[R] need help with a time series plotting problem
Dear R Users, I am a beginner in R programming and need some help with a simple plotting problem that i am having. My dataset consist of three columns: first one has data_id, second is the date and third is the actual data itself corresponding to each date. The date ranges from 1/1/2000-12/31/2009. I am trying to plot my data versus the dates as a long term time series but what's happening is that R is plotting each year on top of the previous year. so instead of getting 1 line (dated 2000-2009) in the plot i am getting 9 lines (1 line for each year). i tried to look for solutions online but found nothing. can someone suggest how can i make a plot with x-axis ranging from 2000-2009. my code is copied below: setwd(J:/Rstuff/flow) flow=read.delim(flow.dat,header=TRUE,sep=\t) plot(flow$usgs1500~as.Date(flow$date, %m/%d/%y),type=l,xlab=date,ylab=daily discharge (m3/s) ,main=USGS1500,yaxs=i, xaxs=i,) any help would be appreciated regards vibhava -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4230672.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] Estimate of smooth.spline
thank you vey much Jean. On 12/21/11, Jean V Adams [via R] ml-node+s789695n4221894...@n4.nabble.com wrote: ali_protocol wrote on 12/21/2011 05:39:19 AM: Hi everyone, I have: s= smooth.spline (cbind(1,2,3,4,3,3),cbind (4,2,4,6,5,6)) how may I obtain s hat (s^)? Thanks a lot. Read the help file on the smooth.spline() function, ?smooth.spline The section titled Value tells you what the function returns ... An object of class smooth.spline with components xthe distinct x values in increasing order, see the ?Details? above. y the fitted values corresponding to x. ... The y component is what you're interested in, s$y Jean [[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. ___ If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Estimate-of-smooth-spline-tp4221253p4221894.html To unsubscribe from Estimate of smooth.spline, visit http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4221253code=bW9oYW1tYWRpYW5hbGltb2hhbW1hZGlhbkBnbWFpbC5jb218NDIyMTI1M3wtMTkwNDgxNTk3NQ== -- View this message in context: http://r.789695.n4.nabble.com/Estimate-of-smooth-spline-tp4221253p4230616.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.
[R] readLines errors
Hi All, I met a problem using readLines function to return the data from Google maps. readLines(url( http://ditu.google.cn/maps/geo?q=+qianshuichong,+guichi+anhui,+CNoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg;), n=1, warn=FALSE) [1] 200,4,30.6528380,117.4872250 The above code is ok because the address is English, but it cannot work when i change the address into zh-hans (Chinese simplified). See below. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CNregion=zh_Hans), n=1, warn=FALSE) [1] 400,0,0,0 The returned value is wrong. I also tried several other options, but still have the same problem. The interesting thing is if i paste the URL into the IE explorer directly, it do return the correct values. Anybody can give me some suggestions or hints to solve this? Thanks a lot. P.S. See below for the other possibilities i tried. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CNregion=zh), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CN), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CN), n=1, warn=FALSE,encoding=UTF-8) [1] 400,0,0,0 -- - Jane Chang Queen's [[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.
[R] Fw:
http://secretosdeunaprincesa.com/next2012.php?aid=11sywa=76wud=764rax=68 __ 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] if statement problem
This is almost Circle 8.1.7 of
'The R Inferno':
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
but is making the mistake in the
other direction.
On 23/12/2011 22:40, reena wrote:
Hello,
I want to do fisher test for the rows in data file which has value less than
5 otherwise chi square test .The p values from both test should be stored in
one resulted file. but there is some problem with bold if statement. I don't
know how
implement this line properly.
x = cbind(obs1,obs2,exp1,exp2)
a = matrix(c(0,0,0,0), ncol=2, byrow =TRUE)#matrix with initialized
values
for (i in 1: length(x[,1]))
{
*if((x[i,1] || x[i,2] || x[i,3] || x[i,4]) 5)*
{
a[1,1]- x[i,1];
a[1,2]- x[i,2];
a[2,1]- x[i,3];
a[2,2]- x[i,4];
result- fisher.test(a)
write.table(result[[p.value]],file=results.txt,
sep=\n, append=TRUE, col.names=FALSE, row.names=FALSE);
}
else
{
a[1,1]- x[i,1];
a[1,2]- x[i,2];
a[2,1]- x[i,3];
a[2,2]- x[i,4];
result- chisq.test(a)
write.table(result[[p.value]],file=results.txt,
sep=\n, append=TRUE, col.names=FALSE, row.names=FALSE);}
}
Regards
R
--
View this message in context:
http://r.789695.n4.nabble.com/if-statement-problem-tp4230026p4230026.html
Sent from the R help mailing list archive at Nabble.com.
__
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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.
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')
__
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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.
[R] Turning my subscription to mailing list back on
Please respond directly to me as I am not receiving Email from the R-Help mailing list: jsor...@grecc.umaryland.edu Before I left for vacation, I turned delivery of mail from R-Help off. I cannot find the proper web-page to use to turn mail back on. Can someone send me the URL, or perhaps an Email address and a command I can sent via Email? Thanks, John Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ 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] need help with a time series plotting problem
you need to supply a subset of your data since the problem is probably related to its representation. please follow the posting guidelines. Sent from my iPad On Dec 24, 2011, at 2:21, vibhava vibhavasrivast...@gmail.com wrote: Dear R Users, I am a beginner in R programming and need some help with a simple plotting problem that i am having. My dataset consist of three columns: first one has data_id, second is the date and third is the actual data itself corresponding to each date. The date ranges from 1/1/2000-12/31/2009. I am trying to plot my data versus the dates as a long term time series but what's happening is that R is plotting each year on top of the previous year. so instead of getting 1 line (dated 2000-2009) in the plot i am getting 9 lines (1 line for each year). i tried to look for solutions online but found nothing. can someone suggest how can i make a plot with x-axis ranging from 2000-2009. my code is copied below: setwd(J:/Rstuff/flow) flow=read.delim(flow.dat,header=TRUE,sep=\t) plot(flow$usgs1500~as.Date(flow$date, %m/%d/%y),type=l,xlab=date,ylab=daily discharge (m3/s) ,main=USGS1500,yaxs=i, xaxs=i,) any help would be appreciated regards vibhava -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4230672.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@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] Help creating a symmetric matrix?
