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 0.0
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 implemen
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?
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.n
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
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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
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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
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
... 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, D
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 r
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
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
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 0.0277505
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,
0.194
It didn't work. :(
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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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 pos
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
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 ob
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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 conferenc
On Sat, Dec 24, 2011 at 8:38 AM, William Revelle 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 interpretati
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
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)
di
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,+CN&output=csv&key=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTk
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 wrote:
>
> Matt Considine wrote
>>
>> Hi,
>> I am trying to work with the output of the MINE analy
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 wrote:
> Dear R Users,
> I am a beginner in R programming and need some help
> wi
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 E
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 te
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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,+CN&output=csv&key=ABQIq8Fnd_oUka-7RdS6BrD7GBTqeABoQuNTXS36G_rkiwQnKRW6GBTkns8JpKz6y6dScgB8827dlddUlg";),
n=1, warn=FALSE)
[1
thank you vey much Jean.
On 12/21/11, Jean V Adams [via R]
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
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
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