Re: [R] pairwise deletion in regression models
Hi Yes, I am sure. Have a look here for SPSS e.g.: http://www-01.ibm.com/support/docview.wss?uid=swg21475199 and here http://www.ats.ucla.edu/stat/spss/modules/missing.htm and for STATA here, http://www.ats.ucla.edu/stat/stata/modules/missing.html I know that R allows for pairwise deletion in the cor() function, but I need it for regression analysis. The default is listwise (casewise) deletion. Would be grateful for further input on this. Best, Adel -- Adel Daoud, PhD, Researcher *Newly published:* * Daoud, Adel and Kohl, Sebastian, *How Much Do Sociologists Write About Economic Topics? Using Big-Data to Test Some Conventional Views in Economic Sociology, 1890 to 2014*. Max Planck Institute for the Study of Societies, Discussion Paper 16/7 <http://www.mpifg.de/pu/mpifg_dp/dp16-7.pdf> * Daoud, Adel, Björn Halleröd, and Debarati Guha-Sapir, (2016) “What Is the Association between Absolute Child Poverty, Poor Governance, and Natural Disasters? A Global Comparison of Some of the Realities of Climate Change”, *PLoS ONE* 11(4) <http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0153296> *Shailen Nandy, Adel Daoud, David Gordon, (2016), Examining the changing profile of undernutrition in the context of food price rises and greater inequality, *Social Science & Medicine*, Volume 149, Pages 153–163 <http://www.sciencedirect.com/science/article/pii/S0277953615302446> *Department of Sociology and Work Science,* *University of Gothenburg* Box 720 405 30, Göteborg, Sweden Email: adel.da...@sociology.gu.se Website: http://adeldaoud.se/ <http://adeldaoud.se/> On Wed, Jul 13, 2016 at 10:53 AM, PIKAL Petr <petr.pi...@precheza.cz> wrote: > Hi > > > > Hm. Are you 100% sure that other software packages can do pairwise > deletion in OLS or GLM? I am not at all familiar with them but > > > > http://www.ats.ucla.edu/stat/spss/modules/stats.htm > > > > suggests that option pairwise is available with corr and I believe the > same option exists in cor function in R. > > > > My statistical knowledge is inferior but I just cannot imagine how whole > model could be computed when one value is missing. > > > > Cheers > > Petr > > > > *From:* adelda...@gmail.com [mailto:adelda...@gmail.com] *On Behalf Of *Adel > Daoud > *Sent:* Wednesday, July 13, 2016 10:00 AM > *To:* PIKAL Petr <petr.pi...@precheza.cz> > *Cc:* r-help <r-help@r-project.org> > *Subject:* Re: [R] pairwise deletion in regression models > > > > Thanks Petr for the suggestion. > > > > I just took the regtools package for a quick test drive. It looks > promising, but it still needs further development to make it a viable > option. You will not get a standard regression output (as in lm or glm), > only the regression coefficients (without standard errors). I will be happy > to try it out once it is more robust. > > > > I am a bit puzzled that the R universe seems to lack a robust package that > allows for pairwise deletion (which is standard in the otherwise poorer > software packages STATA or SPSS…). Would be very happy if anyone can show > me otherwise. > > > Best, > > Adel > > -- > > Adel Daoud, PhD, Researcher > > *Newly published:* > > * Daoud, Adel and Kohl, Sebastian, *How Much Do Sociologists Write About > Economic Topics? Using Big-Data to Test Some Conventional Views in Economic > Sociology, 1890 to 2014*. Max Planck Institute for the Study of > Societies, Discussion Paper 16/7 > <http://www.mpifg.de/pu/mpifg_dp/dp16-7.pdf> > > * Daoud, Adel, Björn Halleröd, and Debarati Guha-Sapir, (2016) “What Is > the Association between Absolute Child Poverty, Poor Governance, and > Natural Disasters? A Global Comparison of Some of the Realities of Climate > Change”, *PLoS ONE* 11(4) > <http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0153296> > > *Shailen Nandy, Adel Daoud, David Gordon, (2016), Examining the changing > profile of undernutrition in the context of food price rises and greater > inequality, *Social Science & Medicine*, Volume 149, Pages 153–163 > <http://www.sciencedirect.com/science/article/pii/S0277953615302446> > > > > *Department of Sociology and Work Science,* > > *University of Gothenburg* > > Box 720 > > 405 30, Göteborg, Sweden > > Email: adel.da...@sociology.gu.se > > Website: http://adeldaoud.se/ <http://adeldaoud.se/> > > > > On Wed, Jul 13, 2016 at 8:21 AM, PIKAL Petr <petr.pi...@precheza.cz> > wrote: > > Hi > > > http://stats.stackexchange.com/questions/158366/fit-multiple-regression-model-with-pairwise-deletion-or-on-a-correlation-covari > > The package is probably not available on CRAN
Re: [R] pairwise deletion in regression models
Thanks Petr for the suggestion. I just took the regtools package for a quick test drive. It looks promising, but it still needs further development to make it a viable option. You will not get a standard regression output (as in lm or glm), only the regression coefficients (without standard errors). I will be happy to try it out once it is more robust. I am a bit puzzled that the R universe seems to lack a robust package that allows for pairwise deletion (which is standard in the otherwise poorer software packages STATA or SPSS…). Would be very happy if anyone can show me otherwise. Best, Adel -- Adel Daoud, PhD, Researcher *Newly published:* * Daoud, Adel and Kohl, Sebastian, *How Much Do Sociologists Write About Economic Topics? Using Big-Data to Test Some Conventional Views in Economic Sociology, 1890 to 2014*. Max Planck Institute for the Study of Societies, Discussion Paper 16/7 <http://www.mpifg.de/pu/mpifg_dp/dp16-7.pdf> * Daoud, Adel, Björn Halleröd, and Debarati Guha-Sapir, (2016) “What Is the Association between Absolute Child Poverty, Poor Governance, and Natural Disasters? A Global Comparison of Some of the Realities of Climate Change”, *PLoS ONE* 11(4) <http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0153296> *Shailen Nandy, Adel Daoud, David Gordon, (2016), Examining the changing profile of undernutrition in the context of food price rises and greater inequality, *Social Science & Medicine*, Volume 149, Pages 153–163 <http://www.sciencedirect.com/science/article/pii/S0277953615302446> *Department of Sociology and Work Science,* *University of Gothenburg* Box 720 405 30, Göteborg, Sweden Email: adel.da...@sociology.gu.se Website: http://adeldaoud.se/ <http://adeldaoud.se/> On Wed, Jul 13, 2016 at 8:21 AM, PIKAL Petr <petr.pi...@precheza.cz> wrote: > Hi > > > http://stats.stackexchange.com/questions/158366/fit-multiple-regression-model-with-pairwise-deletion-or-on-a-correlation-covari > > The package is probably not available on CRAN but seems to be still > maintained on github. > > Cheers > Petr > > > -Original Message- > > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of adel > > daoud > > Sent: Tuesday, July 12, 2016 8:28 PM > > To: r-help <r-help@r-project.org> > > Subject: [R] pairwise deletion in regression models > > > > Dear R users, > > > > > > > > I would like to use a pairwise deletion of missing values in linear > regression > > (lm or glm preferably). I want to replicate some studies done in STATA > that > > uses this type of deletion. What options do we have in R to work with > > pairwise deletion? Most packages I have found do not have this option, it > > seems (lm, glm, plm, psych, sampleSelection). > > > > > > > > This question has been raised here > > < > http://r.789695.n4.nabble.com/set-the-bahavior-that-R-deal-with-missing- > > values-td803840.html> > > and here > > <http://r.789695.n4.nabble.com/Pairwise-deletion-in-a-linear-regression- > > and-in-a-GLM-td4653004.html>, > > but without any clear answer. > > > > > > > > Any input is welcomed > > > > > > > > Thanks in advance > > > > > > Adel > > > > [[alternative HTML version deleted]] > > > > __ > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > > 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. > > > Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou > určeny pouze jeho adresátům. > Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě > neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie > vymažte ze svého systému. > Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email > jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. > Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi > či zpožděním přenosu e-mailu. > > V případě, že je tento e-mail součástí obchodního jednání: > - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření > smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu. > - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; > Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany > příjemce s dodatkem či odchylkou. > - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve > výslovným dosažením shody na všech
[R] pairwise deletion in regression models
Dear R users, I would like to use a pairwise deletion of missing values in linear regression (lm or glm preferably). I want to replicate some studies done in STATA that uses this type of deletion. What options do we have in R to work with pairwise deletion? Most packages I have found do not have this option, it seems (lm, glm, plm, psych, sampleSelection). This question has been raised here <http://r.789695.n4.nabble.com/set-the-bahavior-that-R-deal-with-missing-values-td803840.html> and here <http://r.789695.n4.nabble.com/Pairwise-deletion-in-a-linear-regression-and-in-a-GLM-td4653004.html>, but without any clear answer. Any input is welcomed Thanks in advance Adel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Alpha not working in geom_rect
