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jim holtman

2016-Jul-17 23:14 UTC

[R] means by year, month and day

Here is an example of using dplyr.  Please provide a reasonable subset of
data.  Your was all for the same date.  Use 'dput' to put in your email.
> x <- read.table(text = "     X.YY MM DD hh WDI R.WSP D.GST   PRES 
ATMPDEWP
+   2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+   2015  2  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  2  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  2  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  2  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  2  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  2  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  2  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  2  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  2  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  2  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+   2015  3  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  3  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  3  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  3  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  3  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  3  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  3  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  3  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  3  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  3  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+  ",
+  header = TRUE,
+  as.is = TRUE)>
>  library(dplyr)
>  by_year <- x %>%+             group_by(X.YY) %>%
+             summarise_each(funs(mean))>
>  by_ym <- x %>%+             group_by(X.YY, MM ) %>%
+             summarise_each(funs(mean))>
>  by_ymd <- x %>%+             group_by(X.YY, MM, DD) %>%
+             summarise_each(funs(mean))>
> by_yearSource: local data frame [1 x 10]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <dbl> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6> by_ymSource: local data frame [3 x 10]
Groups: X.YY [?]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <int> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6> by_ymdSource: local data frame [3 x 10]
Groups: X.YY, MM [?]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <int> <int> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6>

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 Sun, Jul 17, 2016 at 5:42 PM, Jianling Fan <fanjianling at gmail.com>
wrote:
> Hello Tom,
>
> try aggregate() or cast(). Both works.I prefer the latter.
>
>
> library(reshape)
> desc<-melt(mydata, measure.vars=c("WDI","R.WSP",
"D.GST", "PRES",
> "ATMP", "DEWP"),
>            id.vars=c("X.YY","MM","DD"))
> summary<-cast(desc, X.YY+MM+DD~variable, mean)
>
>
>
>
>
>
>
>
>
> On 17 July 2016 at 06:22, Tom Mosca <tom at vims.edu> wrote:
> > Hello Good Folk,
> >
> > My dataframe looks like this:
> >> mydata
> >      X.YY MM DD hh WDI R.WSP D.GST   PRES  ATMP  DEWP
> > 1    2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> > 2    2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> > 3    2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> > 4    2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> > 5    2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> > 6    2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> > 7    2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> > 8    2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> > 9    2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> > 10   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3 ?
> >
> > The first four columns are year, month, day, hour (0 ? 23).  I wish to
> take the means of the next six columns (WDIR, WSPD, GST, PRES, ATMP and
> DEWP) by year, month and day.  That is, I want daily averages.
> >
> > Please help.  Thank you.
> >
> > Tom
> >
> >         [[alternative HTML version deleted]]
> >
> >
> > ______________________________________________
> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Jianling Fan
> ???
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
	[[alternative HTML version deleted]]

Tom Mosca

2016-Jul-18 13:18 UTC

head link

[R] means by year, month and day

Dear Jim,



I'm very inexperienced with R.



I'm sorry I failed to recognize the flaw in my example.  I just clipped the
first few lines of data, and should have realized that they were for the first
few hours of a single day.



My first problem was to take the means by day.  As all the data come from a
single year, I accomplished the goal by using:
> datATMP<-aggregate(mydata, by=mydata[c("MM","DD")],
FUN=mean)


My second problem was to sort the results by month, for which I used:
> attach(datATMP)
> datATMP1 <- datATMP[order(MM),]
> datATMP1
> detach(datATMP)


I have read that I should avoid using attach(), and should use with() instead. 
However, I have not yet figured out how to do this with with().  I'm only
running short segments of code, and think I'll be alright with attach() for
the time being, but do want to develop better form.  So, I'll keep working
on it.



Thank you for your kind response and examples.  I will study them.



Sincerely, Tom





________________________________
From: jim holtman [jholtman at gmail.com]
Sent: Sunday, July 17, 2016 7:14 PM
To: Jianling Fan
Cc: Tom Mosca; r-help at r-project.org
Subject: Re: [R] means by year, month and day

