[R] overwritten plots in pdf file
i am having a problem in saving plots in pdf. i have this code below and it
only shows me the last plot. i tried keeping my devices open by removing
dev.off() from the code but the pdf file won't open
for (i in 1:n) {
.
.
.
pdf("D:/research/plot.pdf")
plot(mon, mu, type ='o')
dev.off()
}
my for loop generates 22 plots (to be saved in pdf).. what i want is to save
all these 22 plots in just one pdf file, like 3 plots per page.
can anybody help me?
thanks in advance!
cheers,
CJ
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[R] neural network arguments
hi everyone! my inquiry with neural network is rather basic. i am just learning neural network, particularly the VR bundle. i read the documentations for the said bundle but still is struggling on understanding some arguments - x is the matrix or data frame of x values for example does this mean data frame for training? - y is the matrix or data frame of target values for example does this mean data frame for testing? - would fitting the single-hidden-layer NN train and test my data? what does "fitting" really do? i know these are very basic questions but i just started exploring NN packages. thanks in advance for your help! -- View this message in context: http://www.nabble.com/neural-network-arguments-tp25787734p25787734.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] acf for a univariate time series in a data frame
hi everyone!
i want to check the autocorrelation function for a univariate time series
(streamflow) in a data frame as below:
< DF <- read.table("D:/file path")
< DF
year jan feb mar apr .. dec
1966 0.504 0.406 0.740 0.241 0.429
1967 0.683 0.529 0.780 0.443 0.503
.
.
.
.
what i first tried is:
acf (DF, plot = TRUE)
the resulting plot is correlation between each month ( jan, jan and feb, feb
and march).
what i need is a correlation that would help me find out which lag would be
the most significant. i know i am missing out something simple here. please
bear with me.
should i transform my data frame into a vector of series:
< DF
01/1966 0.504
02/1966 0.406
03/1966 0.740
before i do the command
acf(DF, plot = TRUE)???
thanks in advance!
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Re: [R] acf for a univariate time series in a data frame
*additional: the lags i am expecting is in months.. since i am trying to predict monthly streamflow. thanks again -- View this message in context: http://www.nabble.com/acf-for-a-univariate-time-series-in-a-data-frame-tp25799751p25799784.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] simulating AR() using a
good day everyone!
i have a time series (andong.ts) and fitted and AR() model using the
following code
andong.ts <- ts(read.table("D:/.../andong.csv", header = TRUE), start =
c(1966,1), frequency = 1)
ar(andong.ts)
Call:
ar(x = andong)
Coefficients:
1 2 3
0.3117 0.0607 0.0999
Order selected 3 sigma^2 estimated as 0.8443
I am aware that my model is now
x(t) = 0.3117x(t-1) + 0.0607x(t-2) + 0.0999x(t-3) + e(t)
eqn(1)
but i am not sure of how to perform my simulation for the new series. i
tried to look at arima.sim,
arima.sim(ar = c(0.3117,0.0607, 0.0999), n = 36)
but it doesn't seem to use the old series that should be simulated (or maybe
it is but i am not aware)
Could anyone please brief me on how to perform my simulation properly which
would perform eqn (1) with e(t) being the residual series?
thanks in advance!!!
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[R] help in simulating AR models
good day everyone!
i have a time series (andong.ts) and fitted and AR() model using the
following code
andong.ts <- ts(read.table("D:/.../andong.csv", header = TRUE), start =
c(1966,1), frequency = 1)
ar(andong.ts)
Call: ar(x = andong)
Coefficients:
1 2 3
0.3117 0.06070.0999
Order selected 3 sigma^2 estimated as 0.8443
I am aware that my model is now
x(t) = 0.3117x(t-1) + 0.0607x(t-2) + 0.0999x(t-3) + e(t)
eqn(1)
but i am not sure of how to perform my simulation for the new series. i
tried to look at arima.sim,
arima.sim(ar = c(0.3117,0.0607, 0.0999), n = 36)
but it doesn't seem to use the old series that should be simulated (or maybe
it is but i am not aware) Could anyone please brief me on how to perform my
simulation properly which would perform eqn (1) with e(t) being the residual
series?
thanks in advance!!!
