Re: [R] package mgcv - predict with bam: Error in X[ind, ] : subscript out of bounds
Hi Simon, many thanks for looking into this and making me understand the problem! I'll adjust my factor levels right away... Best, Katharina On 3 February 2014 12:42, Simon Wood s.w...@bath.ac.uk wrote: Hi Katharina, Thanks for sending this. The problem is that the prediction data for site contain levels not available in the (useable non-NA) fit data... levels(m$model$site) [1] KRB NP.FOR WKS.FRE WKS.KRE WKS.RIE WKS.WUE levels(gapData$site) [1] KRB NP.FOR RIE.2 WKS.BBR WKS.FRE WKS.HOE WKS.KRE [8] WKS.RIE WKS.WUE predict.lm has a check for this, and so fails with a rather more informative error message. e.g. m0 - lm(sensor1 ~ sensor2 + site + site:NthSampling, data=xylemRohWeekXnn2011,na.action=na.omit) predict(m0,gapData) ... factor site has new levels RIE.2, WKS.BBR I'll add a better check to predict.gam. best, Simon ps. if you want predictions with the random effects for site set to zero then one trick is to use terms like s(site,bs=re,by=dum) in fitting with dum set to 1. Then in prediction you can set 'site' to any existing level, and dum to zero, in order to get a prediction for the missing level, with the 'site' effect set to zero. On 02/02/14 17:52, Katharina May wrote: Hi Simon, thank you for your reply, I really appreciate any help to understand the problem here... Unluckily the package upgrade didn't help with this issue. An example reproducing the error, and a current sessionInfo() Output can be found below. Many thanks once again, Katharina R Code Example snip library(RCurl) library(mgcv) #retrieve xylemRohWeekXnn2011 test data frame eval( expr = parse( text = getURL(https://webdisk.ads.mwn.de/Handlers/AnonymousDownload.ashx?folder=1a7cbaa4path=xylemRohWeekXnn2011.R;) )) xylemRohWeekXnn.fit.bam - bam(sensor1 ~ sensor2 + s(site, bs=re) + s(site, NthSampling, bs=re) , data=xylemRohWeekXnn2011, na.action=na.omit) #subset data containing gaps for predicting gapData - xylemRohWeekXnn2011[is.na(xylemRohWeekXnn2011[,2]) !is.na(xylemRohWeekXnn2011[,11]),c(2:3,6:7, 11)] xylemRohWeekXnnSite.fit - predict.gam(xylemRohWeekXnn.fit.bam,gapData, type=response, se=F) /snap My current Session Information (sessionInfo() Output - also confirming that the problem exists on both Windows and Mac OS X): snip R version 3.0.2 (2013-09-25) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] mgcv_1.7-28nlme_3.1-113 RCurl_1.95-4.1 bitops_1.0-6 loaded via a namespace (and not attached): [1] grid_3.0.2 lattice_0.20-24 Matrix_1.1-2tools_3.0.2 /snap On 31/01/14 12:57, Simon Wood wrote: Hi Katharina, Could you try upgrading to mgcv_1.7-28, please? There was an occasional problem to do with matching factor levels, which is fixed, but I'm not very confident that is what is going on. If upgrading doesn't work, is there any chance you could send me a small example dataset and code that produces the error, and I'll look at it? best, Simon -- Simon Wood, Mathematical Science, University of Bath BA2 7AY UK +44 (0)1225 386603 http://people.bath.ac.uk/sw283 -- Simon Wood, Mathematical Science, University of Bath BA2 7AY UK +44 (0)1225 386603 http://people.bath.ac.uk/sw283 -- Katharina May BSc. Forest Science and Resource Management IT Specialist (CCI) Prinz-Ludwig-Str. 7 85354 Freising Germany Mobile: +49-176-24031809 www: m3y.de __ 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] package mgcv - predict with bam: Error in X[ind, ] : subscript out of bounds
Hi Simon, thank you for your reply, I really appreciate any help to understand the problem here... Unluckily the package upgrade didn't help with this issue. An example reproducing the error, and a current sessionInfo() Output can be found below. Many thanks once again, Katharina R Code Example snip library(RCurl) library(mgcv) #retrieve xylemRohWeekXnn2011 test data frame eval( expr = parse( text = getURL(https://webdisk.ads.mwn.de/Handlers/AnonymousDownload.ashx?folder=1a7cbaa4path=xylemRohWeekXnn2011.R;) )) xylemRohWeekXnn.fit.bam - bam(sensor1 ~ sensor2 + s(site, bs=re) + s(site, NthSampling, bs=re) , data=xylemRohWeekXnn2011, na.action=na.omit) #subset data containing gaps for predicting gapData - xylemRohWeekXnn2011[is.na(xylemRohWeekXnn2011[,2]) !is.na(xylemRohWeekXnn2011[,11]),c(2:3,6:7, 11)] xylemRohWeekXnnSite.fit - predict.gam(xylemRohWeekXnn.fit.bam,gapData, type=response, se=F) /snap My current Session Information (sessionInfo() Output - also confirming that the problem exists on both Windows and Mac OS X): snip R version 3.0.2 (2013-09-25) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] mgcv_1.7-28nlme_3.1-113 RCurl_1.95-4.1 bitops_1.0-6 loaded via a namespace (and not attached): [1] grid_3.0.2 lattice_0.20-24 Matrix_1.1-2tools_3.0.2 /snap On 31/01/14 12:57, Simon Wood wrote: Hi Katharina, Could you try upgrading to mgcv_1.7-28, please? There was an occasional problem to do with matching factor levels, which is fixed, but I'm not very confident that is what is going on. If upgrading doesn't work, is there any chance you could send me a small example dataset and code that produces the error, and I'll look at it? best, Simon -- Simon Wood, Mathematical Science, University of Bath BA2 7AY UK +44 (0)1225 386603 http://people.bath.ac.uk/sw283 __ 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] package mgcv - predict with bam: Error in X[ind, ] : subscript out of bounds
Dear R-Community, I`m trying to apply the mgcv package to fill gaps in sensor data from different sites (9 sites, 2 sensors per site) and do the filling on a site-wise level. Based on http://r.789695.n4.nabble.com/mgcv-gamm-predict-to-reflect-random-s-effects-td3622738.html my model looks like this: xylemRohWeekXnn.fit.bam - bam(sensor1 ~ sensor2 + s(site, bs=re) + s(site, NthSampling, bs=re) , data=xylemRohWeekXnn2011, na.action=na.omit) However, than I try to use predict, I get an error: gapData - xylemRohWeekXnn2011[is.na(xylemRohWeekXnn2011[,2]) !is.na(xylemRohWeekXnn2011[,11]),c(2:3,6:7, 11)] xylemRohWeekXnnSite.fit - predict.gam(xylemRohWeekXnn.fit.bam,gapData, type=response, se=F) Error in X[ind, ] : subscript out of bounds I was hoping that someone might be able to provide a quick hint on if there is an obvious problem or mistake within my model declaration/approach? I attached the sessionInfo() Output below and the xylemRohWeekXnn2011 dump can be downloaded here: https://webdisk.ads.mwn.de/Handlers/AnonymousDownload.ashx?folder=1a7cbaa4path=xylemRohWeekXnn2011.txt I`m appreciating any help and hints! Thank you very much, Katharina - sessionInfo() - R version 3.0.2 (2013-09-25) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252 LC_MONETARY=German_Germany.1252 [4] LC_NUMERIC=CLC_TIME=German_Germany.1252 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] mgcv_1.7-27 plyr_1.8ggplot2_0.9.3.1 lattice_0.20-24 gdata_2.13.2nlme_3.1-113 [7] zoo_1.7-10 xlsx_0.5.5 xlsxjars_0.5.0 rJava_0.9-6 loaded via a namespace (and not attached): [1] colorspace_1.2-4 dichromat_2.0-0digest_0.6.4 grid_3.0.2 gtable_0.1.2 [6] gtools_3.2.1 labeling_0.2 MASS_7.3-29 Matrix_1.1-2 munsell_0.4.2 [11] proto_0.3-10 RColorBrewer_1.0-5 reshape2_1.2.2 scales_0.2.3 stringr_0.6.2 [16] tools_3.0.2 __ 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] Package zoo - na.rm ignored?
Dear R Users, I've got a strange problem, which I do not really understand: when I use na.spline from the zoo package, the option na.rm is just being ignored, see this adjusted example from the na.spline help page: d0 - as.Date(2000-01-01) z - zoo(c(NA, 11, 13, NA, 15, NA), d0 + 1:6) na.spline(z) na.spline(z, na.rm = FALSE) na.spline(z, na.rm = T) ### In all 3 cases, the output looks like this with the trailing NA being replaced: 2000-01-02 2000-01-03 2000-01-04 2000-01-05 2000-01-06 2000-01-07 8.33 11.00 13.00 14.33 15.00 15.00 Am I missing something here? Any help is very much appreciated... Thanks, Katharina ### Background Info: ### sessionInfo() R version 3.0.1 (2013-05-16) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252 LC_MONETARY=German_Germany.1252 [4] LC_NUMERIC=CLC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] nlme_3.1-109 zoo_1.7-10 xlsx_0.5.1 xlsxjars_0.5.0 rJava_0.9-5 loaded via a namespace (and not attached): [1] grid_3.0.1 lattice_0.20-15 tools_3.0.1 [[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] Package zoo - na.rm ignored?