Or the slightly shorter: z-diag(6) z[row(z) col(z)] - v which is what lower.tri() does, and z - diag(6) z[lower.tri(z)] - v also works. Sarah On Fri, Dec 23, 2011 at 9:31 PM, Rui Barradas ruipbarra...@sapo.pt wrote: Matt Considine wrote Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal - and getting it into a matrix. I have tried using lower.tri as found here https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html but it appears to only partially fill in the matrix. My code and an example of the output is below. Can anyone point me to an example that shows how to create a matrix with this sort of input? Thank you in advance, Matt #v-newx[,3] #or, for the sake of this example v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) ind - lower.tri(z) z[ind] - t(v)[ind] z [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1.0 0.0 0 0 0 0 [2,] 0.26657 1.0 0 0 0 0 [3,] 0.23388 0.19237 1 0 0 0 [4,] 0.23122 0.18633 NA 1 0 0 [5,] 0.21476 0.17298 NA NA 1 0 [6,] 0.20829 0.17174 NA NA NA 1 Hello, Aren't you complicating? In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix, not the vector? z-diag(6) ind - lower.tri(z) z[ind] - v #This works z Rui Barradas -- Sarah Goslee http://www.functionaldiversity.org __ 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] readLines errors
On Dec 24, 2011, at 4:12 AM, rusers.sh wrote:
Hi All,
I met a problem using readLines function to return the data from
Google
maps.
readLines(url(
http://ditu.google.cn/maps/geo?q=+qianshuichong,+guichi+anhui,+CNoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
),
n=1, warn=FALSE)
[1] 200,4,30.6528380,117.4872250
The above code is ok because the address is English, but it cannot
work
when i change the address into zh-hans (Chinese simplified). See
below.
readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5
%86%B2
,+¹ó³Ø+°²»Õ,
+
ÖÐ
¹
ú
output
=
csv
key
=
ABQIq8Fnd_oUka
-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
language=zh-CNregion=zh_Hans),
n=1, warn=FALSE)
When I plug that URL into my browser (Firefox) I get this text on a
white background:
602,0,0,0
When I execute that code in R I get:
[1] {
And when I remove the lines restriction I get:
[1] {
[2] \name\: \ Ç°Ë®³å\\u003chttp://ditu.google.cn/maps/geo?
q= 前水冲\\u003e\,
[3] \Status\: {
[4] \code\: 602,
[5] \request\: \geocode\
[6] }
[7] }
I don't read Chinese , however it does appear that there was some sort
of extra processing step that Firefox carried out to arrive at the
displayed text.
[1] 400,0,0,0
The returned value is wrong. I also tried several other options, but
still have the same problem. The interesting thing is if i paste the
URL
into the IE explorer directly, it do return the correct values.
Anybody can give me some suggestions or hints to solve this?
Thanks a lot.
P.S. See below for the other possibilities i tried.
readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5
%86%B2
,+¹ó³Ø+°²»Õ,
+
ÖÐ
¹
ú
output
=
csv
key
=
ABQIq8Fnd_oUka
-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
language=zh-CNregion=zh),
n=1, warn=FALSE)
[1] 400,0,0,0
readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5
%86%B2
,+¹ó³Ø+°²»Õ,
+
ÖÐ
¹
ú
output
=
csv
key
=
ABQIq8Fnd_oUka
-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
language=zh-CN),
n=1, warn=FALSE)
[1] 400,0,0,0
readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5
%86%B2
,+¹ó³Ø+°²»Õ,
+
ÖÐ
¹
ú
output
=
csv
key
=
ABQIq8Fnd_oUka
-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
language=zh-CN),
n=1, warn=FALSE,encoding=UTF-8)
[1] 400,0,0,0
--
-
Jane Chang
Queen's
[[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.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
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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.
Re: [R] Help creating a symmetric matrix?
Dear Matt, Sarah and Rui, To answer the original question for creating a symmetric matrix v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) z[row(z) col(z)] - v z - z + t(z) diag(z) - 0 z [,1][,2][,3][,4][,5][,6] [1,] 0.0 0.33740 0.26657 0.23388 0.23122 0.21476 [2,] 0.33740 0.0 0.20829 0.20486 0.19439 0.19237 [3,] 0.26657 0.20829 0.0 0.18633 0.17298 0.17174 [4,] 0.23388 0.20486 0.18633 0.0 0.16822 0.16480 [5,] 0.23122 0.19439 0.17298 0.16822 0.0 0.15027 [6,] 0.21476 0.19237 0.17174 0.16480 0.15027 0.0 Bill On Dec 24, 2011, at 6:04 AM, Sarah Goslee wrote: Or the slightly shorter: z-diag(6) z[row(z) col(z)] - v which is what lower.tri() does, and z - diag(6) z[lower.tri(z)] - v also works. Sarah On Fri, Dec 23, 2011 at 9:31 PM, Rui Barradas ruipbarra...@sapo.pt wrote: Matt Considine wrote Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal - and getting it into a matrix. I have tried using lower.tri as found here https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html but it appears to only partially fill in the matrix. My code and an example of the output is below. Can anyone point me to an example that shows how to create a matrix with this sort of input? Thank you in advance, Matt #v-newx[,3] #or, for the sake of this example v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) ind - lower.tri(z) z[ind] - t(v)[ind] z [,1][,2] [,3] [,4] [,5] [,6] [1,] 1.0 0.00000 [2,] 0.26657 1.00000 [3,] 0.23388 0.192371000 [4,] 0.23122 0.18633 NA100 [5,] 0.21476 0.17298 NA NA10 [6,] 0.20829 0.17174 NA NA NA1 Hello, Aren't you complicating? In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix, not the vector? z-diag(6) ind - lower.tri(z) z[ind] - v#This works z Rui Barradas -- Sarah Goslee http://www.functionaldiversity.org __ 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. William Revellehttp://personality-project.org/revelle.html Professor http://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern Universityhttp://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 6 minutes to midnighthttp://www.thebulletin.org __ 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] readLines errors
I just found another strange thing. When i paste the URL into the Google
Chrome, i can get the correct result 200,4,30.5892200,117.4286680. But i
paste it into the IE explorer, the result is wrong 400,0,0,0.
I guess that the R uses the IE as default. Could i tell R to use Google
Chrome as a default explorer? Or how can i find the possible solution for
IE to return the same result as Google Chrome .
Confusing.
