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

Hallo you can extract POSIX object tv <- as.POSIXct(index(dt_train)) and use cut together with aggregate cut(tv, "hour") aggregate(dt_train, list(cut(tv, "hour")), mean) 2014-10-06 21:00:00 9.807692 2014-10-06 22:00:00 8.666667 Cheers. Petr ?t 3. 10. 2024 v 17:25 odes?latel roslinazairimah zakaria < roslinaump at gmail.com> napsal: > Dear all, > > My

Dear all, My data is by minutes and I can see it has seasonal trend by daily and weekly. How do I decompose the minute data into daily and weekly some data: > dput(tail(dt_train,100))structure(c(11L, 11L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 10L,

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 Enrico, yes it works. I can also plot the graph. Next question, how do I delete the date from 16 December until 31 December 2014? Thank you very much. > dt$time <- as.POSIXct(dt$time)> > dt_ts <- xts(dt$count, dt$time)> str(dt_ts)An xts object on 2014-01-01 / 2014-12-31 23:59:00 containing: Data: integer [525600, 1] Index: POSIXct,POSIXt [525600] (TZ: "")

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 all, I tried to rerun the examples given by Hyndman in otexts but keep on getting errors and I have searched through google but no solution yet. Thank you in advance for any help given. library(fpp3) library(lubridate) library(xts) library(fabletools) library(ggplot2) library(dplyr) library(gridExtra) > dput(head(new_us_retail_employment,100))structure(list(Month = structure(c(-11323,

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:

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]

?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

Hi, In perl, to get a substring matching a particular pattern can be implemented like the following example: $x = "AAAA.txt"; if ($x=~ /(.*?)\.txt/){ $prefix = $1; } So how to do the same thing in R? Can someone provide me the code sample? Thanks much in advance. -- Waverley @ Palo Alto

Hi, Can someone help as how to use R to program google search in the R code? I know that other languages can allow or have the google search API If someone can give me some links or sample code I would greatly appreciate. Thanks. -- Waverley @ Palo Alto