(3,20, labels=paste("lev", 1:3, sep=""))
date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5),
by=60*60*24))
date1 <- date1[-c(3,4,15,34,38,40)]
df <- data.frame(lev=lev, date1=date1, y=y)
I would like to produce a new data.frame with missing days in df$date1
in each df$lev, like this:
lev
2002 Mar 08
1
1st January isn't (PR#1370)
In today's pre1.5.0
> ISOdate(2002,1,1)
[1] "2002-03-01 04:00:00 PST"
> ISOdate(2002,1,1)==ISOdate(2002,3,1)
[1] TRUE
It doesn't seem to happen for days other than 1/1
-thomas
> version
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status Under development (unstable)
major 1
minor 5.0
year 2002
month
2010 Jul 15
2
How to plot a histogram of weekday frequencies in a list of dates?
Question from an [R] novice...
Hi,
I have a vector of date/times like the one shown below (a truncated
sample of a much longer list...)
> dates[1:4]
[1] "2006-03-16 08:41:00" "2006-03-16 10:28:00" "2006-03-16 11:03:00"
[4] "2006-03-16 11:04:00"
I would like to generate a weekday histogram showing the frequency of
dates falling on each weekday. I know