2011/12/24 David Winsemius dwinsem...@comcast.net
On Dec 24, 2011, at 4:12 AM, rusers.sh wrote:
Hi All,
I met a problem using readLines function to return the data from Google
maps.
readLines(url(
http://ditu.google.cn/maps/**geo?q=+qianshuichong,+guichi+**
anhui,+CNoutput=csvkey=**ABQIq8Fnd_oUka-**
7RdS6BrD7GBTqeABoQuNTXS36G_**rkiwQnKRW6GBTkns8JpKz6y6dScgB8**827dlddUlghttp://ditu.google.cn/maps/geo?q=+qianshuichong,+guichi+anhui,+CNoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg
),
n=1, warn=FALSE)
[1] 200,4,30.6528380,117.4872250
The above code is ok because the address is English, but it cannot work
when i change the address into zh-hans (Chinese simplified). See below.
readLines(url(http://ditu.**google.cn/maps/geo?q=+Ãhttp://ditu.google.cn/maps/geo?q=+%C3%87
°Ã®³åh**ttp://ditu.google.cn/maps/geo?**q=+%E5%89%8D%E6%B0%B4%E5%86%B2http://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2
**
,+¹ó³Ã+°²»Ã,+Ãùúoutput=csv**key=ABQIq8Fnd_oUka-**
7RdS6BrD7GBTqeABoQuNTXS36G_**rkiwQnKRW6GBTkns8JpKz6y6dScgB8**
827dlddUlglanguage=zh-CN**region=zh_Hans),
n=1, warn=FALSE)
When I plug that URL into my browser (Firefox) I get this text on a white
background:
602,0,0,0
When I execute that code in R I get:
[1] {
And when I remove the lines restriction I get:
[1] {
[2] \name\: \
ðî³å\\u003chttp://ditu.**google.cn/maps/geo?q=http://ditu.google.cn/maps/geo?q=åæ°´å²\\u003e\,
[3] \Status\: {
[4] \code\: 602,
[5] \request\: \geocode\
[6] }
[7] }
I don't read Chinese , however it does appear that there was some sort of
extra processing step that Firefox carried out to arrive at the displayed
text.
[1] 400,0,0,0
The returned value is wrong. I also tried several other options, but
still have the same problem. The interesting thing is if i paste the URL
into the IE explorer directly, it do return the correct values.
Anybody can give me some suggestions or hints to solve this?
Thanks a lot.
P.S. See below for the other possibilities i tried.
readLines(url(http://ditu.**google.cn/maps/geo?q=+Ãhttp://ditu.google.cn/maps/geo?q=+%C3%87
°Ã®³åh**ttp://ditu.google.cn/maps/geo?**q=+%E5%89%8D%E6%B0%B4%E5%86%B2http://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2
**
,+¹ó³Ã+°²»Ã,+Ãùúoutput=csv**key=ABQIq8Fnd_oUka-**
7RdS6BrD7GBTqeABoQuNTXS36G_**rkiwQnKRW6GBTkns8JpKz6y6dScgB8**
827dlddUlglanguage=zh-CN**region=zh),
n=1, warn=FALSE)
[1] 400,0,0,0
readLines(url(http://ditu.**google.cn/maps/geo?q=+Ãhttp://ditu.google.cn/maps/geo?q=+%C3%87
°Ã®³åh**ttp://ditu.google.cn/maps/geo?**q=+%E5%89%8D%E6%B0%B4%E5%86%B2http://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2
**
,+¹ó³Ã+°²»Ã,+Ãùúoutput=csv**key=ABQIq8Fnd_oUka-**
7RdS6BrD7GBTqeABoQuNTXS36G_**rkiwQnKRW6GBTkns8JpKz6y6dScgB8**
827dlddUlglanguage=zh-CN),
n=1, warn=FALSE)
[1] 400,0,0,0
readLines(url(http://ditu.**google.cn/maps/geo?q=+Ãhttp://ditu.google.cn/maps/geo?q=+%C3%87
°Ã®³åh**ttp://ditu.google.cn/maps/geo?**q=+%E5%89%8D%E6%B0%B4%E5%86%B2http://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2
**
,+¹ó³Ã+°²»Ã,+Ãùúoutput=csv**key=ABQIq8Fnd_oUka-**
7RdS6BrD7GBTqeABoQuNTXS36G_**rkiwQnKRW6GBTkns8JpKz6y6dScgB8**
827dlddUlglanguage=zh-CN),
n=1, warn=FALSE,encoding=UTF-8)
[1] 400,0,0,0
--
-
Jane Chang
Queen's
[[alternative HTML version deleted]]
__**
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/**
posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
--
-
Jane Chang
Queen's
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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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.
Re: [R] Help creating a symmetric matrix?
On Sat, Dec 24, 2011 at 8:38 AM, William Revelle li...@revelle.net wrote: Dear Matt, Sarah and Rui, To answer the original question for creating a symmetric matrix I read the original question as *only* wanting the complete lower triangle, with diagonal of 1 and 0 in the upper triangle. If your interpretation is correct, there's also this convenience function: library(ecodist) z - full(v) Sarah v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) z[row(z) col(z)] - v z - z + t(z) diag(z) - 0 z [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.0 0.33740 0.26657 0.23388 0.23122 0.21476 [2,] 0.33740 0.0 0.20829 0.20486 0.19439 0.19237 [3,] 0.26657 0.20829 0.0 0.18633 0.17298 0.17174 [4,] 0.23388 0.20486 0.18633 0.0 0.16822 0.16480 [5,] 0.23122 0.19439 0.17298 0.16822 0.0 0.15027 [6,] 0.21476 0.19237 0.17174 0.16480 0.15027 0.0 Bill On Dec 24, 2011, at 6:04 AM, Sarah Goslee wrote: Or the slightly shorter: z-diag(6) z[row(z) col(z)] - v which is what lower.tri() does, and z - diag(6) z[lower.tri(z)] - v also works. Sarah On Fri, Dec 23, 2011 at 9:31 PM, Rui Barradas ruipbarra...@sapo.pt wrote: Matt Considine wrote Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal - and getting it into a matrix. I have tried using lower.tri as found here https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html but it appears to only partially fill in the matrix. My code and an example of the output is below. Can anyone point me to an example that shows how to create a matrix with this sort of input? Thank you in advance, Matt #v-newx[,3] #or, for the sake of this example v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) ind - lower.tri(z) z[ind] - t(v)[ind] z [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1.0 0.0 0 0 0 0 [2,] 0.26657 1.0 0 0 0 0 [3,] 0.23388 0.19237 1 0 0 0 [4,] 0.23122 0.18633 NA 1 0 0 [5,] 0.21476 0.17298 NA NA 1 0 [6,] 0.20829 0.17174 NA NA NA 1 Hello, Aren't you complicating? In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix, not the vector? z-diag(6) ind - lower.tri(z) z[ind] - v #This works z Rui Barradas -- __ 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] Call for Abstracts for useR! 2012
The international R User Conference useR! 2012 will be held June 12-15 at Vanderbilt University in Nashville, Tennessee, USA, with a pre-conference full-day course offered on June 11. Participants are encouraged to submit a one-page abstract for oral or poster presentation at the conference. Half-day tutorials will be held in the morning and afternoon of June 12th. For tutorials, please submit a one-page proposal with course outline and narrative description along with the target audience. Note that the deadline for tutorial proposals is January 15. All innovative and interesting applications of R are suitable. A list of topics from past conferences is available at the conference website: http://biostat.mc.vanderbilt.edu/wiki/Main/useR-2012#Call_for_Abstracts_and_Tutorial Contributed oral presentations will be allowed 17 minutes, followed by 3 minutes discussion. Following last year's success, all participants are invited to present a Lightning Talk, for which no abstract is required. The format for Lightning Talks is a 15-slide version of Pecha Kucha. Participants wishing to give a Lightning Talk must provide an informative title on their registration form. For regular talks and posters, please use the recently updated LaTeX or Microsoft Word/LibreOffice Writer