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[R] how to create normal qqplot with the 95% confidence interval
hi everyone! season's greetings! is there any way that i can create a normal qqplot showing, aside from the qqline, the 95% confidence limits? thank you very much.. happy holidays! -- View this message in context: http://n4.nabble.com/how-to-create-normal-qqplot-with-the-95-confidence-interval-tp977727p977727.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
Re: [R] how to create normal qqplot with the 95% confidence interval
thank you very much! this is just what my professor was asking for. Walmes Zeviani wrote: > > Rubio, > > > Look at library(fBasics) the function qqnormPlot(). Below an example: > > qqnormPlot(rnorm(100)) > > Best's > > Walmes Zeviani, Brasil. > > > > > > CJ Rubio wrote: >> >> hi everyone! >> >> season's greetings! >> >> is there any way that i can create a normal qqplot showing, aside from >> the qqline, the 95% confidence limits? thank you very much.. >> >> happy holidays! >> > > -- View this message in context: http://n4.nabble.com/how-to-create-normal-qqplot-with-the-95-confidence-interval-tp977727p978298.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] help in identifying the argument "formula" in the package nnet
hi everyone! I have a 40-year monthy streamflow record. And i want to fit a neural network. I already fitted an autoregressive model and found out that an AR(3) model fits my time series (time.series) the best. I am currently having problems on how to express the argument "formula" for my neural network.. I am aware that my model should be Q(t) ~ Q(t-1) + Q(t-2) + Q(t-3). But I can't seem to figure out how to express this in a formula that R would recognize. Any help, suggestions and comments to enlighten me will be truly appreciated. CJ Rubio -- View this message in context: http://n4.nabble.com/help-in-identifying-the-argument-formula-in-the-package-nnet-tp1401467p1401467.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] how to express time series linear model Q(t) ~ Q(t-1)+.. Q(t-n) as a formula
For example I have a time series Q(t) ~ Q(t-1) + Q(t-2) + Q(t-3) meaning that my current value is dependent to the 3 previous values. Can anybody help me express this in a formula that I can use for my neural network model (I am planning to use packages "nnet" and "MASS") -- View this message in context: http://n4.nabble.com/how-to-express-time-series-linear-model-Q-t-Q-t-1-Q-t-n-as-a-formula-tp1401479p1401479.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
[R] annual maximum value
hi everyone! hope you can help me here. i am a new R user. what i am trying to do is to find the maximum annual discharge from a daily record. i have a data.frame which includes date and the discharge. somewhat like this.. 10/1/1989 2410 10/2/1989 2460 10/3/1989 2890 ... ... ... 12/31/2005 5730 i have been browsing through the archives and fount out about the aggregate function and the zoo package. here's one of the codes i've tried DF <- read.table(data[i], sep =",") ##i have several stations to assimilate Date <- as.Date(as.character(DF[,3]), "%m/%d/%Y") #the date is at the 3rd column, obviously library(zoo) z <- aggregate(zoo(DF[,4]), cut(Date, "y"), max) max.discharge <- coredata(z) date <- time(z) the result should somehow look like this 11/21/1926 32600 4/24/1927 66500 ... ... .. 4/26/2005 111000 thanks for your time reading my questions,, any suggestions will be truly appreciated... -- View this message in context: http://www.nabble.com/annual-maximum-value-tp22049205p22049205.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@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.
Re: [R] annual maximum value
thank you very much for your reply.
i studied the code you gave me and tried to make adjustments to fit the
requirements that i need, thank you again.. i used the following codes:
> m <- read.table("D:/documents/5 stations/01014000.csv", sep =",")
> z <- zoo(m[,4],as.Date(as.character(DF[,3]), "%m/%d/%Y"))
> x <- aggregate(z, cut(time(z),"y"), max)
i got the maximum values right using coredata(), my question now is, how can
i call for the complete date (-mm-dd) and the year () when the
maximum observation for that year was observed?
