[R] Leaflet maps. Nudging co-incident markers
I have a dataset showing points, with a category for each point and its
location.
I simply want to display my points, in a way that users can toggle the points
on and off by category.
Where I have two objects in the same category I'd like to display them nudged
to appear as two distinct, but very close points.
I have made reproduceable example (the places are not real), which is loosely
based on a tutorial I found recently
(https://allthisblog.wordpress.com/2016/10/12/r-311-with-leaflet-tutorial/)
I have three categories of things (cafes, libraries and galleries), at three
locations but have four objects in my set. This is because on of my locations
has two functions - there is a cafe at a gallery (North St Gallery and the
Gallery Cafe on the same site)
If I make a selection that includes galleries and cafes there are just two
points. I would like to nudge the point for the North St Galley and the Gallery
Cafe so they appear as two (very close) points on the map and display the name
when clicked on.
Also if anyone has any suggestions for generally tidying up the code I'd be
grateful as my real version is much more complex with many more points and
marker categories. I believe there are Java libraries out there for managing
markers and how they behave, but I'm hoping this can be done in the Leaflet R
library somehow.
The only solution I could think of was to interrogate the entire dataframe,
identify points that were had the same co-ords and move them diagonally apart
by adding and subtracting a fixed amount of longitude and latitude from the
co-ordinates.
Thanks in advance.
GavinR
Here is the code:
library(leaflet)
#make data frame of points
idno=c(1,2,3,4)
x=c(-1.9116, -1.9116,-1.9237,-1.91848)
y=c(52.4898,52.4898,52.5015,52.4851)
cat=c('Gallery','Cafe','Library','Cafe')
n=c('North St Gallery','Gallery cafe', 'South St Library', 'Coffee 2 go')
d<-data.frame(idno,x,y,cat,n)
#get a map and zoom into approx area of interest
m=leaflet()%>% setView(lng = -1.935, lat=52.485, zoom=12)
m=addTiles(m)
m
#create groups of objects
c= subset(d,cat=="Cafe")
l= subset(d, cat=="Library")
g= subset(d, cat=="Gallery")
#add markers
m=addCircleMarkers(m,
lng = c$x,
lat = c$y,
popup = c$n,
radius = 5,
stroke =FALSE,
fillOpacity = 0.75,
group = "1 - Cafes")
m=addCircleMarkers(m,
lng = g$x,
lat = g$y,
popup = g$n,
radius = 5,
stroke =FALSE,
fillOpacity = 0.75,
group = "2 - Galleries")
m=addCircleMarkers(m,
lng = l$x,
lat = l$y,
popup = l$n,
radius = 5,
stroke =FALSE,
fillOpacity = 0.75,
group = "3 - libraries")
m = addLayersControl(m, overlayGroups = c("1 - Cafes","2 - Galleries","3 -
libraries"))
m
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[R] Compiling R2.15.1 on ubuntu with x86-64 architecture and shared library
I am sure I am providing insufficient information, please ask for more. I installed R 2.14.2 on my Ubuntu laptop with and AMD64 processor and also installed RStudio and everything worked fine. Now, I tried to build R 2.15.1 from source and installed it using defaults. RStudio now complained that R was not built as a shared library. Went back and uninstalled, and configured with -enable-R-shlib Now make fails at: /usr/bin/ld: CConverters.o: relocation R_X86_64_325 against '.rodata' can not be used when making a shared object; recompile with -fPIC Being new to Linux, I have no idea how to recompile with -fPIC Any help would be appreciated. Mike W. Michael Conklin Executive Vice President | GfK Marketing Science | Consumer Experiences North America GfK Custom Research, LLC | 8401 Golden Valley Road | Minneapolis, MN, 55427 T +1 763 417 4545 | M +1 612 567 8287 www.gfk.comhttp://www.gfk.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] Cannot write a dataframe to xls or csv Windows 7
Thank you so much, Yes I need to look up how to make a reproducible example, William I will try your advice, I believe this will be my salvation here, once I get my computer. Pancho On Sep 13, 2012, at 23:59, William Dunlap wdun...@tibco.com wrote: -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R. Michael Weylandt Sent: Thursday, September 13, 2012 2:52 PM To: pmulonge Cc: r-help@r-project.org Subject: Re: [R] Cannot write a dataframe to xls or csv Windows 7 On Thu, Sep 13, 2012 at 4:05 PM, pmulonge resea...@namibia.pharmaccess.org wrote: Hello, I have a similar issue , but in my case I am using Windows 7 i try the following command to write a dataframe to xls using the xlsReadWrite package or the write.csv function write.xls(DATA,'Reg_IDcleaned.xls') or I will write.csv replacin the suffix with .csv I get absolutely no error message and the setwd appears at top of my code with absolute path Immediately after the write.xls(..., file=something) run normalizePath(something, mustWork=TRUE) and it should give a full path to the file called something. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com However when I look in the relevant folder ,the xls or cdv outfile is nowhere to be found. Is there a gremlin in the computer or what? It seems unlikely, but I suppose it can't be ruled out Before, this write.xls was working fine and now it stopped, i used the write.csv to test if it was an error in the package. So since write.csv also fails to produce the oufile in .csv in the relevant folder, I am at my witts end Firstly, please do quote context: most of us don't read nabble and I have no idea what similar problem you're having. (Well, actually, I clicked the link and I do, but I'm not letting on) Secondly, can you whip up a reproducible example? http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible- example I see no reason why setwd(~) # Replace with appropriate Windows-ism write.csv(data.frame(1:5, letters[1:5])) wouldn't work if you have the right permissions, but perhaps you are doing something funny. Cheers, Michael __ 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] Tendonitis and R users
Hello This request asks something beyond the technicalities of the R language, I would like to ask you wonderful people if you have ever suffered as programmers ( or de facto programmers like myself though I am a 'research assistant') from tendonitis and how you coped with it, i have golfer's elbow on both sides. Any resources? Pancho Mulongeni Namibia __ 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] Exporting data to spss