abstract templates available at the conference website: http://biostat.mc.vanderbilt.edu/wiki/Main/useR-2012#Templates The deadlines for abstract and tutorial proposals are as follows: Tutorial Submission: Jan 1 - Jan 15 Tutorial Acceptance Notification: Jan 16 - Jan 23 Abstract Submission: Jan 1 - Mar 12 Abstract Acceptance Notification: Mar 13 - Apr 15 The preferred method of abstract and tutorial proposal submission is via the web form at the conference website: http://biostat.mc.vanderbilt.edu/wiki/Main/useR-2012#Online_Submission Participants may register online beginning January 23rd. If earlier registration is preferred, please contact Stephania McNeal-Goddard using the contact information here: http://biostat.mc.vanderbilt.edu/wiki/Main/useR-2012#Contact See http://journal.r-project.org/archive/2011-2/RJournal_2011-2.pdf for an article about the conference with more details, including a list of some of the major Nashville area entertainment events occurring around the time of the conference. We look forward to seeing you at useR! 2012 in Nashville! -- Frank E Harrell Jr Professor and Chairman School of Medicine Department of Biostatistics Vanderbilt University __ 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] Inverse problems in R package
Dear all, I have a general relationship, which can be easily solved using forward engineering model, y=f(x1, x2, x3) Now I know a distribution of y, how can I get the corresponding distributions for x1, x2, and x3? In another situation, if I know the distributions of y and x1, how can I obtain the corresponding distributions of x2 and x3? Are there some R packages for this inverse problem (or called back calculation)? Thanks a lot, Terry [[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] readLines errors
Hi All, I just tried Firefox that David referred and found it can also return the correct result as Google Chrome. So now it is more clear now that readLines() function uses IE as the default explorer, so it return the wrong result. Then the possible solutions are, 1. Could we change the options in readLines() to set another explorer as the default explore, e.g. Google Chrome? 2. Could it be possible to set IE explorer in a rational way, so that it can return the same results as Google Chrome and Firefox? I have tried to modify the encoding method in IE, but it doesnot work. Any suggestions or help? ÔÚ 2011Äê12ÔÂ24ÈÕ ÏÂÎç5:12£¬rusers.sh rusers...@gmail.comдµÀ£º Hi All, I met a problem using readLines function to return the data from Google maps. readLines(url( http://ditu.google.cn/maps/geo?q=+qianshuichong,+guichi+anhui,+CNoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg;), n=1, warn=FALSE) [1] 200,4,30.6528380,117.4872250 The above code is ok because the address is English, but it cannot work when i change the address into zh-hans (Chinese simplified). See below. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CNregion=zh_Hans), n=1, warn=FALSE) [1] 400,0,0,0 The returned value is wrong. I also tried several other options, but still have the same problem. The interesting thing is if i paste the URL into the IE explorer directly, it do return the correct values. Anybody can give me some suggestions or hints to solve this? Thanks a lot. P.S. See below for the other possibilities i tried. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CNregion=zh), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CN), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ,+Öйúoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlglanguage=zh-CN), n=1, warn=FALSE,encoding=UTF-8) [1] 400,0,0,0 -- - Jane Chang Queen's -- - Jane Chang Queen's [[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] readLines errors
On Dec 24, 2011, at 9:55 AM, rusers.sh wrote: Hi All, I just tried Firefox that David referred and found it can also return the correct result as Google Chrome. So now it is more clear now that readLines() function uses IE as the default explorer, so it return the wrong result. Then the possible solutions are, 1. Could we change the options in readLines() to set another explorer as the default explore, e.g. Google Chrome? ?options # and look at the HTTPUserAgent choice I believe this: options()$HTTPUserAgent [1] R (2.14.0 x86_64-apple-darwin9.8.0 x86_64 darwin9.8.0) ... is telling me that the choice is passed off to the OS, and only because you keep referring to (MS)IE do we know that you are running one of that Redmond company's products. (You should evidently do more study of the relationship of R to the OS platform you use.) 2. Could it be possible to set IE explorer in a rational way, so that it can return the same results as Google Chrome and Firefox? I have tried to modify the encoding method in IE, but it doesnot work. Any suggestions or help? That is not an appropriate question for this mailing list. You persist in posting in HTML. Maybe you should also spend more time learning to use you mail client correctly? -- David. ÔÚ 2011Äê12ÔÂ24ÈÕ ÏÂÎç5:12£¬rusers.sh rusers...@gmail.comдµÀ£º Hi All, I met a problem using readLines function to return the data from Google maps. readLines(url( http://ditu.google.cn/maps/geo?q=+qianshuichong,+guichi+anhui,+CNoutput=csvkey=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg ), n=1, warn=FALSE) [1] 200,4,30.6528380,117.4872250 The above code is ok because the address is English, but it cannot work when i change the address into zh-hans (Chinese simplified). See below. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ, + ÖÐ ¹ ú output = csv key = ABQIq8Fnd_oUka -7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg language=zh-CNregion=zh_Hans), n=1, warn=FALSE) [1] 400,0,0,0 The returned value is wrong. I also tried several other options, but still have the same problem. The interesting thing is if i paste the URL into the IE explorer directly, it do return the correct values. Anybody can give me some suggestions or hints to solve this? Thanks a lot. P.S. See below for the other possibilities i tried. readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ, + ÖÐ ¹ ú output = csv key = ABQIq8Fnd_oUka -7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg language=zh-CNregion=zh), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ, + ÖÐ ¹ ú output = csv key = ABQIq8Fnd_oUka -7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg language=zh-CN), n=1, warn=FALSE) [1] 400,0,0,0 readLines(url(http://ditu.google.cn/maps/geo?q=+Ç°Ë®³åhttp://ditu.google.cn/maps/geo?q=+%E5%89%8D%E6%B0%B4%E5%86%B2 ,+¹ó³Ø+°²»Õ, + ÖÐ ¹ ú output = csv key = ABQIq8Fnd_oUka -7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg language=zh-CN), n=1, warn=FALSE,encoding=UTF-8) [1] 400,0,0,0 -- David Winsemius, MD West Hartford, CT __ 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] if statement problem
It didn't work. :( -- View this message in context: http://r.789695.n4.nabble.com/if-statement-problem-tp4230026p4230933.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] Help creating a symmetric matrix?