Thanks for the info Jeff. I will stick to using annotate() -- Adel Daoud, PhD, Researcher The New School for Social Research, Visiting Scholar in the Economics Department, 6 East 16th Street New York, NY 10003, dao...@newschool.edu University of Gothenburg Department of Sociology and Work Science, Box 720 405 30, Göteborg, Sweden Visiting address: Sprängkullsgatan 25, room F411 Sprängkullsgatan 25, room K109 +46 031-786 41 73 adel.da...@sociology.gu.se On Mon, Mar 9, 2015 at 9:42 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: I have run into this a couple of times ... If you generate the rectangles once per row of your data, the fill gets more and more dense so your alpha seems to not work. The annotate call only paints the rectangle once so you don't have this problem. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On March 9, 2015 3:24:23 PM PDT, adel daoud dao...@newschool.edu wrote: Hi Jim, Thanks for the input but that did not work. I am suing Rstudio by the way and I guess that has a better device that would support ggplot output. The annotate options works but that does not explain why the geom_area does not work: annotate(rect, xmin=2, xmax=10, ymin=0, ymax=1, fill=black, alpha=0.5) Best Adel -- Adel Daoud, PhD, Researcher The New School for Social Research, Visiting Scholar in the Economics Department, 6 East 16th Street New York, NY 10003, dao...@newschool.edu University of Gothenburg Department of Sociology and Work Science, Box 720 405 30, Göteborg, Sweden Visiting address: Sprängkullsgatan 25, room F411 Sprängkullsgatan 25, room K109 +46 031-786 41 73 adel.da...@sociology.gu.se On Sun, Mar 8, 2015 at 12:46 AM, Jim Lemon drjimle...@gmail.com wrote: Hi Adel, Almost certainly because the device you were using doesn't support transparency.Try it with a PDF device and check the resulting file in a PDF reader: pdf(ad.pdf) print(p) dev.off() Jim On Sun, Mar 8, 2015 at 4:39 AM, Adel adel.da...@socav.gu.se wrote: Hi I am trying to activate the alpha argument to work, but for some reason it does not to play with me. Anybody has an idea why? p - ggplot(data = prediction_df, aes(x=x, y=prediction, fill=threshold)) + geom_area(colour=black, size=.2, alpha=.4) + scale_fill_brewer(palette=Set1, breaks=rev(levels(prediction_df$threshold))) p + geom_rect(aes(xmin=2, xmax=10, ymin=(0), ymax=(1)), fill=black, alpha=0.5) prediction_df x prediction threshold 1 -10 0.5694161 noAF 2 -9 0.5700513 noAF 3 -8 0.5706863 noAF 4 -7 0.5713211 noAF 5 -6 0.5719556 noAF 6 -5 0.5725899 noAF 7 -4 0.5732240 noAF 8 -3 0.5738578 noAF 9 -2 0.5744914 noAF 10 -1 0.5751247 noAF 11 0 0.5757578 noAF 12 1 0.5763906 noAF 13 2 0.5770232 noAF 14 3 0.5776556 noAF 15 4 0.5782876 noAF 16 5 0.5789195 noAF 17 6 0.5795510 noAF 18 7 0.5801823 noAF 19 8 0.5808134 noAF 20 9 0.5814441 noAF 21 10 0.5820747 noAF 22 -10 0.2359140 singleAF 23 -9 0.2356847 singleAF 24 -8 0.2354550 singleAF 25 -7 0.2352249 singleAF 26 -6 0.2349943 singleAF 27 -5 0.2347634 singleAF 28 -4 0.2345321 singleAF 29 -3 0.2343003 singleAF 30 -2 0.2340682 singleAF 31 -1 0.2338356 singleAF 32 0 0.2336027 singleAF 33 1 0.2333694 singleAF 34 2 0.2331357 singleAF 35 3 0.2329016 singleAF 36 4 0.2326671 singleAF 37 5 0.2324322 singleAF 38 6 0.2321969 singleAF 39 7 0.2319613 singleAF 40 8 0.2317253 singleAF 41 9 0.2314889 singleAF 42 10 0.2312522 singleAF 43 -10 0.1946699 multipleAF 44 -9 0.1942640 multipleAF 45 -8 0.1938587 multipleAF 46 -7 0.1934540 multipleAF 47 -6 0.1930500 multipleAF 48 -5 0.1926467 multipleAF 49 -4 0.1922440 multipleAF 50 -3 0.1918419 multipleAF 51 -2 0.1914404 multipleAF 52 -1 0.1910397 multipleAF 53 0 0.1906395 multipleAF 54 1 0.1902400 multipleAF 55 2 0.1898411 multipleAF 56 3 0.1894429 multipleAF 57 4 0.1890453 multipleAF 58 5 0.1886483 multipleAF 59 6 0.1882520 multipleAF 60 7 0.1878564 multipleAF 61 8 0.1874613 multipleAF
Re: [R] Alpha not working in geom_rect
Hi Jim, Thanks for the input but that did not work. I am suing Rstudio by the way and I guess that has a better device that would support ggplot output. The annotate options works but that does not explain why the geom_area does not work: annotate(rect, xmin=2, xmax=10, ymin=0, ymax=1, fill=black, alpha=0.5) Best Adel -- Adel Daoud, PhD, Researcher The New School for Social Research, Visiting Scholar in the Economics Department, 6 East 16th Street New York, NY 10003, dao...@newschool.edu University of Gothenburg Department of Sociology and Work Science, Box 720 405 30, Göteborg, Sweden Visiting address: Sprängkullsgatan 25, room F411 Sprängkullsgatan 25, room K109 +46 031-786 41 73 adel.da...@sociology.gu.se On Sun, Mar 8, 2015 at 12:46 AM, Jim Lemon drjimle...@gmail.com wrote: Hi Adel, Almost certainly because the device you were using doesn't support transparency.Try it with a PDF device and check the resulting file in a PDF reader: pdf(ad.pdf) print(p) dev.off() Jim On Sun, Mar 8, 2015 at 4:39 AM, Adel adel.da...@socav.gu.se wrote: Hi I am trying to activate the alpha argument to work, but for some reason it does not to play with me. Anybody has an idea why? p - ggplot(data = prediction_df, aes(x=x, y=prediction, fill=threshold)) + geom_area(colour=black, size=.2, alpha=.4) + scale_fill_brewer(palette=Set1, breaks=rev(levels(prediction_df$threshold))) p + geom_rect(aes(xmin=2, xmax=10, ymin=(0), ymax=(1)), fill=black, alpha=0.5) prediction_df x prediction threshold 1 -10 0.5694161 noAF 2 -9 0.5700513 noAF 3 -8 0.5706863 noAF 4 -7 0.5713211 noAF 5 -6 0.5719556 noAF 6 -5 0.5725899 noAF 7 -4 0.5732240 noAF 8 -3 0.5738578 noAF 9 -2 0.5744914 noAF 10 -1 0.5751247 noAF 11 0 0.5757578 noAF 12 1 0.5763906 noAF 13 2 0.5770232 noAF 14 3 0.5776556 noAF 15 4 0.5782876 noAF 16 5 0.5789195 noAF 17 6 0.5795510 noAF 18 7 0.5801823 noAF 19 8 0.5808134 noAF 20 9 0.5814441 noAF 21 10 0.5820747 noAF 22 -10 0.2359140 singleAF 23 -9 0.2356847 singleAF 24 -8 0.2354550 singleAF 25 -7 0.2352249 singleAF 26 -6 0.2349943 singleAF 27 -5 0.2347634 singleAF 28 -4 0.2345321 singleAF 29 -3 0.2343003 singleAF 30 -2 0.2340682 singleAF 31 -1 0.2338356 singleAF 32 0 0.2336027 singleAF 33 1 0.2333694 singleAF 34 2 0.2331357 singleAF 35 3 0.2329016 singleAF 36 4 0.2326671 singleAF 37 5 0.2324322 singleAF 38 6 0.2321969 singleAF 39 7 0.2319613 singleAF 40 8 0.2317253 singleAF 41 9 0.2314889 singleAF 42 10 0.2312522 singleAF 43 -10 0.1946699 multipleAF 44 -9 0.1942640 multipleAF 45 -8 0.1938587 multipleAF 46 -7 0.1934540 multipleAF 47 -6 0.1930500 multipleAF 48 -5 0.1926467 multipleAF 49 -4 0.1922440 multipleAF 50 -3 0.1918419 multipleAF 51 -2 0.1914404 multipleAF 52 -1 0.1910397 multipleAF 53 0 0.1906395 multipleAF 54 1 0.1902400 multipleAF 55 2 0.1898411 multipleAF 56 3 0.1894429 multipleAF 57 4 0.1890453 multipleAF 58 5 0.1886483 multipleAF 59 6 0.1882520 multipleAF 60 7 0.1878564 multipleAF 61 8 0.1874613 multipleAF 62 9 0.1870669 multipleAF 63 10 0.1866732 multipleAF -- View this message in context: http://r.789695.n4.nabble.com/Alpha-not-working-in-geom-rect-tp4704291.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Alpha not working in geom_rect
Hi I am trying to activate the alpha argument to work, but for some reason it does not to play with me. Anybody has an idea why? p - ggplot(data = prediction_df, aes(x=x, y=prediction, fill=threshold)) + geom_area(colour=black, size=.2, alpha=.4) + scale_fill_brewer(palette=Set1, breaks=rev(levels(prediction_df$threshold))) p + geom_rect(aes(xmin=2, xmax=10, ymin=(0), ymax=(1)), fill=black, alpha=0.5) prediction_df x prediction threshold 1 -10 0.5694161 noAF 2 -9 0.5700513 noAF 3 -8 0.5706863 noAF 4 -7 0.5713211 noAF 5 -6 0.5719556 noAF 6 -5 0.5725899 noAF 7 -4 0.5732240 noAF 8 -3 0.5738578 noAF 9 -2 0.5744914 noAF 10 -1 0.5751247 noAF 11 0 0.5757578 noAF 12 1 0.5763906 noAF 13 2 0.5770232 noAF 14 3 0.5776556 noAF 15 4 0.5782876 noAF 16 5 0.5789195 noAF 17 6 0.5795510 noAF 18 7 0.5801823 noAF 19 8 0.5808134 noAF 20 9 0.5814441 noAF 21 10 0.5820747 noAF 22 -10 0.2359140 singleAF 23 -9 0.2356847 singleAF 24 -8 0.2354550 singleAF 25 -7 0.2352249 singleAF 26 -6 0.2349943 singleAF 27 -5 0.2347634 singleAF 28 -4 0.2345321 singleAF 29 -3 0.2343003 singleAF 30 -2 0.2340682 singleAF 31 -1 0.2338356 singleAF 32 0 0.2336027 singleAF 33 1 0.2333694 singleAF 34 2 0.2331357 singleAF 35 3 0.2329016 singleAF 36 4 0.2326671 singleAF 37 5 0.2324322 singleAF 38 6 0.2321969 singleAF 39 7 0.2319613 singleAF 40 8 0.2317253 singleAF 41 9 0.2314889 singleAF 42 10 0.2312522 singleAF 43 -10 0.1946699 multipleAF 44 -9 0.1942640 multipleAF 45 -8 0.1938587 multipleAF 46 -7 0.1934540 multipleAF 47 -6 0.1930500 multipleAF 48 -5 0.1926467 multipleAF 49 -4 0.1922440 multipleAF 50 -3 0.1918419 multipleAF 51 -2 0.1914404 multipleAF 52 -1 0.1910397 multipleAF 53 0 0.1906395 multipleAF 54 1 0.1902400 multipleAF 55 2 0.1898411 multipleAF 56 3 0.1894429 multipleAF 57 4 0.1890453 multipleAF 58 5 0.1886483 multipleAF 59 6 0.1882520 multipleAF 60 7 0.1878564 multipleAF 61 8 0.1874613 multipleAF 62 9 0.1870669 multipleAF 63 10 0.1866732 multipleAF -- View this message in context: http://r.789695.n4.nabble.com/Alpha-not-working-in-geom-rect-tp4704291.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Using gregexpr and regmatches but getting Iconv error
Hi I have stumbled upon a problem when using gregexpr and regmatches, with the following error-message: Error in iconv(x, latin1, ASCII) : 'x' must be a list of NULL or raw vectors The data: (1) I have two journal articles and after some regex manipulation I am at the following situation: # manipluat only two full text articles author.test - articles1[1:2] # extract author informaiton r - gregexpr((\authors\:(.*?)\(.*?)\)|(\authors\: \\[(.*?)\\],), author.test) authors.raw - regmatches(author.test, r) authors.raw [[1]] [1] \authors\: [\Allan G. KING\, \B. Lindsay LOWELL\, \Frank D. BEAN\], [[2]] [1] \authors\: \Chris Baldry\, \ (2) Now, if I want to conduct additional regex manipulation I get the Error stated above. r - gregexpr(([^(\authors\:)])(.*?)(\(.*?)\), authors.raw) authors.raw - regmatches(authors.raw, r) Error in iconv(x, latin1, ASCII) : 'x' must be a list of NULL or raw vectors (3) One of the ways to avoid this is to unlist(authors.raw) - see below - but the problem with this is that I lose some information which was contained in the list. The first element contains three character elements and which are the authors of the first paper. I want to keep them in that list format. authors.raw - unlist(regmatches(authors.raw, r)) authors.raw [1] [\Allan G. KING\ , \B. Lindsay LOWELL\ , \Frank D. BEAN\ \Chris Baldry\ (4) So what I want to do is to avoid unlis() and apply the gregex() multiple times in a row. Any ideas? Thanks in advance Adel -- View this message in context: http://r.789695.n4.nabble.com/Using-gregexpr-and-regmatches-but-getting-Iconv-error-tp4700677.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] list.files accessing subdirectory as relative path?
Thanks for the reply Don and Frede, Your suggestions works perfectly! Best Adel -- View this message in context: http://r.789695.n4.nabble.com/list-files-accessing-subdirectory-as-relative-path-tp4689997p4690048.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] list.files accessing subdirectory as relative path?
