http://blog.gmane.org/gmane.comp.lang.r.general hourly 1 1901-01-01T00:00+00:00 http://gmane.org/img/gmane-25t.png http://gmane.org http://comments.gmane.org/gmane.comp.lang.r.general/293495 <pre>Hi, I would be very graitful if someone could help me to figure out my problem. I used mixed-effects models to analyse my data and AIC approach for model selection. I am studying an effect on Labrador tea on basal diameter of spruce in 2 different habitats (wet and dry zones) during 3 years. This is one of example of my AIC table: Candidate models K AICc $B&amp;$(B AICc AICc Wt Zone + Labrador tea + Year 9 -17.75 0.00 0.80 Zone + Labrador tea + Year + Zone $B!_(B Labrador tea 10 -14.69 3.06 0.17 Zone + Labrador tea + Year + Year $B!_(B Labrador tea 12 -11.21 6.53 0.03 Zone + Labrador tea 6 71.14 88.88 0.00 Zone + Labrador tea + Zone $B!_(B Labrador tea 7 73.85 91.59 0.00 I interpreted the main effect of zone, Labrador tea and Year. My question is should I interpret the interaction term Zone $B!_(B Labrador tea also? Normally I interpreted the effect of variables that have been in the models with $B&amp;$(B AICc &lt; 4. One professor said I should not interpred interaction term if the main effect is stronger. But at the same time I saw articles where author interpreted the interaction term where Akaike weight was still high. Thank you in advance. Galina [[alternative HTML version deleted]] </pre> Galina Kamorina 2013-05-23T22:21:46 http://comments.gmane.org/gmane.comp.lang.r.general/293494 <pre>Hi, I am trying to understand why creating Date variables does not work if I subset to avoid NAs. I had problems creating these Date variables in my code and I thought that the presence of NAs was the cause. So I used a condition to avoid NAs. It turns out that NAs are not a problem and I do not need to subset, but I'd like to understand why subsetting causes the problem. The strange numbers I start with are what I get when I read an Excel sheet with the function read.xls() from package gdata. dat1 = c(41327, 41334, 41341, 41348, 41355, 41362, 41369, 41376, 41383, 41390, 41397) dat2 = dat1 dat2[c(5,9)]=NA Data = data.frame(dat1,dat2) keep1 = !is.na(Data$dat1) keep2 = !is.na(Data$dat2) Data$Dat1a = as.Date(Data[,"dat1"], origin="1899-12-30") Data$Dat1b[keep1] = as.Date(Data[keep1,"dat1"], origin="1899-12-30") Data$Dat2a = as.Date(Data[,"dat2"], origin="1899-12-30") Data$Dat2b[keep2] = as.Date(Data[keep2,"dat2"], origin="1899-12-30") Data dat1 dat2 Dat1a Dat1b Dat2a Dat2b 1 41327 41327 2013-02-22 15758 2013-02-22 15758 2 41334 41334 2013-03-01 15765 2013-03-01 15765 3 41341 41341 2013-03-08 15772 2013-03-08 15772 4 41348 41348 2013-03-15 15779 2013-03-15 15779 5 41355 NA 2013-03-22 15786 &lt;NA&gt; NA 6 41362 41362 2013-03-29 15793 2013-03-29 15793 7 41369 41369 2013-04-05 15800 2013-04-05 15800 8 41376 41376 2013-04-12 15807 2013-04-12 15807 9 41383 NA 2013-04-19 15814 &lt;NA&gt; NA 10 41390 41390 2013-04-26 15821 2013-04-26 15821 11 41397 41397 2013-05-03 15828 2013-05-03 15828 So variables Dat1b and Dat2b are not converted to Date class. sessionInfo() R version 2.15.2 (2012-10-26) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] fr_CA.UTF-8/fr_CA.UTF-8/fr_CA.UTF-8/C/fr_CA.UTF-8/fr_CA.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] gdata_2.12.0 loaded via a namespace (and not attached): [1] gtools_2.7.0 Thanks in advance, Denis </pre> Denis Chabot 2013-05-23T21:35:49 http://comments.gmane.org/gmane.comp.lang.r.general/293492 <pre>I am creating a few dozen multi-panel time series plots using lattice graphics in the lme4 package. Each panel in a given plot represents a tree. each multipanel plot is a particular treatment. Here's my issue: when you use xyplot() to plot this data it orders the panels alphabetically. I would prefer to have them in order of increasing slope of the regression line plotted in each panel. I've read everything I can find regarding the index.cond argument, and the best I can come up with is to manually order them to have the correct increasing slope order, i.e. index.cond=c(7,6,23,4,15,8,...). This would take inordinate amounts of time and I'm sure there is a better, more eloquent solution. Please help! Sorry for the long dataset below, I'm unsure of how to create a reproducible example otherwise. example.plot = xyplot(ht ~ time|tree, data=data, type = c("r", "g", "p"), par.settings=simpleTheme(col="blue"), main="abc", ) example.plot data= tree treat site plot rx mr rxl t w l spp time dia ht 1 1 1 C-H 2002 1 1 Mn N N 14.55 ac 1 9.6 74.5 2 2 1 C-H 2002 1 1 Mn N N 14.55 ac 1 7.4 69.5 3 3 1 C-H 2003 1 