Thank you all for your help and best wishes for the holiday season. Matt Considine On 12/24/2011 8:38 AM, William Revelle wrote: Dear Matt, Sarah and Rui, To answer the original question for creating a symmetric matrix v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) z[row(z) col(z)]- v z- z + t(z) diag(z)- 0 z [,1][,2][,3][,4][,5][,6] [1,] 0.0 0.33740 0.26657 0.23388 0.23122 0.21476 [2,] 0.33740 0.0 0.20829 0.20486 0.19439 0.19237 [3,] 0.26657 0.20829 0.0 0.18633 0.17298 0.17174 [4,] 0.23388 0.20486 0.18633 0.0 0.16822 0.16480 [5,] 0.23122 0.19439 0.17298 0.16822 0.0 0.15027 [6,] 0.21476 0.19237 0.17174 0.16480 0.15027 0.0 Bill On Dec 24, 2011, at 6:04 AM, Sarah Goslee wrote: Or the slightly shorter: z-diag(6) z[row(z) col(z)]- v which is what lower.tri() does, and z- diag(6) z[lower.tri(z)]- v also works. Sarah On Fri, Dec 23, 2011 at 9:31 PM, Rui Barradasruipbarra...@sapo.pt wrote: Matt Considine wrote Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal - and getting it into a matrix. I have tried using lower.tri as found here https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html but it appears to only partially fill in the matrix. My code and an example of the output is below. Can anyone point me to an example that shows how to create a matrix with this sort of input? Thank you in advance, Matt #v-newx[,3] #or, for the sake of this example v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) ind- lower.tri(z) z[ind]- t(v)[ind] z [,1][,2] [,3] [,4] [,5] [,6] [1,] 1.0 0.00000 [2,] 0.26657 1.00000 [3,] 0.23388 0.192371000 [4,] 0.23122 0.18633 NA100 [5,] 0.21476 0.17298 NA NA10 [6,] 0.20829 0.17174 NA NA NA1 Hello, Aren't you complicating? In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix, not the vector? z-diag(6) ind- lower.tri(z) z[ind]- v#This works z Rui Barradas -- Sarah Goslee http://www.functionaldiversity.org __ 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. William Revellehttp://personality-project.org/revelle.html Professor http://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern Universityhttp://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 6 minutes to midnighthttp://www.thebulletin.org __ 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] need help with a time series plotting problem
thanks for the reply. here is subset of the data that i want to plot: dateusgs700 1 10/1/2000 0.050970325 2 10/2/2000 0.041059428 3 10/3/2000 0.032564374 4 10/4/2000 0.02775051 . . . . . 40149/27/2011 0 40159/28/2011 0 40169/29/2011 0 40179/30/2011 0 my script again is: setwd(J:/Rstuff/flow) flow=read.delim(flow.dat,header=TRUE,sep=\t) plot(flow$usgs1500~as.Date(flow$date, %m/%d/%y),type=l,xlab=date,ylab=daily discharge (m3/s) ,main=USGS1500,yaxs=i, xaxs=i,) i wish to plot all this data as a single time series but my program is doing some weird stuff and the x-axis has labels Jan-dec and the long term series is broken and plotted on yearly basis. regards vibhava -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4231285.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] need help with a time series plotting problem
On Dec 24, 2011, at 9:24 AM, vibhava wrote: thanks for the reply. here is subset of the data that i want to plot: dateusgs700 1 10/1/2000 0.050970325 2 10/2/2000 0.041059428 3 10/3/2000 0.032564374 4 10/4/2000 0.02775051 . . Instead of posting print-output why not post the results of dput(head(flow)) further useless output deleted. my script again is: setwd(J:/Rstuff/flow) flow=read.delim(flow.dat,header=TRUE,sep=\t) plot(flow$usgs1500 There's no column with that name. ~as.Date(flow$date, %m/%d/%y) For another thing, you generally need to use %Y for 4 digit years. ,type=l,xlab=date,ylab=daily discharge (m3/s) ,main=USGS1500,yaxs=i, xaxs=i,) Not sure if that empty argument is going to be handled gracefully. i wish to plot all this data as a single time series but my program is doing some weird stuff and the x-axis has labels Jan-dec and the long term series is broken and plotted on yearly basis. Your code is not reproducible. -- David Winsemius, MD West Hartford, CT __ 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] Nested model - singularities not defined
I am using a nested model in R and the lm output shows 47 not defined because of singularities and I have no idea why. Any help on why this is happening or how to fix this problem would be very much appreciated. Below is the output I received from R. Thanks and happy holidays! Call: lm(formula = Dist ~ Treatment/SiteL/Territory) Residuals: Min 1Q Median 3QMax -6.646 -1.443 0.080 1.440 5.613 Coefficients: (47 not defined because of singularities) Estimate Std. Error t value Pr(|t|) (Intercept) 9.470001.24231 7.623 9.48e-13 *** TreatmentReference -4.055001.75689 -2.308 0.02201 * TreatmentSingle Tree -2.430002.15175 -1.129 0.26010 TreatmentTypical -2.035711.55732 -1.307 0.19263 TreatmentIntensive:SiteLB -2.255711.55732 -1.448 0.14904 TreatmentReference:SiteLB -2.253331.60382 -1.405 0.16156 TreatmentSingle Tree:SiteLB 0.432861.99213 0.217 0.82821 TreatmentTypical:SiteLB 0.643711.45484 0.442 0.65863 TreatmentIntensive:SiteLC -3.890002.15175 -1.808 0.07212 . TreatmentReference:SiteLC 0.913001.66673 0.548 0.58445 TreatmentSingle Tree:SiteLC-1.598002.07878 -0.769 0.44296 TreatmentTypical:SiteLC 0.11.32809 0.075 0.94005 TreatmentIntensive:SiteLA:TerritoryQ -3.940001.75689 -2.243 0.02601 * TreatmentReference:SiteLA:TerritoryQ -1.617001.66673 -0.970 0.33313 TreatmentSingle Tree:SiteLA:TerritoryQ 0.010002.15175 0.005 0.99630 TreatmentTypical:SiteLA:TerritoryQ NA NA NA NA TreatmentIntensive:SiteLB:TerritoryQ NA NA NA NA TreatmentReference:SiteLB:TerritoryQ1.596911.38232 1.155 0.24936 TreatmentSingle Tree:SiteLB:TerritoryQ -1.152861.32809 -0.868 0.38639 TreatmentTypical:SiteLB:TerritoryQ NA NA NA NA TreatmentIntensive:SiteLC:TerritoryQ NA NA NA NA TreatmentReference:SiteLC:TerritoryQ0.742001.81451 0.409 0.68303 TreatmentSingle Tree:SiteLC:TerritoryQ 2.329671.32254 1.762 0.07967 . TreatmentTypical:SiteLC:TerritoryQ -0.370001.32809 -0.279 