many thanks for the enlightenment - I guess I completely misunderstood the na.rm option here... On 18 June 2013 00:50, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Mon, Jun 17, 2013 at 6:23 PM, Katharina May may.kathar...@googlemail.com wrote: Dear R Users, I've got a strange problem, which I do not really understand: when I use na.spline from the zoo package, the option na.rm is just being ignored, see this adjusted example from the na.spline help page: d0 - as.Date(2000-01-01) z - zoo(c(NA, 11, 13, NA, 15, NA), d0 + 1:6) na.spline(z) na.spline(z, na.rm = FALSE) na.spline(z, na.rm = T) ### In all 3 cases, the output looks like this with the trailing NA being replaced: 2000-01-02 2000-01-03 2000-01-04 2000-01-05 2000-01-06 2000-01-07 8.33 11.00 13.00 14.33 15.00 15.00 Am I missing something here? Any help is very much appreciated... If the result of the spline interpolation still results in NAs then na.rm=TRUE will remove any leading NAs. It may be that all currently supported methods produce no NAs in which case this argument would only be there for compatability with na.approx and in case future additional spline methods do produce NAs. [[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] round Date object to 10 minutes intervals?
Hello *, I'm trying to round time stamps, consisting of a date and time part, to 10 minutes intervals The timestamps have the following format: %d.%m.%Y %H:%M *timeStamp - c(31.03.2011 09:30, 31.03.2011 09:41, 31.03.2011 10:04, 31.03.2011 10:10, 31.03.2011 10:28, 31.03.2011 10:35, 31.03.2011 10:49, 31.03.2011 23:59)* should look like this: *timeStamp2 - c(31.03.2011 09:30, 31.03.2011 09:40, 31.03.2011 10:00, 31.03.2011 10:10, 31.03.2011 10:30, 31.03.2011 10:40, 31.03.2011 10:55, 01.04.2011 00:00)* I've found this topic on the web: http://stats.stackexchange.com/questions/5305/how-to-re-sample-an-xts-time-series-in-r library(xts) align.time(as.POSIXct(timeStamp, format=%d.%m.%Y %H:%M), 10 * 60) which uses the align.time function from the library xts BUT as far as I can see only supports either all timestamps being either rounded up or down (not both, depending on the minutes 0-9 being either or 5 line the way I need it) As I'm really struggling with this topic, any help and hint is awarded with greatest thankfulness from my side Best regards, Katharina [[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 model: Test difference between coefficients and given values (t.test?)
Thanks to somebody I got the hint to use offset for the purpose of validating if there's a difference between the intercept and slope of a model and some provided values for the coefficients intercept and slope. I read ?model.offset and I'm still struggling to use it for my purpose. If I understood the concepts correctly, offset can be used to define a known coefficient of a model, so therefore I could create two models based on the same data like my original one with an additional offset term (one model with intercept specified and one with the slope specified)? Sorry for troubling you: I struggling with the data analysis of my bachelor thesis and just want to compare the coefficients of my linear regression to published ones... a method often used I guess, but still I cannot find any appropriate documentations. Thanks, Katharina On Aug 8, 2009, Katharina May may.kathar...@googlemail.com wrote: Hi there, I've got a question which is really trivial for sure but still I have to ask as I'm not making any progress solving it by myself (please be patient with an undergraduate student): I've got a linear model (lm and lmer fitted with method=ML). Now I want to compare the coefficients (slope, intercept, not the random effects) of both models with a given value (e.g. intercept=0.5, slope=2) to see if the values estimated with the models are significantly different from the given values. I heard about t.test, but I'm sorry to say that I'm not quite understanding what I have to provide as arguments. I would be more than happy if somebody can point out an example similar to the comparison I have to do... Thanks, Katharina __ 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] linear model: Test difference between coefficients and given values (t.test?)
Hi there, I've got a question which is really trivial for sure but still I have to ask as I'm not making any progress solving it by myself (please be patient with an undergraduate student): I've got a linear model (lm and lmer fitted with method=ML). Now I want to compare the coefficients (slope, intercept, not the random effects) of both models with a given value (e.g. intercept=0.5, slope=2) to see if the values estimated with the models are significantly different from the given values. I heard about t.test, but I'm sorry to say that I'm not quite understanding what I have to provide as arguments. I would be more than happy if somebody can point out an example similar to the comparison I have to do... Thanks, Katharina __ 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] package lmodel2: p-value RMA fitting?
