similar to: Sum data according to date in sequence

Hi, I tried this: # extract date from the time stamp dt1 <- cbind(as.Date(dt$EndDate, format="%m/%d/%Y"), dt$EnergykWh) head(dt1) colnames(dt1) <- c("date", "EnergykWh") and my dt1 becomes these, the dates are replace by numbers. dt1 <- cbind(as.Date(dt$EndDate, format="%m/%d/%Y"), dt$EnergykWh) dput(head(dt1)) colnames(dt1) <-

Is this what you are after? library(tidyverse) library(lubridate) input <- structure(list(StationName = c("PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE

?s 01:49 de 03/11/2023, roslinazairimah zakaria escreveu: > Hi all, > > This is the data: > >> dput(head(dt1,20))structure(list(StationName = c("PALO ALTO CA / CAMBRIDGE #1", > "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", > "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", > "PALO ALTO

How about send a 'dput' of some sample data. My guess is that your date is 'character' and not 'Date'. Thanks Jim Holtman *Data Munger Guru* *What is the problem that you are trying to solve?Tell me what you want to do, not how you want to do it.* On Thu, Nov 2, 2023 at 4:24?PM roslinazairimah zakaria <roslinaump at gmail.com> wrote: > Dear all, > > I

Hi all, This is the data: > dput(head(dt1,20))structure(list(StationName = c("PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE

Hi r-users, I would like to draw line plots. However, the plot starts from 11121 data and plot data ENTRY last in the plot. Here is the code and data. datn <- read.table(header=TRUE, text=' LEVEL STATUS CGPA DIPLOMA ENTRY 3.32 DIPLOMA 11121 2.91 DIPLOMA 11122 2.90 DIPLOMA 12131 2.89 DIPLOMA 12132 2.89 DIPLOMA 13141 2.93 DIPLOMA 13142 2.96 DIPLOMA 14151 2.76 DIPLOMA 14152 2.73 STPM

Hi, Thanks for the reproducible example. Looking at str(datn) would give you a clue. STATUS is a factor because it contains character values. Factor levels by default are alphabetical with numbers first, but you can change those. > str(datn) 'data.frame': 36 obs. of 3 variables: $ LEVEL : Factor w/ 4 levels "DIPLOMA","MATRIC",..: 1 1 1 1 1 1 1 1 1 4 ... $

Dear r-users, I would like to construct 3-day moving average for block maxima series. I tried this: bmthree <- lapply(split(dt, dt$Year), function(x) max(sapply(1:(nrow(x)-2), function(i) with(x, mean(Amount[i:(i+2)],na.rm=TRUE))))) bmthree and got the following output. $`1971` [1] 70.81667 $`1972` [1] 68.94553 $`1973` [1] 102.7236 $`1974` [1] 73.6625 $`1975` [1]

I got advice here that I didn't understand! Can I ask to explain me the meaning of this procedure: first get the structure, and then assign it back. For what? Thanks!? (Great thanks to Moderator/Admin!) You should learn to post in plain text and use dput to present your data structures. At your console do this dput(treat) # and this will appear. Copy it to your plain-text message:

?s 05:11 de 13/01/2023, roslinazairimah zakaria escreveu: > Hi, > > I would like to customise my date series on the plot. I tried this: > > dt_ts <- ts(dt) > autoplot(dt_ts[,2]) + ylab("Charge counts") + xlab("Daily") > > but the label is not the date series. > > Tqvm for any help given. > > >> dput(dt) > structure(list(time

Hi, I'm Roslina, PhD student of University of South Australia, Australia from school Maths and Stats. I use S-Plus before and now has started using R-package. I used to analyse rainfall data using julian date. Is there any similar function that you can suggest to me to be used in R-package? Thank you so much for your attention and help [[alternative HTML version deleted]]

Good morning RGuru's I have a data frame of 575 columns.? I want to extract only those columns that are numeric(double) or integer to do some machine learning with.? I have searched the web for a couple of days (off and on) and have not found anything that shows how to do this.?? Lots of ways to extract rows, but not columns.? I have attempted to use "(x == y)" indices extraction

Can lme or lmer fit a plain regular fixed effects anova? Ie a model without a random effect, or have there be at least one random effect in order for these functions to work? Trying to run such, (1) without specifying a random effect produces an error, (2) specifying that there is no random effect does not produce the same output as an anova run in lm(); (2b) specifying that there is no

> dt1[ vapply(dt1, FUN=is.numeric, FUN.VALUE=NA) ] a c 1 1 1.1 2 2 1.0 ... 10 10 0.2 Bill Dunlap TIBCO Software wdunlap tibco.com On Fri, Apr 29, 2016 at 9:19 AM, Carl Sutton via R-help < r-help at r-project.org> wrote: > Good morning RGuru's > I have a data frame of 575 columns. I want to extract only those columns > that are numeric(double) or integer to do

There seems to be a bug in seq.Date such that it will not allow the user to pass in length.out =3D NULL, despite the fact that this is the = default argument. For example: > dt1 <- as.Date("2004-12-31") > dt2 <- as.Date("2005-12-31") > seq.Date(dt1, dt2, length.out =3D NULL, by =3D "month") Error in seq.Date(dt1, dt2, length.out =3D NULL, by =3D

Dear Friends, I have been trying to save multiple 3x3 (mfrow=c(3,3) graphics inside a loop using tiff figure format (not using PDF or savePlot functions) with no success. Could you please help? Here is a simplified example code: dat=data.frame (ID=rep(1:10,each=10),IDV=rep(seq(1:10),times=10)) dat$DV <- with(dat, 50+15*IDV) dat=dat[order(dat$ID,dat$IDV),] for(i in 1:10){ dt1 =

Hello I have a string which contains microseconds, can anyone help on constructing this in to a time object, with the microseconds, that I can take to a ZOO file? Thanks Sean > UK[1,3] [1] "17:09:53.824" > UK[1,1] [1] "2007-12-11 00:00:00" > mydates <- paste( substr(UK[,1], 1, 10), UK[,3]) > mydates[1] [1] "2007-12-11 17:09:53.824" >

R-Devel, I store and retrieve a large amount of financial data (millions of rows) in a PostgreSQL database keyed by date (and represented in R by class Date). Unfortunately, I frequently find that a great deal of processing time is spent converting dates from character representations to Date class representations in R, presumably because strptime is not fast for large vectors (>10,000

Hi R Community: I want to take the difference in two dates: dt2 - dt1. But, I want the answer in months between those 2 dates. Can you advise me? Please respond to: pzs6 at cdc.gov Thank you! Phil Smith Centers for Disease Control and Prevention

Dear all, I would like to change the "fill" pattern of a histogram using histogram() in the lattice package. I know how to do so using hist(), but would prefer to stay within lattice. dt1 <- rnorm(100,0,1) hist(dt1, density=3, angle=45) library(lattice) histogram(dt1, xlab = "Histogram of rnorm(0,1)", type = "count",