Re: [R] convert real valued matrix to binary matrix
On 01/10/2014 06:56 AM, email wrote: Hi: I am trying to analyze an yeast gene expression data http://arep.med.harvard.edu/biclustering/yeast.matrix I need to convert the real-valued data matrix to a binary (0,1) matrix. Is there any package available? How can it be done? Hi John, It looks like you have been told to perform dichotomous clustering on the yeast data from Tavazoie et al. 1999. You should probably look more closely at the cutoff expression level you are to use and think about creating a new matrix of logical (0,1) values using a logical condition. This may be as much help as you will get on the list for this sort of request. Jim __ 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] locate pattern in matrix
you can find it by use of indexing with setting 1=T and 0=False in your global environment of R. On Fri, Jan 10, 2014 at 12:32 PM, email email8...@gmail.com wrote: Dear all, I have a binary matrix 0 0 0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 I want to find the location of all the square and rectangular 1 blocks, like First block in row=2, col=1 to row=3, col=3. Second block in row=5, col=4, to row=6, col=5. How can I find such blocks of 1? Thanks: John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using lapply to get function values
On Jan 10, 2014, at 2:04 AM, Long Vo wrote: Hi R users, I need to apply a function on a list of vectors. This is simple when I use functions that returns only one numerical value such as 'mean' or 'variance'. Things get complex when I use functions returning a list of value, such as 'acf'. In the following example I first create a list of vectors with equal length 10. Then I try to get the first order autocorrelation coefficient of each vector using lapply: # library(stats) X=rnorm(1:100) Y=split(X,as.numeric(gl(length(X),10,length(X# to divide X into 10 sub-vectors app=lapply(Y,FUN=acf,lag.max=1,plot=F) app # I need to know how to display other values of the function e.g. 'type', 'lag' not just 'acf'. Because using app=lapply(Y,FUN=acf$type,lag.max=1,plot=F) acf$type is not a function. The $ function is a wrapper for [[ so try: lapply(app, [[, 'type') To do it in one step, try: lapply(Y, function(x) acf(x)$type) -- David. prompts the error object of type 'closure' is not subsettable Note that this is only an illustrative example. The function I am studying is not 'acf', but it also return a list of values and I need to call out specific values. Helps and comments are welcome. Long -- View this message in context: http://r.789695.n4.nabble.com/using-lapply-to-get-function-values-tp4683373.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. David Winsemius, MD Alameda, CA, USA __ 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 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] How to determine the order of a time series object
I have monthly claims data from 2009 to end of 2013 and would like to forecast for 2014. How is this done in R. Charles. [[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] Fwd: barchar and box on the same figure! is it possible
On 01/10/2014 09:46 PM, Adel ESSAFI wrote:
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
Hi Adel,
Try this. I have used barp in plotrix as your algorithm labels need
staggering.
aedf-read.table(text=V1 V2 V3 V4
HEFT-AC 1 402499.9 460543.4
HEFT-AC 2 402499.9 470316.8
HEFT-ACU 1 420814.2 433203.7
HEFT-ACU 2 420814.2 453475.8
LPT 1 402499.9 460543.4
LPT 2 402499.9 470316.8
SPT 1 453262.3 459070.3
SPT 2 453262.3 483893.6
MIN-MIN 1 447286.4 477632.5
MIN-MIN 2 447286.4 488146.3
MAX-MIN 1 457122.6 498265.2
MAX-MIN 2 457122.6 491143.3,
header=TRUE)
library(plotrix)
barp(aedf$V3[!duplicated(aedf$V1)],ylim=c(0,60),
names.arg=unique(aedf$V1),staxx=TRUE)
boxat-1
for(alg in unique(aedf$V1)) {
boxplot(aedf$V4[aedf$V1==alg],range=0,at=boxat,add=TRUE)
boxat-boxat+1
}
Jim
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[R] MuMIn Z-Test
Dear list, I'm using MuMIn for model averaging and I had a question about the z-test that MuMIn performs when using the summary call. e.g.: model.1 - subset(model.sel(model.1.glmm.list, rank = AICc), delta3) summary(model.avg(model.1)) ... Model-averaged coefficients: Estimate Std. Error Adjusted SE z value Pr(|z|) 0|1 0.094832 0.3586490.360952 0.263 0.792760 1|2 1.193099 0.3667610.369124 3.232 0.001228 ** age 1.163131 0.4167350.419355 2.774 0.005544 ** distribution 0.063250 0.3210250.323071 0.196 0.844784 left 0.158854 0.2478160.249294 0.637 0.523985 right-1.317514 0.3281740.330271 3.989 6.63e-05 *** income 0.705691 0.1906890.191843 3.678 0.000235 *** population -1.026989 0.3199140.321969 3.190 0.001424 ** weight -0.178042 0.1359620.136761 1.302 0.192967 Signif. codes: 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1 What exactly is the z-test testing? I read a post saying that p0.05 means the confidence intervalsdon't span zero, however you can get the confidence intervals usingthe confint() call. What is the z-test telling me that the confidence intervals are not? Also - the MuMIn vignette says that the p-value assumes a normalerror distribution. A normal distribution of what? The models usedin model averaging? Or the original data? Any clarification would be much appreciated. Tom [[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 riddle
Hi, Use ?reshape() dat1- read.table(text=V1 V2 V3 V4 V5 1 210505 ARS B A 1 210505 BFGL A B 1 210505 NGS B B 1 210506 ARS B B 1 210506 BFGL A A 1 210506 NGS B B 1 210507 ARS B B 1 210507 BFGL A B 1 210507 NGS A B,sep=,header=TRUE,stringsAsFactors=FALSE) res - reshape(dat1[,-1],timevar=V2,idvar=V3,direction=wide) colnames(res)[-1] -gsub(.*\\.,,colnames(res)[-1]) res # V3 210505 210505 210506 210506 210507 210507 #1 ARS B A B B B B #2 BFGL A B A A A B #3 NGS B B B B A B A.K. Dear R users I have a question. I have example file: V1 V2 V3 V4 V5 1 210505 ARS B A 1 210505 BFGL A B 1 210505 NGS B B 1 210506 ARS B B 1 210506 BFGL A A 1 210506 NGS B B 1 210507 ARS B B 1 210507 BFGL A B 1 210507 NGS A B from which I want to get file which looks like this: V2 210505 210505 210506 210506 210507 210507 ARS B A B B B B BFGL A B A A A B NGS B B B B A B I think this can be done with a loop but unfortunatelly I'm not experienced with loops. Thank you very much for help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Package TM dataframes
Hi, I am trying to use the package TM on a dataframe get the following error: complaints - tm_map(complaints, tolower) Error in UseMethod(tm_map, x) : no applicable method for 'tm_map' applied to an object of class data.frame Tm doesn't work on dataframes? My data frame consists of 1 text variable containing customer complaints. What do I have to change it to in order to use the TM package for text mining? Thanks ahead for your help/thoughts. -- View this message in context: http://r.789695.n4.nabble.com/Package-TM-dataframes-tp4683392.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] Package TM dataframes
On Jan 10, 2014, at 10:11 AM, ramoss wrote: Hi, I am trying to use the package TM on a dataframe get the following error: complaints - tm_map(complaints, tolower) Error in UseMethod(tm_map, x) : no applicable method for 'tm_map' applied to an object of class data.frame Tm doesn't work on dataframes? My data frame consists of 1 text variable containing customer complaints. What do I have to change it to in order to use the TM package for text mining? Thanks ahead for your help/thoughts. As the Posting Guide says: Sometimes the appropriate answer is to direct you to the manual. In this case ?tm::tm_map shows that the first argument is expected to be a corpus object. -- David Winsemius, MD Alameda, CA, USA __ 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] How do you transform a dataframe to a corpus?