CJ Rubio wrote:
>
> hi everyone!
>
> hope you can help me here.
>
> i am a new R user. what i am trying to do is to find the maximum annual
> discharge from a daily record. i have a data.frame which includes date and
> the discharge. somewhat like this..
>
> 10/1/1989 2410
> 10/2/1989 2460
> 10/3/1989 2890
> ...
> ...
> ...
> 12/31/2005 5730
>
> i have been browsing through the archives and fount out about the
> aggregate function and the zoo package. here's one of the codes i've tried
>
> DF <- read.table(data[i], sep =",") ##i have several stations to
> assimilate
> Date <- as.Date(as.character(DF[,3]), "%m/%d/%Y") #the date is at the 3rd
> column, obviously
> library(zoo)
> z <- aggregate(zoo(DF[,4]), cut(Date, "y"), max)
> max.discharge <- coredata(z)
> date <- time(z)
>
> the result should somehow look like this
> 11/21/1926 32600
> 4/24/1927 66500
> ...
> ...
> ..
> 4/26/2005 111000
>
> thanks for your time reading my questions,, any suggestions will be truly
> appreciated...
>
>
>
>
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[R] problem with dates in zoo package
i have the following code - assimilating the maximum annual discharge each
year ffrom a daily discharge record from year 1989-2005.
m <- read.table("D:/documents/5 stations/01014000.csv", sep =",")
z <- zoo(m[,4],as.Date(as.character(m[,3]), "%m/%d/%Y"))
x <- aggregate(z, floor(as.numeric(as.yearmon(time(z, max) #code
suggested by Gabor Grothendieck
which gives me the maximum discharge each year and the year itself.. what i
am trying now is to produce the date when the maximum discharge was observed
in the pattern -mm-dd, like:
1988 11/9/1988 18600
1989 5/8/1989 49000
...
2005 4/26/2005111000
thank you all in advance...
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[R] rbind: number of columns of result is not a multiple of vector length (arg 1)
i have the following constructed and running very well,, thanks to Gabor
Grothendieck for his help.
>data.info <- c("station.id", "year", "date", "max.discharge")
>
> for(i in 1:num.files) {
+ station.id <- substring(data[i], 1,8)
+ DF <- read.table(data[i], sep=",", blank.lines.skip = TRUE)
+ z <- zoo(DF[,4], as.Date(as.character(DF[,3]), "%m/%d/%Y"))
+ f <- function(x) time(x) [which.max(x)]
+ ix <- tapply(z, floor(as.yearmon(time(z))),f)
+ year <- (1988:2005)
+ date <- time(z[ix])
+ max.discharge <- coredata(z[ix])
+ data.info <- rbind(data.info, c(station.id, year, date, max.discharge))
+ }
my problem with my code occurs in the part where I arrange my results..
after running this code, i get this warning:
Warning message:
In rbind(data.info, c(station.id, year, date, max.discharge)) :
number of columns of result is not a multiple of vector length (arg 1)
i can't figure out what to do to produce the result i wanted:
(for each station, it should look like this:)
data.info "station.id" "year" "date" "max.discharge"
"01014000" 1988 "1988-11-07" 4360
"01014000" 1989 "1989-05-13" 2
"01014000" 1990 "1990-10-25" 9170
"01014000" 1991 "1991-04-22" 12200
"01014000" 1992 "1992-03-29" 11800
"01014000" 2005 "2005-04-04" 22100
thanks in advence for your help..
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Re: [R] rbind: number of columns of result is not a multiple of vector length (arg 1)
it works perfectly thank you for your help
what if i want to seperate each stations data and save in a .csv file??
while i'm asking you these i'm also finding some ways to do so..
thank you again.
jholtman wrote:
>
> try using:
>
> data.info <- rbind(data.info, cbind(station.id, year, date,
> max.discharge))
>
> On Tue, Feb 17, 2009 at 9:26 PM, CJ Rubio wrote:
>>
>> i have the following constructed and running very well,, thanks to Gabor
>> Grothendieck for his help.