0.78084 TreatmentIntensive:SiteLA:TerritoryR NA NA NA NA TreatmentReference:SiteLA:TerritoryR3.375001.66673 2.025 0.04419 * TreatmentSingle Tree:SiteLA:TerritoryR NA NA NA NA TreatmentTypical:SiteLA:TerritoryR NA NA NA NA TreatmentIntensive:SiteLB:TerritoryR NA NA NA NA TreatmentReference:SiteLB:TerritoryR NA NA NA NA TreatmentSingle Tree:SiteLB:TerritoryR NA NA NA NA TreatmentTypical:SiteLB:TerritoryR NA NA NA NA TreatmentIntensive:SiteLC:TerritoryR NA NA NA NA TreatmentReference:SiteLC:TerritoryR NA NA NA NA TreatmentSingle Tree:SiteLC:TerritoryR NA NA NA NA TreatmentTypical:SiteLC:TerritoryR NA NA NA NA TreatmentIntensive:SiteLA:TerritoryT NA NA NA NA TreatmentReference:SiteLA:TerritoryT1.921001.66673 1.153 0.25046 TreatmentSingle Tree:SiteLA:TerritoryT NA NA NA NA TreatmentTypical:SiteLA:TerritoryT NA NA NA NA TreatmentIntensive:SiteLB:TerritoryT NA NA NA NA TreatmentReference:SiteLB:TerritoryT NA NA NA NA TreatmentSingle Tree:SiteLB:TerritoryT NA NA NA NA TreatmentTypical:SiteLB:TerritoryT NA NA NA NA TreatmentIntensive:SiteLC:TerritoryT NA NA NA NA TreatmentReference:SiteLC:TerritoryT NA NA NA NA TreatmentSingle Tree:SiteLC:TerritoryT NA NA NA NA TreatmentTypical:SiteLC:TerritoryT NA NA NA NA TreatmentIntensive:SiteLA:TerritoryW -3.385001.75689 -1.927 0.05542 . TreatmentReference:SiteLA:TerritoryW -0.285001.89766 -0.150 0.88077 TreatmentSingle Tree:SiteLA:TerritoryW -0.235001.96426 -0.120 0.90489 TreatmentTypical:SiteLA:TerritoryW NA NA NA NA TreatmentIntensive:SiteLB:TerritoryW NA NA NA NA TreatmentReference:SiteLB:TerritoryW NA NA NA NA TreatmentSingle Tree:SiteLB:TerritoryW NA NA NA NA TreatmentTypical:SiteLB:TerritoryW NA NA NA NA TreatmentIntensive:SiteLC:TerritoryW NA NA NA NA TreatmentReference:SiteLC:TerritoryW NA NA NA NA TreatmentSingle Tree:SiteLC:TerritoryW 3.336331.50451 2.218 0.02770 * TreatmentTypical:SiteLC:TerritoryW 0.285711.99213 0.143 0.88610 TreatmentIntensive:SiteLA:TerritoryX -3.170001.75689 -1.804 0.07267 .
Re: [R] Nested model - singularities not defined
On Dec 24, 2011, at 12:07 PM, Kristen Mancuso wrote: I am using a nested model in R and the lm output shows 47 not defined because of singularities and I have no idea why. Any help on why this is happening or how to fix this problem would be very much appreciated. Below is the output I received from R. not defined because of singularities This question should be a FAQ if it's not already: http://lmgtfy.com/?q=%22not+defined+because+of+singularities%22 -- David Winsemius, MD West Hartford, CT __ 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] Nested model - singularities not defined
... and John Nash's comment (regarding another issue) may also be appropriate: ... As with many tools in this domain, for effective use they require more knowledge than many of their users possess, and can be dangerous because they seem to work. -- Bert On Sat, Dec 24, 2011 at 9:44 AM, David Winsemius dwinsem...@comcast.net wrote: On Dec 24, 2011, at 12:07 PM, Kristen Mancuso wrote: I am using a nested model in R and the lm output shows 47 not defined because of singularities and I have no idea why. Any help on why this is happening or how to fix this problem would be very much appreciated. Below is the output I received from R. not defined because of singularities This question should be a FAQ if it's not already: http://lmgtfy.com/?q=%22not+defined+because+of+singularities%22 -- David Winsemius, MD West Hartford, CT __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] need help with a time series plotting problem
sorry about the change in variable name (data has column named usgs700 but code has usgs700). actually i am trying to get familiar with R as i need to make more complex time series plots in near future (stackplots, scatterplot etc.). Let me try to explain what i am intending to achieve here. below i have copied first few lines of a large dataset (daily flow records for 4017 days at 8 usgs locations). first column is date and rest of them are flow data measured at these 8 different location. now all i wish to do is to read them in and make time series plots for the entire period of record for each stations. i have attached some figures that i made in excel and i wish to do something like this using R. I know my code might have some errors and that's the reason why i am requesting for help from people who know R better than I do. i am R user for less than a day but i know what i am trying to do is really simple and all i need is to read 9 columns and make a simple time series plot. i would appreciate if anyone can correct the code that i have written below or if they have some alternate way of doing this i would be happy to learn something new regards vibhava DateUSGS700 USGS1000USGS1500USGS 1898 USGS1975 USGS2500USGS2700USGS2800 10/1/2000 0.050.572.322.274.1129.45 29.45 29.45 10/2/2000 0.040.542.121.704.0529.17 29.17 29.17 10/3/2000 0.030.481.931.983.9628.88 28.88 28.88 10/4/2000 0.030.451.761.423.9128.60 28.60 28.60 10/5/2000 0.030.421.641.273.8228.32 28.32 28.32 10/6/2000 0.030.421.531.133.7428.26 28.26 28.26 10/7/2000 0.110.511.595.663.6828.23 28.23 28.23 10/8/2000 0.160.451.703.403.6227.84 27.84 27.84 10/9/2000 0.100.421.782.553.5426.56 26.56 26.56 setwd(J:/Rstuff/flow) # defining the working directory flow=read.delim(flow.dat,header=TRUE,sep=\t) # opening the above tab separated data file plot(flow$USGS1500~as.Date(flow$Date, %m/%d/%y),type=l,xlab=date,ylab=daily discharge (m3/s) ,main=USGS1500,yaxs=i, xaxs=i) #just to test my code, i am trying to make time series #plot of 1 variable USGS1500 http://r.789695.n4.nabble.com/file/n4231737/flow.docx flow.docx -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4231737.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] need help with a time series plotting problem
This should plot all the columns for you:
flow - read.table(text = DateUSGS700 USGS1000USGS1500
USGS1898 USGS1975USGS2500USGS2700
USGS2800
10/1/2000 0.050.572.322.274.1129.45 29.45 29.45
10/2/2000 0.040.542.121.704.0529.17 29.17 29.17
