Hi Rod,
How about this?
scenarios <- expand.grid(A = c("pass", "fail"), B = c("pass", "fail"), C =
c("pass", "fail"), D = c("pass", "fail"), E = c("pass", "fail"))
scenarios$F<-ifelse(scenarios$E=="pass","fail","pass")
Jim
On Thu, Aug 2, 2018 at 11:20 AM, R Stafford wrote:
> I have 6
On Wed, 1 Aug 2018 17:40:54 +0200
Edoardo Silvestri wrote:
> I have a database based on hourly data and I need to forecast next
> 24h of a single variable. I was thinking about applying an ARIMA
> model with some exogenous variables but I don't succeed to configure
> the hourly frequency,
Hello,
I am interested to apply an econometric technique of Latent Variable
framework on Environmental Kuznets Curve for 164 countries for a span of 25
years.
The methodology and the code are from Simulation exercise from an
unpublished paper "Two Examples of Convex-Programming-Based
I have 6 variables, (A,B,C,D,E,F) that can either pass or fail (i.e., true
or false).
I can get a table of all pass/fail combinations with this:
scenarios <- expand.grid(A = c("pass", "fail"), B = c("pass", "fail"), C =
c("pass", "fail"), D = c("pass", "fail"), E = c("pass", "fail"), F =
Statistics issues are generally off topic here; and we generally prefer
posters to show us their own efforts rather than expecting us to solve the
problem for them.
However, this CRAN time series task view may be useful to you:
https://cran.r-project.org/web/views/TimeSeries.html
Cheers,
Bert
I have a database based on hourly data and I need to forecast next 24h of a
single variable. I was thinking about applying an ARIMA model with some
exogenous variables but I don't succeed to configure the hourly frequency,
estimate ARIMA parameters, pdq ( exists some tests to check which
Try this:
> library(lubridate)
> library(tidyverse)
> input <- read.csv(text = "date,str1,str2,str3
+ 10/1/1998 0:00,0.6,0,0
+ 10/1/1998 1:00,0.2,0.2,0.2
+ 10/1/1998 2:00,0.6,0.2,0.4
+ 10/1/1998 3:00,0,0,0.6
+ 10/1/1998
Hi
see in line
From: Diego Avesani
Sent: Wednesday, August 1, 2018 2:30 PM
To: PIKAL Petr
Cc: r-help mailing list
Subject: Re: [R] read txt file - date - no space
Dear Pikal, Deal all,
again really thank.
it seems not working.
Some specifications: My non data are -999, but I could change
Dear Pikal, DEar all,
I do not if it could help:
if I print MyData%date, I get (at some point)
[281] "1998-12-10 16:00:00 CET" "1998-12-10 17:00:00 CET"
"1998-12-10 18:00:00 CET" "1998-12-10 19:00:00 CET"
[285] "1998-12-10 20:00:00 CET" "1998-12-10 21:00:00 CET"
"1998-12-10 22:00:00 CET"
Dear Pikal, Deal all,
again really thank.
it seems not working.
Some specifications: My non data are -999, but I could change it.
My final results is:
11 -55.86242 -55.84764660 -277.4775
22 -55.47554 -94.58921682 -277.4845
33 -55.47095 -99.20239198 -277.4709
4
Hi
I did not get through all answers you already got and you probably obtained
similar advice as mine.
# read data (if you have csv file just use read.csv)
> test<-read.table("clipboard", header=T, sep=",")
# control your object(s)
> str(test)
'data.frame': 8 obs. of 4 variables:
$ date:
Dear all,
I am sorry, I did a lot of confusion. I am sorry, I have to relax and stat
all again in order to understand.
If I could I would like to start again, without mixing strategy and waiting
for your advice.
I am really appreciate you help, really really.
Here my new file, a *.csv file (buy
I am trying to build an R package with Rtools version 3.5.0.4 along with
R-3.5.1
using the following sequence of commands:
File -> Open Project -> Build -> Build Binary Package
I received the following error message:
zip I/O error: No such file or directory
zip error: Temporary file failure
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