>>
>>>data.info <- c("station.id", "year", "date", "max.discharge")
>>>
>>> for(i in 1:num.files) {
>> + station.id <- substring(data[i], 1,8)
>> + DF <- read.table(data[i], sep=",", blank.lines.skip = TRUE)
>> + z <- zoo(DF[,4], as.Date(as.character(DF[,3]), "%m/%d/%Y"))
>> + f <- function(x) time(x) [which.max(x)]
>> + ix <- tapply(z, floor(as.yearmon(time(z))),f)
>> + year <- (1988:2005)
>> + date <- time(z[ix])
>> + max.discharge <- coredata(z[ix])
>> + data.info <- rbind(data.info, c(station.id, year, date, max.discharge))
>> + }
>>
>> my problem with my code occurs in the part where I arrange my results..
>> after running this code, i get this warning:
>>
>> Warning message:
>> In rbind(data.info, c(station.id, year, date, max.discharge)) :
>> number of columns of result is not a multiple of vector length (arg 1)
>>
>>
>> i can't figure out what to do to produce the result i wanted:
>> (for each station, it should look like this:)
>>
>> data.info "station.id" "year" "date" "max.discharge"
>> "01014000" 1988 "1988-11-07" 4360
>> "01014000" 1989 "1989-05-13" 2
>> "01014000" 1990 "1990-10-25" 9170
>> "01014000" 1991 "1991-04-22" 12200
>> "01014000" 1992 "1992-03-29" 11800
>>
>> "01014000" 2005 "2005-04-04" 22100
>>
>> thanks in advence for your help..
>> --
>> View this message in context:
>> http://www.nabble.com/rbind%3A-number-of-columns-of-result-is-not-a-multiple-of-vector-length-%28arg-1%29-tp22070942p22070942.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
> __
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>
>
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[R] basic inquiry regarding write.csv
i have a loop which looks likes this:
> data.info <- rbind(data.info, cbind(station.id, year, date,
> max.discharge))
+ y <- split(data.info, data.info[station.id])
+ for (i in names(y))
{write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
i am wondering, where the file (which i am about to write in .csv format)
will be saved? i looked at ?write.csv and it says there that :
file either a character string naming a file or a
connection open for writing. "" indicates output to the console.
correct me if i'm wrong but the way i undestand it is, i should have a file
or a working directory where the .csv will be written. if for example i
have a working directory "E:/my_work_directory", how can i save this
splitted files in the same directory?
can anybody please enlighten me more with write.csv and the argument file?
thanks..
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Re: [R] basic inquiry regarding write.csv
thanks for your reply.. is there something wrong with the code i have?
because it doesn't write the file in the directory that i am using...
for (i in names(y))
> {write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
thanks again..
ronggui-2 wrote:
>
> If the "file" is a relative path, then it should be in the working
> directory. Say, the working directory is E:/my_work_directory (of
> course, you can get it by getwd()), and you export a data frame "a" to
> csv by:
> write.csv(a, file="a.csv"), then the file should be
> "E:/my_work_directory/a.csv".
>
> Best
>
> 2009/2/18 CJ Rubio :
>>
>> i have a loop which looks likes this:
>>
>>
>>> data.info <- rbind(data.info, cbind(station.id, year, date,
>>> max.discharge))
>> + y <- split(data.info, data.info[station.id])
>> + for (i in names(y))
>> {write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
>>
>> i am wondering, where the file (which i am about to write in .csv format)
>> will be saved? i looked at ?write.csv and it says there that :
>>
>> file either a character string naming a file or a
>> connection open for writing. "" indicates output to the console.
>>
>> correct me if i'm wrong but the way i undestand it is, i should have a
>> file
>> or a working directory where the .csv will be written. if for example i
>> have a working directory "E:/my_work_directory", how can i save this
>> splitted files in the same directory?
>>
>> can anybody please enlighten me more with write.csv and the argument
>> file?
>>
>> thanks..