Hi; I have a data frame complains w/ dimensions 11335291 ( 1.13m obs 1 col) I am trying to transform it into a corpus using the following code: myCorpus -Corpus(VectorSource(complaints$text)) Error in .Source(readPlain, encoding, length(x), FALSE, names(x), 0, TRUE, : vectorized sources must have positive length Does anybody understand the error message? Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-do-you-transform-a-dataframe-to-a-corpus-tp4683396.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] How to determine the order of a time series object
There are many ways, but this is not a statistical theory support forum for helping you choose among them. Your question becomes more appropriate when you ask something along the lines of how do I regress data using AR model with R? (Though reading the Time Series Task View on CRAN would be even more appropriate than asking that question here.) --- 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. Charles Thuo tcmui...@gmail.com wrote: I have monthly claims data from 2009 to end of 2013 and would like to forecast for 2014. How is this done in R. Charles. [[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.
Re: [R] locate pattern in matrix
What would you like to get from the following? 0 1 0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 0 1 1 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of email Sent: Thursday, January 09, 2014 11:03 PM To: r-help@r-project.org Subject: [R] locate pattern in matrix Dear all, I have a binary matrix 0 0 0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 I want to find the location of all the square and rectangular 1 blocks, like First block in row=2, col=1 to row=3, col=3. Second block in row=5, col=4, to row=6, col=5. How can I find such blocks of 1? Thanks: John __ 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.
Re: [R] How do you transform a dataframe to a corpus?
The column length is 4000 bytes long if that helps. -- View this message in context: http://r.789695.n4.nabble.com/How-do-you-transform-a-dataframe-to-a-corpus-tp4683396p4683402.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] ggplot2; svg; text point size specification
library(ggplot2) a - data.frame(x=rnorm(10), y=rnorm(10)) #conversion of mm to points 2.83464567 plot.demo - qplot(x, y, data=a, xlab=test.x, ylab=test.y)+theme(panel.border=element_rect(fill=NA, size=1), axis.title.x = element_text(size=10*2.83464567, family=Arial), axis.title.y = element_text(size=12*2.83464567, family=Arial), axis.text.y = element_text(size=10*2.83464567, family=Arial), axis.text.x = element_text(size=10*2.83464567, family=Arial)) print(plot.demo) Happy New Year Everyone! I would like to preserve point size with different figure sizes using the svg graphics device. #for instance #ggsave(~/Desktop/test2.svg, p ,height=3, width=3) #ggsave(~/Desktop/test2.svg, p ,height=10, width=10) #both look very different to me. What am I missing? Thank you for all of the help in advance. Stephen -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman [[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.
Re: [R] embedd fonts in pdf generated by R
On 10/01/2014 19:29, Adel ESSAFI wrote: 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. If you have suitable licenses for those fonts, see ?embedFonts (surely an obvious thing to search for!). Otherwise, select fonts you do have a license to embed. And see the posting guide: a reproducible example would have helped a lot here. [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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] Maximum of Maximum
Hi,
May be this helps:
set.seed(42)
vec1 - sapply(1:1000,function(x) max(rnorm(1000,0,1)))
#or
set.seed(42)
vec2 - replicate(1000,max(rnorm(1000,0,1)))
identical(vec1,vec2)
#[1] TRUE
set.seed(598)
res - sample(vec1,100,replace=FALSE)
max(res)
#[1] 4.408794
A.K.
Hello I need some help in programming:
I want to after take a rnorm(1000,0,1) sample want to save the
max of it in a vector, this was difficult to me, then I need to repeat
this process 1000 times, so I shall have 1000 maximum from the normal of
samples of size 1000.
Then I want to sample from this maximum series a new sample, this time of size
100.
I drive a programm:
sample-rnorm(1000,0,1)
for(i in 1:length(sample)){
maxim-vector(mode=numeric,length(sample))
maxim[i]-max(sample)
}
then my result was a vector with n-1 zeroes and the last one
value was a maximum. I think it took the last value, but I don't know
how to repair it.
I need most help, thanks
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[R] SE for R-squared
In R package psychometrics an estimate of SE of R squared of /sersq - sqrt((4*rsq*(1-rsq)^2*(n-k-1)^2)/((n^2-1)*(n+3))) with n sample size, and k number of parameters if sample size greater than 60 is found. Does anyone have a formula for smaller sample size or an exact formula? I have been through sos - but only found the above? All best wishes Troels Ring Denmark / [[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] Conditional inclusion of an element in an R object
Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[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] Conditional inclusion of an element in an R object
Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[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] Conditional inclusion of an element in an R object
Apologies, but all that ifelse() stuff is too hard to follow. What I would do is compute a character vector of column names to keep, then do a1[ , names.to.keep] -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B), each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[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.
Re: [R] Maximum of Maximum
It probably doesn't make much difference in this small example, but
maybe it's worth noting that **if possible** (it typically is not),
things can be speeded up by generating all the random numbers at once
then applying a vectorized operation to the entire ensemble. In this
case, this becomes:
ran - do.call(pmax,data.frame(matrix(rnorm(1e6),nr=1000)))
pmax() is the vectorized operation. The do.call(...,data.frame(..)))
construction must be used to satisfy the argument list requirements
for pmax and do.call.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Fri, Jan 10, 2014 at 12:31 PM, arun smartpink...@yahoo.com wrote:
Hi,
May be this helps:
set.seed(42)
vec1 - sapply(1:1000,function(x) max(rnorm(1000,0,1)))
#or
set.seed(42)
vec2 - replicate(1000,max(rnorm(1000,0,1)))
identical(vec1,vec2)
#[1] TRUE
set.seed(598)
res - sample(vec1,100,replace=FALSE)
max(res)
#[1] 4.408794
A.K.
Hello I need some help in programming:
I want to after take a rnorm(1000,0,1) sample want to save the
max of it in a vector, this was difficult to me, then I need to repeat
this process 1000 times, so I shall have 1000 maximum from the normal of
samples of size 1000.
Then I want to sample from this maximum series a new sample, this time of
size 100.
