search for: unqs

Hi R users, I still have the problem about plotting. I wanted to put the datasets on one figure, x-axis represents values B, y-axis represents values C, while different colors label column A. Each record uses a circle on the figure, while hollow circles represent DF=1 and solid circles represent DF=2. I put my code below, but the A labels do not correspond to the true record, so I don't know

Hi Jim, For true color, I meant that the points in the figure do not correspond to the values from the dataframe. Also, why to use rainbow(9) here? And the legend is straight in the middle, is it possible to reformat it to the very bottom? Thanks again. On Sun, Jul 16, 2017 at 2:50 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi lily, > As I have no idea of what the "true

Hi lily, As I have no idea of what the "true record" is, I can only guess. Maybe this will help: # get some fairly distinct colors rainbow_colors<-rainbow(9) # this should sort the numbers in dfm$A dfm$Acolor<-factor(dfm$A) plot(dfm$B,dfm$C,pch=ifelse(dfm$DF==1,1,19), col=rainbow_colors[as.numeric(dfm$Acolor)]) legend("bottom",legend=sort(unique(dfm$A)),

For more than 10 records, how to reformat the colors? Also, how to show the first legend only, but at the bottom, while the second legend in your code is not necessary? In all, the same A values have the same color, but different symbols in DF==1 and DF==2. Thanks for your help. On Sun, Jul 16, 2017 at 9:28 AM, lily li <chocold12 at gmail.com> wrote: > Hi Jim, > > For true color,

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 I have a rectangular matrix of size 920 by 85. I'd like to coerce it into a square matrix such that all row/col names are present in the new matrix and the additional values are zero. As an example: A B C D A 1 2 3 4 E 5 6 7 8 F 9 10 11 12 Would be coerced to: A B C D E F A 1 2 3 4 5 9 B 2 0 0 0 6 10 C 3 0 0

Dear All - The problem: comment lines in an R function (lines beginning with # ) are *sometimes* removed on leaving the R default editor (same with notepad). I'm working on a Windows machine with R version 2.14.1. An example is below. Couldn't find anything that seemed to relate to this in the Changelog. I don't recall encountering this behavior with previous versions of R. Any

I have a binary matrix of size N x 300. I then create the following: > set.seed(1234) > (key_file <- sample(letters[1:4], 300, replace=TRUE)) [1] "a" "c" "c" "c" "d" "c" "a" "a" "c" "c" "c" "c" "b" "d" "b" "d" "b"

Hi, I searched the archives and did not find a good solution to that. assume I have 10 sets and I want to have the common character elements of them. how could i do that? -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

folks, if i have three matrices, a, b, cc with some colnames in common, and i want to create a matrix which consists of the common columns added up, and the other columns tacked on, what's a good way to do it? i've got the following roundabout code for two matrices, but if the number of matrices increases, then i'm a bit stymied. > a <- matrix(1:20,ncol=4); colnames(a) <-

...ling(length(data$x)/4)),each=2) b=rep(c(0,1),ceiling(length(data$x)/4)) rk.up=c(1,(a*4+b))[1:ceiling(length(data$x)/2)] rk.down=rev(c(a*4+b-2)[1:floor(length(data$x)/2)]) cat("Performing Siegel-Tukey rank transformation...","\n","\n") rks=c(rk.up,rk.down) unqs=unique(sort(data$x)) corr.rks=tapply(rks,data$x,mean) cbind(unqs,corr.rks) rks.data=data.frame(unqs,corr.rks) names(rks.data)=c("unique values of x","tie-adjusted Siegel-Tukey rank") print(rks.data,row.names=F) names(rks.data)=c("unqs","corr.rks")...

If you want colors mapped to the _values_ in DF1$C, there are a number of ways to do it: Color_unq<-color.scale(DF1$C,c(1,0),c(0,0,c(0,1)) This will produce colors from the lowest values (red) through the highest (blue). See the help page for color.scale to get different colors. With this you can use color.legend to add a mapping of the values and colors. If you just want different colors,

Hi, I am just wondering if there is an easy way to count in a numeric vector how many numbers don't have replicates. For example, a=c(1,1,2,2,3,4,5), how can I know there are three numbers (3, 4 and 5) without replicates? Thank you! Jun =====

Hi Neil / Thorsten, The full test result is attached. Thanks and Regards, Anil ------------------------------------------------------------------------------------------------------------------------------- root at genericarmv8:~/anilss/AArch64/mksh_44-1/mksh# ./mksh test.sh -p $PWD/mksh Testing mksh for conformance: # $MirOS: src/bin/mksh/check.t,v 1.599 2013/02/24 14:22:41 tg Exp $

On Wed, 9 Oct 2013 10:44:27 +0000 (UTC) Thorsten Glaser <tg at mirbsd.de> wrote: > Anil Singhar dixit: > > >Manual testing as provided within the package has been done with all > >tests passing. This includes the units tests available under > >usr/klibc/tests, usr/utils, usr/dash and usr/gzip. For dash and > >gzip, only sanity testing has been done. > >

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Dear Everyone, The document of `tsdiag? says > These tests are sometimes applied to the residuals from an ARMA(p, q) fit, in which case the references suggest a better approximation to the null-hypothesis distribution is obtained by setting fitdf = p+q, provided of course that lag > fitdf. This implies that we should pass the `fitdf' parameter when applying `Box.test' to