Hi *, is there a way to obtain some kind of p-value for a model fitted with RMA using the lmodel2 package? I know that p-values are discussed and criticized a lot and as you can image from my question I'm not very much of a statistican (only writing my bachelor thesis). As fare as I understood the confidence interval statistic correctly, a coefficient is regarded as statistically significant if the corresponding CI does not include 0 (null hypothesis). But can I obtain some kind of a p-value to say that it is highly significant ( 0.01), significant (0.05),... like in the output of lm? Sorry for bothering everybody with this, well, probably rather idiotic question, but I don't know where to continue from this point... Thanks, Katharina Here the output of my lmodel2 regression: Model II regression Call: lmodel2(formula = log(AGB) ~ log(BM_roots), data = biomass_data, range.y = interval, range.x = interval, nperm = 99) n = 1969 r = 0.9752432 r-square = 0.9510993 Parametric P-values: 2-tailed = 01-tailed = 0 Angle between the two OLS regression lines = 1.433308 degrees Permutation tests of OLS, MA, RMA slopes: 1-tailed, tail corresponding to sign A permutation test of r is equivalent to a permutation test of the OLS slope P-perm for SMA = NA because the SMA slope cannot be tested Regression results Method Intercept Slope Angle (degrees) P-perm (1-tailed) 1OLS 0.6122146 1.038792 46.09002 0.01 2 MA 0.5787299 1.066868 46.85300 0.01 3SMA 0.5807645 1.065162 46.80725 NA 4RMA 0.5792123 1.066463 46.84216 0.01 Confidence intervals Method 2.5%-Intercept 97.5%-Intercept 2.5%-Slope 97.5%-Slope 1OLS 0.5779465 0.64648281.0283761.049207 2 MA 0.5659033 0.59142031.0562271.077622 3SMA 0.5682815 0.59312601.0547971.075628 4RMA 0.5663916 0.59189891.0558261.077213 Eigenvalues: 19.83213 0.2475542 H statistic used for computing C.I. of MA: 2.502866e-05 [[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] package lmodel2: p-value RMA fitting?
sorry for spaming, but IU just had an idea not sure if that may be a way of doing it: 1. calculate the standrad error of e.g. intercept as mean of the standards errors obtained from the upper/lower confidence intervals of the intercept error_intercept1 = (allo.lmodel2$confidence.intervals[4,2] - mean(log(biomass_data$BM_roots)))/1.96 error_intercept2 = (allo.lmodel2$confidence.intervals[4,3] - mean(log(biomass_data$BM_roots)))/1.96 stderr_intercept = round((error_intercept1+error_intercept2)/2, digits=8) 2. Calculate the t-value as intercept estimate divided by the standard error from 1. and using the following for calculating a two-tailed p-value p_intercept = 2 * (1 - pt(abs(intercept/stderr_intercept), df=length(biomass_data)-1)) Might this a reasonable approach for a 'rough' estimation of an p-value? I glad for every suggestion... 2009/7/20 Katharina May may.kathar...@googlemail.com Hi *, is there a way to obtain some kind of p-value for a model fitted with RMA using the lmodel2 package? I know that p-values are discussed and criticized a lot and as you can image from my question I'm not very much of a statistican (only writing my bachelor thesis). As fare as I understood the confidence interval statistic correctly, a coefficient is regarded as statistically significant if the corresponding CI does not include 0 (null hypothesis). But can I obtain some kind of a p-value to say that it is highly significant ( 0.01), significant (0.05),... like in the output of lm? Sorry for bothering everybody with this, well, probably rather idiotic question, but I don't know where to continue from this point... Thanks, Katharina Here the output of my lmodel2 regression: Model II regression Call: lmodel2(formula = log(AGB) ~ log(BM_roots), data = biomass_data, range.y = interval, range.x = interval, nperm = 99) n = 1969 r = 0.9752432 r-square = 0.9510993 Parametric P-values: 2-tailed = 01-tailed = 0 Angle between the two OLS regression lines = 1.433308 degrees Permutation tests of OLS, MA, RMA slopes: 1-tailed, tail corresponding to sign A permutation test of r is equivalent to a permutation test of the OLS slope P-perm for SMA = NA because the SMA slope cannot be tested Regression results Method Intercept Slope Angle (degrees) P-perm (1-tailed) 1OLS 0.6122146 1.038792 46.09002 0.01 2 MA 0.5787299 1.066868 46.85300 0.01 3SMA 0.5807645 1.065162 46.80725 NA 4RMA 0.5792123 1.066463 46.84216 0.01 Confidence intervals Method 2.5%-Intercept 97.5%-Intercept 2.5%-Slope 97.5%-Slope 1OLS 0.5779465 0.64648281.0283761.049207 2 MA 0.5659033 0.59142031.0562271.077622 3SMA 0.5682815 0.59312601.0547971.075628 4RMA 0.5663916 0.59189891.0558261.077213 Eigenvalues: 19.83213 0.2475542 H statistic used for computing C.I. of MA: 2.502866e-05 -- Time flies like an arrow, fruit flies like bananas. [[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] distinguish regression lines in grouped, black and white lattice xyplot
Hi,
I've got the following problem which I cannot think of a solution right now:
if got a lattice xyplot in black and white and a grouping variable
with many (more than 8
values) and I plot it as regression lines (type=r), just like this
one (not reproducable but that's
I guess not the point here):
xyplot(log(AGWB) ~ log(BM_roots), data=sub_agwb_data, groups=species,
type=r, lty=c(1:6),panel=allo.panel.5)
The problem is that I've got 26 different values for the grouping
variable species and only 6 default values for the line type
lty (and according to the par {graphics} help page customizable to up
to 8 different line types).