10/3/2000 0.030.481.931.983.9628.88 28.88 28.88
10/4/2000 0.030.451.761.423.9128.60 28.60 28.60
10/5/2000 0.030.421.641.273.8228.32 28.32 28.32
10/6/2000 0.030.421.531.133.7428.26 28.26 28.26
10/7/2000 0.110.511.595.663.6828.23 28.23 28.23
10/8/2000 0.160.451.703.403.6227.84 27.84 27.84
10/9/2000 0.100.421.782.553.5426.56 26.56 26.56
, as.is = TRUE
, header = TRUE
)
# setup margins
par(mar = c(5, 5, 3, 3))
# plot each column of data
for (i in names(flow)[-1]){ # ignore the Date column
plot(as.Date(flow$Date, %m/%d/%y)
, flow[[i]] # column to plot
,type=l
,xlab=date
,ylab=expression(daily discharge ( * m^3/s * ))
,main=i
,yaxs=i
, xaxs=i
)
}
On Sat, Dec 24, 2011 at 1:20 PM, vibhava vibhavasrivast...@gmail.com wrote:
sorry about the change in variable name (data has column named usgs700 but
code has usgs700). actually i am trying to get familiar with R as i need to
make more complex time series plots in near future (stackplots, scatterplot
etc.). Let me try to explain what i am intending to achieve here. below i
have copied first few lines of a large dataset (daily flow records for 4017
days at 8 usgs locations). first column is date and rest of them are flow
data measured at these 8 different location. now all i wish to do is to read
them in and make time series plots for the entire period of record for each
stations. i have attached some figures that i made in excel and i wish to do
something like this using R.
I know my code might have some errors and that's the reason why i am
requesting for help from people who know R better than I do. i am R user for
less than a day but i know what i am trying to do is really simple and all i
need is to read 9 columns and make a simple time series plot.
i would appreciate if anyone can correct the code that i have written below
or if they have some alternate way of doing this i would be happy to learn
something new
regards
vibhava
Date USGS700 USGS1000 USGS1500 USGS 1898 USGS1975
USGS2500 USGS2700 USGS2800
10/1/2000 0.05 0.57 2.32 2.27 4.11 29.45 29.45 29.45
10/2/2000 0.04 0.54 2.12 1.70 4.05 29.17 29.17 29.17
10/3/2000 0.03 0.48 1.93 1.98 3.96 28.88 28.88 28.88
10/4/2000 0.03 0.45 1.76 1.42 3.91 28.60 28.60 28.60
10/5/2000 0.03 0.42 1.64 1.27 3.82 28.32 28.32 28.32
10/6/2000 0.03 0.42 1.53 1.13 3.74 28.26 28.26 28.26
10/7/2000 0.11 0.51 1.59 5.66 3.68 28.23 28.23 28.23
10/8/2000 0.16 0.45 1.70 3.40 3.62 27.84 27.84 27.84
10/9/2000 0.10 0.42 1.78 2.55 3.54 26.56 26.56 26.56
setwd(J:/Rstuff/flow)
# defining the working directory
flow=read.delim(flow.dat,header=TRUE,sep=\t)
# opening the above tab separated data file
plot(flow$USGS1500~as.Date(flow$Date,
%m/%d/%y),type=l,xlab=date,ylab=daily discharge (m3/s)
,main=USGS1500,yaxs=i, xaxs=i)
#just to test my code, i am trying to make time series
#plot of 1 variable USGS1500
http://r.789695.n4.nabble.com/file/n4231737/flow.docx flow.docx
--
View this message in context:
http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4231737.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
--
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.
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[R] Fwd:
http://www.xn--altnilekhap-p9a47gja.gen.tr/unread_message.php?gid=28udihehog=819fyzuvomud=96 __ 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] p values in lmer
This takes me back to listening to a professor lament about the researchers that would spend years collecting their data, then negate all that effort because they insist on using tools that are quick rather than correct. So, before dismissing the use of pvals.fnc you might ask how long it takes to run relative to how long it took to collect the data and the importance of the answer. If you feel the need to compute p-values multiple times, then you may need to rethink your approach (model selection based on repeated p-values results in p-values that are meaningless at best). If you consider the above and still feel the need for a quick p-value rather than a correct one then you can use the SnowsCorrectlySizedButOtherwiseUselessTestOfAnything function from the TeachingDemos package. It is quick (but be sure to fully read the documentation). -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of arunkumar Sent: Thursday, December 22, 2011 9:13 PM To: r-help@r-project.org Subject: [R] p values in lmer hi How to get p-values for lmer funtion other than pvals.fnc(), since it takes long time for execution - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/p-values-in-lmer-tp4227434p4227434.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@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] (no subject)
http://kinderlandia.edu.co/unread_message.php?gid=89yripukin=434cjwalal=57 __ 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] linear interpolation of time series
Dear R users, I have two irregular time series say x and y. Each series is supposed to cover 20 years. The data looks, for instance: x-c(200,178, 330, 127, 420) ## only 5 observations out of the expected 20 annual values y-c(0.35,-0.18,-0.54,0.78,1.7,-1.1,0.2,1.9,0.49)### only 9 observations of the expected 20 I need to intepolate each of the series into equally spaced 20 points. Is there a function and library in R to do this please? Many thanks Tsegaye [[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] linear interpolation of time series
zoo:::na.approx will do nicely. Though you are going to have to supply some sort of time metric or it won't know where interpolations should happen. Something like this is my usual route: x - zoo(1:5, Sys.Date() + 2*(1:5)) x.new - zoo(NA, seq(min(time(x)), to = max(time(x)), by = day)) x.new[time(x)] - x na.approx(x.new) Michael On Sat, Dec 24, 2011 at 5:39 PM, Alemtsehai Abate abatea...@gmail.com wrote: Dear R users, I have two irregular time series say x and y. Each series is supposed to cover 20 years. The data looks, for instance: x-c(200,178, 330, 127, 420) ## only 5 observations out of the expected 20 annual values y-c(0.35,-0.18,-0.54,0.78,1.7,-1.1,0.2,1.9,0.49)### only 9 observations of the expected 20 I need to intepolate each of the series into equally spaced 20 points. Is there a function and library in R to do this please? Many thanks Tsegaye [[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. __ 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] extract factor scores post-varimax