1 Mn N N 13.34 ac 1 6.0 66.7 4 4 1 C-H 2003 1 1 Mn N N 13.34 ac 1 7.1 75.4 5 5 1 C-H 2003 1 1 Mn N N 13.34 ac 1 7.5 57.5 6 6 1 C-H 2008 2 1 Mc N N 11.63 ac 1 5.7 71.5 7 7 1 C-H 2008 2 1 Mc N N 11.63 ac 1 5.2 50.0 8 8 1 C-H 2011 2 1 Mc N N 13.04 ac 1 6.3 62.0 9 9 1 C-H 2011 2 1 Mc N N 13.04 ac 1 6.7 60.5 10 10 1 C-H 2017 3 1 H N N 11.38 ac 1 10.7 82.0 11 11 1 C-H 2017 3 1 H N N 11.38 ac 1 4.4 27.0 12 12 1 C-H 2018 3 1 H N N 11.08 ac 1 5.8 49.0 13 13 1 C-H 2018 3 1 H N N 11.08 ac 1 4.3 64.2 14 14 1 C-H 2013 4 1 McH N N 15.09 ac 1 11.4 86.0 15 15 1 C-H 2013 4 1 McH N N 15.09 ac 1 7.6 87.5 16 16 1 C-H 2014 4 1 McH N N 14.17 ac 1 5.8 60.1 17 17 1 C-H 2014 4 1 McH N N 14.17 ac 1 11.5 100.5 18 18 1 C-H 2014 4 1 McH N N 14.17 ac 1 4.7 53.2 19 19 1 C-H 2019 5 1 MnH N N 11.72 ac 1 8.1 56.0 20 20 1 C-H 2019 5 1 MnH N N 11.72 ac 1 7.1 56.0 21 21 1 C-H 2019 5 1 MnH N N 11.72 ac 1 7.1 56.0 22 22 1 C-H 2020 5 1 MnH N N 14.71 ac 1 7.0 78.2 23 23 1 C-H 2020 5 1 MnH N N 14.71 ac 1 5.2 47.2 24 24 1 C-H 2020 5 1 MnH N N 14.71 ac 1 7.0 83.5 595 1 1 C-H 2002 1 1 Mn N N 14.55 ac 2 9.6 96.0 596 2 1 C-H 2002 1 1 Mn N N 14.55 ac 2 6.0 72.0 597 3 1 C-H 2003 1 1 Mn N N 13.34 ac 2 5.7 75.0 598 4 1 C-H 2003 1 1 Mn N N 13.34 ac 2 7.5 101.0 599 5 1 C-H 2003 1 1 Mn N N 13.34 ac 2 6.9 58.0 600 6 1 C-H 2008 2 1 Mc N N 11.63 ac 2 6.0 84.0 601 7 1 C-H 2008 2 1 Mc N N 11.63 ac 2 6.3 72.0 602 8 1 C-H 2011 2 1 Mc N N 13.04 ac 2 7.4 101.0 603 9 1 C-H 2011 2 1 Mc N N 13.04 ac 2 5.6 62.0 604 10 1 C-H 2017 3 1 H N N 11.38 ac 2 10.7 110.0 605 11 1 C-H 2017 3 1 H N N 11.38 ac 2 4.7 60.0 606 12 1 C-H 2018 3 1 H N N 11.08 ac 2 6.4 48.0 607 13 1 C-H 2018 3 1 H N N 11.08 ac 2 5.6 70.0 608 14 1 C-H 2013 4 1 McH N N 15.09 ac 2 11.0 116.0 609 15 1 C-H 2013 4 1 McH N N 15.09 ac 2 7.5 104.0 610 16 1 C-H 2014 4 1 McH N N 14.17 ac 2 6.5 61.0 611 17 1 C-H 2014 4 1 McH N N 14.17 ac 2 10.9 110.0 612 18 1 C-H 2014 4 1 McH N N 14.17 ac 2 5.9 50.0 613 19 1 C-H 2019 5 1 MnH N N 11.72 ac 2 8.1 76.0 614 20 1 C-H 2019 5 1 MnH N N 11.72 ac 2 7.1 82.0 615 21 1 C-H 2019 5 1 MnH N N 11.72 ac 2 7.1 82.0 616 22 1 C-H 2020 5 1 MnH N N 14.71 ac 2 7.6 98.0 617 23 1 C-H 2020 5 1 MnH N N 14.71 ac 2 6.1 70.0 618 24 1 C-H 2020 5 1 MnH N N 14.71 ac 2 8.4 95.0 1189 1 1 C-H 2002 1 1 Mn N N 14.55 ac 3 13.0 109.0 1190 2 1 C-H 2002 1 1 Mn N N 14.55 ac 3 9.8 77.0 1191 3 1 C-H 2003 1 1 Mn N N 13.34 ac 3 8.0 80.0 1192 4 1 C-H 2003 1 1 Mn N N 13.34 ac 3 13.0 113.0 1193 5 1 C-H 2003 1 1 Mn N N 13.34 ac 3 NA NA 1194 6 1 C-H 2008 2 1 Mc N N 11.63 ac 3 7.7 89.0 1195 7 1 C-H 2008 2 1 Mc N N 11.63 ac 3 9.5 84.0 1196 8 1 C-H 2011 2 1 Mc N N 13.04 ac 3 6.2 122.0 1197 9 1 C-H 2011 2 1 Mc N N 13.04 ac 3 NA NA 1198 10 1 C-H 2017 3 1 H N N 11.38 ac 3 11.5 104.0 1199 11 1 C-H 2017 3 1 H N N 11.38 ac 3 6.1 62.0 1200 12 1 C-H 2018 3 1 H N N 11.08 ac 3 NA NA 1201 13 1 C-H 2018 3 1 H N N 11.08 ac 3 NA NA 1202 14 1 C-H 2013 4 1 McH N N 15.09 ac 3 15.4 100.0 1203 15 1 C-H 2013 4 1 McH N N 15.09 ac 3 4.8 113.0 1204 16 1 C-H 2014 4 1 McH N N 14.17 ac 3 NA NA 1205 17 1 C-H 2014 4 1 McH N N 14.17 ac 3 NA NA 1206 18 1 C-H 2014 4 1 McH N N 14.17 ac 3 NA NA 1207 19 1 C-H 2019 5 1 MnH N N 11.72 ac 3 12.4 92.0 1208 20 1 C-H 2019 5 1 MnH N N 11.72 ac 3 9.6 88.0 1209 21 1 C-H 2019 5 1 MnH N N 11.72 ac 3 9.6 88.0 1210 22 1 C-H 2020 5 1 MnH N N 14.71 ac 3 11.0 94.0 1211 23 1 C-H 2020 5 1 MnH N N 14.71 ac 3 9.0 75.0 1212 24 1 C-H 2020 5 1 MnH N N 14.71 ac 3 10.5 100.0 1783 1 1 C-H 2002 1 1 Mn N N 14.55 ac 4 13.5 102.0 1784 2 1 C-H 2002 1 1 Mn N N 14.55 ac 4 NA 76.0 1785 3 1 C-H 2003 1 1 Mn N N 13.34 ac 4 11.0 79.0 1786 4 1 C-H 2003 1 1 Mn N N 13.34 ac 4 11.8 115.0 1787 5 1 C-H 2003 1 1 Mn N N 13.34 ac 4 NA NA 1788 6 1 C-H 2008 2 1 Mc N N 11.63 ac 4 NA 84.0 1789 7 1 C-H 2008 2 1 Mc N N 11.63 ac 4 10.5 84.0 1790 8 1 C-H 2011 2 1 Mc N N 13.04 ac 4 10.4 125.0 1791 9 1 C-H 2011 2 1 Mc N N 13.04 ac 4 NA NA 1792 10 1 C-H 2017 3 1 H N N 11.38 ac 4 10.2 108.0 1793 11 1 C-H 2017 3 1 H N N 11.38 ac 4 7.2 66.0 1794 12 1 C-H 2018 3 1 H N N 11.08 ac 4 NA NA 1795 13 1 C-H 2018 3 1 H N N 11.08 ac 4 NA NA 1796 14 1 C-H 2013 4 1 McH N N 15.09 ac 4 18.5 121.0 1797 15 1 C-H 2013 4 1 McH N N 15.09 ac 4 15.0 106.0 1798 16 1 C-H 2014 4 1 McH N N 14.17 ac 4 NA NA 1799 17 1 C-H 2014 4 1 McH N N 14.17 ac 4 NA NA 1800 18 1 C-H 2014 4 1 McH N N 14.17 ac 4 NA NA 1801 19 1 C-H 2019 5 1 MnH N N 11.72 ac 4 11.4 114.0 1802 20 1 C-H 2019 5 1 MnH N N 11.72 ac 4 10.6 83.0 1803 21 1 C-H 2019 5 1 MnH N N 11.72 ac 4 10.6 83.0 1804 22 1 C-H 2020 5 1 MnH N N 14.71 ac 4 8.2 94.0 1805 23 1 C-H 2020 5 1 MnH N N 14.71 ac 4 9.1 30.0 1806 24 1 C-H 2020 5 1 MnH N N 14.71 ac 4 11.0 