Here is an example of using dplyr.  Please provide a reasonable subset of data. 
Your was all for the same date.  Use 'dput' to put in your email.
> x <- read.table(text = "     X.YY MM DD hh WDI R.WSP D.GST   PRES 
ATMP  DEWP+   2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+   2015  2  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  2  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  2  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  2  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  2  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  2  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  2  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  2  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  2  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  2  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+   2015  3  1  0 328   3.6   4.5 1028.0   3.8  -3.5
+   2015  3  1  1 300   2.1   2.7 1027.9   3.7  -4.4
+   2015  3  1  2 264   2.4   2.9 1027.7   3.6  -4.5
+   2015  3  1  3 230   4.1   4.5 1027.4   4.2  -3.8
+   2015  3  1  4 242   8.1   9.2 1026.6   4.4  -3.1
+   2015  3  1  5 262   9.3  10.1 1026.6   4.1  -3.8
+   2015  3  1  6 267   8.6   9.6 1026.3   4.2  -3.8
+   2015  3  1  7 264   9.3   9.9 1026.1   3.9  -2.8
+   2015  3  1  8 268   8.2   9.1 1026.1   3.5  -3.0
+   2015  3  1  9 272   8.8   9.6 1025.4   3.2  -3.3
+  ",
+  header = TRUE,
+  as.is<http://as.is> = TRUE)>
>  library(dplyr)
>  by_year <- x %>%+             group_by(X.YY) %>%
+             summarise_each(funs(mean))>
>  by_ym <- x %>%+             group_by(X.YY, MM ) %>%
+             summarise_each(funs(mean))>
>  by_ymd <- x %>%+             group_by(X.YY, MM, DD) %>%
+             summarise_each(funs(mean))>
> by_yearSource: local data frame [1 x 10]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <dbl> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6> by_ymSource: local data frame [3 x 10]
Groups: X.YY [?]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <int> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6> by_ymdSource: local data frame [3 x 10]
Groups: X.YY, MM [?]
   X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
  <int> <int> <int> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86 
-3.6>

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 Sun, Jul 17, 2016 at 5:42 PM, Jianling Fan <fanjianling at
gmail.com<mailto:fanjianling at gmail.com>> wrote:
Hello Tom,

try aggregate() or cast(). Both works.I prefer the latter.


library(reshape)
desc<-melt(mydata, measure.vars=c("WDI","R.WSP",
"D.GST", "PRES",
"ATMP", "DEWP"),
           id.vars=c("X.YY","MM","DD"))
summary<-cast(desc, X.YY+MM+DD~variable, mean)









On 17 July 2016 at 06:22, Tom Mosca <tom at vims.edu<mailto:tom at
vims.edu>> wrote:> Hello Good Folk,
>
> My dataframe looks like this:
>> mydata
>      X.YY MM DD hh WDI R.WSP D.GST   PRES  ATMP  DEWP
> 1    2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> 2    2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> 3    2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> 4    2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> 5    2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> 6    2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> 7    2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> 8    2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> 9    2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> 10   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3 ?K
>
> The first four columns are year, month, day, hour (0 ?V 23).  I wish to
take the means of the next six columns (WDIR, WSPD, GST, PRES, ATMP and DEWP) by
year, month and day.  That is, I want daily averages.
>
> Please help.  Thank you.
>
> Tom
>
>         [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at r-project.org<mailto:R-help at r-project.org> mailing list
-- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


--
Jianling Fan
??????

______________________________________________
R-help at r-project.org<mailto:R-help at r-project.org> mailing list -- To
UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


	[[alternative HTML version deleted]]

William Dunlap

2016-Jul-18 16:14 UTC

head link

[R] means by year, month and day

If you are very inexperienced with R you still have time to forget you ever
heard
of the attach function.  Your code
    > attach(datATMP)
    > datATMP1 <- datATMP[order(MM),]
    > detach(datATMP)
can be replaced by
    > datATMP1 <- datATMP[order(datATMP[["MM"]]),]


