[R] means by year, month and day
Tom Mosca
tom at vims.edu
Mon Jul 18 18:33:48 CEST 2016
Dear William,
The line of code you composed works perfectly, as you knew it would. Thank you for your kind response. I will now endeavor to forget that other function exists.
Sincerely, Tom
________________________________
From: William Dunlap [wdunlap at tibco.com]
Sent: Monday, July 18, 2016 12:14 PM
To: Tom Mosca
Cc: jim holtman; Jianling Fan; r-help at r-project.org
Subject: Re: [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<http://tibco.com>
On Mon, Jul 18, 2016 at 6:18 AM, Tom Mosca <tom at vims.edu<mailto: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<mailto:jholtman at gmail.com>]
Sent: Sunday, July 17, 2016 7:14 PM
To: Jianling Fan
Cc: Tom Mosca; r-help at r-project.org<mailto: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><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><mailto: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><mailto: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]]
>
>
> ______________________________________________
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--
Jianling Fan
¼Ô«Øâ
______________________________________________
R-help at r-project.org<mailto:R-help at r-project.org><mailto:R-help at r-project.org<mailto:R-help at r-project.org>> mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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______________________________________________
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.
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