111.0 2377 1 1 C-H 2002 1 1 Mn N N 14.55 ac 5 14.2 110.0 2378 2 1 C-H 2002 1 1 Mn N N 14.55 ac 5 10.7 92.0 2379 3 1 C-H 2003 1 1 Mn N N 13.34 ac 5 10.1 91.0 2380 4 1 C-H 2003 1 1 Mn N N 13.34 ac 5 12.3 128.0 2381 5 1 C-H 2003 1 1 Mn N N 13.34 ac 5 NA NA 2382 6 1 C-H 2008 2 1 Mc N N 11.63 ac 5 8.5 86.0 2383 7 1 C-H 2008 2 1 Mc N N 11.63 ac 5 9.6 89.0 2384 8 1 C-H 2011 2 1 Mc N N 13.04 ac 5 11.3 130.0 2385 9 1 C-H 2011 2 1 Mc N N 13.04 ac 5 NA NA 2386 10 1 C-H 2017 3 1 H N N 11.38 ac 5 11.2 110.0 2387 11 1 C-H 2017 3 1 H N N 11.38 ac 5 8.5 83.0 2388 12 1 C-H 2018 3 1 H N N 11.08 ac 5 NA NA 2389 13 1 C-H 2018 3 1 H N N 11.08 ac 5 NA NA 2390 14 1 C-H 2013 4 1 McH N N 15.09 ac 5 21.1 156.0 2391 15 1 C-H 2013 4 1 McH N N 15.09 ac 5 17.5 132.0 2392 16 1 C-H 2014 4 1 McH N N 14.17 ac 5 NA NA 2393 17 1 C-H 2014 4 1 McH N N 14.17 ac 5 NA NA 2394 18 1 C-H 2014 4 1 McH N N 14.17 ac 5 NA NA 2395 19 1 C-H 2019 5 1 MnH N N 11.72 ac 5 15.1 132.0 2396 20 1 C-H 2019 5 1 MnH N N 11.72 ac 5 9.0 105.0 2397 21 1 C-H 2019 5 1 MnH N N 11.72 ac 5 9.9 90.0 2398 22 1 C-H 2020 5 1 MnH N N 14.71 ac 5 13.0 122.0 2399 23 1 C-H 2020 5 1 MnH N N 14.71 ac 5 12.3 91.0 2400 24 1 C-H 2020 5 1 MnH N N 14.71 ac 5 14.5 145.0 [[alternative HTML version deleted]] </pre> Belair, Ethan D 2013-05-23T20:21:03 http://comments.gmane.org/gmane.comp.lang.r.general/293490 <pre>Dear useRs, i have the following data arranged in three columns structure(c(0.492096635764151, 0.433332688044914, 0.521585941816778, 1.66472272302545, 2.61878329527404, 2.19154489521664, 0.493876245329722, 0.4915787202584, 0.889477365620806, 0.609135860199222, 0.739201878930367, 0.854663750519518, 2.06195904001605, 1.41493262330451, 1.35748791897328, 1.19490680241894, 0.702488756183322, 0.338258418490199, 0.123398398622741, 0.138548982660226, 0.16170889185798, 0.414543218677095, 1.84629295875002, 2.24547399004563), .Dim = c(12L, 2L)) The distance is to be calculated by subtracting each value of each column from the corresponding column value in the following way =&gt;The column values are cyclic. For example, after row 12 there is once again row 1. So, in a way, row 3 is more closer to row 12 than to row 8. =&gt; The peak value is the maximum value for any column. the values falling in the range of 80% of the maximum values can also be considered as maximum value provided they are not falling immediatly next to eachother. =&gt; If we plot column 1 and column 2 the peak value of column 1 is at 5th grade of x-axis and for column 2 its in 12th. For column 2 at x=1 the value is very close to that of the value at x=12 (in 80% range of it), but still it can considered as peak value as it is immediatly falling next to the maximum value. Now The peaks are moved towards eachother in a shortest possible way unless maximum values are under eachother more precisely, column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 1 2 3 4 5 6 7 8 9 10 11 12(max) Now distance is measured in the following way column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 12(max) 1 2 3 4 5 6 7 8 9 10 11 a&gt;sum(abs(col1-col2)) ==column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 11 12(max) 1 2 3 4 5 6 7 8 9 10 b&gt;sum(abs(col1-col2))==column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 10 11 12(max) 1 2 3 4 5 6 7 8 9 c&gt;sum(abs(col1-col2))==column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 9 10 11 12(max) 1 2 3 4 5 6 7 8 d&gt;sum(abs(col1-col2))==column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 8 9 10 11 12(max) 1 2 3 4 5 6 7 e&gt;sum(abs(col1-col2)) total distance= a+b+c+d+e For the following two column it should work the following way structure(c(0.948228727226247, 1.38569091844218, 0.910510759802679, 1.25991218521949, 0.993123416952421, 0.553640392997634, 0.357487763503204, 0.368328033777003, 0.344255688489322, 0.423679560916755, 1.32093576037521, 3.13420679229785, 0.766278117577654, 0.751997501086888, 0.836280758630117, 1.188156460303, 1.56771616670373, 1.15928168139479, 0.522523036011874, 0.561678840701488, 1.11155735914479, 1.26467106348848, 1.09378883406298, 1.17607018089421), .Dim = c(12L, 2L)) column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 1 2 3 4 5(max) 6 7 8 9 10(max) 11 12 Now as for column 2, 10th value is closer to col1 maximum value, therefore distance is measured in the following way column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 12 1 2 3 4 5 6 7 8 9 10(max) 11 a&gt;sum(abs(col1-col2)) --- column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 11 12 1 2 3 4 5 6 7 8 9 10(max) b&gt;sum(abs(col1-col2)) total distance=a+b How can i do