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...data points. e.g. Remove the data points with the 20 lowest density values. Or remove the data points with the lowest 5% of density values. I'll let you decide whether that is a good idea or a bad idea. And if it's a good idea, then how much to trim. On Sat, Oct 10, 2020 at 5:47 AM Ana Marija <sokovic.anamarija at gmail.com> wrote: > > Hi Bert, > > Another confrontational response from you... > > You might have noticed that I use the word "outlier" carefully in this > post and only in relation to the plotted ellipses. I do not know the > underlyin...

...ata points with the 20 lowest density values. > Or remove the data points with the lowest 5% of density values. > > I'll let you decide whether that is a good idea or a bad idea. > And if it's a good idea, then how much to trim. > > > On Sat, Oct 10, 2020 at 5:47 AM Ana Marija <sokovic.anamarija at gmail.com> wrote: > > > > Hi Bert, > > > > Another confrontational response from you... > > > > You might have noticed that I use the word "outlier" carefully in this > > post and only in relation to the plotted ellipse...

Hello, I have a data frame with one column: > remove V1 1 ABAFT_g_4RWG569_BI_SNP_A10_35096 2 ABAFT_g_4RWG569_BI_SNP_B12_35130 3 ABAFT_g_4RWG569_BI_SNP_E09_35088 4 ABAFT_g_4RWG569_BI_SNP_E12_35136 5 ABAFT_g_4RWG569_BI_SNP_F11_35122 6 ABAFT_g_4RWG569_BI_SNP_F12_35138 7 ABAFT_g_4RWG569_BI_SNP_G07_35060 8 ABAFT_g_4RWG569_BI_SNP_G12_35140 I want to remove these 8

...provide. That's why I believe you need local expertise. Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Fri, Oct 9, 2020 at 8:25 AM Ana Marija <sokovic.anamarija at gmail.com> wrote: > Hi Abby, > > thank you for getting back to me and for this useful information. > > I'm trying to detect the outliers in my distribution based of mean and > variance. Can I see that from the plot I provided? Would outliers be &gt...

...ou need local expertise. > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Fri, Oct 9, 2020 at 8:25 AM Ana Marija <sokovic.anamarija at gmail.com> wrote: >> >> Hi Abby, >> >> thank you for getting back to me and for this useful information. >> >> I'm trying to detect the outliers in my distribution based of mean and >> variance. Can I see that from the plot I...

On Wed, 21 Oct 2020 16:15:22 -0500 Ana Marija <sokovic.anamarija at gmail.com> wrote: > Hello, > > I have a data frame with one column: > > > remove > > V1 > > 1 ABAFT_g_4RWG569_BI_SNP_A10_35096 > 2 ABAFT_g_4RWG569_BI_SNP_B12_35130 > 3 ABAFT_g_4RWG569_BI_SNP_E09_350...

Hello, This is probably because basename keeps the file extension, try instead filename <- sub("(^[^\\.]*)\\..+$", "\\1", basename(celFiles)) celFiles[!filename %in% as.character(remove$V1)] Hope this helps, Rui Barradas ?s 22:15 de 21/10/20, Ana Marija escreveu: > Hello, > > I have a data frame with one column: > >> remove > > V1 > > 1 ABAFT_g_4RWG569_BI_SNP_A10_35096 > 2 ABAFT_g_4RWG569_BI_SNP_B12_35130 > 3 ABAFT_g_4RWG569_BI_SNP_E09_35088 > 4 ABAFT_g_4RWG569_BI_SNP_E12_...

...basename keeps the file extension, try instead > > > filename <- sub("(^[^\\.]*)\\..+$", "\\1", basename(celFiles)) > celFiles[!filename %in% as.character(remove$V1)] > > > Hope this helps, > > Rui Barradas > > ?s 22:15 de 21/10/20, Ana Marija escreveu: >> Hello, >> >> I have a data frame with one column: >> >>> remove >> >> ???????????????????????????????? V1 >> >> 1 ABAFT_g_4RWG569_BI_SNP_A10_35096 >> 2 ABAFT_g_4RWG569_BI_SNP_B12_35130 >> 3 ABAFT_g_4RWG569_BI_SNP_E09_...