Dear list members, I would like to access a subdirectory given where the work directory has been set. So I have: getwd() [1] C:/Users/Lord Adellus/Dropbox/I8child1/Data list.files() # give three folders [1] Least Developed Countries [4] Low middle income grouping More advanced developing countries and territories list.files(path = ../Least Developed Countries) # I want to access now one of the subdirectories character(0) But as you can see R does not want to look into the specified subdirectory. I have tried several combination and searched the list but without any great success. Actually: list.files(path = ../...) #goes up one level in the folder structure so I cannot see what the problem is. Thanks in advance. Adel -- View this message in context: http://r.789695.n4.nabble.com/list-files-accessing-subdirectory-as-relative-path-tp4689997.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] barchar or barp on multiple data
here is the solution if anyone is interrested barchart(V5 ~ V3 | V1 * V2 , data = t,groups = V4, layout = c(1,6), auto.key = list(space = right), ylab = Makespan) Regards 2014/1/14 Adel ESSAFI adeless...@gmail.com Hello list I have the following data in file in attachment. in want to draw bars for every value of V5 in finction of V3 grouped by V2*V1. however, each unique value of V4, V2 and V1 I want to draw a bar. This example is very close to want I want to do. but here, for each value of V3, barchart draws 2 superimposed bar. How to do to make them one beside another. Thanks t=read.table(actual) barchart( V5 ~ V3 | V2 * V1,t) -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] barchar or barp on multiple data
Hello list I have the following data in file in attachment. in want to draw bars for every value of V5 in finction of V3 grouped by V2*V1. however, each unique value of V4, V2 and V1 I want to draw a bar. This example is very close to want I want to do. but here, for each value of V3, barchart draws 2 superimposed bar. How to do to make them one beside another. Thanks t=read.table(actual) barchart( V5 ~ V3 | V2 * V1,t) __ 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] barchar or barp on multiple data
Ok, I put here the data exported with dput(). structure(list(V1 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), V2 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), V3 = structure(c(1L, 1L, 2L, 2L, 8L, 8L, 7L, 7L, 9L, 9L, 10L, 10L, 4L, 4L, 3L, 3L, 5L, 5L, 6L, 6L, 1L, 1L, 2L, 2L, 8L, 8L, 7L, 7L, 9L, 9L, 10L, 10L, 4L, 4L, 3L, 3L, 5L, 5L, 6L, 6L, 1L, 1L, 2L, 2L, 8L, 8L, 7L, 7L, 9L, 9L, 10L, 10L, 4L, 4L, 3L, 3L, 5L, 5L, 6L, 6L, 1L, 1L, 2L, 2L, 8L, 8L, 7L, 7L, 9L, 9L, 10L, 10L, 4L, 4L, 3L, 3L, 5L, 5L, 6L, 6L), .Label = c(C/L, C/S, D/F, D/I, D/L, D/S, I/F, I/I, I/L, I/S), class = factor), V4 = c(0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L), V5 = c(671115, 1028526, 734174, 1327998, 838787, 1422540, 803142, 1206913, 1981495, 1856112, 1981495, 1856112, 689313.033281565, 1167279.07984257, 666335.015648723, 1087554.01538312, 670777.019860625, 1028317.0531472, 730457.025310636, 1298133.01414597, 690879, 996712, 755213, 1293494, 871769, 1478186, 830802, 1169810, 2364168, 2364168, 2364168, 2364168, 710988.088020086, 1119450.09723687, 685209.028554797, 1041768.02318358, 689760.04900682, 997493.021301806, 752893.050110579, 1286333.03097144, 671437, 1029284, 732144, 1317596, 844051, 1422540, 803142, 1205147, 1981495, 1856112, 1981495, 1856112, 690552.014781833, 1168328.05673718, 14.011364698, 1089523.05270934, 670944.00994724, 1029246.06969887, 728912.037932515, 1294319.04904819, 691311, 996821, 753640, 1299848, 871769, 1443884, 830802, 1169810, 2364168, 2364168, 2364168, 2364168, 711121.048051834, 1115219.00324583, 685480.015842438, 1042424.03302765, 690023.032608509, 998007.038496614, 751040.029144168, 1277138.01515984), V6 = c(631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 631830L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 651129L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 632125L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L, 651458L), V7 = c(471L, 900L, 423L, 830L, 4L, 4L, 192L, 157L, 477L, 900L, 428L, 827L, 59L, 72L, 6221L, 5409L, 11752L, 17060L, 10766L, 17546L, 462L, 830L, 417L, 774L, 4L, 4L, 182L, 142L, 472L, 824L, 421L, 771L, 58L, 66L, 5799L, 4819L, 11581L, 16325L, 10860L, 17255L, 472L, 901L, 424L, 835L, 5L, 5L, 192L, 160L, 477L, 900L, 428L, 826L, 58L, 72L, 6176L, 5390L, 11590L, 17028L, 10782L, 17561L, 465L, 827L, 422L, 777L, 4L, 4L, 180L, 142L, 470L, 827L, 422L, 774L, 57L, 68L, 5781L, 4798L, 11558L, 16348L, 10885L, 17221L), V8 = c(4.71, 9, 4.23, 8.3, 0.04, 0.04, 1.92, 1.57, 4.77, 9, 4.28, 8.27, 0.59, 0.72, 62.21, 54.09, 117.52, 170.6, 107.66, 175.46, 4.62, 8.3, 4.17, 7.74, 0.04, 0.04, 1.82, 1.42, 4.72, 8.24, 4.21, 7.71, 0.58, 0.66, 57.99, 48.19, 115.81, 163.25, 108.6, 172.55, 4.72, 9.01, 4.24, 8.35, 0.05, 0.05, 1.92, 1.6, 4.77, 9, 4.28, 8.26, 0.58, 0.72, 61.76, 53.9, 115.9, 170.28, 107.82, 175.61, 4.65, 8.27, 4.22, 7.77, 0.04, 0.04, 1.8, 1.42, 4.7, 8.27, 4.22, 7.74, 0.57, 0.68, 57.81, 47.98, 115.58, 163.48, 108.85, 172.21)), row.names = c(NA, -80L ), .Names = c(V1, V2, V3, V4, V5, V6, V7, V8), class = data.frame) 2014/1/14 Adel ESSAFI adeless...@gmail.com Hello list I have the following data in file in attachment. in want to draw bars for every value of V5 in finction of V3 grouped by V2*V1. however, each unique value of V4, V2 and V1 I want to draw a bar. This example is very close to want I want to do. but here, for each value of V3, barchart draws 2 superimposed bar. How
[R] barchar and box on the same figure! is it possible
Hello list I have this data frame which represent values grouped by algorithm. remark that for each algorithm, we have the same value in V3 but differnt values in V4. I want to make a plot with algorithms in axes, and for each algorithm I draw a bar (for V3) and a bwplot for V4. is thart possible. Thanks for any entry. t[t[,2]3,] V1 V2 V3 V4 1HEFT-AC 1 402499.9 460543.4 2HEFT-AC 2 402499.9 470316.8 31 HEFT-ACU 1 420814.2 433203.7 32 HEFT-ACU 2 420814.2 453475.8 60 LPT 1 402499.9 460543.4 61 LPT 2 402499.9 470316.8 90 SPT 1 453262.3 459070.3 91 SPT 2 453262.3 483893.6 119 MIN-MIN 1 447286.4 477632.5 120 MIN-MIN 2 447286.4 488146.3 149 MAX-MIN 1 457122.6 498265.2 150 MAX-MIN 2 457122.6 491143.3 -- *PhD in Computer ScienceAddress * *Avenue Taha Hussein Montfleury, 1008 Tunistél : +216 71 49 60 66 fax: +216 71 39 11 66* [[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] Fwd: barchar and box on the same figure! is it possible
Hello list I have this data frame which represent values grouped by algorithm. remark that for each algorithm, we have the same value in V3 but differnt values in V4. I want to make a plot with algorithms in axes, and for each algorithm I draw a bar (for V3) and a bwplot for V4. is thart possible. Thanks for any entry. t[t[,2]3,] V1 V2 V3 V4 1HEFT-AC 1 402499.9 460543.4 2HEFT-AC 2 402499.9 470316.8 31 HEFT-ACU 1 420814.2 433203.7 32 HEFT-ACU 2 420814.2 453475.8 60 LPT 1 402499.9 460543.4 61 LPT 2 402499.9 470316.8 90 SPT 1 453262.3 459070.3 91 SPT 2 453262.3 483893.6 119 MIN-MIN 1 447286.4 477632.5 120 MIN-MIN 2 447286.4 488146.3 149 MAX-MIN 1 457122.6 498265.2 150 MAX-MIN 2 457122.6 491143.3 -- *PhD in Computer ScienceAddress * *Avenue Taha Hussein Montfleury, 1008 Tunis tél : +216 71 49 60 66 %2B216%2071%2049%2060%2066 fax: +216 71 39 11 66 %2B216%2071%2039%2011%2066* -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] embedd fonts in pdf generated by R
Hello list I generated pdf files with R that I integrated in .tex file and compiled with pdflatex. however, all the fonts of my R figure not embadded. so the file is rejected from EDAS web site. Could you help please. [adel@localhost hcw]$ pdffonts hcw.pdf name type encoding emb sub uni object ID - --- --- --- - ZIMYVR+NimbusRomNo9L-MediType 1Custom yes yes no 7 0 EYECKV+NimbusRomNo9L-ReguType 1Custom yes yes no 8 0 WKYGSZ+CMSY8 Type 1Builtin yes yes no 9 0 TEABSF+NimbusRomNo9L-ReguItalType 1Custom yes yes no 10 0 MPLOPB+NimbusRomNo9L-MediItalType 1Custom yes yes no 11 0 XSKVTD+CMMI10Type 1Builtin yes yes no 19 0 NKXZUI+CMSY10Type 1Builtin yes yes no 20 0 PVJEEI+CMR10 Type 1Builtin yes yes no 21 0 TEORRU+CMR7 Type 1Builtin yes yes no 22 0 NGESNP+CMMI7 Type 1Builtin yes yes no 23 0 ZDXXKY+CMMI5 Type 1Builtin yes yes no 24 0 VGZAHY+CMR5 Type 1Builtin yes yes no 25 0 NOPBSI+CMSY5 Type 1Builtin yes yes no 26 0 ZYGQZM+CMMI6 Type 1Builtin yes yes no 27 0 QRZPBZ+CMMI8 Type 1Builtin yes yes no 28 0 TZYURE+MSBM10Type 1Builtin yes yes no 29 0 HIBZGW+Times-Roman Type 1C WinAnsi yes yes no 37 0 QLAAJV+CMSY7 Type 1Builtin yes yes no 46 0 RMTRPB+CMEX10Type 1Builtin yes yes no 47 0 BNPUIJ+Courier Type 1C WinAnsi yes yes no 60 0 GZSZHC+Times-Roman Type 1C WinAnsi yes yes no 61 0 QBHKPM+Times-Roman Type 1C WinAnsi yes yes no 70 0 Symbol Type 1Custom no no no 81 0 ZapfDingbats Type 1Custom no no yes 82 0 HelveticaType 1Standard no no no 83 0 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] ifelse statement with two vectors of different length
Dear list-members, I have the following problem: I have a vector (countrydiff) with length 72 and another vector (long_df$country_name) which is about 12000 long. Basically what I want to do is to if the factor level (or string name) in long_df$country_name appears on the countrydiff, then long_df$povdat should be equal to 1, if it does not appear on the countrydiff vector then long_df$povdat should be equal to zero. I have tried different combinations and read some. The following code should in my mind do it, but it doesn’t: long_df$povdat-ifelse(long_df$country_name == countrydiff, 1, 0) long_df$povdat-ifelse(long_df$country_name %in% countrydiff, 1, 0) Additional information: the factor vector countrydiff contains unique country names (Albania, Zimbabwe etc.), whereas long_df$country_name also contains country names albeit not unique since it is in longform. The unique names that appear in long_df$country_name is around 200. Any suggestions? Thanks in advance. Best Adel -- View this message in context: http://r.789695.n4.nabble.com/ifelse-statement-with-two-vectors-of-different-length-tp4682401.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] ifelse statement with two vectors of different length
Dear Arun Thanks for your reply, it made me realize that the problem was not in the code but in the levels() of the factors. Some countries had some extra spacing which made the ifelse() function not work. So if I modify your code (added space to countrydiff), it will then look something like this: countrydiff - c(Albania, Algeria, Belarus, Canada , Germany ) long_df - data.frame(country_name = c(Algeria, Guyana, Hungary, Algeria, Canada, Iran, Iran, Norway,Uruguay, Zimbabwe) ) I had to use the gsub to fix this first. Interestingly, the setdiff() function did not react on spacing difference which I used before coming to the ifelse statement and therefore I did not react on this in the first place #no reaction from R on spacing diff. setdiff(countrydiff, long_df$country_name) Nevertheless, thanks again for being helpful! Adel -- View this message in context: http://r.789695.n4.nabble.com/ifelse-statement-with-two-vectors-of-different-length-tp4682401p4682403.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] multiple bar for barchart
Hello list, I have the following data on dm table dm Group.1 V1 V2 V3V4 V5 V6 V7 V8 1 C/L NA 15.5 732179 875270.6 -143091.46 1107270 1088300 18964.40 2 C/S NA 15.5 803926 850352.1 -46426.03 1395710 1312310 83403.30 3 D/D NA 15.5 751660 857828.2 -106168.17 1340360 1322790 17569.30 4 D/F NA 15.5 724924 969418.7 -244494.67 1181280 1160760 20519.20 5 D/I NA 15.5 755841 842130.5 -86289.48 1264250 1241750 22495.20 6 D/L NA 15.5 731904 875340.0 -143435.84 1107600 1087940 19657.30 7 D/S NA 15.5 798289 844102.0 -45812.85 1399840 1305000 94832.10 8 I/F NA 15.5 871670 1074136.3 -202466.58 1304290 1249006 55286.59 9 I/I NA 15.5 897718 1029579.0 -131861.35 1542810 1398716 144100.07 10 I/L NA 15.5 2628110 862466.8 1765645.67 2628110 1073510 1554610.00 11 I/S NA 15.5 2628110 831486.8 1796627.33 2475450 1282100 1193350.00 barchart (dm[,4] ~ dm[,1]) For each value of Group.1 I want to draw 4 bars (v3,v4, v6 and v7). Can you suggest me a solution please barchat draws only one value. regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] the wilcox.test() and pairwise.wilcox.test are producing different results
Thanks Arun! It was simple as that. You suggestion solved it. Best Adel -- View this message in context: http://r.789695.n4.nabble.com/the-wilcox-test-and-pairwise-wilcox-test-are-producing-different-results-tp4681375p4681423.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] the wilcox.test() and pairwise.wilcox.test are producing different results
Dear list member, I want to compare if the rank order is significantly different for seven different measures. So we have same sample but different measures which reduces the problem to a paired one sample Wilcox test if I understood the test correctly. In constructed toy examples for my sake of understanding, but things are not adding up. Basically, the wilcox.test() and pairwise.wilcox.test are producing different results when they should not (according to my understanding of course): #take a vector daily.intake - c(5260,5470,5640,6180,6390,6515, + 6805,7515,7515,8230,8770) #I get desired results when I do the following daily-data.frame(pre=daily.intake, post=daily.intake) #add som differences daily[1,1]-5000 daily[2,1]-5100 #reshape the data for pairwise comparison library(reshape2) daily_long-melt(daily, id=) #conduct simple test wilcox.test(daily$pre,daily$post, paired=T) #produces desired results #do the test again but now in a pairwise, which produces the same p-value as in the simple test above pairwise.wilcox.test(daily_long$value,daily_long$variable, paired=T) #But now I the issues arise when testing more than two vectors. #take three vectors this time daily-data.frame(pre=daily.intake, post=daily.intake, posttwo=daily.intake) #add some differences daily[1,1]-5000 daily[2,1]-5100 daily[10,3]-9000 daily[11,3]-9100 #the wilcox.test() produces a set of p-values wilcox.test(daily$pre,daily$posttwo, paired=T) wilcox.test(daily$pre,daily$post, paired=T) wilcox.test(daily$post,daily$posttwo, paired=T) #and the pairwise.wilcox.test produces another set pairwise.wilcox.test(daily_long$value,daily_long$variable, paired=T) ##And from the manual for pairwise.wilcox we get similar issues #produces a given set of p-values attach(airquality) Month - factor(Month, labels = month.abb[5:9]) ## These give warnings because of ties : pairwise.wilcox.test(Ozone, Month) pairwise.wilcox.test(Ozone, Month, p.adj = bonf) detach() #but if we want to test the rank difference between the 6th and 7th month we get a p-value of 0.5775 testar skillnaden mellan 6e och 7e mÃ¥naden â observ however that this data is not paired which makes it different to the example I gave above. #p-värdet är 0.5775 air-subset(airquality, airquality$Month 7) #p-value is now 0.1925 wilcox.test(air$Ozone~air$Month) What am I doing wrong here? Best Adel -- Adel Daoud, PhD Visiting researcher (post-doc) Max Planck Institute for the Study of Societies / Max-Planck-Institut für Gesellschaftsforschung Paulstr. 3 | 506 76 Köln | Germany Tel.: + 49 (0) 221 2767-534 da...@mpifg.de Department of Sociology and Work Science, University of Gothenburg Box 720 405 30 Göteborg, Sweden Visiting address: Sprängkullsgatan 25, room K109 +46 031-786 41 73 adel.da...@sociology.gu.se -- View this message in context: http://r.789695.n4.nabble.com/the-wilcox-test-and-pairwise-wilcox-test-are-producing-different-results-tp4681375.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] legend for bar plot ?