it?? Thankyou very very much in advance Elisa i have the following data arranged in three columns structure(c(0.492096635764151, 0.433332688044914, 0.521585941816778, 1.66472272302545, 2.61878329527404, 2.19154489521664, 0.493876245329722, 0.4915787202584, 0.889477365620806, 0.609135860199222, 0.739201878930367, 0.854663750519518, 2.06195904001605, 1.41493262330451, 1.35748791897328, 1.19490680241894, 0.702488756183322, 0.338258418490199, 0.123398398622741, 0.138548982660226, 0.16170889185798, 0.414543218677095, 1.84629295875002, 2.24547399004563), .Dim = c(12L, 2L)) The distance is to be calculated by subtracting each value of each column from the corresponding column value in the following way =&gt;The column values are cyclic. For example, after row 12 there is once again row 1. So, in a way, row 3 is more closer to row 12 than to row 8. =&gt; The peak value is the maximum value for any column. the values falling in the range of 80% of the maximum values can also be considered as maximum value provided they are not falling immediatly next to eachother. =&gt; If we plot column 1 and column 2 the peak value of column 1 is at 5th grade of x-axis and for column 2 its in 12th. For column 2 at x=1 the value is very close to that of the value at x=12 (in 80% range of it), but still it can considered as peak value as it is immediatly falling next to the maximum value. Now The peaks are moved towards eachother in a shortest possible way unless maximum values are under eachother more precisely, column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 1 2 3 4 5 6 7 8 9 10 11 12(max) Now distance is measured in the following way column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 12(max) 1 2 3 4 5 6 7 8 9 10 11 a&gt;sum(abs(col1-col2)) == column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 11 12(max) 1 2 3 4 5 6 7 8 9 10 b&gt;sum(abs(col1-col2)) == column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 10 11 12(max) 1 2 3 4 5 6 7 8 9 c&gt;sum(abs(col1-col2)) == column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 9 10 11 12(max) 1 2 3 4 5 6 7 8 d&gt;sum(abs(col1-col2)) == column 1 1 2 3 4 5(max) 6 7 8 9 10 11 12 column 2 8 9 10 11 12(max) 1 2 3 4 5 6 7 e&gt;sum(abs(col1-col2)) total distance= a+b+c+d+e For the following two column it should work the following way structure(c(0.948228727226247, 1.38569091844218, 0.910510759802679, 1.25991218521949, 0.993123416952421, 0.553640392997634, 0.357487763503204, 0.368328033777003, 0.344255688489322, 0.423679560916755, 1.32093576037521, 3.13420679229785, 0.766278117577654, 0.751997501086888, 0.836280758630117, 1.188156460303, 1.56771616670373, 1.15928168139479, 0.522523036011874, 0.561678840701488, 1.11155735914479, 1.26467106348848, 1.09378883406298, 1.17607018089421), .Dim = c(12L, 2L)) column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 1 2 3 4 5(max) 6 7 8 9 10(max) 11 12 Now as for column 2, 10th value is closer to col1 maximum value, therefore distance is measured in the following way column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 12 1 2 3 4 5 6 7 8 9 10(max) 11 a&gt;sum(abs(col1-col2)) --- column 1 1 2 3 4 5 6 7 8 9 10 11 12(max) column 2 11 12 1 2 3 4 5 6 7 8 9 10(max) b&gt;sum(abs(col1-col2)) total distance=a+b How can i do it.. Thankyou very very much in advance Elisa </pre> eliza botto 2013-05-23T20:29:19 http://comments.gmane.org/gmane.comp.lang.r.general/293487 <pre>Hi List, I have a little confused when to glmnet() vs cv.glmnet(). I know that, glmnet(): gives the fit cv.glment(): does the cv after the fit I just want to get the beta coefficients after the fit, that's it! But of all the glmnet examples I've seen, the beta coefficient is obtained ONLY AFTER cv.glmnet(). Why is that? Also, why is there so many more extra beta's after the fit? Thanks, Mike </pre> C W 2013-05-23T19:44:18 http://comments.gmane.org/gmane.comp.lang.r.general/293481 <pre>I have two files containing words. I want to print the are in file 1 but NOT in file 2. How do I go about? file 1: ABL1 1 ALKBH1 2 ALKBH2 3 ALKBH3 4 ANKRD17 5 APEX1 6 APEX2 7 APTX 8 ASF1A 9 ASTE1 10 ATM 11 ATR 12 ATRIP 13 ATRX 14 ATXN3 15 BCCIP 16 BLM 17 BRCA1 18 BRCA2 file2: ALKBH2 1 ALKBH3 2 APEX1 3 APEX2 4 APLF 5 APTX 6 ATM 7 ATR 8 ATRIP 9 BLM 10 BRCA1 11 BRCA2 12 BRIP1 13 BTBD12 14 CCNH [[alternative HTML version deleted]] </pre> Robin Mjelle 2013-05-23T18:04:34 http://comments.gmane.org/gmane.comp.lang.r.general/293474 <pre>Hello all, I would like to use the Nadaraya-Watson estimator assuming a Gaussian kernel: So far I sued the