I have a dataframe of 80+ columns and over 700 rows. I use write.foreigin(data,C:/filename.dat,codefile.sps) and it does write out the .dat file and the code file. Problem is that when I open the codefile in SPSS 20, I can an error message saying there are too many variables and something about the formatting (this is not an SPSS list so the details of the error are not germane). Question: is there another way of reading data into SPSS without having to first create a text file, excel file that is then opened in SPSS manually. Is there any other function apart from write.foreign? Thank you [[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] analysing a three level reponse
Hello, I am struggling to figure out how to analyse a dataset I have inherited (please note this was conducted some time ago, so the data is as it is, and I know it isn't perfect!). A brief description of the experiment follows: Pots of grass were grown in 1l pots of standad potting medium for 1 month with a regular light and watering regime. At this point they were randomly given 1l of one of 4 different pesticides at one of 4 different concentrations (100%, 75%, 50% or 25% in water). There were 20 pots of grass for each pesticide/concentration giving 320 pots. There were no control (untreated) pots. The response was measured after 1 week and recorded as either: B1 - grass dead B2 - grass affected but not dead B3 - no visible effect I could analyse this as lethal effect vs non-lethal effect (B1 vs B2+B3) or some effect vs no effect (B1+B2 vs B3) binomial model, but I can't see how to do it with three levels. Any pointing in the right direction greatly appreciated! Thanks Matt -- Disclaimer: This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you have received this email in error please notify me at matt.el...@basc.org.uk then delete it. BASC may monitor email traffic. By replying to this e-mail you consent to BASC monitoring the content of any email you send or receive from BASC. Any views expressed in this message are those of the individual sender, except where the sender specifies with authority, states them to be the views of the British Association for Shooting and Conservation. BASC can confirm that this email message and any attachments have been scanned for the presence of computer viruses but recommends that you make your own virus checks. Registered Industrial and Provident Society No.: 28488R. Registered Office: Marford Mill, Rossett, Wrexham, LL12 0HL. -- [[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] Odp: Help me with prediction in linear model
Thanks Murphy and pikal, I need another help,for fitting first fourier transformation ,i used following thing .Please advise on this beer_monthl has 400+ records EXample: head(beer_monthly) beer 1 93.2 2 96.0 3 95.2 4 77.1 5 70.9 6 64.8 time-seq(1956,1995.2,length=length(beer_monthly)) sin.t-sin(2*pi*time) cos.t-cos(2*pi*time) beer_fit_fourier=lm(beer_monthly[,1]~poly(time,2)+sin.t+cos.t) #this is not working beer_fit_fourier=lm(beer_monthly[,1]~time+time2+sin.t+cos.t) #it is working #prediction is not workinng tpred_four - data.frame(time = seq(1995, 1998, length = 20)) predict(beer_fit_fourier, newdata = tpred_four) Is there any way to fit first fourier frequency , Please assist. Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/Help-me-with-prediction-in-linear-model-tp2297313p2300991.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 me with holt-winter model
Please help me with this i need to submit my thesis . Thanks In advance -- View this message in context: http://r.789695.n4.nabble.com/help-me-with-holt-winter-model-tp2295552p2298464.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] Please advise acf and pacf in order to determine order of Arima
I have data as below.Please let me know how the ACF and Pacf used to determine the order od arima model. Is there any rules need to be followed to determine order.Please advise turkey.price.ts Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 2001 1.58 1.75 1.63 1.45 1.56 2.07 1.81 1.74 1.54 1.45 0.57 1.15 2002 1.50 1.66 1.34 1.67 1.81 1.60 1.70 1.87 1.47 1.59 0.74 0.82 2003 1.43 1.77 1.47 1.38 1.66 1.66 1.61 1.74 1.62 1.39 0.70 1.07 2004 1.48 1.48 1.50 1.27 1.56 1.61 1.55 1.69 1.49 1.32 0.53 1.03 2005 1.62 1.63 1.40 1.73 1.73 1.80 1.92 1.77 1.71 1.53 0.67 1.09 2006 1.71 1.90 1.68 1.46 1.86 1.85 1.88 1.86 1.62 1.45 0.67 1.18 2007 1.68 1.74 1.70 1.49 1.81 1.96 1.97 1.91 1.89 1.65 0.70 1.17 2008 1.76 1.78 1.53 1.90 acf(turkey.price.ts,plot=FALSE) Autocorrelations of series ‘turkey.price.ts’, by lag 0. 0.0833 0.1667 0.2500 0. 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333 1.000 0.465 -0.019 -0.165 -0.145 -0.219 -0.215 -0.122 -0.136 -0.200 -0.016 0.9167 1. 1.0833 1.1667 1.2500 1. 1.4167 1.5000 1.5833 0.368 0.723 0.403 -0.013 -0.187 -0.141 -0.180 -0.226 -0.130 pacf(turkey.price.ts,plot=FALSE) Partial autocorrelations of series ‘turkey.price.ts’, by lag 0.0833 0.1667 0.2500 0. 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333 0.9167 0.465 -0.300 -0.020 -0.060 -0.218 -0.054 -0.061 -0.211 -0.180 0.098 0.299 1. 1.0833 1.1667 1.2500 1. 1.4167 1.5000 1.5833 0.571 -0.122 -0.077 -0.075 0.119 0.064 -0.149 -0.061 Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/Please-advise-acf-and-pacf-in-order-to-determine-order-of-Arima-tp2298474p2298474.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] KALMAN
Hello, I am looking for some very simple, step by step, hands on application/examples/notes etc. on setting up a multivariate time series Kalman filter model in R. Any help/pointers much appreciated. Best regards, Costas __ 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] List of zoo multivariate series
Hi,
I have a list which is basically a database of zoo multivariate objects:
x.Date - as.Date(2003-02-01) + c(1:100) - 1
x - zoo(rnorm(100), x.Date)
z - zoo(rnorm(100), x.Date)
y - zoo(rnorm(100), x.Date)
k - zoo(rnorm(100), x.Date)
a-merge(x,z)
b-merge(y,k)
c-merge(z,x,y)
test-list(a=a, b=b, c=c)
Can I access the elements of the list in a for/next do loop?
for example something like:
for (i in 1:3)
{
x11()
plot(test[i])
}
So far I can only recover the zoo objects as test$a, test$b etc. etc...
