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Henry

2012-Feb-17 19:56 UTC

[R] convert zoo object to "standard" R object so I can plot and output to csv file

Another newbie question
I got the 1 minute spine interpolation and 15 mean aggregation working with
many thanks to Gabor Grothendieck using Zoo functions.  I got a tip from
Hasan Diwan to look at xts but it seemed I would make better progress using
code from Gabor.

Now I'm having trouble plotting this zoo object.  I'm thinking I want a
function to "split" the zoo object back to a regular R object with x
time
values and y values so I can plot using plot functions I'm familiar with.

What is the vector name that has the values (e.g. the first value is
432.2189)

I also got the 15 minute aggregation mean working - I'm happy about that
also.

In addition next I'll want to write out a csv. file of the 15min block
aggregated mean data.

Here is the trial data and code I used so far.


Lines <- "10/11/2011 23:30:01     432.22
10/11/2011 23:31:17     432.32
10/11/2011 23:35:00     432.32
10/11/2011 23:36:18     432.22
10/11/2011 23:37:18     432.72
10/11/2011 23:39:19     432.23
10/11/2011 23:40:02     432.23
10/11/2011 23:45:00     432.23
10/11/2011 23:45:20     429.75
10/11/2011 23:46:20     429.65
10/11/2011 23:50:00     429.65
10/11/2011 23:51:22     429.75
10/11/2011 23:55:01     429.75
10/11/2011 23:56:23     429.55
10/12/2011 0:00:07      429.55
10/12/2011 0:01:24      429.95
10/12/2011 0:05:00      429.95
10/12/2011 0:06:25      429.85
10/12/2011 0:10:00      429.85
10/12/2011 0:11:26      428.85
10/12/2011 0:15:00      428.85
10/12/2011 0:20:03      428.85
10/12/2011 0:21:29      428.75
10/12/2011 0:25:01      428.75
10/12/2011 0:30:01      428.75
10/12/2011 0:31:31      428.75"

library(zoo)
library(chron)

fmt <- "%m/%d/%Y %H:%M:%S"
toChron <- function(d, t) as.chron(paste(d, t), format = fmt)

z <- read.zoo(text = Lines, index = 1:2, FUN = toChron)

# 1 minute spline fit
m1 <- times("00:01:00")
g <- seq(trunc(start(z), m1), end(z), by = m1)
na.spline(z, xout = g) # this did what I want but what is the vector name?

# 15 minute aggregates
m15 <- times("00:15:00")
ag15 <- aggregate(z, trunc(time(z), m15), mean)

the results of the na.spline(z, xout = g) function below
(10/11/11 23:30:00) (10/11/11 23:31:00) (10/11/11 23:32:00) (10/11/11
23:33:00) (10/11/11 23:34:00) (10/11/11 23:35:00) 
           432.2189            432.2950            432.3869           
432.4584            432.4545            432.3200 
(10/11/11 23:36:00) (10/11/11 23:37:00) (10/11/11 23:38:00) (10/11/11
23:39:00) (10/11/11 23:40:00) (10/11/11 23:41:00) 
           432.1639            432.5834            432.7443           
432.3624            432.2095            433.8208 
(10/11/11 23:42:00) (10/11/11 23:43:00) (10/11/11 23:44:00) (10/11/11
23:45:00) (10/11/11 23:46:00) (10/11/11 23:47:00) 
           436.3606            438.0969            437.2974           
432.2300            428.9265            430.6503 
(10/11/11 23:48:00) (10/11/11 23:49:00) (10/11/11 23:50:00) (10/11/11
23:51:00) (10/11/11 23:52:00) (10/11/11 23:53:00) 
           430.8493            430.2351            429.6500           
429.6715            429.8502            429.9054 
(10/11/11 23:54:00) (10/11/11 23:55:00) (10/11/11 23:56:00) (10/11/11
23:57:00) (10/11/11 23:58:00) (10/11/11 23:59:00) 
           429.8623            429.7522            429.6074           
429.4636            429.3664            429.3678 
(10/12/11 00:00:00) (10/12/11 00:01:00) (10/12/11 00:02:00) (10/12/11
00:03:00) (10/12/11 00:04:00) (10/12/11 00:05:00) 
           429.5200            429.8310            430.0703           
430.1312            430.0707            429.9500 
(10/12/11 00:06:00) (10/12/11 00:07:00) (10/12/11 00:08:00) (10/12/11
00:09:00) (10/12/11 00:10:00) (10/12/11 00:11:00) 
           429.8495            429.9134            430.0879           
430.1446            429.8500            429.1407 
(10/12/11 00:12:00) (10/12/11 00:13:00) (10/12/11 00:14:00) (10/12/11
00:15:00) (10/12/11 00:16:00) (10/12/11 00:17:00) 
           428.6042            428.4933            428.6367           
428.8500            428.9834            429.0264 
(10/12/11 00:18:00) (10/12/11 00:19:00) (10/12/11 00:20:00) (10/12/11
00:21:00) (10/12/11 00:22:00) (10/12/11 00:23:00) 
           429.0031            428.9376            428.8542           
428.7773            428.7315            428.7217 
(10/12/11 00:24:00) (10/12/11 00:25:00) (10/12/11 00:26:00) (10/12/11
00:27:00) (10/12/11 00:28:00) (10/12/11 00:29:00) 
           428.7330            428.7498            428.7594           
428.7615            428.7588            428.7541 
(10/12/11 00:30:00) (10/12/11 00:31:00) 
           428.7500            428.7491 