library(sm) library(sm) x&lt;-runif(5000) y&lt;-rnorm(5000) plot(x,y,col='black') h1&lt;-h.select(x,y,method='aicc') lines(ksmooth(x,y,bandwidth=h1)) which works fine. What if my data were clustered requiring a bandwidth that varies with x? How can I do that? Thanks in advance, Ioanna </pre> IOANNA 2013-05-23T16:10:58 http://comments.gmane.org/gmane.comp.lang.r.general/293468 <pre>Hi, I use R 2.14.0 on Win XP Pro SP3 and it behaves same - some times. After I draw a lot of plots (more then 200, 2 concurrent rgui processes running in parallel) to png then I get same error message. Bmp(), jpg(), png() - same error. Restart of Rgui helps nothing. Solutin - restart system and voila everything is ok. I suspect that there might be something wrong with allocation/deallocation of Windows resources in windows() function. Ondrej Novak [[alternative HTML version deleted]] </pre> Ondrej Novak 2013-05-23T15:06:12 http://comments.gmane.org/gmane.comp.lang.r.general/293467 <pre>Dear r-helpers, I am trying to extract quantities of interest from my iTunes library xml file.  For example, i'd like to be able to run a simple regression of playcount on track number, under the theory that tracks near the beginning of albums get played more (either because they are "better" or because people listen to the beginnings of albums) I have an xml file that is of the following form: &lt;key&gt;13162&lt;/key&gt; &lt;dict&gt; &lt;key&gt;Track ID&lt;/key&gt;&lt;integer&gt;13162&lt;/integer&gt; &lt;key&gt;Name&lt;/key&gt;&lt;string&gt;I'm A Wheel&lt;/string&gt; &lt;key&gt;Artist&lt;/key&gt;&lt;string&gt;Wilco&lt;/string&gt; &lt;key&gt;Composer&lt;/key&gt;&lt;string&gt;Jeff Tweedy&lt;/string&gt; &lt;key&gt;Album&lt;/key&gt;&lt;string&gt;A Ghost is Born&lt;/string&gt; &lt;key&gt;Genre&lt;/key&gt;&lt;string&gt;Rock&lt;/string&gt; &lt;key&gt;Kind&lt;/key&gt;&lt;string&gt;Matched AAC audio file&lt;/string&gt; &lt;key&gt;Size&lt;/key&gt;&lt;integer&gt;6248701&lt;/integer&gt; &lt;key&gt;Total Time&lt;/key&gt;&lt;integer&gt;154648&lt;/integer&gt; &lt;key&gt;Disc Number&lt;/key&gt;&lt;integer&gt;1&lt;/integer&gt; &lt;key&gt;Disc Count&lt;/key&gt;&lt;integer&gt;1&lt;/integer&gt; &lt;key&gt;Track Number&lt;/key&gt;&lt;integer&gt;9&lt;/integer&gt; &lt;key&gt;Track Count&lt;/key&gt;&lt;integer&gt;12&lt;/integer&gt; &lt;key&gt;Year&lt;/key&gt;&lt;integer&gt;2004&lt;/integer&gt; &lt;key&gt;Date Modified&lt;/key&gt;&lt;date&gt;2012-07-26T22:29:15Z&lt;/date&gt; &lt;key&gt;Date Added&lt;/key&gt;&lt;date&gt;2010-01-27T00:02:21Z&lt;/date&gt; &lt;key&gt;Bit Rate&lt;/key&gt;&lt;integer&gt;256&lt;/integer&gt; &lt;key&gt;Sample Rate&lt;/key&gt;&lt;integer&gt;44100&lt;/integer&gt; &lt;key&gt;Play Count&lt;/key&gt;&lt;integer&gt;3&lt;/integer&gt; &lt;key&gt;Play Date&lt;/key&gt;&lt;integer&gt;3434905791&lt;/integer&gt; &lt;key&gt;Play Date UTC&lt;/key&gt;&lt;date&gt;2012-11-05T00:29:51Z&lt;/date&gt; &lt;key&gt;Artwork Count&lt;/key&gt;&lt;integer&gt;1&lt;/integer&gt; &lt;key&gt;Sort Album&lt;/key&gt;&lt;string&gt;Ghost is Born&lt;/string&gt; &lt;key&gt;Persistent ID&lt;/key&gt;&lt;string&gt;A8B0E5CF2E86A4C6&lt;/string&gt; &lt;key&gt;Track Type&lt;/key&gt;&lt;string&gt;File&lt;/string&gt; &lt;key&gt;Location&lt;/key&gt;&lt;string&gt;file://localhost/Users/Alex/Music/iTunes/iTunes%20Media/Music/Wilco/A%20Ghost%20is%20Born/09%20I'm%20A%20Wheel.m4a&lt;/string&gt; &lt;key&gt;File Folder Count&lt;/key&gt;&lt;integer&gt;5&lt;/integer&gt; &lt;key&gt;Library Folder Count&lt;/key&gt;&lt;integer&gt;1&lt;/integer&gt; &lt;/dict&gt;  From each entry, i'd like to extract: Track ID, Track Number and Play Count.  In this case, it would be  13162, 9, 3 my guess is that this can be done using library(XML). If anyone has any guidance, it would be appreciated.  Please note:  a) I do not understand XML data structures, so please explain what you mean by "children" etc… b) Not every entry in my database has a track number and a play count -- i'd like to have NAs associated with the appropriate Track ID, which all entries have. c) it'd also be OK if this XML database just got turned into a normal r data frame. Thanks! [[alternative HTML version deleted]] </pre> Alexander Coppock 2013-05-23T14:31:21 http://comments.gmane.org/gmane.comp.lang.r.general/293463 <pre>Hi, I have a few graph objects created by some graphic package (say, ggplot2, which I use frequently). Because of the existent relation between the graphs, I'd like to index them in two dimensions as p[1,1], p[1,2], p[2,1], p[2,2] for convenience. To my knowledge, the only data type capable of storing graph objects (and any R object) is list, but unfortunately it is available in only