Does anybody have any idea how these 26 different lines can be made
distinguishable from each other without the use
of colors?
Thanks,
Katharina
__
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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] distinguish regression lines in grouped, black and white lattice xyplot
That's a point. I justed wanted to provide an overview for myself to
see the tendencies in a direct comparement
and with an easy way to distinct them, but maybe the text panel can
help me with that...
Well anyway, is it right that a grouped black and white plot can
contain a maxinum of 8 distinguishable lines or might
there be a way to increase that?
I know some graphics from papers containing lines with equally
distance points on the lines (one type of point per line)
as a form of distinction. Can this be realised using grouped lattice
plots with regression lines?
2009/6/24 Bert Gunter gunter.ber...@gene.com:
Don't be silly. They can't be made distinguishable by any number of line
types and/or colors. The brain can't keep that many different symbol
representations straight. Referring back and forth to a legend is also
similarly useless. You need to think more creatively about how to make a
more meaningful display to provide viewers interpretable information.
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Katharina May
Sent: Wednesday, June 24, 2009 12:28 PM
To: r-help@r-project.org
Subject: [R] distinguish regression lines in grouped,black and white lattice
xyplot
Hi,
I've got the following problem which I cannot think of a solution right now:
if got a lattice xyplot in black and white and a grouping variable
with many (more than 8
values) and I plot it as regression lines (type=r), just like this
one (not reproducable but that's
I guess not the point here):
xyplot(log(AGWB) ~ log(BM_roots), data=sub_agwb_data, groups=species,
type=r, lty=c(1:6),panel=allo.panel.5)
The problem is that I've got 26 different values for the grouping
variable species and only 6 default values for the line type
lty (and according to the par {graphics} help page customizable to up
to 8 different line types).
Does anybody have any idea how these 26 different lines can be made
distinguishable from each other without the use
of colors?
Thanks,
Katharina
__
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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.
--
Time flies like an arrow, fruit flies like bananas.
__
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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] lattice logaritmic scale (basis e ), rewriting labels using xscale.component
thanks Deepayan, that works great!
2009/6/19 Deepayan Sarkar deepayan.sar...@gmail.com:
On 6/18/09, Katharina May may.kathar...@googlemail.com wrote:
Hi there,
sorry for troubling everybody once again, I've got a problem rewriting
Sarkar's function for
rewriting the tick locations in a logaritmic way (s.
http://lmdvr.r-forge.r-project.org/code/Chapter08.R):
His example works for log 2 but I need log e (natural logarithm). My
problem is that if I replace
2 with e (using paste()), I get the error message that the location
isn't a numeric value.
R doesn't have a constant that represents e, but
tick.at - logTicks(exp(lim), loc = c(1, 3))
instead of
tick.at - logTicks(paste(e^,lim,sep=), loc = c(1, 3))
should give you e^lim. Or, if it makes it easier,
e - exp(1)
tick.at - logTicks(e^lim, loc = c(1, 3))
I don't think you need to change the logTicks function.
-Deepayan
Is there any way to get this working somehow or do I have to take a
different approach?
Thanks, Katharina
Here my failing approach:
require(lattice)
data(Earthquake, package = MEMSS)
xscale.components.log - function(lim, ...) {
ans - xscale.components.default(lim = lim, ...)
tick.at - logTicks(paste(e^,lim,sep=), loc = c(1, 3))
ans$bottom$ticks$at - log(tick.at, 2)
ans$bottom$labels$at - log(tick.at, 2)
ans$bottom$labels$labels - as.character(tick.at)
ans
}
logTicks - function (lim, loc = c(1, 5)) {
ii - floor(log(range(lim))) + c(-1, 2)
main - paste(e^,(ii[1]:ii[2]),sep=)
r - as.numeric(outer(loc, main, *))
r[lim[1] = r r = lim[2]]
}
xyplot(accel ~ distance, data=Earthquake, scales = list(log = e),
xscale.components = xscale.components.log,
Here is the original code of Sarkar:
logTicks - function (lim, loc = c(1, 5)) {
ii - floor(log10(range(lim))) + c(-1, 2)
main - 10^(ii[1]:ii[2])
r - as.numeric(outer(loc, main, *))
r[lim[1] = r r = lim[2]]
}
xscale.components.log2 - function(lim, ...) {
ans - xscale.components.default(lim = lim, ...)
tick.at - logTicks(2^lim, loc = c(1, 3))
ans$bottom$ticks$at - log(tick.at, 2)
ans$bottom$labels$at - log(tick.at, 2)
ans$bottom$labels$labels - as.character(tick.at)
ans
}
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Re: [R] lattice: axis ticks, axis alignment and remove axis from plot
on either side. (May not
always work as expected.)