the results from when I enter ag15 in the R command line - which looks
correct

(10/11/11 23:30:00) (10/11/11 23:45:00) (10/12/11 00:00:00) (10/12/11
00:15:00) (10/12/11 00:30:00) 
           432.3229            430.0471            429.6667           
428.8000            428.7500 




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R. Michael Weylandt

2012-Feb-17 20:05 UTC

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[R] convert zoo object to "standard" R object so I can plot and output to csv file

?write.zoo

Perhaps with sep = "," if you have multiple columns.

Michael

On Fri, Feb 17, 2012 at 2:56 PM, Henry <hccoles at lbl.gov>
wrote:> Another newbie question
> I got the 1 minute spine interpolation and 15 mean aggregation working with
> many thanks to Gabor Grothendieck using Zoo functions. ?I got a tip from
> Hasan Diwan to look at xts but it seemed I would make better progress using
> code from Gabor.
>
> Now I'm having trouble plotting this zoo object. ?I'm thinking I
want a
> function to "split" the zoo object back to a regular R object
with x time
> values and y values so I can plot using plot functions I'm familiar
with.
>
> What is the vector name that has the values (e.g. the first value is
> 432.2189)
>
> I also got the 15 minute aggregation mean working - I'm happy about
that
> also.
>
> In addition next I'll want to write out a csv. file of the 15min block
> aggregated mean data.
>
> Here is the trial data and code I used so far.
>
>
> Lines <- "10/11/2011 23:30:01 ? ? 432.22
> 10/11/2011 23:31:17 ? ? 432.32
> 10/11/2011 23:35:00 ? ? 432.32
> 10/11/2011 23:36:18 ? ? 432.22
> 10/11/2011 23:37:18 ? ? 432.72
> 10/11/2011 23:39:19 ? ? 432.23
> 10/11/2011 23:40:02 ? ? 432.23
> 10/11/2011 23:45:00 ? ? 432.23
> 10/11/2011 23:45:20 ? ? 429.75
> 10/11/2011 23:46:20 ? ? 429.65
> 10/11/2011 23:50:00 ? ? 429.65
> 10/11/2011 23:51:22 ? ? 429.75
> 10/11/2011 23:55:01 ? ? 429.75
> 10/11/2011 23:56:23 ? ? 429.55
> 10/12/2011 0:00:07 ? ? ?429.55
> 10/12/2011 0:01:24 ? ? ?429.95
> 10/12/2011 0:05:00 ? ? ?429.95
> 10/12/2011 0:06:25 ? ? ?429.85
> 10/12/2011 0:10:00 ? ? ?429.85
> 10/12/2011 0:11:26 ? ? ?428.85
> 10/12/2011 0:15:00 ? ? ?428.85
> 10/12/2011 0:20:03 ? ? ?428.85
> 10/12/2011 0:21:29 ? ? ?428.75
> 10/12/2011 0:25:01 ? ? ?428.75
> 10/12/2011 0:30:01 ? ? ?428.75
> 10/12/2011 0:31:31 ? ? ?428.75"
>
> library(zoo)
> library(chron)
>
> fmt <- "%m/%d/%Y %H:%M:%S"
> toChron <- function(d, t) as.chron(paste(d, t), format = fmt)
>
> z <- read.zoo(text = Lines, index = 1:2, FUN = toChron)
>
> # 1 minute spline fit
> m1 <- times("00:01:00")
> g <- seq(trunc(start(z), m1), end(z), by = m1)
> na.spline(z, xout = g) # this did what I want but what is the vector name?
>
> # 15 minute aggregates
> m15 <- times("00:15:00")
> ag15 <- aggregate(z, trunc(time(z), m15), mean)
>
> the results of the na.spline(z, xout = g) function below