one dimension. Could the graphs be stored in any two-dimensional data type? One remedy that comes to my mind is to build a function f so that f(1,1)=1 f(1,2)=2 f(2,1)=3 f(2,2)=4 With functions f and f^{-1} (inverse function of f) , the two-dimensional indices could be mapped to and from a set of one-dimensional indices, and the functions are exactly the way R numbers elements in a matrix. Does R have this built-in function for a m by n matrix or more generally, m*n*p array? (I know this function is easy to write, but just want to make sure whether it exists already) Thanks, Miao [[alternative HTML version deleted]] </pre> jpm miao 2013-05-23T15:30:11 http://comments.gmane.org/gmane.comp.lang.r.general/293459 <pre>Dear All, I wrote a program using R 2.15.2 but this error message "cannot allocate vector of size 200Mb" appeared. I want to ask in general how to handle this situation. I try to run the same program on other computers. It is perfectly fine. Can anybody help? Thank you very much in advance. Best Regards, Ray [[alternative HTML version deleted]] </pre> Ray Cheung 2013-05-23T14:53:31 http://comments.gmane.org/gmane.comp.lang.r.general/293457 <pre>Hello, I have a column called max_date in my data frame and I only want to keep the bigger values for the same activity. How can I do that? data frame: activity max_dt A 2013-03-05 B 2013-03-28 A 2013-03-28 C 2013-03-28 B 2013-03-01 Thank you for your help -- View this message in context: http://r.789695.n4.nabble.com/Removing-rows-w-smaller-value-from-data-frame-tp4667816.html Sent from the R help mailing list archive at Nabble.com. </pre> ramoss 2013-05-23T14:23:58 http://comments.gmane.org/gmane.comp.lang.r.general/293443 <pre>Dear R Gurus, I am trying to run a multigroup SEM using Prof. John Fox's SEM package. The two groups are "Ready to Eat" denoted by RTE and "Ready to Cook" denoted by RTC. I ran a omnibus CFA on the data of consumer perceptions &amp; preferences and am satisfied with what I got. When I tried to do a multigroup SEM - my understanding is limited to the SEM manual in CRAN - using the code below, I get the following message: Error in summary.msemObjectiveML(sem.MG) : no 'dimnames' attribute for array Execution halted The relevant part of my code follows: mod.mg &lt;- multigroupModel(sbmod.cfa,groups=c("RTC","RTE")) sem.MG &lt;- sem(mod.mg,data=srt,group="RTind", formula = ~ inv + imp + tch + emo + loy + usg + sig + dif + ndif + vda + vdb + vdc + vdd + vde + vdf + vdg + vdh + riskT + riskP + riskS + riskFi + riskFu + riskPs ) summary(sem.MG) I was expecting two sets of fit indices for RTE &amp; RTC and want to do an ANOVA across the models; as well as possibly check for loading equivalence. Can somebody please throw some light on where I am making a mistake? Thanks Amarnath Bose </pre> Amarnath Bose 2013-05-23T10:53:38 http://comments.gmane.org/gmane.comp.lang.r.general/293438 <pre>Dear all, I have an ASCII-Grid for Switzerland in the Swiss National Coordinate System of CH1903. Now for a Webapplication of the ASCII-Grid, I need to deliver the ASCII-Grid in the WGS84 System. Via coordinates(ascii) I can "export" the coordinates and convert them with a formula into WGS84. My problem is now, how can I implement these into the ASCII-Grid, so that the whole grid-structure is from now on gonna be saved in the WGS84-coordinate format? (important: I don't want to change the projection, I want to actually change the numeric format of the coordinates) Thank you so much for your help, jas -- View this message in context: http://r.789695.n4.nabble.com/Transform-Coordinate-System-of-an-ASCII-Grid-tp4667786.html Sent from the R help mailing list archive at Nabble.com. </pre> jas 2013-05-23T07:44:46 http://comments.gmane.org/gmane.comp.lang.r.general/293431 <pre>Hi, From time to time I need to do the aggregation. To illustrate, I present a toy example as below. In this example, the task is to aggregate x and y by z with the function mean. Could I call the aggregation function with x_test, where x_test=c("x","y")? Thanks Miao x y z 1 1 1 1 2 2 2 0 3 3 3 1 4 4 0 0 5 5 1 1 6 6 2 0 7 7 3 1 8 8 0 0 9 9 1 1 10 10 2 0 11 11 3 1 12 12 0 0 z x y 1 0 7 1 2 1 6 2 Error in model.frame.default(formula = cbind(x_test) ~ z, data = dftest) : variable lengths differ (found for 'z') a1aggregate(cbind(factor(x_test))~z, data=dftest, FUN=mean) Error in model.frame.default(formula = cbind(factor(x_test)) ~ z, data = dftest) : variable lengths differ (found for 'z') Error in model.frame.default(formula = factor(x_test) ~ z, data = dftest) : variable lengths differ (found for 