I drive a programm:
sample-rnorm(1000,0,1)
for(i in 1:length(sample)){
maxim-vector(mode=numeric,length(sample))
maxim[i]-max(sample)
}
then my result was a vector with n-1 zeroes and the last one
value was a maximum. I think it took the last value, but I don't know
how to repair it.
I need most help, thanks
__
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.
Re: [R] Maximum of Maximum
... In fact, on my computer, the pmax construction took longer. I
would guess that's due to the data.frame construction. Nevertheless,
the comment is still germane.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Fri, Jan 10, 2014 at 1:03 PM, Bert Gunter bgun...@gene.com wrote:
It probably doesn't make much difference in this small example, but
maybe it's worth noting that **if possible** (it typically is not),
things can be speeded up by generating all the random numbers at once
then applying a vectorized operation to the entire ensemble. In this
case, this becomes:
ran - do.call(pmax,data.frame(matrix(rnorm(1e6),nr=1000)))
pmax() is the vectorized operation. The do.call(...,data.frame(..)))
construction must be used to satisfy the argument list requirements
for pmax and do.call.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Fri, Jan 10, 2014 at 12:31 PM, arun smartpink...@yahoo.com wrote:
Hi,
May be this helps:
set.seed(42)
vec1 - sapply(1:1000,function(x) max(rnorm(1000,0,1)))
#or
set.seed(42)
vec2 - replicate(1000,max(rnorm(1000,0,1)))
identical(vec1,vec2)
#[1] TRUE
set.seed(598)
res - sample(vec1,100,replace=FALSE)
max(res)
#[1] 4.408794
A.K.
Hello I need some help in programming:
I want to after take a rnorm(1000,0,1) sample want to save the
max of it in a vector, this was difficult to me, then I need to repeat
this process 1000 times, so I shall have 1000 maximum from the normal of
samples of size 1000.
Then I want to sample from this maximum series a new sample, this time of
size 100.
I drive a programm:
sample-rnorm(1000,0,1)
for(i in 1:length(sample)){
maxim-vector(mode=numeric,length(sample))
maxim[i]-max(sample)
}
then my result was a vector with n-1 zeroes and the last one
value was a maximum. I think it took the last value, but I don't know
how to repair it.
I need most help, thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: Conditional inclusion of an element in an R object
My intention is to include certain columns if they meet certain criteria. For example, if b is one of the columns of a1, then keep otherwise don't. HTH.. santosh On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don macque...@llnl.gov wrote: Apologies, but all that ifelse() stuff is too hard to follow. What I would do is compute a character vector of column names to keep, then do a1[ , names.to.keep] -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B), each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Tinn-R eBook (free)
Dear users, An eBook (2. Ed.) of Tinn-R was published today. Dowload (free): - http://www.uesc.br/editora/index.php?item=conteudo_livros_digitais.php - http://www.uesc.br/editora/livrosdigitais2/tredit.pdf - Http://nbcgib.uesc.br/lec/software/editores/tinn-r/en # h10-ebook All the best, -- ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ Jose Claudio Faria Estatistica UESC/DCET/Brasil joseclaudio.faria at gmail.com Telefones: 55(73)3680.5545 - UESC 55(73)9100.7351 - TIM 55(73)8817.6159 - OI ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ __ 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] Fwd: Conditional inclusion of an element in an R object
Don's response seems apropos to me. Do you understanding indexing, i.e. the [ operator? If not, you should read An Introduction to R or other tutorial (there are many good ones on the web). If that is not the issue, you need to explain more clearly why his answer does not suffice. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, Jan 10, 2014 at 1:59 PM, Santosh santosh2...@gmail.com wrote: My intention is to include certain columns if they meet certain criteria. For example, if b is one of the columns of a1, then keep otherwise don't. HTH.. santosh On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don macque...@llnl.gov wrote: Apologies, but all that ifelse() stuff is too hard to follow. What I would do is compute a character vector of column names to keep, then do a1[ , names.to.keep] -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B), each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] Fwd: Conditional inclusion of an element in an R object
I don't think apropos or indexing would help. I am open to your suggestions/tips. I usually get multiple versions of a dataset (even with the same column names). In the source data, I occasionally notice inconsistencies... formatting issues, column naming issues etc.. As shown In the a1 example,.. the values that are supposed to be in column Qr are sometimes in column b. Such differences between versions crop up due to various unknown reasons, e.g. when different programmers prepare the data set or if the existing practices/processes change. Likewise, formats of certain date-time columns (not shown in the example) also vary, the time date format may be in %m/%d/%Y %H:%M, %d %b %Y %H:%M or %d%b%Y %H:%M So, I would like use programming methods to pick the right one if available or don't pick at all. Besides, is there an R equivalent of %m[/][.]%d[/][,]%y [%H[:%M[:%S[.%N]]][%p][[(]%3Z[)]]] available in Splus (?class.timeDate) for tackling time-date format inconsistencies as mentioned above. Thanks, Santosh On Fri, Jan 10, 2014 at 2:32 PM, Bert Gunter gunter.ber...@gene.com wrote: Don's response seems apropos to me. Do you understanding indexing, i.e. the [ operator? If not, you should read An Introduction to R or other tutorial (there are many good ones on the web). If that is not the issue, you need to explain more clearly why his answer does not suffice. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, Jan 10, 2014 at 1:59 PM, Santosh santosh2...@gmail.com wrote: My intention is to include certain columns if they meet certain criteria. For example, if b is one of the columns of a1, then keep otherwise don't. HTH.. santosh On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don macque...@llnl.gov wrote: Apologies, but all that ifelse() stuff is too hard to follow. What I would do is compute a character vector of column names to keep, then do a1[ , names.to.keep] -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b, NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B), each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b ,NULL)),C3=Ra) a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't Is there a way to dynamically include columns in a dataframe? Regards, santosh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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,
Re: [R] Fwd: Conditional inclusion of an element in an R object