> (10/11/11 23:30:00) (10/11/11 23:31:00) (10/11/11 23:32:00) (10/11/11
> 23:33:00) (10/11/11 23:34:00) (10/11/11 23:35:00)
> ? ? ? ? ? 432.2189 ? ? ? ? ? ?432.2950 ? ? ? ? ? ?432.3869
> 432.4584 ? ? ? ? ? ?432.4545 ? ? ? ? ? ?432.3200
> (10/11/11 23:36:00) (10/11/11 23:37:00) (10/11/11 23:38:00) (10/11/11
> 23:39:00) (10/11/11 23:40:00) (10/11/11 23:41:00)
> ? ? ? ? ? 432.1639 ? ? ? ? ? ?432.5834 ? ? ? ? ? ?432.7443
> 432.3624 ? ? ? ? ? ?432.2095 ? ? ? ? ? ?433.8208
> (10/11/11 23:42:00) (10/11/11 23:43:00) (10/11/11 23:44:00) (10/11/11
> 23:45:00) (10/11/11 23:46:00) (10/11/11 23:47:00)
> ? ? ? ? ? 436.3606 ? ? ? ? ? ?438.0969 ? ? ? ? ? ?437.2974
> 432.2300 ? ? ? ? ? ?428.9265 ? ? ? ? ? ?430.6503
> (10/11/11 23:48:00) (10/11/11 23:49:00) (10/11/11 23:50:00) (10/11/11
> 23:51:00) (10/11/11 23:52:00) (10/11/11 23:53:00)
> ? ? ? ? ? 430.8493 ? ? ? ? ? ?430.2351 ? ? ? ? ? ?429.6500
> 429.6715 ? ? ? ? ? ?429.8502 ? ? ? ? ? ?429.9054
> (10/11/11 23:54:00) (10/11/11 23:55:00) (10/11/11 23:56:00) (10/11/11
> 23:57:00) (10/11/11 23:58:00) (10/11/11 23:59:00)
> ? ? ? ? ? 429.8623 ? ? ? ? ? ?429.7522 ? ? ? ? ? ?429.6074
> 429.4636 ? ? ? ? ? ?429.3664 ? ? ? ? ? ?429.3678
> (10/12/11 00:00:00) (10/12/11 00:01:00) (10/12/11 00:02:00) (10/12/11
> 00:03:00) (10/12/11 00:04:00) (10/12/11 00:05:00)
> ? ? ? ? ? 429.5200 ? ? ? ? ? ?429.8310 ? ? ? ? ? ?430.0703
> 430.1312 ? ? ? ? ? ?430.0707 ? ? ? ? ? ?429.9500
> (10/12/11 00:06:00) (10/12/11 00:07:00) (10/12/11 00:08:00) (10/12/11
> 00:09:00) (10/12/11 00:10:00) (10/12/11 00:11:00)
> ? ? ? ? ? 429.8495 ? ? ? ? ? ?429.9134 ? ? ? ? ? ?430.0879
> 430.1446 ? ? ? ? ? ?429.8500 ? ? ? ? ? ?429.1407
> (10/12/11 00:12:00) (10/12/11 00:13:00) (10/12/11 00:14:00) (10/12/11
> 00:15:00) (10/12/11 00:16:00) (10/12/11 00:17:00)
> ? ? ? ? ? 428.6042 ? ? ? ? ? ?428.4933 ? ? ? ? ? ?428.6367
> 428.8500 ? ? ? ? ? ?428.9834 ? ? ? ? ? ?429.0264
> (10/12/11 00:18:00) (10/12/11 00:19:00) (10/12/11 00:20:00) (10/12/11
> 00:21:00) (10/12/11 00:22:00) (10/12/11 00:23:00)
> ? ? ? ? ? 429.0031 ? ? ? ? ? ?428.9376 ? ? ? ? ? ?428.8542
> 428.7773 ? ? ? ? ? ?428.7315 ? ? ? ? ? ?428.7217
> (10/12/11 00:24:00) (10/12/11 00:25:00) (10/12/11 00:26:00) (10/12/11
> 00:27:00) (10/12/11 00:28:00) (10/12/11 00:29:00)
> ? ? ? ? ? 428.7330 ? ? ? ? ? ?428.7498 ? ? ? ? ? ?428.7594
> 428.7615 ? ? ? ? ? ?428.7588 ? ? ? ? ? ?428.7541
> (10/12/11 00:30:00) (10/12/11 00:31:00)
> ? ? ? ? ? 428.7500 ? ? ? ? ? ?428.7491
>
> the results from when I enter ag15 in the R command line - which looks
> correct
>
> (10/11/11 23:30:00) (10/11/11 23:45:00) (10/12/11 00:00:00) (10/12/11
> 00:15:00) (10/12/11 00:30:00)
> ? ? ? ? ? 432.3229 ? ? ? ? ? ?430.0471 ? ? ? ? ? ?429.6667
> 428.8000 ? ? ? ? ? ?428.7500
>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/convert-zoo-object-to-standard-R-object-so-I-can-plot-and-output-to-csv-file-tp4398302p4398302.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.