Hello; I have the following table m V2 V1 V3 1 C/L 0 732179 3 C/S 0 803926 19 D/F 0 724924 17 D/I 0 755841 13 D/L 0 731904 15 D/S 0 798289 11 I/F 0 871670 9 I/I 0 897718 5 I/L 0 2628113 7 I/S 0 2628113 2 C/L 1 1107269 4 C/S 1 1395714 20 D/F 1 1181282 18 D/I 1 1264249 14 D/L 1 1107595 16 D/S 1 1399836 12 I/F 1 1304294 10 I/I 1 1542813 6 I/L 1 2628113 8 I/S 1 2475448 as you can see, the table is sorted by the second and the first column. with this command : barplot(m$V3,names.arg=m$V2,col=rainbow(10)) I succeded to print the figure attached. Now, I need to indicate that the first 10 bar are for beta=0 and the second 10 bars are for beta=1. Could you help please. Best regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] legend for bar plot ?
Hello; I have the following table m V2 V1 V3 1 C/L 0 732179 3 C/S 0 803926 19 D/F 0 724924 17 D/I 0 755841 13 D/L 0 731904 15 D/S 0 798289 11 I/F 0 871670 9 I/I 0 897718 5 I/L 0 2628113 7 I/S 0 2628113 2 C/L 1 1107269 4 C/S 1 1395714 20 D/F 1 1181282 18 D/I 1 1264249 14 D/L 1 1107595 16 D/S 1 1399836 12 I/F 1 1304294 10 I/I 1 1542813 6 I/L 1 2628113 8 I/S 1 2475448 as you can see, the table is sorted by the second and the first column. with this command : barplot(m$V3,names.arg=m$V2,col=rainbow(10)) I succeded to print the figure attached. Now, I need to indicate that the first 10 bar are for beta=0 and the second 10 bars are for beta=1. Could you help please. Best regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 plot.pdf Description: Adobe PDF document __ 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] scatter plot matrix with different x-y variables
Greg, the pairs2 function was exactly what I wanted. Thanks for a very useful function. May I ask a follow up question, is it possible to draw a correlation line with R2 values for each graph on each graph, or on the side of the graph. I am sure this is possible somehow, but I am new to R programming so please just direct me into the right direction. John, the grid.arrange() function is useful too but not exactly what I needed. Thanks any way. Best Adel On Sat, May 11, 2013 at 4:52 PM, Greg Snow 538...@gmail.com wrote: The pairs2 function in the TeachingDemos package does what you describe. You give it 2 matricies instead of just one and it creates the plots. On Wed, May 8, 2013 at 10:49 AM, Adel adelda...@gmail.com wrote: Dear list-members, I wonder if there is a way of creating a scatter plot table/grid with different variables on the y-axis compared to the x-axis? Something like this: A*** B*** C*** -XYZ I know that you can create scatter plot matrix with the same variables on the y-axis as on the x-axis, like this: A* *B **C*** ***X** Y* *Z But this is not what I want. I basically want to get a visual representation of different dependent variables (y-axis) on various independent variables (x-axis). So I want it to look like a scatter plot matrix or grid, but with my own specified variables on each axis. Secondly, to add to this, can I add a correlation line (linear and quadratic) through all scatter plots with both correlation and R2 values? All help appreciated. Adel -- Adel Daoud, PhD Department of Sociology, University of Gothenburg Box 720 405 30 Göteborg, SWEDEN Visiting address: Sprängkullsgatan 25, room F411 +46 031-786 47 84 adel.da...@sociology.gu.se -- View this message in context: http://r.789695.n4.nabble.com/scatter-plot-matrix-with-different-x-y-variables-tp4666598.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com -- Adel Daoud, PhD Department of Sociology, University of Gothenburg Box 720 405 30 Göteborg, SWEDEN Visiting address: Sprängkullsgatan 25, room F411 +46 031-786 47 84 adel.da...@sociology.gu.se [[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] R prints empty group on my figure!!
2013/5/9 PIKAL Petr petr.pi...@precheza.cz Hi Thank you, that was the option I look for Use drop=TRUE argument in interaction for removing unused levels. Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Adel ESSAFI Sent: Wednesday, May 08, 2013 12:48 PM To: r-help Subject: Re: [R] R prints empty group on my figure!! hello, I attach the figure generated by R: 2013/5/8 Adel ESSAFI adeless...@gmail.com Hello list I am trying to solve a problem when drawing a figure related to the dataframe below. I draw V6 in Y axes and V3 as X axes. The data are grouped by V1 AND V2. I use this commande to make it: xyplot(cm[,6]~cm[,3],type=b,group=interaction(cm[,1],cm[,2],sep=/) , auto.key =list( title=Makespan en fonction de beta, points = FALSE, columns=2, lines = TRUE) ,data=cm,xlab=expression(beta),ylab=Ratio) The problem is that, R puts (for example) the entry C/D in the legend, however, I have no such entry. Note that R it does not draw a line for that entry. So, the question : how to do to remove the empty groups from the legend. Cordially cm V1 V2 V3V4 V5 V6 1 C L 0.0 732179.0 686983 1.065789 2 C L 0.2 900956.4 686983 1.311468 3 C L 0.4 964725.0 686983 1.404292 4 C L 0.6 1017984.0 686983 1.481818 5 C L 0.8 1064745.8 686983 1.549887 6 C L 1.0 1107269.0 686983 1.611785 7 C S 0.0 803926.0 686983 1.170227 8 C S 0.2 1020492.0 686983 1.485469 9 C S 0.4 1134818.0 686983 1.651887 10 C S 0.6 1223596.0 686983 1.781115 11 C S 0.8 1322548.8 686983 1.925155 12 C S 1.0 1395714.0 686983 2.031657 13 I L 0.0 2628113.0 686983 3.825587 14 I L 0.2 2628113.0 686983 3.825587 15 I L 0.4 2628113.0 686983 3.825587 16 I L 0.6 2628113.0 686983 3.825587 17 I L 0.8 2628113.0 686983 3.825587 18 I L 1.0 2628113.0 686983 3.825587 19 I S 0.0 2628113.0 686983 3.825587 20 I S 0.2 2628113.0 686983 3.825587 21 I S 0.4 2475448.0 686983 3.603361 22 I S 0.6 2475448.0 686983 3.603361 23 I S 0.8 2475448.0 686983 3.603361 24 I S 1.0 2475448.0 686983 3.603361 25 I F 0.0 871670.0 686983 1.268838 26 I F 0.2 1066742.0 686983 1.552792 27 I F 0.4 1136213.0 686983 1.653917 28 I F 0.6 1210547.0 686983 1.762121 29 I F 0.8 1256777.0 686983 1.829415 30 I F 1.0 1304294.0 686983 1.898583 31 I I 0.0 897718.0 686983 1.306754 32 I I 0.2 1123968.0 686983 1.636093 33 I I 0.4 1301097.0 686983 1.893929 34 I I 0.6 1373663.0 686983 1.999559 35 I I 0.8 1424121.0 686983 2.073008 36 I I 1.0 1542813.0 686983 2.245780 37 I D 0.0 937621.0 686983 1.364839 38 I D 0.2 1205063.0 686983 1.754138 39 I D 0.4 1341839.0 686983 1.953235 40 I D 0.6 1431970.0 686983 2.084433 41 I D 0.8 1538944.0 686983 2.240149 42 I D 1.0 1670073.0 686983 2.431025 43 D L 0.0 731904.0 686983 1.065389 44 D L 0.2 900183.0 686983 1.310342 45 D L 0.4 964870.1 686983 1.404504 46 D L 0.6 1017462.8 686983 1.481060 47 D L 0.8 1064435.2 686983 1.549435 48 D L 1.0 1107595.0 686983 1.612260 49 D S 0.0 798289.0 686983 1.162022 50 D S 0.2 1014643.2 686983 1.476955 51 D S 0.4 1125711.8 686983 1.638631 52 D S 0.6 1220923.0 686983 1.777224 53 D S 0.8 1306571.2 686983 1.901897 54 D S 1.0 1399836.0 686983 2.037657 55 D F 0.0 724924.0 686983 1.055229 56 D F 0.2 935413.5 686983 1.361625 57 D F 0.4 1011621.9 686983 1.472557 58 D F 0.6 1071081.6 686983 1.559109 59 D F 0.8 1139325.8 686983 1.658448 60 D F 1.0 1181282.1 686983 1.719522 61 D I 0.0 755841.0 686983 1.100232 62 D I 0.2 964335.0 686983 1.403725 63 D I 0.4 1051233.4 686983 1.530218 64 D I 0.6 1110842.0 686983 1.616986 65 D I 0.8 1193509.0 686983 1.737320 66 D I 1.0 1264249.0 686983 1.840292 67 D D 0.0 751660.0 686983 1.094146 68 D D 0.2 1019735.0 686983 1.484367 69 D D 0.4 1101191.0 686983 1.602938 70 D D 0.6 1191547.0 686983 1.734464 71 D D 0.8 1293433.0 686983 1.882773 72 D D 1.0 1340360.0 686983 1.951082 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] R prints empty group on my figure!!