Thanks,
Costas
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[R] merge.zoo and fill
Hello again, I merge different zoo time series with prices at different dates. This returns a multivariate zoo object with NA's at various points i.e., 2010-02-28 NA NA NA NA 850.2 2444.4 NA NA NA NA NA NA NA 2010-03-01 61.1 55.361.581.24 NA NA 1712.2 3.3 11139.3 163.7 2242.4 9015.6 109.791 2010-03-31 NA NA NA NA 846.5 2439.0 NA NA NA NA NA NA NA 2010-04-01 66.9 49.465.784.48 NA NA 1700.6 3.8 11203.0 164.4 2245.7 9064.5 109.887 2010-04-30 NA NA NA NA 838.0 2440.0 NA NA NA NA NA NA NA 2010-05-01 66.6 45.665.773.84 NA NA 1705.3 4.0 11252.0 165.1 2248.4 9113.9 110.065 Will the fill parameter of the zoo.merge() allow a missing value for each zoo object merged to take the last available (previous in time) price for that particular time series or one has to do that using the na.locf() command afterwards? I want ot keep the series as zoo and not convert to other formats or interpolate (as in the FAQ) Thanks, Costas __ 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] Duplicate dates in zoo objects
Hello, I have a zoo time series read from an excel file which has some dates the same, such as the following example: 02/10/1995 4925.5 30/10/1995 4915.9 23/01/1996 4963.5 23/01/1996 5009.2 04/03/1996 5031.9 # here 04/03/1996 5006.5 # here 03/04/1996 5069.2 03/05/1996 5103.7 31/05/1996 5107.1 01/07/1996 5153.1 02/08/1996 5151.7 Is there a simple way to keep the last price of the ones that have the same dates? 04/03/19965031.9 04/03/19965006.5 i.e., keep only the 04/03/19965006.5 price and discard the previous one... Is there an implicit function that does that or do I need some sort of recursive algorithm? You can try a solution on this example (for convenience): x.Date - as.Date(2003-02-01) + c(1, 3, 7, 7, 14) - 1 x - zoo(rnorm(5), x.Date) Zoo object has 2 prices with same dates. Many thanks in advance, Costas __ 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] Time problems (POSIXct)
Thanks Gabor, I suppose that in the example below, you can convet Date to POSIXct as well, can't you? Best, Costas On 20/05/2010 18:08, Gabor Grothendieck wrote: The warning message does tell you exactly what was wrong. You are trying to merge zoo objects that have two different index classes. The first is POSIXct and the second is Date class. Next time please provide your data in reproducible form as shown here: library(zoo) z1- zoo(1:6, as.Date(c(1950-01-05, 1950-01-06, 1950-01-09, + 1950-01-10, 1950-01-11, 1950-01-12))) z2- zoo(c(16.66, 16.85, 16.93, 16.98, 17.08, 17.03), + as.POSIXct(c(1950-01-03, GMT, 1950-01-04, GMT, 1950-01-05, GMT, + 1950-01-06, GMT, 1950-01-09 GMT, 1950-01-10 GMT))) merge(z1, z2 = aggregate(z2, as.Date, identity)) z1z2 1950-01-03 NA 16.66 1950-01-04 NA 16.85 1950-01-05 1 16.93 1950-01-06 2 16.98 1950-01-09 3 17.08 1950-01-10 4 17.03 1950-01-11 5NA 1950-01-12 6NA # or z2.Date- z2 time(z2.Date)- as.Date(time(z2.Date)) merge(z1, z2.Date) z1 z2.Date 1950-01-03 NA 16.66 1950-01-04 NA 16.85 1950-01-05 1 16.93 1950-01-06 2 16.98 1950-01-09 3 17.08 1950-01-10 4 17.03 1950-01-11 5 NA 1950-01-12 6 NA On Thu, May 20, 2010 at 10:48 AM, Researchrisk2...@ath.forthnet.gr wrote: Hello, I have a zoo time series object (say a) with the following time stamp/format [1] 1950-01-03 GMT 1950-01-04 GMT 1950-01-05 GMT 1950-01-06 GMT [5] 1950-01-09 GMT 1950-01-10 GMT and another (say b) with [1] 1950-01-05 1950-01-06 1950-01-09 1950-01-10 1950-01-11 [6] 1950-01-12 I want to merge these series but when I try: head(merge(a, b )) a b 1950-01-03 02:00:00 16.66 NA 1950-01-04 02:00:00 16.85 NA 1950-01-05 02:00:00 16.93 NA 1950-01-06 02:00:00 16.98 NA 1950-01-09 02:00:00 17.08 NA 1950-01-10 02:00:00 17.03 NA Warning message: In merge.zoo(a, b) : Index vectors are of different classes: POSIXt Date Anybody can help? I tried as.POSIXct(a,tz=GMT) but it doesn't seem to work Many thanks in advance! Costas _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 10.1 year 2009 month 12 day14 svn rev50720 language R version.string R version 2.10.1 (2009-12-14) __ 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. __ Information from ESET Smart Security, version of virus signature database 5133 (20100520) __ The message was checked by ESET Smart Security. http://www.eset.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] Time problems (POSIXct)
Hello, I have a zoo time series object (say a) with the following time stamp/format [1] 1950-01-03 GMT 1950-01-04 GMT 1950-01-05 GMT 1950-01-06 GMT [5] 1950-01-09 GMT 1950-01-10 GMT and another (say b) with [1] 1950-01-05 1950-01-06 1950-01-09 1950-01-10 1950-01-11 [6] 1950-01-12 I want to merge these series but when I try: head(merge(a, b )) a b 1950-01-03 02:00:00 16.66 NA 1950-01-04 02:00:00 16.85 NA 1950-01-05 02:00:00 16.93 NA 1950-01-06 02:00:00 16.98 NA 1950-01-09 02:00:00 17.08 NA 1950-01-10 02:00:00 17.03 NA Warning message: In merge.zoo(a, b) : Index vectors are of different classes: POSIXt Date Anybody can help? I tried as.POSIXct(a,tz=GMT) but it doesn't seem to work Many thanks in advance! Costas _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 10.1 year 2009 month 12 day14 svn rev50720 language R version.string R version 2.10.1 (2009-12-14) __ 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] Histogram Bin