>>
>> --
>> View this message in context:
>> http://www.nabble.com/basic-inquiry-regarding-write.csv-tp22073053p22073053.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@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.
>>
>
>
>
> --
> HUANG Ronggui, Wincent
> Tel: (00852) 3442 3832
> PhD Candidate
> Dept of Public and Social Administration
> City University of Hong Kong
> Homepage: http://ronggui.huang.googlepages.com/
>
> __
> R-help@r-project.org mailing list
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>
>
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Re: [R] basic inquiry regarding write.csv
this is the code that i have so far:
> data.path <- file.path ("D:/documents/research/5 stations")
> setwd(data.path)
> getwd()
> data <- dir(".")
> num.files <- length(data)
> for(i in 1:num.files) {
+ station.id <- substring(data[i], 1,8)
+ DF <- read.table(data[i], sep=",", blank.lines.skip = TRUE)
+ z <- zoo(DF[,4], as.Date(as.character(DF[,3]), "%m/%d/%Y"))
+ f <- function(x) time(x) [which.max(x)]
+ ix <- tapply(z, floor(as.yearmon(time(z))),f)
+ year <- (1988:2005)
+ date <- time(z[ix])
+ max.discharge <- coredata(z[ix])
+ data.info <- rbind(data.info, cbind(station.id, year, date,
max.discharge))
+ y <- split(data.info, data.info[station.id])
+ for (i in names(y)) {write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
+ }
i have done the suggestion you gave me.. and there were no errors when i
run it.
hope this would further clarify my question.
So far, I cannot see any mistake, though the sep="" will be more
elegant that sep=",". Are you sure your working directory is
"E:/my_work_directory"? and is there any error msg?
BTW, a reproducible example will help to get better response from the list.
2009/2/18 CJ Rubio :
>
> thanks for your reply.. is there something wrong with the code i have?
> because it doesn't write the file in the directory that i am using...
>
> for (i in names(y))
>> {write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
>
> thanks again..
>
>
>
> ronggui-2 wrote:
>>
>> If the "file" is a relative path, then it should be in the working
>> directory. Say, the working directory is E:/my_work_directory (of
>> course, you can get it by getwd()), and you export a data frame "a" to
>> csv by:
>> write.csv(a, file="a.csv"), then the file should be
>> "E:/my_work_directory/a.csv".
>>
>> Best
>>
>> 2009/2/18 CJ Rubio :
>>>
>>> i have a loop which looks likes this:
>>>
>>>
>>>> data.info <- rbind(data.info, cbind(station.id, year, date,
>>>> max.discharge))
>>> + y <- split(data.info, data.info[station.id])
>>> + for (i in names(y))
>>> {write.csv(y[[i]], file=paste(i, ".csv", sep=","))}
>>>
>>> i am wondering, where the file (which i am about to write in .csv
>>> format)
>>> will be saved? i looked at ?write.csv and it says there that :
>>>
>>> file either a character string naming a file or a
>>> connection open for writing. "" indicates output to the console.
>>>
>>> correct me if i'm wrong but the way i undestand it is, i should have a
>>> file
>>> or a working directory where the .csv will be written. if for example i
>>> have a working directory "E:/my_work_directory", how can i save this
>>> splitted files in the same directory?
>>>
>>> can anybody please enlighten me more with write.csv and the argument
>>> file?
>>>
>>> thanks..
>>>
>>> --
>>> View this message in context:
>>> http://www.nabble.com/basic-inquiry-regarding-write.csv-tp22073053p22073053.html
>>> Sent from the R help mailing list archive at Nabble.com.
>>>
>>> __
>>> R-help@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.
>>>
>>
>>
>>
>> --
>> HUANG Ronggui, Wincent
>> Tel: (00852) 3442 3832
>> PhD Candidate
>> Dept of Public and Social Administration
>> City University of Hong Kong
>> Homepage: http://ronggui.huang.googlepages.com/
>>
>> __
>> R-help@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.
>>
>>
>
> --
> View this message in context:
> http://www.nabble.com/basic-inquiry-regarding-write.csv-tp22073053p22073393.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://
