Thank you very much for your help. It is very much appreciated.
On Fri, Nov 3, 2023 at 7:23 AM roslinazairimah zakaria
wrote:
> Dear all,
>
> I have this set of data. I would like to sum the EnergykWh according date
> sequences.
>
> > head(dt1,20) StationName date time
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",
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
date appears to be a character variable, and R is treating it as such.
str(dt1)
might give you some insight. Or the dplyr equivalent
glimpse(dt1)
I think R did what you asked, but if you want to be able to order
records by date, in temporal order, you need to tell R that it is a date:
Dear all,
I have this set of data. I would like to sum the EnergykWh according date
sequences.
> head(dt1,20) StationName date time EnergykWh
1 PALO ALTO CA / CAMBRIDGE #1 1/14/2016 12:09 4.680496
2 PALO ALTO CA / CAMBRIDGE #1 1/14/2016 19:50 6.272414
3 PALO ALTO CA
В Wed, 25 Oct 2023 09:18:26 +0300
"Christian Asseburg" пишет:
> > str(x)
> 'data.frame': 1 obs. of 3 variables:
> $ A: num 1
> $ B: num 1
> $ C:'data.frame': 1 obs. of 1 variable:
> ..$ A: num 1
>
> Why does the print(x) not show "C" as the name of the third element?
I think it is more clear-cut than so, at least if the Poisson situation is
something to go by.
There, you can do either of these and get equivalent results
> fit.lung <- glm(cases ~ age + city, offset=log(pop),
+ family=poisson, data=lungcancer)
> fit.lung2 <- glm(cases/pop ~
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