Gabor Grothendieck

2012-Feb-18 04:34 UTC

head link

[R] convert zoo object to "standard" R object so I can plot and output to csv file

On Fri, Feb 17, 2012 at 2:56 PM, Henry <hccoles at lbl.gov>
wrote:> Another newbie question
> I got the 1 minute spine interpolation and 15 mean aggregation working with
> many thanks to Gabor Grothendieck using Zoo functions. ?I got a tip from
> Hasan Diwan to look at xts but it seemed I would make better progress using
> code from Gabor.
>
> Now I'm having trouble plotting this zoo object. ?I'm thinking I
want a
> function to "split" the zoo object back to a regular R object
with x time
> values and y values so I can plot using plot functions I'm familiar
with.
>
If z is a zoo object then

plot(z)

or

library(lattice)
xyplot(z)

will plot it.  See ?plot.zoo and ?xyplot.zoo




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tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

Henry

2012-Feb-23 02:54 UTC

head link

[R] convert zoo object to "standard" R object so I can plot and output to csv file

Gabor,
Thanks very much.
I have all the zoo functions to get 1 minute aggregation and 15 min. means
working and now able to write out to a file/etc.

One question on the 15 min. mean results.

m1 <- times("00:01:00")
g <- seq(trunc(start(z),m1),end(z),by = m1)
z1<-na.approx(z,xout = g)
m15 <- times("00:15:00")
ag15 <- aggregate(z1,trunc(time(z1),m15),mean)

Does the value of ag15 get assigned to the "center" of the time, front
or
rear?
Can this be defined?
It seems aggregate(z1,trunc(time(z1),m15),mean) is getting the mean and
"assigning" it to the first time stamp, not a time stamp assigned to
7.5
minutes from the first 1 minute data point used to get the mean.

Is there an option to "create" a new regular timestamp on 15 periods
but
centered or reporting on 7.5 minute periods?

Thanks.