Hello list I am trying to solve a problem when drawing a figure related to the dataframe below. I draw V6 in Y axes and V3 as X axes. The data are grouped by V1 AND V2. I use this commande to make it: xyplot(cm[,6]~cm[,3],type=b,group=interaction(cm[,1],cm[,2],sep=/), auto.key =list( title=Makespan en fonction de beta, points = FALSE, columns=2, lines = TRUE) ,data=cm,xlab=expression(beta),ylab=Ratio) The problem is that, R puts (for example) the entry C/D in the legend, however, I have no such entry. Note that R it does not draw a line for that entry. So, the question : how to do to remove the empty groups from the legend. Cordially cm V1 V2 V3V4 V5 V6 1 C L 0.0 732179.0 686983 1.065789 2 C L 0.2 900956.4 686983 1.311468 3 C L 0.4 964725.0 686983 1.404292 4 C L 0.6 1017984.0 686983 1.481818 5 C L 0.8 1064745.8 686983 1.549887 6 C L 1.0 1107269.0 686983 1.611785 7 C S 0.0 803926.0 686983 1.170227 8 C S 0.2 1020492.0 686983 1.485469 9 C S 0.4 1134818.0 686983 1.651887 10 C S 0.6 1223596.0 686983 1.781115 11 C S 0.8 1322548.8 686983 1.925155 12 C S 1.0 1395714.0 686983 2.031657 13 I L 0.0 2628113.0 686983 3.825587 14 I L 0.2 2628113.0 686983 3.825587 15 I L 0.4 2628113.0 686983 3.825587 16 I L 0.6 2628113.0 686983 3.825587 17 I L 0.8 2628113.0 686983 3.825587 18 I L 1.0 2628113.0 686983 3.825587 19 I S 0.0 2628113.0 686983 3.825587 20 I S 0.2 2628113.0 686983 3.825587 21 I S 0.4 2475448.0 686983 3.603361 22 I S 0.6 2475448.0 686983 3.603361 23 I S 0.8 2475448.0 686983 3.603361 24 I S 1.0 2475448.0 686983 3.603361 25 I F 0.0 871670.0 686983 1.268838 26 I F 0.2 1066742.0 686983 1.552792 27 I F 0.4 1136213.0 686983 1.653917 28 I F 0.6 1210547.0 686983 1.762121 29 I F 0.8 1256777.0 686983 1.829415 30 I F 1.0 1304294.0 686983 1.898583 31 I I 0.0 897718.0 686983 1.306754 32 I I 0.2 1123968.0 686983 1.636093 33 I I 0.4 1301097.0 686983 1.893929 34 I I 0.6 1373663.0 686983 1.999559 35 I I 0.8 1424121.0 686983 2.073008 36 I I 1.0 1542813.0 686983 2.245780 37 I D 0.0 937621.0 686983 1.364839 38 I D 0.2 1205063.0 686983 1.754138 39 I D 0.4 1341839.0 686983 1.953235 40 I D 0.6 1431970.0 686983 2.084433 41 I D 0.8 1538944.0 686983 2.240149 42 I D 1.0 1670073.0 686983 2.431025 43 D L 0.0 731904.0 686983 1.065389 44 D L 0.2 900183.0 686983 1.310342 45 D L 0.4 964870.1 686983 1.404504 46 D L 0.6 1017462.8 686983 1.481060 47 D L 0.8 1064435.2 686983 1.549435 48 D L 1.0 1107595.0 686983 1.612260 49 D S 0.0 798289.0 686983 1.162022 50 D S 0.2 1014643.2 686983 1.476955 51 D S 0.4 1125711.8 686983 1.638631 52 D S 0.6 1220923.0 686983 1.777224 53 D S 0.8 1306571.2 686983 1.901897 54 D S 1.0 1399836.0 686983 2.037657 55 D F 0.0 724924.0 686983 1.055229 56 D F 0.2 935413.5 686983 1.361625 57 D F 0.4 1011621.9 686983 1.472557 58 D F 0.6 1071081.6 686983 1.559109 59 D F 0.8 1139325.8 686983 1.658448 60 D F 1.0 1181282.1 686983 1.719522 61 D I 0.0 755841.0 686983 1.100232 62 D I 0.2 964335.0 686983 1.403725 63 D I 0.4 1051233.4 686983 1.530218 64 D I 0.6 1110842.0 686983 1.616986 65 D I 0.8 1193509.0 686983 1.737320 66 D I 1.0 1264249.0 686983 1.840292 67 D D 0.0 751660.0 686983 1.094146 68 D D 0.2 1019735.0 686983 1.484367 69 D D 0.4 1101191.0 686983 1.602938 70 D D 0.6 1191547.0 686983 1.734464 71 D D 0.8 1293433.0 686983 1.882773 72 D D 1.0 1340360.0 686983 1.951082 [[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] R prints empty group on my figure!!
hello, I attach the figure generated by R: 2013/5/8 Adel ESSAFI adeless...@gmail.com Hello list I am trying to solve a problem when drawing a figure related to the dataframe below. I draw V6 in Y axes and V3 as X axes. The data are grouped by V1 AND V2. I use this commande to make it: xyplot(cm[,6]~cm[,3],type=b,group=interaction(cm[,1],cm[,2],sep=/), auto.key =list( title=Makespan en fonction de beta, points = FALSE, columns=2, lines = TRUE) ,data=cm,xlab=expression(beta),ylab=Ratio) The problem is that, R puts (for example) the entry C/D in the legend, however, I have no such entry. Note that R it does not draw a line for that entry. So, the question : how to do to remove the empty groups from the legend. Cordially cm V1 V2 V3V4 V5 V6 1 C L 0.0 732179.0 686983 1.065789 2 C L 0.2 900956.4 686983 1.311468 3 C L 0.4 964725.0 686983 1.404292 4 C L 0.6 1017984.0 686983 1.481818 5 C L 0.8 1064745.8 686983 1.549887 6 C L 1.0 1107269.0 686983 1.611785 7 C S 0.0 803926.0 686983 1.170227 8 C S 0.2 1020492.0 686983 1.485469 9 C S 0.4 1134818.0 686983 1.651887 10 C S 0.6 1223596.0 686983 1.781115 11 C S 0.8 1322548.8 686983 1.925155 12 C S 1.0 1395714.0 686983 2.031657 13 I L 0.0 2628113.0 686983 3.825587 14 I L 0.2 2628113.0 686983 3.825587 15 I L 0.4 2628113.0 686983 3.825587 16 I L 0.6 2628113.0 686983 3.825587 17 I L 0.8 2628113.0 686983 3.825587 18 I L 1.0 2628113.0 686983 3.825587 19 I S 0.0 2628113.0 686983 3.825587 20 I S 0.2 2628113.0 686983 3.825587 21 I S 0.4 2475448.0 686983 3.603361 22 I S 0.6 2475448.0 686983 3.603361 23 I S 0.8 2475448.0 686983 3.603361 24 I S 1.0 2475448.0 686983 3.603361 25 I F 0.0 871670.0 686983 1.268838 26 I F 0.2 1066742.0 686983 1.552792 27 I F 0.4 1136213.0 686983 1.653917 28 I F 0.6 1210547.0 686983 1.762121 29 I F 0.8 1256777.0 686983 1.829415 30 I F 1.0 1304294.0 686983 1.898583 31 I I 0.0 897718.0 686983 1.306754 32 I I 0.2 1123968.0 686983 1.636093 33 I I 0.4 1301097.0 686983 1.893929 34 I I 0.6 1373663.0 686983 1.999559 35 I I 0.8 1424121.0 686983 2.073008 36 I I 1.0 1542813.0 686983 2.245780 37 I D 0.0 937621.0 686983 1.364839 38 I D 0.2 1205063.0 686983 1.754138 39 I D 0.4 1341839.0 686983 1.953235 40 I D 0.6 1431970.0 686983 2.084433 41 I D 0.8 1538944.0 686983 2.240149 42 I D 1.0 1670073.0 686983 2.431025 43 D L 0.0 731904.0 686983 1.065389 44 D L 0.2 900183.0 686983 1.310342 45 D L 0.4 964870.1 686983 1.404504 46 D L 0.6 1017462.8 686983 1.481060 47 D L 0.8 1064435.2 686983 1.549435 48 D L 1.0 1107595.0 686983 1.612260 49 D S 0.0 798289.0 686983 1.162022 50 D S 0.2 1014643.2 686983 1.476955 51 D S 0.4 1125711.8 686983 1.638631 52 D S 0.6 1220923.0 686983 1.777224 53 D S 0.8 1306571.2 686983 1.901897 54 D S 1.0 1399836.0 686983 2.037657 55 D F 0.0 724924.0 686983 1.055229 56 D F 0.2 935413.5 686983 1.361625 57 D F 0.4 1011621.9 686983 1.472557 58 D F 0.6 1071081.6 686983 1.559109 59 D F 0.8 1139325.8 686983 1.658448 60 D F 1.0 1181282.1 686983 1.719522 61 D I 0.0 755841.0 686983 1.100232 62 D I 0.2 964335.0 686983 1.403725 63 D I 0.4 1051233.4 686983 1.530218 64 D I 0.6 1110842.0 686983 1.616986 65 D I 0.8 1193509.0 686983 1.737320 66 D I 1.0 1264249.0 686983 1.840292 67 D D 0.0 751660.0 686983 1.094146 68 D D 0.2 1019735.0 686983 1.484367 69 D D 0.4 1101191.0 686983 1.602938 70 D D 0.6 1191547.0 686983 1.734464 71 D D 0.8 1293433.0 686983 1.882773 72 D D 1.0 1340360.0 686983 1.951082 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 cmax.pdf Description: Adobe PDF document __ 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] scatter plot matrix with different x-y variables
Dear list-members, I wonder if there is a way of creating a scatter plot table/grid with different variables on the y-axis compared to the x-axis? Something like this: A*** B*** C*** -XYZ I know that you can create scatter plot matrix with the same variables on the y-axis as on the x-axis, like this: A* *B **C*** ***X** Y* *Z But this is not what I want. I basically want to get a visual representation of different dependent variables (y-axis) on various independent variables (x-axis). So I want it to look like a scatter plot matrix or grid, but with my own specified variables on each axis. Secondly, to add to this, can I add a correlation line (linear and quadratic) through all scatter plots with both correlation and R2 values? All help appreciated. Adel -- Adel Daoud, PhD Department of Sociology, University of Gothenburg Box 720 405 30 Göteborg, SWEDEN Visiting address: Sprängkullsgatan 25, room F411 +46 031-786 47 84 adel.da...@sociology.gu.se -- View this message in context: http://r.789695.n4.nabble.com/scatter-plot-matrix-with-different-x-y-variables-tp4666598.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] xyplot legend : simple question
thank you 2013/5/6 Gerrit Eichner gerrit.eich...@math.uni-giessen.de Hello, Adel, maybe the argument sep of interaction() helps you. Try interaction( g[,1], g[,2], sep = /) Regards -- Gerrit I am using this command to draw the figure attached to this mail. xyplot(g[,4]~g[,3],type=b,**group=interaction(g[,1],g[,2])**, auto.key =list( title=Evolution de la stabilité , points = FALSE, columns=2, lines = TRUE,space=right) ,data=g,xlab=expression(beta),** ylab=stabilité) I want to change name of the legend that were attributed automaticalle (for example from C.D to C/D)? Can you help me to solve this? Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD in Computer Science Address * *Avenue Taha Hussein Montfleury, 1008 Tunis tél : +216 71 49 60 66 fax: +216 71 39 11 66* [[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] xyplot legend : simple question
Thank you maybe the argument sep of interaction() helps you. Try interaction( g[,1], g[,2], sep = /) Regards -- Gerrit I am using this command to draw the figure attached to this mail. xyplot(g[,4]~g[,3],type=b,**group=interaction(g[,1],g[,2])**, auto.key =list( title=Evolution de la stabilité , points = FALSE, columns=2, lines = TRUE,space=right) ,data=g,xlab=expression(beta),** ylab=stabilité) I want to change name of the legend that were attributed automaticalle (for example from C.D to C/D)? Can you help me to solve this? Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] ggplot and colors
Hello, I have a problem with ggplot and colors I used this function to draw somes lines and I want them to be all black (just to test) however, I dont get any black line in the figure. Do you have any experience with ggplot and manual colors?? ggplot(cmax, aes(cmax[,3], cmax[,6],colour=interaction(cmax[,1],cmax[,2]))) + geom_line() + geom_point() + scale_fill_manual(values=c(black, black , black, black, black, black, black, black, black, black, black, black )) best regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 attachment: test.png__ 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] monte carlo simulation on R
Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] mean and kurtosis
Need help! I know how to get the mean and kurtosis for a single variable but I am reading in an excel file that has several columns representing variables. I need a way to find descriptive statistics across ALL the variables? [[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] Problems getting slope and intercept to change when do multiple reps.
library(ROCR)
n - 1000
fitglm - function(iteration,intercept,sigma,tau,beta){
x - rnorm(n,0,sigma)
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
*int*-coef(model)[1]
*slope*-coef(model)[2] # when add error you are suppose to get slightly
bias slope. However when I change the beta in the original X, I am not
getting the save average slope as output? strange?