Hello, Is there a function that returns the number of the bin (or quantile, or percentile etc. etc.) that a value of a variable may belong to? Tor example: breaks-hist(variable, 18, plot=FALSE) If the following breaks are 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 the boundaries of successive bins of a histogram, then value 6 belongs to the 2nd bin. Best regards, Costas __ 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] mvpart : Printing response values at terminal nodes
I have created a multivariate regression tree using mvpart, with 3-4 responses. Though the plot shows bargraphs for each response, I would like to have the VALUES of the responses printed or indicated (via a scale or something) alongside the bargraph. Is this possible ?? Thanks, Manjunath [[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] N'th of month working day problem
Dear Gabor,
Thanks for your reply. however:
tail(DJd)
^DJI.Close
2010-04-01 10927.07
2010-04-05 10973.55
2010-04-06 10969.99
2010-04-07 10897.52
2010-04-08 10927.07
*2010-04-09 10997.35*
tail(ag)
2009-11-30 10344.84
2009-12-31 10428.05
2010-01-31 10067.33
2010-02-28 10325.26
2010-03-31 10856.63
*2010-04-30 10997.35
*
It seems the script makes up dates (?)
Best,
Costas
On 09/04/2010 14:55, Gabor Grothendieck wrote:
The function seems to be working properly. You are asking for a day
of the month which does not exist. I assume this was written a very
long time ago since there are easier ways to do this now. yearmon
class gives an object representing the year and month of a date and if
ym is such an object then as.Date(ym) gives the first of the month and
as.Date(ym, frac = 1) gives the last of the month so:
# nth day of month or last day of month if less
nth.of.month- function(date, n) {
ym- as.yearmon(date)
pmin(as.Date(ym) + n - 1, as.Date(ym, frac = 1))
}
ag- aggregate(DJd, nth.of.month(time(DJd), 31), tail, 1)
On Fri, Apr 9, 2010 at 7:01 AM, Researchrisk2...@ath.forthnet.gr wrote:
Dear all,
Some time ago I received some very kind help (special thanks to Gabor) to
construct a function that isolates the n'th working day of each month for
zoo object (time series) to create monthly data from daily observations.
I found out that the code works fine except for the 29 till 31st dates of
each month as it skips some months (February for example).
If you could help me isolate the problem I would be grateful as I can not
find a way to explain to R to keep the last working day of month if I
choose the 29th, 30th or 31st dates...
I enclose a working version of the function and a script for demo purposes.
Many thanks in advance,
Costas
library(fImport)
library(zoo)
DJ-yahooSeries(^DJI, frequency=daily, nDaysBack=1)
DJd-as.zoo(DJ[,4])
### Choose number of day for month
chooseday-function(z, day)
{
# z.na is same as z but with missing days added using NAs
# Its formed by merging z with a zoo-width series containing all days.
rng- range(time(z))
z.na- merge(z, zoo(, seq(rng[1], rng[2], by = day)))
# form a series that has NAs wherever z.na does but has 1, 2, 3, ...
# instead of z.na's data values and then use na.locf to fill in NAs
idx- na.locf(seq_along(z.na) + (0 * z.na))
# pick off elements of z.na corresponding to i'th of month
noofday- paste(day)
if (day10) noofday-paste(0,day, sep=)
tempdata-z.na[idx[format(time(z.na), %d) == noofday]]
return(tempdata)
}
length(chooseday(DJd,1))
length(chooseday(DJd,2))
length(chooseday(DJd,31))
length(chooseday(DJd,30))
length(chooseday(DJd,29))
length(chooseday(DJd,28))
tail(chooseday(DJd,31))
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[R] Exchanging zoo object dates
Hi, I have two zoo objects (time series of same frequency) say A and B. I need to exchange A's dates for B's. Is there and easy way? I managed to do it by converting A to a vector and pasting it on a zoo vector (all ones) with B's dates but I wonder if there is an easier way by some zoo command I have overlooked... Thanks in advance, Costas __ 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] Playwith with Linux Mint
Hi, I am having problems installing playwith on a Linux Mint ver. 8 (Helena) computer running R-2.10.1. src/base/R-2/R-2.10.1.tar.gz Anybody facing similar problems? Is it a GTK issue? I can not install the cairo related packages/libraries as well. Any pointers very welcome. Thanks + bestregards, Costas [[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] List of zoo dataframes
Hello,
I have various datasets of zoo time series (merged into single frames
via the merge command) such as
datevar0var1var2var3
08/07/1996 652.54 0.2239220.5158190.502638
08/08/1996 662.59 0.9978410.0003830.999806
06/09/1996 655.68 0.9016850.5697630.866333
08/10/1996 700.64 0.2688550.2447010.329285
08/11/1996 730.82 0.4384070.5014270.461374
06/12/1996 739.6 0.4322330.5621750.052423
08/01/1997 748.41 0.6162110.7992115.96E-09
07/02/1997 789.56 0.3784150.6450880.93862
07/03/1997 804.97 0.1427060.4221560.145648
08/04/1997 766.12 0.990.9994350.998576
08/05/1997 820.26 0.8107950.9660440.000427
06/06/1997 858.01 0.9986520.9982870.994222
08/07/1997 918.75 0.9515530.9742510.89633
I am trying to put all these into a sort of a list so I can invoke them
via the list
and not individually so I can run a batch of statistical analyses on them.