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Gabor Grothendieck

2012-Feb-23 04:17 UTC

head link

[R] convert zoo object to "standard" R object so I can plot and output to csv file

On Wed, Feb 22, 2012 at 9:54 PM, Henry <hccoles at lbl.gov>
wrote:> Gabor,
> Thanks very much.
> I have all the zoo functions to get 1 minute aggregation and 15 min. means
> working and now able to write out to a file/etc.
>
> One question on the 15 min. mean results.
>
> m1 <- times("00:01:00")
> g <- seq(trunc(start(z),m1),end(z),by = m1)
> z1<-na.approx(z,xout = g)
> m15 <- times("00:15:00")
> ag15 <- aggregate(z1,trunc(time(z1),m15),mean)
>
> Does the value of ag15 get assigned to the "center" of the time,
front or
> rear?
> Can this be defined?
> It seems aggregate(z1,trunc(time(z1),m15),mean) is getting the mean and
> "assigning" it to the first time stamp, not a time stamp assigned
to 7.5
> minutes from the first 1 minute data point used to get the mean.
>
> Is there an option to "create" a new regular timestamp on 15
periods but
> centered or reporting on 7.5 minute periods?
>
The by argument of aggregate.zoo defines the time index (as discussed
in ?aggregate.zoo) so time index of ag15 is trunc(time(z1), m15);
therefore, to push it ahead by 7.5 minutes just use trunc(time(z1),
m15) + as.numeric(m15)/2 in its place.


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Henry

2012-Feb-23 20:06 UTC

head link

[R] convert zoo object to "standard" R object so I can plot and output to csv file

Another simple question - trying to specify xlim in a zoo plot and getting
error
my plot line is
plot(z1,
ylim=c(-100,3000),xlim=c(chron("10/30/2011","00:00:00"),chron("10/30/2011","00:20:00")),type="b",xlab="",ylab="1
Minute Fit",cex.lab=1.3)
Error in substring(paste("0", v$day, sep = ""), first =
nchar(paste(v$day)))
: 
  invalid substring argument(s)

Most of the complete code pasted below.....
fmt<-"%m/%d/%Y %H:%M:%S"  # describe the date/time format in the
file
tail1<-function(x) tail(x,1)
z<-read.zoo("Cooling-1.txt",FUN=as.chron,format=fmt,sep="\t",header=TRUE,aggregate=tail1)

par(oma=c(6,1,4,2)) # set the outside space boundaries

par(mfrow=c(2,1)) # set the number of graphs (rows,cols) going on one page

# plot the original data
par(mar=c(1.1, 5, .9, 0.5))
plot(z, ylim=c(-100,3000),type="b",xlab="",ylab="Raw
Data",col="red",cex.lab=1.3)
grid(nx=NA,ny=NULL,lwd=.5,lty=1,col="red")

# calculate and plot the 1 minute straight line interpolation 
m1 <- times("00:01:00")
g <- seq(trunc(start(z),m1),end(z),by = m1)
z1<-na.approx(z,xout = g)

#plot the 1 minute linear interpolation fit
par(mar=c(1.1, 5, .9, 0.5))
#the following plot line generates the error
plot(z1,
ylim=c(-100,3000),xlim=c(chron("10/30/2011","00:00:00"),chron("10/30/2011","00:20:00")),type="b",xlab="",ylab="1
Minute Fit",cex.lab=1.3)
grid(nx=NA,ny=NULL,lwd=.5,lty=1,col="darkgrey")



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Gabor Grothendieck

2012-Feb-23 20:21 UTC

head link

[R] convert zoo object to "standard" R object so I can plot and output to csv file

On Thu, Feb 23, 2012 at 3:06 PM, Henry <hccoles at lbl.gov>
wrote:> Another simple question - trying to specify xlim in a zoo plot and getting
> error
> my plot line is
> plot(z1,
>
ylim=c(-100,3000),xlim=c(chron("10/30/2011","00:00:00"),chron("10/30/2011","00:20:00")),type="b",xlab="",ylab="1
> Minute Fit",cex.lab=1.3)
> Error in substring(paste("0", v$day, sep = ""), first =
nchar(paste(v$day)))
> :
> ?invalid substring argument(s)
>
> Most of the complete code pasted below.....
Most???????????

Read the last two lines of every message to r-help.

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