pred-predict(model,type=response)
cutp-.5
result-ifelse(predcutp,1,0)
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
accuracy-length(which(result==z))/length(z)
tn- sum(z==0 result==0) # True Negative
fp- sum(z==0 result==1) # False Positive
tp- sum(z==1 result==1) # True Positive
fn- sum(z==1 result==0) # False Negative
sensitivity- tp/(tp+fn)
specificity-tn/(tn+fp)
output-c(*int,slope*,cutp,accuracy,auc,sensitivity,specificity,iteration)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sensitivity,Specificity,iteration)
return(output)
}
y-fitglm(1,2,1,2,4)
y
Output-t(sapply(1:10, function(x) fitglm(x,intercept=2, sigma=1,tau=2,beta=
4)))
apply(output,2, function(x) mean(unlist(x)))
apply(output,2, function(x) var(unlist(x)))
Output-t(sapply(1:500, function(x) fitglm(x,intercept=0, sigma=1,tau=.25,
beta=1)))
apply(output,2, function(x) mean(unlist(x)))
apply(output,2, function(x) var(unlist(x)))
Output-t(sapply(1:500, function(x) fitglm(x,intercept=2, sigma=1,tau=.25,
beta=6)))
apply(output,2, function(x) mean(unlist(x)))
apply(output,2, function(x) var(unlist(x)))
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] How to extract auc, specificity and sensitivity
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(.05,1)
y
The code runs without the sentiv and specs but when I remove the loop i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract auc, specificity and sensitivity
n - 10; # Sample size
fitglm - function(sigma,tau){
+ x - rnorm(n,0,sigma)
+ intercept - 0
+ beta - 0
+ ystar - intercept+beta*x
+ z - rbinom(n,1,plogis(ystar))
+ xerr - x + rnorm(n,0,tau)
+ model-glm(z ~ xerr, family=binomial(logit))
+ int-coef(model)[1]
+ slope-coef(model)[2]
+ pred-predict(model)
+
+ result-ifelse(pred.5,1,0)
+
+ accuracy-length(which(result==z))/length(z)
+ accuracy
+
+ rocpreds-prediction(result,z)
+ auc-performance(rocpreds,auc)@y.values
+ fp-performance(rocpreds,sens)
+ sentiv-slot(fp,y.values)[[1]]
+ sentiv-sentiv[2]
+ sentiv
+ fp2-performance(rocpreds,spec)
+ specs-slot(fp2,y.values)[[1]]
+ specs-specs[2]
+ specs
+ output-c(int,slope,.5,accuracy,auc,sentiv,specs)
+
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
+ return(output)
+
+ }
y-fitglm(2,1)
y
$Intercept
[1] 1.335284
$Slope
[1] 0.1562984
$CutPoint
[1] 0.5
$Accuracy
[1] 0.8
$AUC
[1] 0.5
$Sentivity
[1] 1
$Specificity
[1] 0
Don't get error message but wrong values
On Thu, Oct 25, 2012 at 4:05 PM, Berend Hasselman b...@xs4all.nl wrote:
Your code is still not runnable.
It gives the error message
Error in fitglm(0.05, 1) : could not find function prediction
Berend
On 25-10-2012, at 21:55, Adel Powell wrote:
I think I have corrected it. Can you tell me are my spec and sens values
correct
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
fp-performance(rocpreds,sens)
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
fp2-performance(rocpreds,spec)
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(.05,1)
y
On Thu, Oct 25, 2012 at 3:43 PM, Berend Hasselman b...@xs4all.nl wrote:
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop
i can
get the sensitivity and spec. values ???
[[alternative HTML version deleted]]
__
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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.
[[alternative HTML version deleted]]
__
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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] How to extract auc, specificity and sensitivity
n - 10; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
fp-performance(rocpreds,sens)
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
fp2-performance(rocpreds,spec)
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
return(output)
}
y-fitglm(2,1)
y
I corrected the code. I am put everything in a loop because I am running
monte carlo reps outside of this later.
It works but the value returned is wrong.
Here is the manual manipulation:
* x - rnorm(10,0,2)*
* *
* intercept - 0*
* *
* beta - 5*
* *
* ystar - intercept+beta*x*
* *
* ystar*
[1] 16.5436337 7.7740329 -10.1653928 -2.8338118 -21.5410780 2.6902171
5.1156558 5.0729933 -10.8556430 0.9208434
* test - plogis(ystar)*
* *
* test*
[1] 9.99e-01 9.995797e-01 3.847772e-05 5.552417e-02 4.413963e-10
9.364469e-01 9.940338e-01 9.937753e-01 1.929504e-05 7.152139e-01
* z - rbinom(10,1,plogis(ystar))*
* *
* z*
[1] 1 1 0 0 0 1 1 1 0 1
* xerr - x + rnorm(10,0,1) *
* *
* xerr*
[1] 0.5610573 3.1741687 -2.3915066 -0.2546224 -4.1790037 -1.4387786
1.4211227
-1.1141176 -1.6230087 0.7595021
* model-glm(z ~ xerr, family=binomial(logit))*
* *
* model*
Call: glm(formula = z ~ xerr, family = binomial(logit))
Coefficients:
(Intercept) xerr
1.5001.309
* int-coef(model)[1]*
* *
* slope-coef(model)[2]*
* *
* pred1-predict(model)*
* *
* pred2-predict(model,type=response)*
* *
* pred1*
1 2 3 4 5
6
7 8 9 10
2.23478077 5.65499178 -1.62972799 1.16716581 -3.96932102 -0.38273530
3.36049056 0.04220225 -0.62386760 2.49451832
* pred2*
1 2 3 4 5 6
7
8 9 10
0.90332965 0.99651221 0.16386763 0.76263234 0.01853617 0.40546735
0.96644669 0.51054900 0.34890234 0.92375664
* result-ifelse(pred2.5,1,0) *
* *
* result*
1 2 3 4 5 6 7 8 9 10
1 1 0 1 0 0 1 1 0 1
* accuracy-length(which(result==z))/length(z)*
* *
* accuracy*
[1] 0.8
* rocpreds-prediction(result,z)*
* *
* rocpreds*
* auc-performance(rocpreds,auc)@y.values*
* *
* auc*
[[1]]
[1] 0.7916667
* fp-performance(rocpreds,sens)*
* *
* sentiv-slot(fp,y.values)[[1]]*
* *
* sentiv-sentiv[2]*
* *
* sentiv*
[1] 0.833
* *
* fp2-performance(rocpreds,spec)*
* *
* specs-slot(fp2,y.values)[[1]]*
* *
* specs*
[1] 1.00 0.75 0.00
* specs-specs[2]*
* *
* specs*
[1] 0.75
On Thu, Oct 25, 2012 at 3:43 PM, Berend Hasselman b...@xs4all.nl wrote:
On 25-10-2012, at 21:28, Adel Powell wrote:
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 0
ystar - intercept+beta*x
z - rbinom(n,1,plogis(ystar))
xerr - x + rnorm(n,0,tau)
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model)
result-ifelse(pred.5,1,0)
accuracy-length(which(result==z))/length(z)
accuracy
rocpreds-prediction(result,z)
auc-performance(rocpreds,auc)@y.values
sentiv-performance(rocpreds,sens)@y.values
sentiv-slot(fp,y.values)[[1]]
sentiv-sentiv[2]
sentiv
specs-performance(rocpreds,spec)@y.values
specs-slot(fp2,y.values)[[1]]
specs-specs[2]
specs
output-c(int,slope,.5,accuracy,auc,sentiv,specs)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Sentivity,Specificity)
A missing before Specificity?
return(output)
}
y-fitglm(.05,1)
y
Running this after correction of the missing one gets en error
Error in fitglm(0.05, 1) : could not find function prediction
How are you using a loop?
Your example is not reproducible.
Berend
The code runs without the sentiv and specs but when I remove the loop i
can
get the sensitivity and spec. values ???
[[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] Logistic regression/Cut point? predict ??
I am new to R and I am trying to do a monte carlo simulation where I
generate data and interject error then test various cut points; however, my
output was garbage (at x equal zero, I did not get .50)
I am basically testing the performance of classifiers.
Here is the code:
n - 1000; # Sample size
fitglm - function(sigma,tau){
x - rnorm(n,0,sigma)
intercept - 0
beta - 5
* ystar - intercept+beta*x*
* z - rbinom(n,1,plogis(ystar))**# I believe plogis accepts the a
+bx augments and return the e^x/(1+e^x) which is then used to generate 0
and 1 data*
xerr - x + rnorm(n,0,tau)# error is added here
model-glm(z ~ xerr, family=binomial(logit))
int-coef(model)[1]
slope-coef(model)[2]
pred-predict(model) #this gives me the a+bx data for new error? I
know I can add type= response to get the probab. but only e^x not *e^x/(1+e^x)
*
pi1hat-length(z[which(z==1)]/length(z)) My cut point is calculated is
the proportion of 0s to 1.
pi0hat-length(z[which(z==0)]/length(z))
cutmid - log(pi0hat/pi1hat)
pred-ifelse(predcutmid,1,0) * I am not sure if I need to compare
these two. I think this is an error.
*
accuracy-length(which(pred==z))/length(z)
accuracy
rocpreds-prediction(pred,z)
auc-performance(rocpreds,auc)@y.values
output-c(int,slope,cutmid,accuracy,auc)
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC)
return(output)
}
y-fitglm(.05,1)
y
nreps - 500;
output-data.frame(matrix(rep(NA),nreps,6,ncol=6))
mysigma-.5
mytau-.1
i-1
for(j in 1:nreps) {
output[j,1:5]-fitglm(mysigma,mytau)
output[j,6]-j
}
names(output)-c(Intercept,Slope,CutPoint,Accuracy,AUC,Iteration)
apply(output,2, mean)
apply(output,2, var)
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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] syntaxe problem
Hello,
I want to create 10 dataframe using a for loop.
I don t know what to do to create 10 different dataframes whose name is
parametrized with variable i.
This syntaxe fails. It create a unique fataframe called dfn.
Any input will help
Thanks
for (i in 1:10){
+ filename=paste(avail4,i,sep = _)
+ dfn=read.table(filename)
+ }
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] transforming a data frame to matrix
Hello orderulcount Group.1 Group.2 Group.3 xV5 7C L 0.0 30 C / L 19 C L 0.2 27 C / L 31 C L 0.4 15 C / L 43 C L 0.6 7 C / L 54 C L 0.8 2 C / L 10 C S 0.0 27 C / S 22 C S 0.2 10 C / S 34 C S 0.4 6 C / S 46 C S 0.6 1 C / S 1D D 0.0 30 D / D 13 D D 0.2 30 D / D 25 D D 0.4 17 D / D 37 D D 0.6 9 D / D 49 D D 0.8 2 D / D 3D F 0.0 30 D / F 15 D F 0.2 26 D / F 27 D F 0.4 13 D / F 39 D F 0.6 3 D / F 51 D F 0.8 3 D / F 5D I 0.0 26 D / I 17 D I 0.2 7 D / I 29 D I 0.4 7 D / I 41 D I 0.6 1 D / I 53 D I 0.8 1 D / I 8D L 0.0 30 D / L 20 D L 0.2 26 D / L 32 D L 0.4 14 D / L 44 D L 0.6 8 D / L 55 D L 0.8 2 D / L 11 D S 0.0 27 D / S 23 D S 0.2 8 D / S 35 D S 0.4 4 D / S 47 D S 0.6 1 D / S 2I D 0.0 30 I / D 14 I D 0.2 30 I / D 26 I D 0.4 14 I / D 38 I D 0.6 9 I / D 50 I D 0.8 2 I / D 4I F 0.0 30 I / F 16 I F 0.2 29 I / F 28 I F 0.4 16 I / F 40 I F 0.6 5 I / F 52 I F 0.8 2 I / F 6I I 0.0 30 I / I 18 I I 0.2 17 I / I 30 I I 0.4 4 I / I 42 I I 0.6 1 I / I 9I L 0.0 30 I / L 21 I L 0.2 27 I / L 33 I L 0.4 17 I / L 45 I L 0.6 8 I / L 56 I L 0.8 2 I / L 12 I S 0.0 28 I / S 24 I S 0.2 11 I / S 36 I S 0.4 6 I / S 48 I S 0.6 3 I / S I have a data frame formed by 5 colomns. How to transform this data frame to a matrix formed as follow: 1. the colomn 5 is the comumn (string) 2. the colomn 5 row 3. the data are the colomn x. I aim to draw barplot later using the matrix. Thanks for any help -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] transforming a data frame to matrix
with simpler expressions: if I have a dataframe A B C is it possible de transform it to a matrix where nrow(A) colomns, nrow(B) row and for each value of C, we put the corresponding value in the matrix. Regards Le 18 février 2012 13:57, Adel ESSAFI adeless...@gmail.com a écrit : Hello orderulcount Group.1 Group.2 Group.3 xV5 7C L 0.0 30 C / L 19 C L 0.2 27 C / L 31 C L 0.4 15 C / L 43 C L 0.6 7 C / L 54 C L 0.8 2 C / L 10 C S 0.0 27 C / S 22 C S 0.2 10 C / S 34 C S 0.4 6 C / S 46 C S 0.6 1 C / S 1D D 0.0 30 D / D 13 D D 0.2 30 D / D 25 D D 0.4 17 D / D 37 D D 0.6 9 D / D 49 D D 0.8 2 D / D 3D F 0.0 30 D / F 15 D F 0.2 26 D / F 27 D F 0.4 13 D / F 39 D F 0.6 3 D / F 51 D F 0.8 3 D / F 5D I 0.0 26 D / I 17 D I 0.2 7 D / I 29 D I 0.4 7 D / I 41 D I 0.6 1 D / I 53 D I 0.8 1 D / I 8D L 0.0 30 D / L 20 D L 0.2 26 D / L 32 D L 0.4 14 D / L 44 D L 0.6 8 D / L 55 D L 0.8 2 D / L 11 D S 0.0 27 D / S 23 D S 0.2 8 D / S 35 D S 0.4 4 D / S 47 D S 0.6 1 D / S 2I D 0.0 30 I / D 14 I D 0.2 30 I / D 26 I D 0.4 14 I / D 38 I D 0.6 9 I / D 50 I D 0.8 2 I / D 4I F 0.0 30 I / F 16 I F 0.2 29 I / F 28 I F 0.4 16 I / F 40 I F 0.6 5 I / F 52 I F 0.8 2 I / F 6I I 0.0 30 I / I 18 I I 0.2 17 I / I 30 I I 0.4 4 I / I 42 I I 0.6 1 I / I 9I L 0.0 30 I / L 21 I L 0.2 27 I / L 33 I L 0.4 17 I / L 45 I L 0.6 8 I / L 56 I L 0.8 2 I / L 12 I S 0.0 28 I / S 24 I S 0.2 11 I / S 36 I S 0.4 6 I / S 48 I S 0.6 3 I / S I have a data frame formed by 5 colomns. How to transform this data frame to a matrix formed as follow: 1. the colomn 5 is the comumn (string) 2. the colomn 5 row 3. the data are the colomn x. I aim to draw barplot later using the matrix. Thanks for any help -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] multiple histograms from a dataframe
Le 11 février 2012 02:33, David Winsemius dwinsem...@comcast.net a écrit : On Feb 10, 2012, at 7:05 PM, Adel ESSAFI wrote: Hi list I need some help for drawing some histograms I have a dataframe , say, X Y Z T I want to draw a histogram Z-T for each value of the couple (X-Y). When I use thus syntax library(lattice) histogram(law[,3] ~ law[,66] | law[,1] ) Perhaps (but untested in the absence of data); histogram( Z ~ T | interaction(X, Y) , data=dfrmname ) Thanks , that helped a lot. now, I have another problem: I want to draw many (two) figures together. The par(new=T) directve does not recognize the ploy provided by lattice library when I tired : xyplot(law[,66] ~ law[,3]| interaction(law[,1],law[,2]),type='l') par(new=T) *Warning message: In par(new = T) : calling par(new=TRUE) with no plot* xyplot(law[,67] ~ law[,3]| interaction(law[,1],law[,2]),type='l') and the second xyplot() draws a new figure. what can I do to draw to figures together using lattice? Thanks it draws multiple histograms but by selecting distinct values of law[,1] The deal is to make the same thing but for a couple of columns Thanks in advance for help Adel -- David Winsemius, MD West Hartford, CT -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] multiple histograms from a dataframe
Hi list I need some help for drawing some histograms I have a dataframe , say, X Y Z T I want to draw a histogram Z-T for each value of the couple (X-Y). When I use thus syntax library(lattice) histogram(law[,3] ~ law[,66] | law[,1] ) it draws multiple histograms but by selecting distinct values of law[,1] The deal is to make the same thing but for a couple of columns Thanks in advance for help Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] generate a random number with rexp ?
dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] generate a number using exponential low
Deal list I want to generate a random number in an interval using exponential low I know how to make that simply using uniform low x1 - runif(1, 5.0, 7.5) this will generate a number between 5 et 7.5 but with uniform low. Could you help pease.? Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[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] Fwd: question about aggregate
Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
Hi, It is not what I want, but, thx anyway. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com try this table(sci[,2]=0,sci[,1]=L) Weidong Gu On Sat, Oct 22, 2011 at 7:51 AM, Adel ESSAFI adel.s...@imag.fr wrote: Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
NO aggregate(si[,7],list(si[,2],si[,1]),mean) Group.1 Group.2 x 1 0.0 D 212.5015448 2 0.2 D 200.5271137 3 0.4 D 191.5387529 4 0.6 D 131.5856218 5 0.8 D 16.4515798 6 1.0 D 0.9698699 7 0.0 F 211.6176036 8 0.2 F 199.5358336 9 0.4 F 179.1472057 10 0.6 F 70.3376311 11 0.8 F 25.0939253 12 1.0 F 0.9759778 13 0.0 I 199.1829674 14 0.2 I 188.8703456 15 0.4 I 147.5194562 16 0.6 I 18.1743204 17 0.8 I 0.9563789 18 1.0 I 0.9531993 19 0.0 L 95.1813009 20 0.2 L 95.1813009 21 0.4 L 92.3533476 22 0.6 L 47.1235041 23 0.8 L 9.4129464 24 1.0 L 0.9336508 25 0.0 S 95.1813009 26 0.2 S 91.4119070 27 0.4 S 81.9875960 28 0.6 S 11.3183831 29 0.8 S 1.8843307 30 1.0 S 0.9310779 for exemple, for the last line, I want aggregate to show me the number of lines with col1=1 and col2=S in stead of the mean. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com Is this waht you want? sci[,5][sci[,2]==0 sci[,1]==L] Weidong On Sat, Oct 22, 2011 at 8:19 AM, Adel ESSAFI adeless...@gmail.com wrote: Hi, It is not what I want, but, thx anyway. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com try this table(sci[,2]=0,sci[,1]=L) Weidong Gu On Sat, Oct 22, 2011 at 7:51 AM, Adel ESSAFI adel.s...@imag.fr wrote: Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[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] question about aggregate
2011/10/22 Adel ESSAFI adel.s...@imag.fr NO aggregate(si[,7],list(si[,2],si[,1]),mean) Group.1 Group.2 x 1 0.0 D 212.5015448 2 0.2 D 200.5271137 3 0.4 D 191.5387529 4 0.6 D 131.5856218 5 0.8 D 16.4515798 6 1.0 D 0.9698699 7 0.0 F 211.6176036 8 0.2 F 199.5358336 9 0.4 F 179.1472057 10 0.6 F 70.3376311 11 0.8 F 25.0939253 12 1.0 F 0.9759778 13 0.0 I 199.1829674 14 0.2 I 188.8703456 15 0.4 I 147.5194562 16 0.6 I 18.1743204 17 0.8 I 0.9563789 18 1.0 I 0.9531993 19 0.0 L 95.1813009 20 0.2 L 95.1813009 21 0.4 L 92.3533476 22 0.6 L 47.1235041 23 0.8 L 9.4129464 24 1.0 L 0.9336508 25 0.0 S 95.1813009 26 0.2 S 91.4119070 27 0.4 S 81.9875960 28 0.6 S 11.3183831 29 0.8 S 1.8843307 30 1.0 S 0.9310779 for exemple, for the last line, I want aggregate to show me the number of lines with col1=1 and col2=S in stead of the mean. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com Is this waht you want? sci[,5][sci[,2]==0 sci[,1]==L] Weidong On Sat, Oct 22, 2011 at 8:19 AM, Adel ESSAFI adeless...@gmail.com wrote: Hi, It is not what I want, but, thx anyway. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com try this table(sci[,2]=0,sci[,1]=L) Weidong Gu On Sat, Oct 22, 2011 at 7:51 AM, Adel ESSAFI adel.s...@imag.fr wrote: Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
Thank you I works as I need. aggregate(tmp[,5],list(tmp[,2],tmp[,1]),function(x) sum(table(x))) Group.1 Group.2 x 1 0.0 D 100 2 0.2 D 100 3 0.4 D 100 4 0.6 D 72 5 0.8 D 9 2011/10/22 Weidong Gu anopheles...@gmail.com Thanks for providing the example data, try this aggregate(si[,7],list(si[,2],si[,1]),function(x) sum(table(x))) Weidong On Sat, Oct 22, 2011 at 9:28 AM, Adel ESSAFI adel.s...@imag.fr wrote: NO aggregate(si[,7],list(si[,2],si[,1]),mean) Group.1 Group.2 x 1 0.0 D 212.5015448 2 0.2 D 200.5271137 3 0.4 D 191.5387529 4 0.6 D 131.5856218 5 0.8 D 16.4515798 6 1.0 D 0.9698699 7 0.0 F 211.6176036 8 0.2 F 199.5358336 9 0.4 F 179.1472057 10 0.6 F 70.3376311 11 0.8 F 25.0939253 12 1.0 F 0.9759778 13 0.0 I 199.1829674 14 0.2 I 188.8703456 15 0.4 I 147.5194562 16 0.6 I 18.1743204 17 0.8 I 0.9563789 18 1.0 I 0.9531993 19 0.0 L 95.1813009 20 0.2 L 95.1813009 21 0.4 L 92.3533476 22 0.6 L 47.1235041 23 0.8 L 9.4129464 24 1.0 L 0.9336508 25 0.0 S 95.1813009 26 0.2 S 91.4119070 27 0.4 S 81.9875960 28 0.6 S 11.3183831 29 0.8 S 1.8843307 30 1.0 S 0.9310779 for exemple, for the last line, I want aggregate to show me the number of lines with col1=1 and col2=S in stead of the mean. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com Is this waht you want? sci[,5][sci[,2]==0 sci[,1]==L] Weidong On Sat, Oct 22, 2011 at 8:19 AM, Adel ESSAFI adeless...@gmail.com wrote: Hi, It is not what I want, but, thx anyway. Regards 2011/10/22 Weidong Gu anopheles...@gmail.com try this table(sci[,2]=0,sci[,1]=L) Weidong Gu On Sat, Oct 22, 2011 at 7:51 AM, Adel ESSAFI adel.s...@imag.fr wrote: Hello, is there any help please. Regards 2011/10/21 Adel ESSAFI adeless...@gmail.com Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
Hi list I am discovering R, and -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] question about aggregate
Hello I am discovering R and I find it is really very powerful. However, I find some newbie difficulties. Here, I have a data frame with manu values that I want to calculate the frequency (the nomber of line) of the some criteria. For exemple here, I want it to print the number of occurence where sci[,2]=0 and sci[,1]=L. In my exemple, he is printing the number of the line in the result data frame. however, I have at least 90 line with sci[,2]=0 and sci[,1]=L. Thank you in advance for any input. aggregate(sci[,5],list(sci[,2],sci[,1]),frequency) Group.1 Group.2 x 1 0.0 L 1 2 0.2 L 1 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] multiple lines with the same data frame?