At the risk of being annoying ... Your original question was, Is there a way to dynamically include columns in a dataframe? The answer is yes. One way, and I think the simplest, is to calculate the names of the columns you want to keep, and then use an expression like I suggested, that is, one like a1[ , names.to.keep] instead of using select(). Or calculate the names you do not want to keep, if that is easier, and use for example a1[, setdiff(names(a1),names.to.drop) ] Based on your most recent email it seems like calculating which columns to keep or drop may be difficult. But I would still suggest a better approach would be to focus on calculating a character vector of column names. Maybe you can convert your lc1 and lc2 objects to vectors of column names. Speaking personally, nested ifelse() expressions make me want to get up and run away. So I've very reluctant to put any effort into trying to figure out what they produce. But that's just my preference; others may feel differently. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 3:24 PM, Santosh santosh2...@gmail.com wrote: I don't think apropos or indexing would help. I am open to your suggestions/tips. I usually get multiple versions of a dataset (even with the same column names). In the source data, I occasionally notice inconsistencies... formatting issues, column naming issues etc.. As shown In the a1 example,.. the values that are supposed to be in column Qr are sometimes in column b. Such differences between versions crop up due to various unknown reasons, e.g. when different programmers prepare the data set or if the existing practices/processes change. Likewise, formats of certain date-time columns (not shown in the example) also vary, the time date format may be in %m/%d/%Y %H:%M, %d %b %Y %H:%M or %d%b%Y %H:%M So, I would like use programming methods to pick the right one if available or don't pick at all. Besides, is there an R equivalent of %m[/][.]%d[/][,]%y [%H[:%M[:%S[.%N]]][%p][[(]%3Z[)]]] available in Splus (?class.timeDate) for tackling time-date format inconsistencies as mentioned above. Thanks, Santosh On Fri, Jan 10, 2014 at 2:32 PM, Bert Gunter gunter.ber...@gene.com wrote: Don's response seems apropos to me. Do you understanding indexing, i.e. the [ operator? If not, you should read An Introduction to R or other tutorial (there are many good ones on the web). If that is not the issue, you need to explain more clearly why his answer does not suffice. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, Jan 10, 2014 at 1:59 PM, Santosh santosh2...@gmail.com wrote: My intention is to include certain columns if they meet certain criteria. For example, if b is one of the columns of a1, then keep otherwise don't. HTH.. santosh On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don macque...@llnl.gov wrote: Apologies, but all that ifelse() stuff is too hard to follow. What I would do is compute a character vector of column names to keep, then do a1[ , names.to.keep] -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: There is a correction to the earlier post.. my apologies... a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10], R=rep(c(A,B),each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b , NULL)),C3=R) lc2 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b , NULL)), C3=ifelse(is.element(Ra,names(a1)),Ra,NULL)) *The error for the above:* Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) : replacement has length zero In addition: Warning message: In rep(no, length.out = length(ans)) : 'x' is NULL so the result will be NULL a2 - subset(a1, sel=unlist(lc1)) # this works a3 - subset(a1, sel=unlist(lc2)) # this doesn't work Is there a way to dynamically include columns in a dataframe? Regards, santosh On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com wrote: Dear Rxperts... I would like to conditionally include an element (as a column) in a dataframe. Please see the sample code below: a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B ), each=5)) lc1 - list(C1 = P,C2 = ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)), b ,NULL)),C3=R) lc2 - list(C1 = P,C2 =
Re: [R] Fwd: Conditional inclusion of an element in an R object
I agree with Don... focus on identifying the names of the columns and then use column name indexing to extract the columns you want. You will probably want to rename them to a standard set of names once you have gone to all this trouble... just assign the new vector of names to the names function of the extracted data frame. Using nested if-else may be messy, but regardless of whether you figure out a better way to identify columns that works for your data, you should be able to using indexing by name both now and later. --- 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. MacQueen, Don macque...@llnl.gov wrote: At the risk of being annoying ... Your original question was, Is there a way to dynamically include columns in a dataframe? The answer is yes. One way, and I think the simplest, is to calculate the names of the columns you want to keep, and then use an expression like I suggested, that is, one like a1[ , names.to.keep] instead of using select(). Or calculate the names you do not want to keep, if that is easier, and use for example a1[, setdiff(names(a1),names.to.drop) ] Based on your most recent email it seems like calculating which columns to keep or drop may be difficult. But I would still suggest a better approach would be to focus on calculating a character vector of column names. Maybe you can convert your lc1 and lc2 objects to vectors of column names. Speaking personally, nested ifelse() expressions make me want to get up and run away. So I've very reluctant to put any effort into trying to figure out what they produce. But that's just my preference; others may feel differently. -Don __ 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] Function inside Sweave
Hi Silvano
To save you typing try paste0 (...) instead of paste(, sep = )
Just checking quickly
In
### code chunk number 3: Serie1
for (i in 1:7){
aux - paste(Q, i, sep=)}
will always end up with Q7
Is the } to go elsewhere?
OR
Is it superfluous?
Duncan
-Original Message-
From: silv...@uel.br [mailto:silv...@uel.br]
Sent: Friday, 10 January 2014 23:50
To: Duncan Mackay
Subject: Re: [R] Function inside Sweave
Dear Duncan,
I get solved the problem. This is the solution:
###
### code chunk number 1: Dados
###
rm(list=ls())
require(foreign)
require(xtable)
require(reshape)
setwd('C:/Colegiado de Veterinária/Dados/A2013/')
Geral = read.epiinfo('A2013.rec')
Professores = read.epiinfo('Professor.rec')
Dados = merge(Geral, Professores, by=c('ALUNO', 'DISCIPLINA'))
###
### code chunk number 2: Leitura
###
serie_1 = subset(Dados, SERIE=='1')
attach(serie_1)
head(serie_1)
round(100*prop.table(table(DISCIPLINA, Q1, PROFESSOR), 1), 2)
###
### code chunk number 3: Serie1
###
DISCIPLINA - serie_1$DISCIPLINA
for (i in 1:7){
aux - paste(Q, i, sep=)}
for (i in 1:7){
aux - paste(nome.Questao, i, sep=)
assign(aux, paste(Nome da Questao, i))
}
nome.Questoes = c(Apresentação da proposta de programa a ser desenvolvida
na disciplina.,
Profundidade dos temas em relação aos objetivos da
disciplina.,
Aplicabilidade dos temas abordados.,
Articulação do conteúdo da disciplina com outras e com a
profissão.,
Estabelecimento de critérios de avaliação claros e
adequados.,
Os resultados das avaliações são discutidos com os
alunos.,
Atendimento da disciplina às suas expectativas.)
cria.tabela - function(Questao, i){
Questao1 - get(Questao)
tab1 - table(DISCIPLINA, Questao1)
tab1.prop = round(100*prop.table(tab1, 1), 2)
capt - nome.Questoes[i]
tab1.txt = xtable(tab1.prop, align=l|r, label=Questao,
caption=paste(capt))
print(tab1.txt, format.args=list(big.mark = ., decimal.mark = ,),
caption.placement='top', table.placement='H')
cat(\n\n\n)
}
Discip - function(){
for (i in 1:7){
x - paste(Q, i, sep=)
cria.tabela(x, i)
}
}
Discip()
Thanks a lot.
Silvano.
-Mensagem Original-
From: Duncan Mackay
Sent: Friday, January 10, 2014 12:18 AM
To: silv...@uel.br
Cc: R
Subject: RE: [R] Function inside Sweave
Hi Silvano
I am not sure exactly what you want as I am not sure of the structure and
format going into xtable
but the structure after you have formed the table is apparently the same
structure.