'z') [[alternative HTML version deleted]] </pre> jpm miao 2013-05-23T07:05:19 http://comments.gmane.org/gmane.comp.lang.r.general/293425 <pre>I'm comparing a variety of datasets with over 4M rows. I've solved this problem 5 different ways using a for/while loop but the processing time is murder (over 8 hours doing this row by row per data set). As such I'm trying to find whether this solution is possible without a loop or one in which the processing time is much faster. Each dataset is a time series as such: DF1: X.DATE X.TIME VALUE VALUE2 1 01052007 0200 37 29 2 01052007 0300 42 24 3 01052007 0400 45 28 4 01052007 0500 45 27 5 01052007 0700 45 35 6 01052007 0800 42 32 7 01052007 0900 45 32 . . . n DF2 X.DATE X.TIME VALUE VALUE2 1 01052007 0200 37 29 2 01052007 0300 42 24 3 01052007 0400 45 28 4 01052007 0500 45 27 5 01052007 0600 45 35 6 01052007 0700 42 32 7 01052007 0800 45 32 . . n+4000 In other words there are 4000 more rows in DF2 then DF1 thus the datasets are of unequal length. I'm trying to ensure that all dataframes have the same number of X.DATE and X.TIME entries. Where they are missing, I'd like to insert a new row. In the above example, when comparing DF2 to DF1, entry 01052007 0600 entry is missing in DF1. The solution would add a row to DF1 at the appropriate index. so new dataframe would be X.DATE X.TIME VALUE VALUE2 1 01052007 0200 37 29 2 01052007 0300 42 24 3 01052007 0400 45 28 4 01052007 0500 45 27 5 01052007 0600 45 27 6 01052007 0700 45 35 7 01052007 0800 42 32 8 01052007 0900 45 32 Value and Value2 would be the same as row 4. Of course this is simple to accomplish using a row by row analysis but with of 4M rows the processing time destroying and rebinding the datasets is very time consuming and I believe highly un-R'ish. What am I missing? Thanks! [[alternative HTML version deleted]] </pre> Adeel Amin 2013-05-23T05:00:50 http://comments.gmane.org/gmane.comp.lang.r.general/293420 <pre>Hello, I have a 'zoo' object containing dates [dd/mm/yr] in the first column and values in the second column. I tried as.Date but did not succeed. Here is an example of date format  01/01/2000 01/02/2000 ... 01/12/2000 01/01/2001 01/02/2001 ... 01/12/2000   etc. I would like to sort all Jans from 2000 to 2010, all Febs from 2000 to 2010 ... all Decs from 2000 to 2010. So, basically I would like to sort by month. Thanks for your help. Atem. [[alternative HTML version deleted]] </pre> Zilefac Elvis 2013-05-23T03:11:40 http://comments.gmane.org/gmane.comp.lang.r.general/293419 <pre>Greetings.  My wife is teaching an introductory stat class at UC Davis.  The class emphasizes the use of simulations, rather than mathematics, to get insight into statistics, and R is the mandated tool.   A student in the class recently inquired about different approaches to sampling from a binomial distribution.  I've appended some code that exhibits the idea, the gist of which is that using sample(c(0, 1), ...) and rbinom(...) should give equivalent results. The surprising (to me) result is that the two approaches DO give the same result, EXCEPT when the probability is exactly 0.5.  See Appendix A for the code and Appendix B for the output.  I don't think this issue is system-dependent, but I've put my session information in Appendix C. Another wrinkle in this is that if I omit the "prob" parameter from the call to sample, meaning to take the default value of 0.5, the two methods DO give the same result. Any thoughts about this?  Thanks. --Mike Appendix A: some R code that exhibits the problem ================================================= ppp &lt;- seq(0, 1, by = 0.01) result &lt;- do.call(rbind, lapply(ppp, function(p) {     set.seed(1)     sampleRes &lt;- sample(c(0, 1), size = 1, replace = TRUE,                         prob=c(1-p, p))          set.seed(1)     rbinomRes &lt;- rbinom(1, size = 1, prob = p)          data.frame(prob = p, equivalent = all(sampleRes == rbinomRes))      })) result Appendix B: the output from the R code ======================================     prob equivalent 1   0.00       TRUE 2   0.01       TRUE 3   0.02       TRUE 4   0.03       TRUE 5   0.04       TRUE 6   0.05       TRUE 7   0.06       TRUE 8   0.07       TRUE 9   0.08       TRUE 10  0.09       TRUE 11  0.10       TRUE 12  0.11       TRUE 13  0.12       TRUE 14  0.13       TRUE 15  0.14       TRUE 16  0.15       TRUE 17  0.16       TRUE 18  0.17       TRUE 19  0.18       TRUE 20  0.19       TRUE 21  0.20       TRUE 22  0.21       TRUE 23  0.22       TRUE 24  0.23       TRUE 25  0.24       TRUE 26  0.25       TRUE 27  0.26       TRUE 28  0.27       TRUE 