Displaying 8 results from an estimated 8 matches for "daynum".
2011 May 19
1
Creating a "shifted" month (one that starts not on the first of each month but on another date)
...some package - that makes it
easier/more elegant?
Thanks a lot!
# Example data:
mydf<-data.frame(mydate=seq(as.Date("2011-01-01"), length = 92, by = "day"))
(mydf)
### Creating a new variable that has one value before
### the 20th of each month and next value after it
mydf$daynum<-as.numeric(format(mydate,"%d"))
library(zoo)
mydf$yearmon<-as.yearmon(mydf$mydate)
(mydf); str(mydf)
mydf$newfactor<-NA
for(i in unique(mydf$yearmon)){ # looping through "yearmon" (important
because true data has many years of data)
tempdf<-mydf[mydf$yearmon == i,...
2024 Dec 02
1
Identify first row of each ID within a data frame, create a variable first =1 for the first row and first=0 of all other rows
...ata;
> by lat_lon daytime; /* lat long is a string giving latitude and longitude */
> run;
>
> 3) For each geographic area determine the records that mark the start and stop of each of the four days, let's say from 00:00 hrs to 23:59 hours, and create a variable, daynum that indicates the day number (valid values 1 to 4). I do not know how to accomplish this in R. This is the part of my analysis that I asked the R community to help me write. I know how this this can be easily done in SAS:
> * Arrange data date and time within each geographic region (i.e. lat_lo...
2020 Feb 17
2
Colocar objeto Dates dentro de un vector.
...s vectores "original" en un data.frame y luego
> convertirlo a fechas.
>
> Una vez tienes estas fechas, puedes hacer cálculos o extraer otras
> variables.
>
>
> > original<-c(2019,308,1700, 25)
> > df <- data.frame(
> + Year = original[1],
> + DayNum = original[2],
> + Hour = original[3]
> + )
> > my_fecha <- strptime( paste(df$Year, df$DayNum, df$Hour, sep = ""),
> "%Y%j%H" )
> > my_fecha
> [1] "2019-11-04 17:00:00 CET"
> > my_fecha <- strptime( paste(original[1], original[2],...
2010 Jun 29
1
Problem with GoToIfTime
Hello list,
why is it that GoToIfTime thinks a date of **|*|29-*|jun *is not valid ??
[Jun 29 14:06:34] -- Executing [s at macro-vac:10]
*GotoIfTime*("SIP/testcorp-00000036", "**|*|29-*|jun*?onvac") in new stack
[Jun 29 14:06:34] WARNING[3076]: pbx.c:4127 get_range: Invalid end day
'*', assuming none
[Jun 29 14:06:34] -- Executing [s at macro-vac:11]
2020 Feb 15
2
Colocar objeto Dates dentro de un vector.
Hola,
Estoy encontrando un problema al intentar poner un objeto Dates en un
vector o dataframe.
Mi ejemplo
# preliminares
install.packages( lubridate ); library( lubridate )
v <- c(0, 0, 0)
original<-c(2019,308,1700, 25) # c(año, día del año, hora, temperatura)
esto sale así de un sensor de temperatura
# convertimos los datos originales en algo que entienda R
a <-
2020 Feb 24
2
Colocar objeto Dates dentro de un vector.
...s.
>> >
>> > Una vez tienes estas fechas, puedes hacer cálculos o extraer otras
>> > variables.
>> >
>> >
>> > > original<-c(2019,308,1700, 25)
>> > > df <- data.frame(
>> > + Year = original[1],
>> > + DayNum = original[2],
>> > + Hour = original[3]
>> > + )
>> > > my_fecha <- strptime( paste(df$Year, df$DayNum, df$Hour, sep = ""),
>> > "%Y%j%H" )
>> > > my_fecha
>> > [1] "2019-11-04 17:00:00 CET"
>> >...
2017 Jul 06
1
Convert date to continuous variable in R
Thanks it worked for me. I wanted to plot days since planting on
x-axis 1 and years on x-axis 3.
LAI_simulation$Date <- as.Date( LAI_simulation$Date, '%Y/%m/%d')
LAI_simulation$Date <- as.integer(LAI_simulation$Date - as.Date("2009-10-07"))
plot(LAI~Date,data=LAI_simulation,xlab="Days since Oct, 7,
2012 Mar 12
8
CentOS Server Backup Options
What do you guys recommend for backing up a small CentOS server in a
business environment. It will have (3) 300gb drives in a raid 5 array but I
don't anticipate more than about 25gb of data that needs to be backed up
each night.
I want a lot of backups with a rotation scheme that included daily, weekly,
and monthly copies. I want the daily copies of the data kept until the next
week, and