--
Sorn Norng
Biometrician
Department of Primary Industries
Knoxfield Centre
621 Burwood Highway, Ferntree Gully 3156
Ph: (03) 9210 9358
Mob: 0428344515
Fax: (03) 9800 3521
Email: sorn.no...@dpi.vic.gov.au
-
*Katharina May may.kathar...@googlemail.com*
Sent by: r-help-boun...@r-project.org
17/06/2009 07:24 PM
To
r-help@r-project.org cc
Subject
[R] lattice: axis ticks, axis alignment and remove axis from plot
Hi there,
I'm a bit confused concerning the axis tck setting in the lattice
package as the ticks on left sided axis aren't
drawn at all with the following setting:
dados - data.frame(varsep = factor(rep(1:2,10)),
i = runif(20))
library(lattice)
my.theme - list(
axis.components = list(left = list(tck = 1, pad1 = 1, pad2 =
2), top = list(tck = 0, pad1 = 1, pad2 = 0), right = list(tck = 0,
pad1 = 1, pad2 = 0), bottom = list(tck = +0.5, pad1 = 1, pad2 = 2)))
trellis.par.set(theme = my.theme)
bwplot(varsep ~ i, dados,
xlab = names(dados)[1],
ylab = names(dados)[2],
panel = function(...) {
panel.grid(v = -1, h = 0)
panel.bwplot(...)
})
Maybe somebody can help me recognise the problem why the left sided
ticks are not drawn.
I'm also wondering how to tell the trellis object not to draw the
right and top axes at all an how to align the y axis
that it goes through x=0?
Thanks,
Katharina
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[R] lattice logaritmic scale (basis e ), rewriting labels using xscale.component
Hi there,
sorry for troubling everybody once again, I've got a problem rewriting
Sarkar's function for
rewriting the tick locations in a logaritmic way (s.
http://lmdvr.r-forge.r-project.org/code/Chapter08.R):
His example works for log 2 but I need log e (natural logarithm). My
problem is that if I replace
2 with e (using paste()), I get the error message that the location
isn't a numeric value.
Is there any way to get this working somehow or do I have to take a
different approach?
Thanks, Katharina
Here my failing approach:
require(lattice)
data(Earthquake, package = MEMSS)
xscale.components.log - function(lim, ...) {
ans - xscale.components.default(lim = lim, ...)
tick.at - logTicks(paste(e^,lim,sep=), loc = c(1, 3))
ans$bottom$ticks$at - log(tick.at, 2)
ans$bottom$labels$at - log(tick.at, 2)
ans$bottom$labels$labels - as.character(tick.at)
ans
}
logTicks - function (lim, loc = c(1, 5)) {
ii - floor(log(range(lim))) + c(-1, 2)
main - paste(e^,(ii[1]:ii[2]),sep=)
r - as.numeric(outer(loc, main, *))
r[lim[1] = r r = lim[2]]
}
xyplot(accel ~ distance, data=Earthquake, scales = list(log = e),
xscale.components = xscale.components.log,
Here is the original code of Sarkar:
logTicks - function (lim, loc = c(1, 5)) {
ii - floor(log10(range(lim))) + c(-1, 2)
main - 10^(ii[1]:ii[2])
r - as.numeric(outer(loc, main, *))
r[lim[1] = r r = lim[2]]
}
xscale.components.log2 - function(lim, ...) {
ans - xscale.components.default(lim = lim, ...)