I also want to have the specification of the analysis on a list of some
sort as well.
For example:
List a has 3 zoo data sets and list b has 2 specifications (calls)
for a function (say, analysis(specification, inputdata) ) that runs
some statistical manipulations (regressions, nonlin models, etc. etc.)
on the elements of the zoo data frames. Thus I want to be able to run
(this is pseudo R code):
for (i in 1:3)
{
output1-analysis(b[1], a[i])
output2-analysis(b[2], a[i])
}
Thanks in advance for any pointers or help.
Best,
Costas
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[R] ZOO object colnames refering to Dates
Hello, I have large zoo objects (about 100 or more time series merged next to eachother). Example: X05.Oct.99 X05.Nov.99 X05.Dec.99 X05.Jan.00 X05.Feb.00 X05.Mar.00 X05.Apr.00 X05.May.00 X05.Jun.00 [1,] 5649.3 5679.4 5679.4 5679.4 5679.4 5679.4 5679.4 5679.4 5679.4 [2,] 5682.7 5719.2 5719.2 5719.2 5719.2 5719.2 5719.2 5719.2 5719.2 [3,] 5697.5 5745.5 5745.5 5745.5 5745.5 5745.5 5745.5 5745.5 5745.5 [4,] 5723.9 5767.8 5767.8 5767.8 5767.8 5767.8 5767.8 5767.8 5767.8 [5,] 5782.1 5829.8 5829.8 5829.8 5829.8 5829.8 5829.8 5829.8 5829.8 [6,] 5815.0 5850.9 5850.9 5850.9 5850.9 5850.9 5850.9 5850.9 5850.9 The column names contain dates, i.e., X05.Oct.99 stands for a time sequence that was obtained on 05/10/1999. Each column is obtained month + 1 from the previous one. Is it possible to access these columns (time series data) via a for/next loop (or a while) that uses dates i/o (i in 1:8) in the above example? Say: if month==Oct and year==1999 choose column X05.Oct.99 ... Could I also generate automatically a variable, appropriately named from the corresponding columns name that would contain the data and access it by that date? Any pointers really appreciated. Thanks in advance, Costas __ Information from ESET Smart Security, version of virus signature database 4857 (20100211) __ The message was checked by ESET Smart Security. http://www.eset.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] Using tcltk or other graphical widgets to view zoo time series objects
Dear Felix, Thanks. That was most useful. I am having trouble though exporting values of variables to the R environment from within the playwith. Say I need to do a claculation by setting up an extra button within playwith(xyplot(dat[,c(1,i)]), parameters = list(i = 1:100, do_something = function(playState) print(playState$env$i), do_something_else = function(playstate) ###Export here mean value of sequence playState$env$i### ) ) Can I pass the value of this to the rest of R ? I could not find something in the help pages. Can you direct me to it please? Thanks + Best regards, Costas Felix Andrews wrote: The playwith package might help, though if I understand the problem correctly, the help(xyplot.zoo) example is not so relevant. If you want to switch between many series you could use a spin-button or somesuch. To execute a function you can create a button. If you have a hundred-column dataset like dat - zoo(matrix(rnorm(100*100),ncol=100), Sys.Date()+1:100) colnames(dat) - paste(Series, 1:100) Then this will give you a spin button to choose the column to plot, and a button to print out the current series number. playwith(xyplot(dat[,c(1,i)]), parameters = list(i = 1:100, do_something = function(playState) print(playState$env$i)) ) Note that the playwith package uses RGtk2, and therefore requires the GTK+ libraries to be installed on your system. On 28 January 2010 23:16, Gabor Grothendieck ggrothendi...@gmail.com wrote: There is an example of using zoo together with the playwith package at the end of the examples section of help(xyplot.zoo) which may address this. On Thu, Jan 28, 2010 at 7:10 AM, Research risk2...@ath.forthnet.gr wrote: Dear all, I am looking at the R-help entry below: http://finzi.psych.upenn.edu/R/Rhelp02/archive/26640.html I have a more complicatedt problem. I have a zoo time series frame with 100+ sequences. I want to cycle through them back and forth and compare them to the 1st column at any time. I need also a button to click when I need the viewed-selected sequence (that is being compared to the 1st column one) to be manipulated (by some algorithm or be saved individually etc. etc.)... I am trying to modify the code at the above link but somehow I can not make it to work with zoo time series objects. Any help would be greatly appreciated. Thanks in advance, Costas __ Information from ESET Smart Security, version of virus signature database 4813 (20100128) __ The message was checked by ESET Smart Security. http://www.eset.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. __ Information from ESET Smart Security, version of virus signature database 4823 (20100201) __ The message was checked by ESET Smart Security. http://www.eset.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] Using tcltk or other graphical widgets to view zoo time series objects
Many thanks Liviu!