This is what I suggested before (modified)
for(j in 1:25){
yourtable - ...
xx - xtable(yourtable)
caption(xx) - paste(Table, j)
print(xx)
if(j 25) cat(\n\n)
}
Pre xtable
If you have different colums to tabulate I suggest you use a formula or
bquote via a list to vary the columns in a loop
and these are the variables to be tabulated. You can then use this for your
column headers in xtable /latex.
durin/Post xtable
If single line header for table and formatting not needed to be changed
everything OK
otherwise inside chunk
head, echo=FALSE, results=tex=
for(j in 1:25){
yourtable - ...
hdx = c(\\begin{table}[h]\n,
\\caption{\\prevt{} and their equivalent \\dcode{}}%\n,
\\label{tab:, j }%\n,
\\begin{tabular}{l|r}\n) # adjust to suit
cat(hdx)
# header from col names or otherwise
out - names(yourtable)
out -
c(,
paste(\\multicolumn{1}{c}{,
out,
rep(c(} %\n, } ), c(length(out)-1, 1)), sep = ), \n
)
cat(out, sep = )
# xtable
xx - xtable(yourtable)
xnd = c(\\end{tabular}\n,
\\end{table}\n)
cat(xnd)
if(j 25) cat(\n\n)
}
@ %% end
You can add booktabs functions within tabular by cat if needed
If you have things that need to be changed they can be accessed by a
vector/list
If you have variable numbers of decimal points see an colour is black
http://tex.stackexchange.com/questions/96982/coloring-text-in-a-dcolumn-alig
ned-cell-of-a-table
Henrik Bengtsson has a package to do this sort of thing but he discontinued
it for ver 3 or 2.15 ca
Regards
Duncan
-Original Message-
From: silv...@uel.br [mailto:silv...@uel.br]
Sent: Friday, 10 January 2014 04:48
To: Duncan Mackay
Subject: Re: [R] Function inside Sweave
Dear Duncan,
I want relatioship two variables: DISCIPLINA and Questions.
I have about 100 different disciplines (in same column) and 25 questions (25
differents columns).
Head is:
ALUNO DISCIPLINA FIELD1 DATA SERIE TURMA Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10
PROFESSOR Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25
Something like this:
[R] Selecting individuals to maximize the correlation of two variables
Hi,
Maybe it is not directly related to R but sine many are statistical experts so
I post it here for help:
I have two variables (say x and y) of length n. Now the cor(x,y) is close to 0.
I need to find the subset in {1,.. n} so that the correlation between x and y
using the subset data is maximized. A trivial choice would be selecting 2
individuals only so that cor(x,y) =1. As the size of the subset increases,
cor(x,y) will go down to 0, but I am assuming the best correlation for each
size of the subsets would not be monotonically decreasing.
Any idea of how to find the solution?
Thanks,
Jing
[[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] Fwd: Conditional inclusion of an element in an R object
?as.POSIXct for time-formatting. This function makes a structured list of
time data, where you specify an input time and format i.e.
as.POSIXct('2014-01-09 01:30:00', format='%Y-%m-%d %H:%M:%S')
On 1/10/14, 15:24 , Santosh santosh2...@gmail.com wrote:
I don't think apropos or indexing would help. I am open to your
suggestions/tips.
I usually get multiple versions of a dataset (even with the same column
names). In the source data, I occasionally notice inconsistencies...
formatting issues, column naming issues etc..
As shown In the a1 example,.. the values that are supposed to be in
column Qr are sometimes in column b. Such differences between versions
crop up due to various unknown reasons, e.g. when different programmers
prepare the data set or if the existing practices/processes change.
Likewise, formats of certain date-time columns (not shown in the example)
also vary, the time date format may be in %m/%d/%Y %H:%M, %d %b %Y
%H:%M or %d%b%Y %H:%M
So, I would like use programming methods to pick the right one if available
or don't pick at all.
Besides, is there an R equivalent of %m[/][.]%d[/][,]%y
[%H[:%M[:%S[.%N]]][%p][[(]%3Z[)]]] available in Splus (?class.timeDate)
for tackling time-date format inconsistencies as mentioned above.
Thanks,
Santosh
On Fri, Jan 10, 2014 at 2:32 PM, Bert Gunter gunter.ber...@gene.com
wrote:
Don's response seems apropos to me. Do you understanding indexing,
i.e. the [ operator? If not, you should read An Introduction to R or
other tutorial (there are many good ones on the web). If that is not
the issue, you need to explain more clearly why his answer does not
suffice.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Fri, Jan 10, 2014 at 1:59 PM, Santosh santosh2...@gmail.com wrote:
My intention is to include certain columns if they meet certain
criteria.
For example, if b is one of the columns of a1, then keep otherwise
don't.
HTH..
santosh
On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don macque...@llnl.gov
wrote:
Apologies, but all that ifelse() stuff is too hard to follow.
What I would do is compute a character vector of column names to keep,
then do
a1[ , names.to.keep]
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/10/14 12:53 PM, Santosh santosh2...@gmail.com wrote:
Dear Rxperts...
I would like to conditionally include an element (as a column) in a
dataframe. Please see the sample code below:
There is a correction to the earlier post.. my apologies...
a1 - data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],
R=rep(c(A,B),each=5))
lc1 - list(C1 = P,C2 =
ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b
,
NULL)),C3=R)
lc2 - list(C1 = P,C2 =
ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b
,
NULL)),
C3=ifelse(is.element(Ra,names(a1)),Ra,NULL))
*The error for the above:*
Error in ifelse(is.element(Ra, names(a1)), Ra, NULL) :
replacement has length zero
In addition: Warning message:
In rep(no, length.out = length(ans)) :
'x' is NULL so the result will be NULL
a2 - subset(a1, sel=unlist(lc1)) # this works
a3 - subset(a1, sel=unlist(lc2)) # this doesn't work
Is there a way to dynamically include columns in a dataframe?
Regards,
santosh
On Fri, Jan 10, 2014 at 12:45 PM, Santosh santosh2...@gmail.com
wrote:
Dear Rxperts...
I would like to conditionally include an element (as a column) in a
dataframe. Please see the sample code below:
a1 -
data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c(A,B)
,
each=5))
lc1 - list(C1 = P,C2 =
ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b
,NULL)),C3=R)
lc2 - list(C1 = P,C2 =
ifelse(is.element(Q,names(a1)),Q,ifelse(is.element(b,names(a1)),b
,NULL)),C3=Ra)
a2 - subset(a1, sel=unlist(lc1)) # this works
a3 - subset(a1, sel=unlist(lc2)) # this doesn't
Is there a way to dynamically include columns in a dataframe?
Regards,
santosh
[[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
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML
[R] Species Distribution Modeling in R, with already extracted environmental values
Hi all, I am attempting to do SDM in R, mostly I have used the 'dismo' package. My problem is that all the methods call for a raster stack as the input representing the environmental layers. However, my processing steps leave me with a table of extracted values for each occurrence (much like section 4.2 in the dismo vignette). I am wondering if there is a way to feed this type of data into these modeling methods, in dismo or any other SDM package. Thanks! Cody Schank [[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] looping through 3D array
It seems like emdist does not like to compare matrices with all 0 values. I
ended up removing those from my 3D array and have ~8000 matrices instead of
13000.