Hello all, I've run a principal component regression using the PLS package. I then applied varimax rotation (i.e., using http://stat.ethz.ch/R-manual/R-patched/library/stats/html/varimax.html). I cannot figure out how to extract the factor loadings post-varimax. Is there a command to do this? scores(x) does not do it. Thanks and happy holidays -- View this message in context: http://r.789695.n4.nabble.com/extract-factor-scores-post-varimax-tp4232192p4232192.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] need help with a time series plotting problem
thanks for your reply. i don't think your solution is what i am looking for. Please look at the attached two plots; first plot is what i have made using excel and the second plot is what i was getting when i was running my script through R yesterday night. I think my script was reading and implementing things nicely but the only problem was that at the end of each year my time series line use to start from the same axis label (instead of keep creating more labels and extending the x-axis. therefore what happened was that unlike the excel plot which has x-axis from oct00-oct2011, i have x-axis labels jan-nov. i am looking for something similar to excel where you select the axis and right click on it and you can extend the x-axis. so to sum up i *just need* to know how to extend my x-axis (see attached figures) so that i get a continuous line (i need one line for each plot representing 11 years of record AND DO NOT want 11 lines representing the same thing). regards vibhava http://r.789695.n4.nabble.com/file/n4232160/flow.docx flow.docx -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-a-time-series-plotting-problem-tp4230672p4232160.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] need help with a time series plotting problem
Try this; this should put the labels on the x-axis like Excel does:
flow - read.table(text = DateUSGS700 USGS1000USGS1500
USGS1898 USGS1975USGS2500USGS2700 USGS2800
10/1/2001 0.050.572.322.274.1129.45 29.45 29.45
10/1/2002 0.040.542.121.704.0529.17 29.17 29.17
10/1/2003 0.030.481.931.983.9628.88 28.88 28.88
10/1/2004 0.030.451.761.423.9128.60 28.60 28.60
10/1/2005 0.030.421.641.273.8228.32 28.32 28.32
10/1/2006 0.030.421.531.133.7428.26 28.26 28.26
10/1/2007 0.110.511.595.663.6828.23 28.23 28.23
10/1/2008 0.160.451.703.403.6227.84 27.84 27.84
10/1/2009 0.100.421.782.553.5426.56 26.56 26.56
, as.is = TRUE
, header = TRUE
)
# setup margins
par(mar = c(5, 5, 3, 3))
# create x-axis with yearly tick marks starting in Oct
x_range - seq(from = as.Date(2001-10-1)
, to = as.Date(2011-10-1)
, by = '1 year'
)
# now create the format of month-year
x_label - sprintf(%s-%02d, format(x_range, %b),
as.POSIXlt(x_range)$year %% 100)
# plot each column of data
for (i in names(flow)[-1]){ # ignore the Date column
plot(as.Date(flow$Date, %m/%d/%Y)
, flow[[i]] # column to plot
,type=l
,xlab=date
,ylab=expression(daily discharge ( * m^3/s * ))
,main=i
,yaxs=i
, xaxs=i
, xaxt = 'n'
, xlim = range(x_range)
)
axis(1, at = x_range, labels = x_label, las = 2)
}
On Sat, Dec 24, 2011 at 3:23 PM, jim holtman jholt...@gmail.com wrote:
This should plot all the columns for you:
flow - read.table(text = Date USGS700 USGS1000 USGS1500
USGS1898 USGS1975 USGS2500 USGS2700
USGS2800
10/1/2000 0.05 0.57 2.32 2.27 4.11 29.45 29.45 29.45
10/2/2000 0.04 0.54 2.12 1.70 4.05 29.17 29.17 29.17
10/3/2000 0.03 0.48 1.93 1.98 3.96 28.88 28.88 28.88
10/4/2000 0.03 0.45 1.76 1.42 3.91 28.60 28.60 28.60
10/5/2000 0.03 0.42 1.64 1.27 3.82 28.32 28.32 28.32
10/6/2000 0.03 0.42 1.53 1.13 3.74 28.26 28.26 28.26
10/7/2000 0.11 0.51 1.59 5.66 3.68 28.23 28.23 28.23
10/8/2000 0.16 0.45 1.70 3.40 3.62 27.84 27.84 27.84
10/9/2000 0.10 0.42 1.78 2.55 3.54 26.56 26.56 26.56
, as.is = TRUE
, header = TRUE
)
# setup margins
par(mar = c(5, 5, 3, 3))
# plot each column of data
for (i in names(flow)[-1]){ # ignore the Date column
plot(as.Date(flow$Date, %m/%d/%y)
, flow[[i]] # column to plot
,type=l
,xlab=date
,ylab=expression(daily discharge ( * m^3/s * ))
,main=i
,yaxs=i
, xaxs=i
)
}
On Sat, Dec 24, 2011 at 1:20 PM, vibhava vibhavasrivast...@gmail.com wrote:
sorry about the change in variable name (data has column named usgs700 but
code has usgs700). actually i am trying to get familiar with R as i need to
make more complex time series plots in near future (stackplots, scatterplot
etc.). Let me try to explain what i am intending to achieve here. below i
have copied first few lines of a large dataset (daily flow records for 4017
days at 8 usgs locations). first column is date and rest of them are flow
data measured at these 8 different location. now all i wish to do is to read
them in and make time series plots for the entire period of record for each
stations. i have attached some figures that i made in excel and i wish to do
something like this using R.
I know my code might have some errors and that's the reason why i am
requesting for help from people who know R better than I do. i am R user for
less than a day but i know what i am trying to do is really simple and all i
need is to read 9 columns and make a simple time series plot.
i would appreciate if anyone can correct the code that i have written below
or if they have some alternate way of doing this i would be happy to learn
something new
regards
vibhava
Date USGS700 USGS1000 USGS1500 USGS 1898 USGS1975
USGS2500 USGS2700 USGS2800
10/1/2000 0.05 0.57 2.32 2.27 4.11 29.45 29.45 29.45
10/2/2000 0.04 0.54 2.12 1.70 4.05 29.17 29.17 29.17
10/3/2000 0.03 0.48 1.93 1.98 3.96 28.88 28.88 28.88
10/4/2000 0.03 0.45 1.76 1.42 3.91 28.60 28.60 28.60
10/5/2000 0.03 0.42 1.64 1.27 3.82 28.32 28.32 28.32
10/6/2000 0.03 0.42 1.53 1.13 3.74 28.26 28.26 28.26
10/7/2000 0.11 0.51 1.59 5.66 3.68 28.23