Liviu Andronic wrote:
On 2/1/10, Research risk2...@ath.forthnet.gr wrote:
do_something_else = function(playstate) ###Export here mean value of
sequence playState$env$i###
Would this do the trick?
playwith(xyplot(dat[,c(1,i)]), parameters = list(i = 1:100,
do_something = function(playState) print(playState$env$i),
do_something_else = function(playState) print(mean(dat[,
playState$env$i])),
do_something_diff = function(playState) {
assign(x, mean(dat[, playState$env$i]), envir = .GlobalEnv)
print(x)
}
)
)
Liviu
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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] Using tcltk or other graphical widgets to view zoo time series objects
Dear all, I am looking at the R-help entry below: http://finzi.psych.upenn.edu/R/Rhelp02/archive/26640.html I have a more complicatedt problem. I have a zoo time series frame with 100+ sequences. I want to cycle through them back and forth and compare them to the 1st column at any time. I need also a button to click when I need the viewed-selected sequence (that is being compared to the 1st column one) to be manipulated (by some algorithm or be saved individually etc. etc.)... I am trying to modify the code at the above link but somehow I can not make it to work with zoo time series objects. Any help would be greatly appreciated. Thanks in advance, Costas __ Information from ESET Smart Security, version of virus signature database 4813 (20100128) __ The message was checked by ESET Smart Security. http://www.eset.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] piecewise regression using nlme function
Is a 95% CI on a breakpoint fixed effect legitimate when the nonlinear
equation is continuous but not differentiable at the breakpoint?
I used the nlme to generate a non-linear mixed-effects piecewise model
with initial slope zero prior to an unknown breakpoint. It did give a
95% CI for the breakpoint.
Thanks
Erik
whole model is complex, but here is the simple equation used:
dr.hockey-function(a,b,x,brk){
a+ifelse(x=brk,0,b*(x-brk))}
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[R] End of Month date capture
Dear R-users, I have the following zoo object: x1x2 x3 x4 x5 x6 1998-08-31 -0.0704375904 NA NA NA NA NA 1998-09-01 0.0379028122 NA NA NA NA 0.00609139 1998-09-02 -0.0038191639 NA NA NA NA NA 1998-09-03 -0.0083235389 NA NA NA NA NA 1998-09-04 -0.0085576782 NA NA 0.0028570541 NA NA 1998-09-07 0.00 NA NA NA NA NA 1998-09-08 0.0496459618 NA NA NA NA NA 1998-09-09 -0.0170081847 NA NA NA NA NA 1998-09-10 -0.0261897076 NA NA NA NA NA 1998-09-11 0.0290280530 NA NA NA NA NA 1998-09-14 0.0202677162 NA NA NA NA NA 1998-09-15 0.0077005314 NA NA NA NA NA 1998-09-16 0.0074886581 NA NA NA -0.002710978 NA 1998-09-17 -0.0257819401 NA NA NA NA NA 1998-09-18 0.0011966887 NA NA NA NA NA 1998-09-21 0.0037182403 NA NA NA NA NA 1998-09-22 0.0055904154 NA NA NA NA NA 1998-09-23 0.0347982355 NA NA NA NA NA 1998-09-24 -0.0221650663 NA NA NA NA NA 1998-09-25 0.0019449387 NA 0.007833611 NA NA NA 1998-09-28 0.0037641439 NA NA NA NA NA 1998-09-29 0.0003146288 NA NA NA NA NA 1998-09-30 -0.0309894451 NA NA NA NA NA 1998-10-01 -0.0305704149 NA NA NA NA 0. 1998-10-02 0.0163000909 -0.03409975 NA 0.0004991463 NA NA 1998-10-05 -0.0141025660 NA NA NA NA NA 1998-10-06 -0.0040240279 NA NA NA NA NA 1998-10-07 -0.0142284540 NA NA NA NA NA 1998-10-08 -0.0116470759 NA NA NA NA NA 1998-10-09 0.0256723791 NA NA NA NA NA 1998-10-12 0.0134404929 NA NA NA NA NA 1998-10-13 -0.0029209410 NA NA NA NA NA 1998-10-14 0.0107283327 NA NA NA NA NA 1998-10-15 0.0408820605 NA NA NA NA NA 1998-10-16 0.0084890073 NA NA NA 0.006675831 NA 1998-10-19 0.0056352536 NA NA NA NA NA 1998-10-20 0.0014485122 NA NA NA NA NA 1998-10-21 0.0056142800 NA NA NA NA NA 1998-10-22 0.0079687631 NA NA NA NA NA 1998-10-23 -0.0072680217 NA NA NA NA NA I need to create second one which contains the end of each month rows only. Is there an easy way of doing this? The data starts at 1998-08-31 and ends at 2009-11-06. Many thanks in advance, Costas __ 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] AMRAtoMA
Hello R Users! I have a question about the output of ARMAtoMA when used to calculate the variance of a model. I have a mixed model of the form ARMA(1,1). The actual model takes the form: X(t) = 0.75X(t-12) + a(t) - 0.4a(t-1) Given that gamma(0) takes the form [(1 + theta^2 - 2*theta*phi)/(1-phi^2)]*sigma(a), I would expect a process variance of 4.02*sigma(a) when I substitute 0.75 for phi and -0.4 for theta. When I run ARMAtoMA, result - ARMAtoMA(ar=c(0.75), ma=(-0.4), lag.max=40) sum(result^2)+1 I get 1.28. If I input 0.4 instead of -0.4 in ARMAtoMA I get the result I expected. Is there a sign dependence in the R function I am overlooking? Thanks in advance. Matt __ 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] A beginner's question about ggplot
Dear R-users, I would have another question about the ggplot() function in the ggplot2 package. All the examples I've read so far in the documentation make use of a single neatly formatted data.frame. However, sometimes, one may be interested in plotting on the same grid information or objects derived from two totally different datasets and customize both displays. I still cannot tell how this can be done using ggplot(). Here's an example. ### ## A very simple data.frame; my.data = data.frame(X1 = as.factor(rep(1:2,c(4,4))),X2=c(4,3,5,2,6,2,3,5),X3=c(1:3,2,2:4,5)) ; ## Let's say I want to add the X^2 line to the plot; squared = data.frame(X=1:12,Y=((1:12)/2)^2) ; ## A scatterplot for my.data ; p = ggplot(my.data,aes(x=X2,y=X3,group=X1)) ; p = p+geom_point(aes(colour=X1)) ; # How can squared be added to the plot? At first, I used p+geom_line(data=squared,aes(x=X,y=Y,group=1,colour=green)) ; but the plotted line is always blue! In fact, I can replace colour by any character value and I will still get a blue line. Although I may be wrong, I think this is pretty straightforward. Can anyone give me a pointer as to how we can add arbitrary curves to a ggplot graph and then customize them? A bit later, I'll have to overlay histograms derived from totally different datasets and, if possible, I'd like to use the ggplot2 library for that too, hence the importance of understanding how ggplot objects can be mixed. Thanks a lot, Luc -- View this message in context: http://www.nabble.com/A-beginner%27s-question-about-ggplot-tp23336793p23336793.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] Overlaying graphs from different datasets with ggplot