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Mon, Jul 18, 2016 at 6:18 AM, Tom Mosca <tom at vims.edu> wrote:
> Dear Jim,
>
>
>
> I'm very inexperienced with R.
>
>
>
> I'm sorry I failed to recognize the flaw in my example.  I just clipped
> the first few lines of data, and should have realized that they were for
> the first few hours of a single day.
>
>
>
> My first problem was to take the means by day.  As all the data come from
> a single year, I accomplished the goal by using:
>
> > datATMP<-aggregate(mydata,
by=mydata[c("MM","DD")], FUN=mean)
>
>
>
> My second problem was to sort the results by month, for which I used:
>
> > attach(datATMP)
> > datATMP1 <- datATMP[order(MM),]
> > datATMP1
> > detach(datATMP)
>
>
>
> I have read that I should avoid using attach(), and should use with()
> instead.  However, I have not yet figured out how to do this with with().
> I'm only running short segments of code, and think I'll be alright
with
> attach() for the time being, but do want to develop better form.  So,
I'll
> keep working on it.
>
>
>
> Thank you for your kind response and examples.  I will study them.
>
>
>
> Sincerely, Tom
>
>
>
>
>
> ________________________________
> From: jim holtman [jholtman at gmail.com]
> Sent: Sunday, July 17, 2016 7:14 PM
> To: Jianling Fan
> Cc: Tom Mosca; r-help at r-project.org
> Subject: Re: [R] means by year, month and day
>
> Here is an example of using dplyr.  Please provide a reasonable subset of
> data.  Your was all for the same date.  Use 'dput' to put in your
email.
>
> > x <- read.table(text = "     X.YY MM DD hh WDI R.WSP D.GST  
PRES  ATMP
> DEWP
> +   2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> +   2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> +   2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> +   2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> +   2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> +   2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> +   2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> +   2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> +   2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> +   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3
> +   2015  2  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> +   2015  2  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> +   2015  2  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> +   2015  2  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> +   2015  2  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> +   2015  2  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> +   2015  2  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> +   2015  2  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> +   2015  2  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> +   2015  2  1  9 272   8.8   9.6 1025.4   3.2  -3.3
> +   2015  3  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> +   2015  3  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> +   2015  3  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> +   2015  3  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> +   2015  3  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> +   2015  3  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> +   2015  3  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> +   2015  3  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> +   2015  3  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> +   2015  3  1  9 272   8.8   9.6 1025.4   3.2  -3.3
> +  ",
> +  header = TRUE,
> +  as.is<http://as.is> = TRUE)
> >
> >  library(dplyr)
> >  by_year <- x %>%
> +             group_by(X.YY) %>%
> +             summarise_each(funs(mean))
> >
> >  by_ym <- x %>%
> +             group_by(X.YY, MM ) %>%
> +             summarise_each(funs(mean))
> >
> >  by_ymd <- x %>%
> +             group_by(X.YY, MM, DD) %>%
> +             summarise_each(funs(mean))
> >
> > by_year
> Source: local data frame [1 x 10]
>    X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
>   <int> <dbl> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
> 1  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> > by_ym
> Source: local data frame [3 x 10]
> Groups: X.YY [?]
>    X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
>   <int> <int> <dbl> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
> 1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> 2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> 3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> > by_ymd
> Source: local data frame [3 x 10]
> Groups: X.YY, MM [?]
>    X.YY    MM    DD    hh   WDI R.WSP D.GST    PRES  ATMP  DEWP
>   <int> <int> <int> <dbl> <dbl> <dbl>
<dbl>   <dbl> <dbl> <dbl>
> 1  2015     1     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> 2  2015     2     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> 3  2015     3     1   4.5 269.7  6.45  7.21 1026.81  3.86  -3.6
> >
>
>
> 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 Sun, Jul 17, 2016 at 5:42 PM, Jianling Fan <fanjianling at gmail.com
> <mailto:fanjianling at gmail.com>> wrote:
> Hello Tom,
>
> try aggregate() or cast(). Both works.I prefer the latter.
>
>
> library(reshape)
> desc<-melt(mydata, measure.vars=c("WDI","R.WSP",
"D.GST", "PRES",
> "ATMP", "DEWP"),
>            id.vars=c("X.YY","MM","DD"))
> summary<-cast(desc, X.YY+MM+DD~variable, mean)
>
>
>
>
>
>
>
>
>
> On 17 July 2016 at 06:22, Tom Mosca <tom at vims.edu<mailto:tom at
vims.edu>>
> wrote:
> > Hello Good Folk,
> >
> > My dataframe looks like this:
> >> mydata
> >      X.YY MM DD hh WDI R.WSP D.GST   PRES  ATMP  DEWP
> > 1    2015  1  1  0 328   3.6   4.5 1028.0   3.8  -3.5
> > 2    2015  1  1  1 300   2.1   2.7 1027.9   3.7  -4.4
> > 3    2015  1  1  2 264   2.4   2.9 1027.7   3.6  -4.5
> > 4    2015  1  1  3 230   4.1   4.5 1027.4   4.2  -3.8
> > 5    2015  1  1  4 242   8.1   9.2 1026.6   4.4  -3.1
> > 6    2015  1  1  5 262   9.3  10.1 1026.6   4.1  -3.8
> > 7    2015  1  1  6 267   8.6   9.6 1026.3   4.2  -3.8
> > 8    2015  1  1  7 264   9.3   9.9 1026.1   3.9  -2.8
> > 9    2015  1  1  8 268   8.2   9.1 1026.1   3.5  -3.0
> > 10   2015  1  1  9 272   8.8   9.6 1025.4   3.2  -3.3 ?
> >
> > The first four columns are year, month, day, hour (0 ? 23).  I wish to
> take the means of the next six columns (WDIR, WSPD, GST, PRES, ATMP and
> DEWP) by year, month and day.  That is, I want daily averages.
> >
> > Please help.  Thank you.
> >
> > Tom
> >
> >         [[alternative HTML version deleted]]
> >
> >
> > ______________________________________________
> > R-help at r-project.org<mailto:R-help at r-project.org> mailing
list -- To
> UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Jianling Fan
> ???
>
> ______________________________________________
> R-help at r-project.org<mailto:R-help at r-project.org> mailing list
-- To
> UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>         [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

R help - Jul 2016 - means by year, month and day

[R] means by year, month and day

[R] means by year, month and day

[R] means by year, month and day