I am using res2 - unlist(mclapply(seq_len(ncol(indx)),function(i) {x1 -
indx[,i]; emd2d(results[x1[1],,],results[x1[2],,]) }) )
But even with mclapply it is taking extremely long. Any way to speed this
up?
On Jan 9, 2014 4:10 PM, arun kirshna [via R]
ml-node+s789695n4683362...@n4.nabble.com wrote:
Hi,
No problem.
You can use ?lower.tri() or ?upper.tri()
res[lower.tri(res)]
res[lower.tri(res,diag=TRUE)]
#Other way would be to use:
?combn
indx - combn(dim(results)[1],m=2)
res2 - sapply(seq_len(ncol(indx)),function(i) {x1 - indx[,i];
emd2d(results[x1[1],,],results[x1[2],,]) })
identical(res[lower.tri(res)], res2)
#[1] TRUE
A.K.
On Thursday, January 9, 2014 4:03 PM, alex padron [hidden email] wrote:
Thanks. This works. I just noticed that half of the matrix repeats. For
example res[1,2] is the same as res[2,1]. any way to get half of the matrix
output (notice the diagonal 0 across the output matrix)?
-Alex
On Thu, Jan 9, 2014 at 12:57 PM, arun [hidden email] wrote:
#or
you can use ?expand.grid() and then loop over:
indx - expand.grid(rep(list(seq(dim(results)[1])),2))
res1 - matrix(sapply(seq_len(nrow(indx)),function(i) {x1 - indx[i,];
emd2d(results[x1[,1],,],results[x1[,2],,]) }),ncol=10)
identical(res,res1)
#[1] TRUE
On Thursday, January 9, 2014 3:46 PM, arun [hidden email] wrote:
Hi,
Try:
library(emdist)
set.seed(435)
results- array(sample(1:400,120,replace=TRUE),dim=c(10,3,4))
res - sapply(seq(dim(results)[1]),function(i) {x1 - results[i,,]; x2
- results; sapply(seq(dim(x2)[1]),function(i) emd2d(x1,x2[i,,]))})
dim(res)
#[1] 10 10
A.K.
On Thursday, January 9, 2014 3:25 PM, alex padron [hidden email]
wrote:
I'll try to be clearer. in your example we have: results-
array(1:120,dim=c(10,3,4))
I want to do the following: compare results[1,,] with every matrix
inside results. I then want to jump to results[2,,] and compare it to all
of the other 10 matrices inside results and so on. so emd2d from
the emdist package outputs a single value when comparing matrices and since
your example has 10 matrices who are all being compared, the output should
be 100 values.
Does that make sense?
-Alex
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Re: [R] Levelplots with non-continuous x-axis breaks
Sorry for bugging you again. I was wondering whether it is possible to include
multiple different colors instead of just two.
In the example below, the color.scale() function goes from yellow to red, but
never passes through plain white. Would it be possible to tweak the function or
would I have to start all over again?
Thanks for any tips.
patrick
On Jan 9, 2014, at 7:54, Pachapep pacha...@gmail.com wrote:
Great, it works perfectly! Thanks so much for the awesome help!
(Happy) patrick
On Jan 9, 2014, at 7:07, Adams, Jean jvad...@usgs.gov wrote:
Patrick,
No error in your code, just two different ways of deriving a range of colors
... heat.colors() and color.scale(). I modified the code to just use
color.scale(). You can tell it what two colors you want it to use for the
extremes, and it will work out the shades in between.
Jean
# fake data
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1), runif(6,
0, 2), runif(6, 0, 2))
mymatrix - fake
dimnames(mymatrix)[[2]] - letters[1:dim(fake)[2]]
# separate the first two columns from the remaining columns for simplicity
mymatrix12 - mymatrix[, 1:2]
mymatrix3plus - mymatrix[, -(1:2)]
# create variables for the dimensions for easy reference later
nrows - dim(mymatrix3plus)[1]
ncols - dim(mymatrix3plus)[2]
# to make sure that the min and max contain all of the data
# use floor(100*...)/100 and ceiling(100*...)/100 instead of round(..., 2)
datalim - c(floor(100*min(mymatrix3plus)),
ceiling(100*max(mymatrix3plus)))/100
# choose two colors to be the end points
colorlim - c(yellow, red)
# create a sequence of 10 colors across the selected range for the legend
lut - color.scale(seq(datalim[1], datalim[2], length=10),
extremes=colorlim, xrange=datalim)
# it's not clear from the help file, but the extremes= argument wants a
range (a vector of length 2)
mycolors - array(color.scale(mymatrix3plus, extremes=colorlim,
xrange=datalim), dim=dim(mymatrix3plus))
ycenter - seq(from=0, to=1, length=(nrows + 2))
yheight - 1/(nrows + 2)
# Plot the color key on the lower part
ColorBar(lut=lut, min=datalim[1], max=datalim[2])
# Plot the actual heatmap
par(new=TRUE)
# can't use title for main= argument, because title only exists inside the
ColorBar() function
plot(c(mymatrix12[1, 1], mymatrix12[nrows, 2]), c(0, 1), col=White,
xlab=, ylab=, main=SOMETHING, axes=F)
for(i in 1:ncols){
RectPlotter(mymatrix12, ycenter[nrows - i + 2], yheight, mycolors[, i])
text(mymatrix12[1,1], ycenter[nrows - i + 2],
colnames(mymatrix3plus)[i], pos=4)
}
On Wed, Jan 8, 2014 at 3:59 PM, Pachapep pacha...@gmail.com wrote:
Hi Jean,
Thanks for the great help. Indeed, it seems that that helps a bit. However,
I included a control column made of ones.
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
runif(6, 0, 2), runif(6, 0, 2), runif(6, 0, 2))
However, the color doesnt correspond to that in the key (it is red-ish in
the plot while the 1 corresponds to yellow-ish in the key), so Im not sure
whether it is the key or the colors that are messed up. Sorry for all these
questions.. and thanks (again) for the help provided.
On Jan 8, 2014, at 16:15, Adams, Jean jvad...@usgs.gov wrote:
Patrick,
Thanks for providing reproducible code!
I think the main problem was that the extremes= argument in the
color.scale() function wants a range (a vector of length 2), and you were
providing with more than that, length(lut) is 10.
In the process of tracking this down, I made a bunch of minor changes to
your code to help me see what was going on. This is what I ended up with.