Bonjour I have this data frame and I am newbie in R. I want to ask if it is possible to draw 10 lines in a plot such that: a line for every colomn, the x - axis is the second column and the y-axis is the third one. Thank you for any input 1 0 1094442 1 0.2 1163576.2 1 0.4 1238539.6 1 0.6 1303510.4 1 0.8 1376573.2 1 1 1454175 2 0 1076068 2 0.2 1139246 2 0.4 1212344 2 0.6 1277591.8 2 0.8 1346156.6 2 1 1410058 3 0 1097901 3 0.2 1173643.2 3 0.4 1258849.8 3 0.6 1343001.6 3 0.8 1427705.8 3 1 1507793 4 0 1197047 4 0.2 1292918.6 4 0.4 1383640.8 4 0.6 1480487.8 4 0.8 1571557.6 4 1 1659578 5 0 1120010 5 0.2 1200076.6 5 0.4 1279653 5 0.6 1360494.6 5 0.8 1437223.2 5 1 1512636 6 0 1082650 6 0.2 1154676.6 6 0.4 1223416.6 6 0.6 1297593.6 6 0.8 1366764.6 6 1 1432705 7 0 1023590 7 0.2 1077783.8 7 0.4 1133154.2 7 0.6 1190296 7 0.8 1244411.4 7 1 1297709 8 0 1042257 8 0.2 1107452.8 8 0.4 1174574.8 8 0.6 1238547.6 8 0.8 1301507.6 8 1 1363338 9 0 1066917 9 0.2 1136411.8 9 0.4 1202822 9 0.6 1273514.8 9 0.8 1340641.8 9 1 1408249 10 0 1069685 10 0.2 1138909 10 0.4 1213284.2 10 0.6 1282508.8 10 0.8 1346093.2 10 1 1410707 [[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] two ecdf in the same figure
Hello, is ot possible to draw two ecdf of vectors (say s1 and s2) on the same figire with R. plot function draws a new plot and there is no function like points or lines to draw a second ecdf on the figure. Regards -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[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] draw points in R
Hello list I have this file 0 121289479 25 0 212599129 1 0 285254098 21 0 444889848 45 0 469197123 30 0 640007403 82 0 718215617 8 0 758534043 56 0 799706577 4 0 814441385 93 0 843545059 37 0.2 121289479 6 0.2 285254098 3 0.2 444889848 6 0.2 469197123 13 0.2 640007403 24 0.2 718215617 3 0.2 758534043 2 0.2 799706577 2 0.2 814441385 70 0.2 843545059 3 0.4 121289479 1 0.4 444889848 14 0.4 469197123 4 0.4 640007403 11 0.4 799706577 2 0.4 814441385 8 I read it with R and next I want to draw a point for each line the X is in col 2 and y is in col 3 but, I need that for each value of the first colomn , to have a color. Thus, I want to get a draw with 3 colors for values 0 , 0.2 and 0.4 The probleme is that when I use points function; it plots to me lines and not simpe points. Thank you in advance for any input. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] draw points in R
You are kind 2011/9/14 Sarah Goslee sarah.gos...@gmail.com Hi, The probleme is that when I use points function; it plots to me lines and not simpe points. That seems unlikely, but without your code it's hard to figure out what you did. As usual, the telepathy is not working. Regardless, this will create a graph with points: x - read.table(clipboard, header=FALSE, as.is=TRUE) # including data is good; including data with dput() is better plot(x[,2], x[,3], type=p, xlab=X, ylab=Y, main=Points) I read it with R and next I want to draw a point for each line the X is in col 2 and y is in col 3 but, I need that for each value of the first colomn , to have a color. Thus, I want to get a draw with 3 colors for values 0 , 0.2 and 0.4 And here's one way to color them: plot(x[,2], x[,3], col=c(red, blue, green)[as.factor(x[,1])], xlab=X, ylab=Y, main=Colored Points) You might be interested in ?par especially pch and also in ?legend as well as in the basic help for ?plot and ?points. Sarah On Wed, Sep 14, 2011 at 1:57 PM, Adel ESSAFI adeless...@gmail.com wrote: Hello list I have this file 0 121289479 25 0 212599129 1 0 285254098 21 0 444889848 45 0 469197123 30 0 640007403 82 0 718215617 8 0 758534043 56 0 799706577 4 0 814441385 93 0 843545059 37 0.2 121289479 6 0.2 285254098 3 0.2 444889848 6 0.2 469197123 13 0.2 640007403 24 0.2 718215617 3 0.2 758534043 2 0.2 799706577 2 0.2 814441385 70 0.2 843545059 3 0.4 121289479 1 0.4 444889848 14 0.4 469197123 4 0.4 640007403 11 0.4 799706577 2 0.4 814441385 8 Thank you in advance for any input. -- Sarah Goslee http://www.functionaldiversity.org -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] printing big real values
Hello, I have a file with very big values. I want to display the real values with classic ways (without exposant) summary(a[,1]); Min. 1st Qu.Median Mean 3rd Qu. Max. 1.198e+09 1.199e+09 1.200e+09 1.200e+09 1.201e+09 1.202e+09 Can you help please? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] basic hist() question
Hi list I loaded the content of a file dureetasks.txt to variable a. This file contains an interger per line. when I print a vector, it displays correctly. however, when I try to print the histogram, I get this error message a=read.table(dureetasks.txt) hist(a) Error in hist.default(a) : 'x' must be numeric Can you help please? regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] basic hist() question
Hi In fact, I searching for a simpler solution. I remember that I have done this without these functions (but I forgot) simply, is there any function that force R to take variable a as table ? Regards Adel a=read.table(dureetasks.txt) summary(a) V1 Min. : 1 1st Qu.: 77 Median : 2658 Mean : 25802 3rd Qu.: 42558 Max. :1575814 hist(a); Error in hist.default(a) : 'x' must be numeric 2010/8/21 Gavin Simpson gavin.simp...@ucl.ac.uk On Sat, 2010-08-21 at 11:37 +0200, Adel ESSAFI wrote: Hi list I loaded the content of a file dureetasks.txt to variable a. This file contains an interger per line. when I print a vector, it displays correctly. however, when I try to print the histogram, I get this error message a=read.table(dureetasks.txt) hist(a) Error in hist.default(a) : 'x' must be numeric Can you help please? class(a) str(a) a - data.frame(x = rnorm(100)) hist(a) Error in hist.default(a) : 'x' must be numeric So I suspect a is a data frame with a single column/variable. You need to pass hist the vector from /inside/ a that you want to plot: with(a, hist(x)) is one way to do that. Note that 'x' is used because that is the name of the variable (component really) in a that I want to plot: names(a) [1] x You will need to work out what the correct name is for the component you want to plot. If there was no header to your text file then IIRC the variable will be named X1, but run names(a) and see what it returns. Because we don't have your dureetasks.txt file, there is no way for me to clarify that this is the problem and nothing else. If you can't send us data at the very least use tools like str(a) and class(a), and head(a) to show us what the data look like. HTH G regards __ 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. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] Fwd: basic hist() question
-- Forwarded message --
From: Adel ESSAFI adeless...@gmail.com
Date: 2010/8/21
Subject: Re: [R] basic hist() question
To: Henrique Dallazuanna www...@gmail.com
It works fine.
Could you explain to me why it did not worked for read.table?
regards
Adel
2010/8/21 Henrique Dallazuanna www...@gmail.com
Try this:
a - scan('dureetasks.txt')
hist(a)
On Sat, Aug 21, 2010 at 10:32 AM, Adel ESSAFI adeless...@gmail.comwrote:
Hi
In fact, I searching for a simpler solution. I remember that I have done
this without these functions (but I forgot)
simply, is there any function that force R to take variable a as table
?
Regards
Adel
a=read.table(dureetasks.txt)
summary(a)
V1
Min. : 1
1st Qu.: 77
Median : 2658
Mean : 25802
3rd Qu.: 42558
Max. :1575814
hist(a);
Error in hist.default(a) : 'x' must be numeric
2010/8/21 Gavin Simpson gavin.simp...@ucl.ac.uk
On Sat, 2010-08-21 at 11:37 +0200, Adel ESSAFI wrote:
Hi list
I loaded the content of a file dureetasks.txt to variable a. This
file
contains an interger per line.
when I print a vector, it displays correctly.
however, when I try to print the histogram, I get this error message
a=read.table(dureetasks.txt)
hist(a)
Error in hist.default(a) : 'x' must be numeric
Can you help please?
class(a)
str(a)
a - data.frame(x = rnorm(100))
hist(a)
Error in hist.default(a) : 'x' must be numeric
So I suspect a is a data frame with a single column/variable. You need
to pass hist the vector from /inside/ a that you want to plot:
with(a, hist(x))
is one way to do that. Note that 'x' is used because that is the name of
the variable (component really) in a that I want to plot:
names(a)
[1] x
You will need to work out what the correct name is for the component you
want to plot. If there was no header to your text file then IIRC the
variable will be named X1, but run names(a) and see what it returns.
Because we don't have your dureetasks.txt file, there is no way for me
to clarify that this is the problem and nothing else. If you can't send
us data at the very least use tools like str(a) and class(a), and
head(a) to show us what the data look like.
HTH
G
regards
__
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.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w]
http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/
http://www.ucl.ac.uk/%7Eucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[alternative HTML version deleted]]
__
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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] basic hist() question
thanks,
I'am not veru used with R!!
regards
2010/8/21 Henrique Dallazuanna www...@gmail.com
Adel,
read.table return a data.frame, Gavin showed, you need pass which column
will be plotted to hist.
scan return a vector.
On Sat, Aug 21, 2010 at 10:59 AM, Adel ESSAFI adeless...@gmail.comwrote:
It works fine.
Could you explain to me why it did not worked for read.table?
regards
Adel
2010/8/21 Henrique Dallazuanna www...@gmail.com
Try this:
a - scan('dureetasks.txt')
hist(a)
On Sat, Aug 21, 2010 at 10:32 AM, Adel ESSAFI adeless...@gmail.comwrote:
Hi
In fact, I searching for a simpler solution. I remember that I have
done
this without these functions (but I forgot)
simply, is there any function that force R to take variable a as
table ?
Regards
Adel
a=read.table(dureetasks.txt)
summary(a)
V1
Min. : 1
1st Qu.: 77
Median : 2658
Mean : 25802
3rd Qu.: 42558
Max. :1575814
hist(a);
Error in hist.default(a) : 'x' must be numeric
2010/8/21 Gavin Simpson gavin.simp...@ucl.ac.uk
On Sat, 2010-08-21 at 11:37 +0200, Adel ESSAFI wrote:
Hi list
I loaded the content of a file dureetasks.txt to variable a. This
file
contains an interger per line.
when I print a vector, it displays correctly.
however, when I try to print the histogram, I get this error
message
a=read.table(dureetasks.txt)
hist(a)
Error in hist.default(a) : 'x' must be numeric
Can you help please?
class(a)
str(a)
a - data.frame(x = rnorm(100))
hist(a)
Error in hist.default(a) : 'x' must be numeric
So I suspect a is a data frame with a single column/variable. You need
to pass hist the vector from /inside/ a that you want to plot:
with(a, hist(x))
is one way to do that. Note that 'x' is used because that is the name
of
the variable (component really) in a that I want to plot:
names(a)
[1] x
You will need to work out what the correct name is for the component
you
want to plot. If there was no header to your text file then IIRC the
variable will be named X1, but run names(a) and see what it returns.
Because we don't have your dureetasks.txt file, there is no way for me
to clarify that this is the problem and nothing else. If you can't
send
us data at the very least use tools like str(a) and class(a), and
head(a) to show us what the data look like.
HTH
G
regards
__
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.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w]
http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/
http://www.ucl.ac.uk/%7Eucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[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] variation
Hi list I am a new user of R. I ask for some beginner question I am searching if there is any function that computes the variation of some discrete values of a vector (mean() and sd() exists, but i have not find variation). Thanks in advance Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] Fwd: How to make R automatic?
-- Forwarded message --
From: Adel ESSAFI adel.s...@imag.fr
Date: 2010/6/5
Subject: Re: [R] How to make R automatic?
To: zhangted001 zhen...@gmail.com
Well, I am new but i will give you an example of script that I run
cat exec2.sh
R --no-saveEOF
fl=list.files(pattern=*.dat)
for( j in 1:length(fl)){
a=read.table(fl[j])
debut=a[,1]
fin=a[,2]
duree=a[,3]
u=matrix(ncol=1,nrow=length(fin)-1)
for( i in 1:length(fin)-1) u[i]=debut[i+1]-fin[i]
if (! (is.na(mean(duree)) || is.na(sd(duree)) || is.na(mean(u)) ||
is.na(sd(u)
) ) ) {
cat (fl[j], ,mean(duree), ,sd(duree), ,mean(u), ,sd(u),
,trunc((mean(duree)/(mean(duree)+mean(u)))*100), ,length(debut),
\n,file=testlogsnavai,append=TRUE)
}
}
This works perfect
2010/6/5 zhangted001 zhen...@gmail.com
Thank you all for the information! That is exactly what I was looking for.
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-make-R-automatic-tp2238541p2244132.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] variation
2010/6/5 Jannis bt_jan...@yahoo.de What exactly do you mean by variation? As I understand it, this term is a broad term for all kinds of different spread measures (like quantile range or standard deviation). Do you mean the Coefficient of Variation? If you found out how to Yes, that what I mean. But I looked for ready function :) compute the mean and the std.dev., it is straightforward to calculate it. Adel ESSAFI schrieb: Hi list I am a new user of R. I ask for some beginner question I am searching if there is any function that computes the variation of some discrete values of a vector (mean() and sd() exists, but i have not find variation). Thanks in advance Adel __ 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. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] variation
2010/6/5 Adel ESSAFI adel.s...@imag.fr 2010/6/5 Jannis bt_jan...@yahoo.de What exactly do you mean by variation? As I understand it, this term is a broad term for all kinds of different spread measures (like quantile range or standard deviation). Do you mean the Coefficient of Variation? If you found out how to Yes, that what I mean. But I looked for ready function :) Yes, that what I mean. But I looked for ready function :) compute the mean and the std.dev., it is straightforward to calculate it. Adel ESSAFI schrieb: Hi list I am a new user of R. I ask for some beginner question I am searching if there is any function that computes the variation of some discrete values of a vector (mean() and sd() exists, but i have not find variation). Thanks in advance Adel __ 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. -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] AIC score
Dear All, I've been using step function to find me the best model.this basically works by using AIC score fucntion that is implemented on step(). The problem I'm facing with lots of variables on the model for example : step(lm(x1~x2,x3,x4,..x13)) sometimes gives me a warning message which is : AIC=- inf Coefficients: (Intercept) wnt3.values wnt6.values wnt10b.values wnt9a.values 2.3462 -0.4689 2.0730 1.2769 -0.2319 sfrp1.valueswnt5b.values sfrp1.1.valuessfrp5.values fzd5.1.values -0.2597 0.3150 0.3811 0.5926 -1.5567 fzd1.values fzd4.values fzd6.values fzd7.values fzd7.1.values 0.6459 -2.3016 0.3636 NA NA fzd8.values NA Warning message: attempting model selection on an essentially perfect fit is nonsense . which stops the search. Does this means that Residual Sum of Squares (RSS) equals to zero that makes AIC goes to -inf .And how would I overcome this problem.Can I for example find those that have strong correlation with x1 first and then use AIC score to find me the best model among them,. Regards Adel, [[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.