Dear R-users, I recently began using the ggplot2 package and I am still in the process of getting used to it. My goal would be to plot on the same grid a number of curves derived from two distinct datasets. The first dataset (called molten.data) looks like this : Column names : Perc, Week, Weight P10 21 333.3554 P90 21 486.0480 P10 22 452.6347 P90 22 563.8263 P10 23 575.0960 P90 23 661.6841 P10 24 700.4449 P90 24 779.4067 P10 25 828.4966 P90 25 917.1222 The second dataset (called skj) looks like this: Column names : Week, Perc, Weight 211 317.5 221 392.5 231 467.5 241 542.5 251 617.5 261 697.5 212 535.0 222 632.5 232 737.5 242 855.0 252 980.0 262 1115.0 213 425.0 223 512.5 233 602.5 243 697.5 253 800.0 263 907.5 Now, I plot my graphs using (with the Perc column in skj being a factor) : p - ggplot(molten.data, aes(x=Week, y=Weight, group=Perc)) ; p - p + geom_line(aes(colour = Perc,size=1,linetype=Perc)) ; p + geom_line(data=skj,mapping=aes(x=Week,y=Weight,group=Perc,linetype=Perc)) ; This yields the following error message: ## Error in data.frame(c(#FF6C91FF, #00C1A9FF), c(solid, 22, 42, : arguments imply differing number of rows: 2, 5 ## If I remove the linetype=Perc argument, I get a graph, but also a warning: ## Warning message: In data$arrow - NULL : Coercing LHS to a list ## So, what am I doing wrong in this situation? I thank you sincerely for your help, Luc -- View this message in context: http://www.nabble.com/Overlaying-graphs-from-different-datasets-with-ggplot-tp23322409p23322409.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] lme between-group and within-group covariance
I will try to make this more precise. In the lme() function, the correlation argument allows the user to specify a within-group correlation structure, i.e. the structure of the Lambda matrix using the mixed model notation in Pineiro and Bates. What I want to do is specify a distinct structure for the Psi matrix (same notation), that is, a correlation structure for the random effects. If lme() doesn't allow for this, is there any other function that I could use? MUHC-Research wrote: Dear R users, I would be interested in using the lme() function to fit a linear mixed model to a longitudinal dataset. I know this function allows for the specification of a within-group covariance structure. However, does it allow for the explicit specification of a between-group covariance structure? Being able to specify both separately would be very important in the context of my project since, as might be expected, they have different implications/interpretations. For instance, the mixed procedure in SAS allows users to specify the two structures separately by adding a value for the type argument after the RANDOM statement and the REPEATED statement. My question is thus if we can do the same with lme(). I thank you most sincerely for your help. -- View this message in context: http://www.nabble.com/lme-between-group-and-within-group-covariance-tp22834748p22868945.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] lme between-group and within-group covariance
Dear R users, I would be interested in using the lme() function to fit a linear mixed model to a longitudinal dataset. I know this function allows for the specification of a within-group covariance structure. However, does it allow for the explicit specification of a between-group covariance structure? Being able to specify both separately would be very important in the context of my project since, as might be expected, they have different implications/interpretations. For instance, the mixed procedure in SAS allows users to specify the two structures separately by adding a value for the type argument after the RANDOM statement and the REPEATED statement. My question is thus if we can do the same with lme(). I thank you most sincerely for your help. -- View this message in context: http://www.nabble.com/lme-between-group-and-within-group-covariance-tp22834748p22834748.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] Reflecting correct color and density values in a barplot legend
Hello- I am trying to build a barplot with a fairly large number or categories (65). I figured out how to plot the bars with solid colors and then with density lines to differentiate the 65 items with the following code: barplot(data,col = seq(2:11)) barplot(data,col = black, density = c(0,seq(10,90,10))) #The above produces one barplot with densities overlaid on my colored columns My question is how do I set up my legend to show the same color/density pattern? Thank you, Matt [[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] Simulated annealing method with restarts
Hello R-Help, I'm using R to do some optimization, specifically using the optim method with method = 'SANN' (simulated annealing). I read the help file, and noticed that this method does not include restarts/reheats, which I think would help my optimization significantly. Does anyone know of an implementation that does this? I searched both the internet and the help archives and was unable to find anything. Thanks in advance. Bernard Gordon This is not an offer (or solicitation of an offer) to buy/sell the securities/instruments mentioned. Morgan Stanley may deal as principal in or own or act as market maker to securities/instruments mentioned or may advise the issuers. This may refer to a research analyst/research report. For additional information, research reports and important disclosures, contact me or see https://secure.ms.com. We do not represent this is accurate or complete and we may not update this. Past performance is not indicative of future returns. This communication is solely for the addressee(s) and may contain confidential information. We do not waive confidentiality by mistransmission. Contact me if you do not wish to receive these communications. This communication is directed in the UK to those persons who are professional and eligible counterparties (as defined in the UK Financial Services Authority's rules). [[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] peak finding
Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x=c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y=somefunction(x) y 4 14 Thanks John [[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] peak finding