Jean
# fake data
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
runif(6, 0, 2), runif(6, 0, 2), runif(6, 0, 2))
mymatrix - fake
dimnames(mymatrix)[[2]] - letters[1:5]
# separate the first two columns from the remaining columns for simplicity
mymatrix12 - mymatrix[, 1:2]
mymatrix3plus - mymatrix[, -(1:2)]
# create variables for the dimensions for easy reference later
nrows - dim(mymatrix3plus)[1]
ncols - dim(mymatrix3plus)[2]
# prepare plotting parameters
lut - rev(heat.colors(10))
# to make sure that the min and max contain all of the data
# use floor(100*...)/100 and ceiling(100*...)/100 instead of round(..., 2)
minimum - floor(100*min(mymatrix3plus))/100
maximum - ceiling(100*max(mymatrix3plus))/100
# it's not clear from the help file, but the extremes= argument wants a
range (a vector of length 2)
mycolors - color.scale(mymatrix3plus, extremes=lut[c(1, length(lut))])
ycenter - seq(from=0, to=1, length=(nrows + 2))
yheight - 1/(nrows + 2)
# Plot the color key on the lower part
ColorBar(lut=lut, min=minimum, max=maximum)
# Plot the actual heatmap
par(new=TRUE)
# can't use title for main= argument, because title only exists inside the
ColorBar() function
plot(c(mymatrix12[1, 1], mymatrix12[nrows,
Re: [R] How do you transform a dataframe to a corpus?
HI, I couldn't reproduce the error. complaints - data.frame(text=c(Do your homework, Provide reproducible example,Read the R-help manual)) library(tm) myCorpus - Corpus(VectorSource(as.character(complaints$text))) myCorpus #A corpus with 3 text documents A.K. On Friday, January 10, 2014 10:57 AM, ramoss ramine.mossad...@finra.org wrote: Hi; I have a data frame complains w/ dimensions 1133529 1 ( 1.13m obs 1 col) I am trying to transform it into a corpus using the following code: myCorpus -Corpus(VectorSource(complaints$text)) Error in .Source(readPlain, encoding, length(x), FALSE, names(x), 0, TRUE, : vectorized sources must have positive length Does anybody understand the error message? Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-do-you-transform-a-dataframe-to-a-corpus-tp4683396.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Time Series STL
Hi, I'm relatively new to R and need to some help in converting some of my data into a ts object that can then be used to run an STL analysis. My initial input file is a .csv output from some atmospheric measurement instruments. The first column is the year fraction and extends over 2 years with a new data point being collected roughly every 2 minutes. Lots of data are recorded in the subsequent columns (such as met info) and concentrations of particular atmospheric species (which I want to run STL on). Here's my initial code: *#Read in file as an object, defined BHD from original file* *BHD - read.csv(C:/Work/PhD/R_Directory/BHD_2009_2012_all_samp_cal.csv, header=T, dec=., sep=,)* *#Define Columns from original files (comment out if using IDL output)* *yearfrac - BHD[,1]* *APO - BHD[,30]* *O2 - BHD [,29]* *CO2 - BHD[,28]* (end of code) What I want to do is take the objects defined above (APO, CO2, O2) and perform stl analyses on them, but I need to convert them to time series objects first based on the information in the yearfrac object. How do I go about this?! Thanks in advance Barney -- Thomas Barningham Centre for Ocean and Atmospheric Sciences School of Environmental Sciences University of East Anglia Norwich Research Park Norwich NR4 7TJ [[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] Levelplots with non-continuous x-axis breaks
ok, I think I got it. Here is the code I used. Thanks for the help guys..
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
c(-10,-10,-10,-10,-10,-10), c(10,10,10,10,10,10), runif(6, 0, 2), runif(6, 0,
2))
datalim - c(-0,10)
matrix3plus - fake[,-(1:2)]
dimnames(fake)[[2]] - letters[1:dim(fake)[2]]
rbPal - colorRampPalette(c('blue', 'white','red'))
gradcol - rbPal(100)
mycolors - cbind(fake[,1:2], array(rbPal(100)[as.numeric(cut(matrix3plus,
breaks=100))], dim=dim(matrix3plus)))
# Plot the color key on the lower part
ColorBar(lut=gradcol, min=datalim[1], max=datalim[2])
par(new=TRUE)
plot(c(fake[1,1],fake[length(fake[,1]),2]), c(0,1), col=White, xlab=,
ylab=, main=title, axes=F)
yheight - 1/length(ycenter)
ycenter - seq(from=0, to=1, length=(length[fake[1,]]))
trackheight - ycenter[2]
for(i in 3:length(fake[1,])){
RectPlotter(matrixval=fake[,c(1,2,i)],
ycoord=ycenter[length(ycenter)-i+2-emptylines], trackheight=ycenter[2],
mycolors=mycolors[,i])
text(fake[1,1], ycenter[length(ycenter)-i+2-emptylines], colnames(fake)[i],
cex=cexsize, pos=4)
}
On Jan 10, 2014, at 15:47, Pachapep pacha...@gmail.com wrote:
Sorry for bugging you again. I was wondering whether it is possible to
include multiple different colors instead of just two.
In the example below, the color.scale() function goes from yellow to red, but
never passes through plain white. Would it be possible to tweak the function
or would I have to start all over again?
Thanks for any tips.
patrick
On Jan 9, 2014, at 7:54, Pachapep pacha...@gmail.com wrote:
Great, it works perfectly! Thanks so much for the awesome help!
(Happy) patrick
On Jan 9, 2014, at 7:07, Adams, Jean jvad...@usgs.gov wrote:
Patrick,
No error in your code, just two different ways of deriving a range of
colors ... heat.colors() and color.scale(). I modified the code to just
use color.scale(). You can tell it what two colors you want it to use for
the extremes, and it will work out the shades in between.
Jean
# fake data
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
runif(6, 0, 2), runif(6, 0, 2))
mymatrix - fake
dimnames(mymatrix)[[2]] - letters[1:dim(fake)[2]]
# separate the first two columns from the remaining columns for simplicity
mymatrix12 - mymatrix[, 1:2]
mymatrix3plus - mymatrix[, -(1:2)]
# create variables for the dimensions for easy reference later
nrows - dim(mymatrix3plus)[1]
ncols - dim(mymatrix3plus)[2]
# to make sure that the min and max contain all of the data
# use floor(100*...)/100 and ceiling(100*...)/100 instead of round(..., 2)
datalim - c(floor(100*min(mymatrix3plus)),
ceiling(100*max(mymatrix3plus)))/100
# choose two colors to be the end points
colorlim - c(yellow, red)
# create a sequence of 10 colors across the selected range for the legend
lut - color.scale(seq(datalim[1], datalim[2], length=10),
extremes=colorlim, xrange=datalim)
# it's not clear from the help file, but the extremes= argument wants a
range (a vector of length 2)
mycolors - array(color.scale(mymatrix3plus, extremes=colorlim,
xrange=datalim), dim=dim(mymatrix3plus))
ycenter - seq(from=0, to=1, length=(nrows + 2))
yheight - 1/(nrows + 2)
# Plot the color key on the lower part
ColorBar(lut=lut, min=datalim[1], max=datalim[2])
# Plot the actual heatmap
par(new=TRUE)
# can't use title for main= argument, because title only exists inside the
ColorBar() function
plot(c(mymatrix12[1, 1], mymatrix12[nrows, 2]), c(0, 1), col=White,
xlab=, ylab=, main=SOMETHING, axes=F)
for(i in 1:ncols){
RectPlotter(mymatrix12, ycenter[nrows - i + 2], yheight, mycolors[, i])
text(mymatrix12[1,1], ycenter[nrows - i + 2],
colnames(mymatrix3plus)[i], pos=4)
}
On Wed, Jan 8, 2014 at 3:59 PM, Pachapep pacha...@gmail.com wrote:
Hi Jean,
Thanks for the great help. Indeed, it seems that that helps a bit. However,
I included a control column made of ones.