tick.at - logTicks(2^lim, loc = c(1, 3))
ans$bottom$ticks$at - log(tick.at, 2)
ans$bottom$labels$at - log(tick.at, 2)
ans$bottom$labels$labels - as.character(tick.at)
ans
}
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Re: [R] lattice: axis ticks, axis alignment and remove axis from plot
Hi Jim, thanks, that sounds like a good idea! The only problem is that axis and pos are a far as I know not for lattice or at least I cannot find an appropriate function and way to do this in lattice. Maybe I have to keep on searching for how to realize that with lattice... Best wishes, Katharina 2009/6/18 Jim Lemon j...@bitwrit.com.au: Katharina May wrote: What I sort of looking for is a xyplot with both the x axis at the bottom and y axis at the left going through 0, but continuing in the positive and negative area (forming a kind of cross like e.g. http://upload.wikimedia.org/wikipedia/commons/thumb/2/2c/Erf_plot.svg/600px-Erf_plot.svg.png I'm not sure if this possible at all or if I just have to add to reference lines at x=0 and y =0 as a workaround even though not fully satisfying...? Hi Katharina, I'm not sure if you can do this in lattice, but if you do the original plot with axes=FALSE, you can specify where the x and y axis are placed with the pos= argument. Jim -- Time flies like an arrow, fruit flies like bananas. __ 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] lattice: axis ticks, axis alignment and remove axis from plot
Hi there,
I'm a bit confused concerning the axis tck setting in the lattice
package as the ticks on left sided axis aren't
drawn at all with the following setting:
dados - data.frame(varsep = factor(rep(1:2,10)),
i = runif(20))
library(lattice)
my.theme - list(
axis.components = list(left = list(tck = 1, pad1 = 1, pad2 =
2), top = list(tck = 0, pad1 = 1, pad2 = 0), right = list(tck = 0,
pad1 = 1, pad2 = 0), bottom = list(tck = +0.5, pad1 = 1, pad2 = 2)))
trellis.par.set(theme = my.theme)
bwplot(varsep ~ i, dados,
xlab = names(dados)[1],
ylab = names(dados)[2],
panel = function(...) {
panel.grid(v = -1, h = 0)
panel.bwplot(...)
})
Maybe somebody can help me recognise the problem why the left sided
ticks are not drawn.
I'm also wondering how to tell the trellis object not to draw the
right and top axes at all an how to align the y axis
that it goes through x=0?
Thanks,
Katharina
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[R] color in grouped lattice xyplot plot
Hi,
I've got a data.frame object which I would like to plot in a lattice
xyplot grouped by a variable and using a value for the color of each
point which is stored in a variable in the same data.frame.
I tried the following code:
my.panel.xy - function(x, y, ...){
panel.xyplot(x, y, ...)
panel.lmline(x, y, lwd = 2)
panel.abline(0, 1, col=red, lty=2)
panel.grid(h = -1, v = -1, lty=1)
}
xyplot(log(AGB) ~ log(BM_roots) | as.character(taxa),
data=sub_agb_data,main=list(AGB in dependence of BM_root [per
taxa!],cex=0.7),
ylab=list(log(AGB),cex=0.7),xlab=list(log(BM_root),cex=0.7),col=sub_agb_data$color,
panel = my.panel.xy)
xyplot(log(AGB) ~ log(BM_roots) | as.character(species),
data=sub_agb_data,main=list(AGB in dependence of BM_root [per
species!],cex=0.7),
ylab=list(log(AGB),cex=0.7),xlab=list(log(BM_root),cex=0.7),col=sub_agb_data$color,
panel = my.panel.xy)
Somehow the colors are not displayed correctly (it works if I use plot
in a for loop for each value for the variables for which I group my
lattice plots. The color variable is of type character so this
shouldn't be the reason)
The object sub_agb_object can be downloaded here:
http://www.wzw.tum.de/oekophys/fileadmin/sub_agb_data.rData
Maybe somebody has any idea how to make this work or what the problem might be?
Thanks,
Katharina
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[R] wrong labels and colors of points in graph/plot
Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(!is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] write values of points beside points in plot
Hi everyone, is it possible to write a certain value beside a point in a plot? I'm plotting the following: plot(coefficient$intercept ~ coefficient$average_BM_leaves_needles, main=intercepts ::: BM_leaves_needles, ylab=intercepts, xlab=average BM_leaves_needles per site [kg]) and I want to have the value of coefficient$site_no written beside each point within the plot (to recognize them) sorry for troubling you with this question, but I somehow can't find any information on it (maybe because I'm not quite sure what to search for) Thanks a lot, Katharina __ 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] remove empty objects from workspace
Hi, how can I remove all empty objects (which are NA or have zero rows) from my workspace? Thanks, Katharina __ 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] Remove objects names like character String
Hi, how can I use rm() on objects named like: paste(site,i,_data,sep=) while looping through i? I tried rm(paste(site,i,_data,sep=)) but I get the error that rm() must contain names or text strings which is confusing me as I thought paste() would create something like that...? Thanks, Katharina -- Time flies like an arrow, fruit flies like bananas. __ 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] remove empty objects from workspace