Hi Thanks for replying. I meant x[4] is the start of a peak shape and x[14] is the end of that peak and x[9] is the maxima of the peak. Thanks, John On Mon, Mar 24, 2008 at 11:09 PM, [EMAIL PROTECTED] wrote: It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] Replacing text
Sorry, another newbie question :-( I loaded a data set with 10 rows and 30 columns. The first column is characters for names of car manufacturers: Jeep Nissan Toyota1 Toyota2 Etc. How can I replace Toyota2 with Scion? Thanks again [[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] combining/deleting rows
Hello List, I'm new to R and have some question on combining (and deleting) rows. I have a dataset that looks like this: msa_id base_sf.2002.2 compltns.2002.2 absorption.2002.2 avrate.2002.2 1 AKRON 27918 0 -1420 19 24 ALBUQU 20469 77 -23 9.2 100 ALLENT 41490 0 -208 19 129 AARBOR 6264 0 62 18 178 ATLANT 395974 1486 638 16.6 254 AUSTIN 34000 31 153 15 330 BALTIM 121080 361 -3447 17.7 How can I add row 1 to row 24? Also, how can I delete rows from the dataframe? Thanks so much in advance, Luciana Luciana Suran | Economist CBRE Torto Wheaton Research | Research 200 High Street, 3rd Floor | Boston, MA 02110 T 617 912 5204 | F 617 912 5240 [EMAIL PROTECTED] | www.tortowheatonresearch.com Please consider the environment before printing this email. This email may contain information that is confidential or attorney-client privileged and may constitute inside information. The contents of this email are intended only for the recipient(s) listed above. If you are not the intended recipient, you are directed not to read, disclose, distribute or otherwise use this transmission. If you have received this email in error, please notify the sender immediately and delete the transmission. Delivery of this message is not intended to waive any applicable privileges. [[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] linear discriminant analysis / search
Dear R help list, I have a training dataset that looks like Table1. I have an unknown dataset that looks like Table2. I want to have a program that should search the training dataset and identify that the unknown sample belongs to which category (type1, type2 or type3) and also if the unknown does not belong to any of the categories, it should let me know. The real dataset has 600 variables and 50 sample types. I tried working with linear discriminant analysis (lda in MASS package) and its predict function. It works great but I think lda is supposed to categorize unknown into one of the types. Most of my unknowns would not be from any category in the training dataset. I don't want to have false positive identification. Table 1: Three types and 10 variables type1type1type1type2type2type2type3type3 type3 var1242825505146182016 var245498910910 var3777121212966 var445410129222 var545410910323 var6545232135 var7545777333 var834310108424 var9343222222 var10333444312 Table 2 unknown var123 var24 var37 var44 var54 var66 var75 var83 var93 var103 Thanks RS [[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] linear discriminant analysis
Dear R help list, I have a training dataset that looks like Table1. I have an unknown dataset that looks like Table2. I want to have a program that should search the training dataset and identify that the unknown sample belongs to which category (type1, type2 or type3) and also if the unknown does not belong to any of the categories, it should let me know. The real dataset has 600 variables and 50 sample types. I tried working with linear discriminant analysis (lda in MASS package) and its predict function. It works great but I think lda is supposed to categorize unknown into one of the types. Most of my unknowns would not be from any category in the training dataset. I don't want to have false positive identification. Table 1: Three types and 10 variables type1type1type1type2type2type2type3type3 type3 var1242825505146182016 var245498910910 var3777121212966 var445410129222 var545410910323 var6545232135 var7545777333 var834310108424 var9343222222 var10333444312 Table 2 unknown var123 var24 var37 var44 var54 var66 var75 var83 var93 var103 Thanks RS [[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] Object Oriented programming in R
You can try this good resource http://www1.maths.lth.se/help/R/R.oo/ RS On Thu, Mar 6, 2008 at 9:30 PM, Davood Tofighi [EMAIL PROTECTED] wrote: Dear all, I was wondering if there a guide/tutorial to the object oriented programming in R for the beginners. Thanks, -- Davood Tofighi Department of Psychology Arizona State University [[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. [[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] forming a linear discriminant function from the output of lda()
Hello all- I am a relatively new user of R and am working through a graduate course in Statistics that uses Minitab, SAS and some Matlab. I like using R but am having some trouble lining up the output from lda() to that of the other programs' results. The dataset below is a modified set of wine data from the Pinot Noir data set as an illustration of the 2 group LDA scenario. MoBa Region 1 0.058 0.225 3 2 0.071 0.105 3 3 0.147 0.301 2 4 0.116 0.166 3 5 0.166 0.132 3 6 0.261 0.078 3 7 0.191 0.085 3 8 0.009 0.072 3 9 0.027 0.094 3 10 0.030 0.349 2 11 0.268 0.099 3 12 0.245 0.071 3 13 0.161 0.181 2 14 0.146 0.328 2 15 0.155 0.081 3 16 0.126 0.299 2 17 0.211 0.206 2 18 0.129 0.281 2 19 0.166 0.292 2 20 0.199 0.292 2 21 0.208 0.087 3 There are various displays from SAS, Minitab and Matlab's Statistics Toolbox, but they all agree with the final linear discriminant function: 0 = -19.51 + 21.47*Mo + 84.08*Ba. When I run the following code in R: library(MASS) wine.lda - lda(Region ~ Mo + Ba, data = wine, prior = c(1,1)/2) wine.lda Call: lda(Region ~ Mo + Ba, data = wine, prior = c(1, 1)/2) Prior probabilities of groups: 2 3 0.5 0.5 Group means: MoBa 2 0.146 0.281 3 0.1479167 0.1079167 Coefficients of linear discriminants: LD1 Mo -5.636024 Ba -22.069187 I am having trouble going from the cofficients derived from the spherical within group covariance to the function I am able to obtain in the other programs. The rest of the problem set up for all programs are: Region = group assignment, prior = equal priors and I do not do any data pretreatment prior to analysis. Any help would be great. Thanks, Matt [[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.