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
runif(6, 0, 2), runif(6, 0, 2), runif(6, 0, 2))
However, the color doesnt correspond to that in the key (it is red-ish in
the plot while the 1 corresponds to yellow-ish in the key), so Im not sure
whether it is the key or the colors that are messed up. Sorry for all these
questions.. and thanks (again) for the help provided.
On Jan 8, 2014, at 16:15, Adams, Jean jvad...@usgs.gov wrote:
Patrick,
Thanks for providing reproducible code!
I think the main problem was that the extremes= argument in the
color.scale() function wants a range (a vector of length 2), and you were
providing with more than that, length(lut) is 10.
In the process of tracking this down, I made a bunch of minor changes to
your code to help me see what was going on. This is what I ended up with.
Jean
# fake data
fake - cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
runif(6, 0, 2),
Re: [R] embedd fonts in pdf generated by R
Hi Adel, You may have a look at the *showtext *package (on CRAN and https://github.com/yixuan/showtext), which could convert text into lines and curves in the pdf file, so it doesn't rely on those fonts after the file is created. Best, Yixuan 2014/1/10 Adel ESSAFI adeless...@gmail.com 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. -- Yixuan Qiu yixuan@cos.name Department of Statistics, Purdue University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with barplot
Hi I’m using bar plot function And I have in my data some zero, but my data are numerics and when i try to use bar plot i have a warning message of « NA was introduced during the automatic conversion » and i can’t see my bar plot. So if somebody have an idea for that? Thanks clémence __ 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] Find the package of a class given classname
How to find the package of a class given classname? For example, there is a class called GAlignments, I want to do something like attr(GAlignments, package) that gives you the package where the class is defined? But of course, attr(GAlignments, package) won't work... Thanks for any help! Yuan [[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: Need to delete my messages in Nabble
Hi, I deleted my account. I need to delete all of messages started by me in here https://stat.ethz.ch/pipermail/r-help/2009-December/221628.html http://comments.gmane.org/gmane.comp.lang.r.general/172827 Can you please guide how to do so? thanks, Sri. [[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] Fwd: Need to delete my messages in Nabble
On 11/01/14 16:54, srinivasa babu wrote: Hi, I deleted my account. I need to delete all of messages started by me in here https://stat.ethz.ch/pipermail/r-help/2009-December/221628.html http://comments.gmane.org/gmane.comp.lang.r.general/172827 Can you please guide how to do so? You can't. Nor should you. It would be unethical. Think before you post, because once you've posted, that's it. The moving finger writes, and having writ/Moves on, nor all thy piety nor wit/Shall lure it back to cancel half a line/Nor all thy tears wash out a word of it. cheers, Rolf Turner __ 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] Find the package of a class given classname
This probably doesn't answer your question, but, amazingly enough, try googling the class name. It worked for GAlignmnets. Incidentally, there's no reason class names have to be unique among R packages.. And they can be hidden in namespaces. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, Jan 10, 2014 at 6:49 PM, Yuan Luo yuan.hypnos@gmail.com wrote: How to find the package of a class given classname? For example, there is a class called GAlignments, I want to do something like attr(GAlignments, package) that gives you the package where the class is defined? But of course, attr(GAlignments, package) won't work... Thanks for any help! Yuan [[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.
Re: [R] Find the package of a class given classname
Hi Yuan, On 01/10/2014 06:49 PM, Yuan Luo wrote: How to find the package of a class given classname? For example, there is a class called GAlignments, I want to do something like attr(GAlignments, package) that gives you the package where the class is defined? But of course, attr(GAlignments, package) won't work... You didn't say whether you wanted to be able to do this programmatically. If that's the case maybe you're lucky and the package where the class is defined is already attached to your session. In that case: attr(class(new(GAlignments)), package) [1] GenomicRanges Otherwise, if the package in question is installed but not attached you can still do: ??`GAlignments-class` If the package is not installed but you have a vague idea that this could be a Bioconductor container, then search the bioconductor mailing list: http://bioconductor.org/help/mailing-list/ If that still doesn't give you the answer, then you're welcome to ask on the list. Cheers, H. PS: If you're using the current devel version of BioC (requires R devel) the GAlignments class has moved to the new GenomicAlignments package. Thanks for any help! Yuan [[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. -- Hervé Pagès Program in Computational Biology Division of Public Health Sciences Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N, M1-B514 P.O. Box 19024 Seattle, WA 98109-1024 E-mail: hpa...@fhcrc.org Phone: (206) 667-5791 Fax:(206) 667-1319 __ 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] How do you transform a dataframe to a corpus?
Not much. Columns are not normally measured in bytes. --- 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. ramoss ramine.mossad...@finra.org wrote: The column length is 4000 bytes long if that helps. -- View this message in context: http://r.789695.n4.nabble.com/How-do-you-transform-a-dataframe-to-a-corpus-tp4683396p4683402.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time Series STL
This might not seem helpful, but there really is a communication gap here. You need to read the Posting Guide, post in plain text so your example code is not mangled by the HTML, and include a bit of data that is representative of the data you are working with. You may find some helpful advice here: http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example, including how to use the dput function. --- 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. Thomas Barningham stbarning...@gmail.com wrote: Hi, I'm relatively new to R and need to some help in converting some of my data into a ts object that can then be used to run an STL analysis. My initial input file is a .csv output from some atmospheric measurement instruments. The first column is the year fraction and extends over 2 years with a new data point being collected roughly every 2 minutes. Lots of data are recorded in the subsequent columns (such as met info) and concentrations of particular atmospheric species (which I want to run STL on). Here's my initial code: *#Read in file as an object, defined BHD from original file* *BHD - read.csv(C:/Work/PhD/R_Directory/BHD_2009_2012_all_samp_cal.csv, header=T, dec=., sep=,)* *#Define Columns from original files (comment out if using IDL output)* *yearfrac - BHD[,1]* *APO - BHD[,30]* *O2 - BHD [,29]* *CO2 - BHD[,28]* (end of code) What I want to do is take the objects defined above (APO, CO2, O2) and perform stl analyses on them, but I need to convert them to time series objects first based on the information in the yearfrac object. How do I go about this?! Thanks in advance Barney __ 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.