Thanks Jim, the removal of objects which are NA works perfectly! For my second problem it didn't express myself correctly: I actually meant objects with rows (attributes?) but no data in it but I solved this adjusting your approach: for(object in objects()) if(is.null(dim(get(object))[1]) || dim((get(object)))[1] == 0) rm(list=object) Thanks a lot! 2009/5/19 Jim Lemon j...@bitwrit.com.au: Katharina May wrote: Hi, how can I remove all empty objects (which are NA or have zero rows) from my workspace? Hi Katharina, To remove objects that are all NA: for(object in objects()) if(all(is.na(get(object rm(list=object) If by zero rows you mean objects that do not have a dimension: for(object in objects()) if(is.null(dim(get(object rm(list=object) Jim -- Time flies like an arrow, fruit flies like bananas. __ 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] Remove objects names like character String
thanks to all your solutions, works out perfectly! 2009/5/19 Henrique Dallazuanna www...@gmail.com: Try this: rm(list=ls(patt=site[0-9]$)) On Tue, May 19, 2009 at 7:47 AM, Katharina May may.kathar...@googlemail.com wrote: Hi, how can I use rm() on objects named like: paste(site,i,_data,sep=) while looping through i? I tried rm(paste(site,i,_data,sep=)) but I get the error that rm() must contain names or text strings which is confusing me as I thought paste() would create something like that...? Thanks, Katharina -- Time flies like an arrow, fruit flies like bananas. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Time flies like an arrow, fruit flies like bananas. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] create string of comma-separated content of vector
Hi, how do I create a string of the comma-separated content of a vector? I've got the vector i with several numeric values as content: str(i) num 99 and want to create a SQL statement to look like the following where the part '(2, 4, 6, 7)' should be the content of the vector i: select * from [biomass_data$] where site_no in (2, 4, 6, 7) Here my approach (which doesn't work): site_all_data = sqlQuery(channel, select * from [biomass_data$] where site_no in (,paste(i,sep=,),) ) sorry for spaming so much today to the mailing list... -Katharina -- Time flies like an arrow, fruit flies like bananas. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create objects in a loop and adding sqlQuery content to them
I finally found the problem within my site_data object...
this works well:
snip
site_data - sqlQuery(channel, select site_no from [biomass_data$]
group by site_no)$site_no
for(i in site_data) {
assign(paste(site,i,_data,sep=), sqlQuery(channel,
paste(select * from [biomass_data$] where site_no = ,i,sep=)))
}
/snap
Thanks for all your solution proposals!
Katharina
On 21.04.2009, at 02:43, David Winsemius wrote:
On Apr 20, 2009, at 8:04 PM, Katharina May wrote:
Hi there,
I've got a database or rather spreadsheet with several columns and
rows.
For one column named sites I want to loop through all possible
values and retrieve
all data out of the database where site = x and write it into an
object named 'sitex_data'.
Somehow I'm really missing something as I'm not able to create
these sitex_data objects with
the database values, neither using list nor assign...
Here some code snipplets:
library (RODBC)
channel - odbcConnectExcel2007 (biomass_data.xlsx)
site_data - sqlQuery(channel, select site_no from
[biomass_data$] group by site_no)
#Here the values I want to loop through
str(site_data)
'data.frame': 44 obs. of 1 variable:
$ site_no: num 4 7 9 10 15 16 17 18 19 20 ...
#Here my first try [error message on the line within the loop,
saying something like:
# 'recursive indexing on level 2 failed']
sites_object_list - vector(list,99)
for(i in site_data) {
sites_object_list[[i]] - sqlQuery(channel, paste(select * from
[biomass_data$] where site_no = ,i,sep=))
}
This does not appear to be a dataset to which we have access, so I
will take an untested stab at this. What happens if you assign the
results of the function unique() around site_data$site_no to your
index, i? That way you would stand a chance of referring to the
sites in a manner that R might understand and to only use the site
numbers once apiece in the loop. I'm not a competent SQL programmer
so i have no comments on that part.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] create objects in a loop and adding sqlQuery content to them
Hi there,
I've got a database or rather spreadsheet with several columns and rows.
For one column named sites I want to loop through all possible values
and retrieve
all data out of the database where site = x and write it into an
object named 'sitex_data'.
Somehow I'm really missing something as I'm not able to create these
sitex_data objects with
the database values, neither using list nor assign...
Here some code snipplets:
library (RODBC)
channel - odbcConnectExcel2007 (biomass_data.xlsx)
site_data - sqlQuery(channel, select site_no from [biomass_data
$] group by site_no)
#Here the values I want to loop through
str(site_data)
'data.frame': 44 obs. of 1 variable:
$ site_no: num 4 7 9 10 15 16 17 18 19 20 ...
#Here my first try [error message on the line within the loop, saying
something like:
# 'recursive indexing on level 2 failed']
sites_object_list - vector(list,99)
for(i in site_data) {
sites_object_list[[i]] - sqlQuery(channel, paste(select * from
[biomass_data$] where site_no = ,i,sep=))
}
#Here my second try [error message on the line within the loop,
saying something like:
# 'only the first element will be used as variable name']
for(i in site_data) {
assign(paste(site,i,_data,sep=), sqlQuery(channel,
paste(select * from [biomass_data$] where site_no = ,i,sep=)))
}
I would be very, very glad if anybody sends me a clue about this, as
I'm a obviuos Newbie using
R...
Thanks,
Katharina
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