29  0.28       TRUE 30  0.29       TRUE 31  0.30       TRUE 32  0.31       TRUE 33  0.32       TRUE 34  0.33       TRUE 35  0.34       TRUE 36  0.35       TRUE 37  0.36       TRUE 38  0.37       TRUE 39  0.38       TRUE 40  0.39       TRUE 41  0.40       TRUE 42  0.41       TRUE 43  0.42       TRUE 44  0.43       TRUE 45  0.44       TRUE 46  0.45       TRUE 47  0.46       TRUE 48  0.47       TRUE 49  0.48       TRUE 50  0.49       TRUE 51  0.50      FALSE 52  0.51       TRUE 53  0.52       TRUE 54  0.53       TRUE 55  0.54       TRUE 56  0.55       TRUE 57  0.56       TRUE 58  0.57       TRUE 59  0.58       TRUE 60  0.59       TRUE 61  0.60       TRUE 62  0.61       TRUE 63  0.62       TRUE 64  0.63       TRUE 65  0.64       TRUE 66  0.65       TRUE 67  0.66       TRUE 68  0.67       TRUE 69  0.68       TRUE 70  0.69       TRUE 71  0.70       TRUE 72  0.71       TRUE 73  0.72       TRUE 74  0.73       TRUE 75  0.74       TRUE 76  0.75       TRUE 77  0.76       TRUE 78  0.77       TRUE 79  0.78       TRUE 80  0.79       TRUE 81  0.80       TRUE 82  0.81       TRUE 83  0.82       TRUE 84  0.83       TRUE 85  0.84       TRUE 86  0.85       TRUE 87  0.86       TRUE 88  0.87       TRUE 89  0.88       TRUE 90  0.89       TRUE 91  0.90       TRUE 92  0.91       TRUE 93  0.92       TRUE 94  0.93       TRUE 95  0.94       TRUE 96  0.95       TRUE 97  0.96       TRUE 98  0.97       TRUE 99  0.98       TRUE 100 0.99       TRUE 101 1.00       TRUE Appendix C: Session information =============================== R version 3.0.0 (2013-04-03) Platform: x86_64-redhat-linux-gnu (64-bit) locale:  [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C                [3] LC_TIME=en_US.UTF-8        LC_COLLATE=en_US.UTF-8      [5] LC_MONETARY=en_US.UTF-8    LC_MESSAGES=en_US.UTF-8     [7] LC_PAPER=C                 LC_NAME=C                   [9] LC_ADDRESS=C               LC_TELEPHONE=C             [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C        attached base packages: [1] stats     graphics  grDevices utils     datasets  methods   base      </pre> Michael Hannon 2013-05-23T02:54:01 http://comments.gmane.org/gmane.comp.lang.r.general/293418 <pre>Dear Bernhard, I am using your R package “var”. I am interested in running impulse response analysis (using ‘irf’) and error variance decomposition (using ‘fevd’). I have two questions: -What decomposition method do you use in your package – Cholesky? -What is the order of entry of the variables – the same as in the data set being analyzed? Thank you very much! Dimitri Liakhovitski [[alternative HTML version deleted]] </pre> Dimitri Liakhovitski 2013-05-22T23:17:23 http://comments.gmane.org/gmane.comp.lang.r.general/293413 <pre>hey, I want to divide my data into three groups based on the value in one column with group name. dat: Var 0 0.2 0.5 1 4 6 I tried: dat &lt;- cbind(dat, group=cut(dat$Var, breaks=c(0.1,0.6))) But it doesnt work, I want to group those &lt;0.1 as group A, 0.1-0.6 as group B, &gt;0.6 as group C Thanks for your help! [[alternative HTML version deleted]] </pre> Ye Lin 2013-05-22T21:40:48 http://comments.gmane.org/gmane.comp.lang.r.general/293408 <pre>Hi everyone, I'm having some difficulty getting the linebreaks I want with the cat() function. I currently have the following function: lab1&lt;-function(iv,dv){ a&lt;-anova(lm(dv~iv)) cat("df(between) is",a[1,1]) cat("df(within) is", a[2,1]) cat("ss(between) is", a[1,2]) cat("ss(within) is", a[2,2]) } And I want my output to be: df(between) is 4 df(within) is 45 ss(between) is 232 ss(within) is 400 but instead it prints: df(between) is 4df(within) is 45ss(between) is 232ss(within) is 400 I have seen other people ask about this issue, but only in the context of character strings. I understand that if i wanted the output: Happy birthday to you the correct input would be However applying the same logic to my string returns an error: Error: unexpected input in "cat("df(between) is",a[1,1],\" Error: unexpected input in "cat("df(between) is",a[1,1]\" Adding the command 'sep="\n"' to the end of each cat() returns: df(between) is 4 df(within) is 45 ss(between) is 232 ss(within) is 400 which is not quite as ugly, but I would still prefer the 4 line format I posted at the beginning. Can anyone help? -- View this message in context: http://r.789695.n4.nabble.com/Linebreaks-in-cat-functions-that-call-other-variables-tp4667748.html Sent from the R help mailing list archive at Nabble.com. </pre> jordanbrace 2013-05-22T20:34:34 Search the mailing list at Gmane query http://search.gmane.org/?group=$group=gmane.comp.lang.r.general