...charts. > > Do not use printouts of your data since it hides important information. > Use str(a11) and dput(a11) or dput(head(a11)) to provide useful information > about your data. > > David L Carlson > Texas A&M University > > > On Tue, Jun 13, 2023 at 4:08?PM Ana Marija <sokovic.anamarija at gmail.com> > wrote: > >> Hello, I have a data frame like this: d11=suppressWarnings(read. >> csv("/Users/anamaria/Downloads/B1. csv", stringsAsFactors=FALSE, >> header=TRUE)) > d11 X Domain. decomp. DD. com. . load Neighbor. search &g...

...kow.com/notes/ggplot2-color-by-density/ and then do something like this: a=SNP[SNP$density>400,] and plot it again: p <- ggplot(a, mapping = aes(x = mean, y = var)) p <- p + geom_density_2d() + geom_point() + my.theme + ggtitle("SNPS_red") On Thu, Oct 8, 2020 at 3:52 PM Ana Marija <sokovic.anamarija at gmail.com> wrote: > > Hello, > > I have a data frame like this: > > > head(SNP) > mean var sd > FQC.10090295 0.0327 0.002678 0.0517 > FQC.10119363 0.0220 0.000978 0.0313 > FQC.10132112 0.0275 0.002088 0.0457 > F...

...e agent be defined in more than one group? Example: group=1 ; queue1 agent => 401,401,Tomislav Parcina agent => 402,402,Katarina Ivanisevic agent => 403,403,Sasa Juginovic group=2 ; queue2 agent => 401,401,Tomislav Parcina agent => 402,402,Katarina Ivanisevic agent => 404,404,Marija Bilic agent => 405,405,Ana Kaliterna Will this work? Will agents 401 and 402 be in both groups? If I join every group to another queue, will one agent be in both queue's? -- Tomislav Par?ina Lama Computers Split Stinice 12, 21000 Split Tel.: +385(21)393447 e-mail: tparcina#lama.hr http:...

Hello, I have a data frame like this: > head(SNP) mean var sd FQC.10090295 0.0327 0.002678 0.0517 FQC.10119363 0.0220 0.000978 0.0313 FQC.10132112 0.0275 0.002088 0.0457 FQC.10201128 0.0169 0.000289 0.0170 FQC.10208432 0.0443 0.004081 0.0639 FQC.10218466 0.0116 0.000131 0.0115 ... and I am creating plot like this: s <- ggplot(SNP, mapping = aes(x = mean, y = var))

Hello, I have a data frame like this: d11=suppressWarnings(read.csv("/Users/anamaria/Downloads/B1.csv", stringsAsFactors=FALSE, header=TRUE)) > d11 X Domain.decomp. DD.com..load Neighbor.search Launch.PP.GPU.ops. Comm..coord. 1 SYCL 2. 1 0 3.7 0. 1 1 .6 2 CUDA 2 0 3. 1 0 1 .0

Hi Abby, thank you for getting back to me and for this useful information. I'm trying to detect the outliers in my distribution based of mean and variance. Can I see that from the plot I provided? Would outliers be outside of ellipses? If so how do I extract those from my data frame, based on which parameter? So I am trying to connect outliers based on what the plot is showing: s <-

I have code like this: data <- read.csv("test1.csv", stringsAsFactors=FALSE, header=TRUE) # Graph myplot=ggplot(data, aes(fill=condition, y=value, x=condition)) + geom_bar(position="dodge", stat="identity", width=0.5) + scale_fill_manual(values=c("#7b3294", "#c2a5cf", "#a6dba0", "#008837"))+

> My understanding is that this represents bivariate normal > approximation of the data which uses the kernel density function to > test for inclusion within a level set. (please correct me) You can fit a bivariate normal distribution by computing five parameters. Two means, two standard deviations (or two variances) and one correlation (or covariance) coefficient. The bivariate normal

Hi, I have a data frame like this: > dput(df) structure(list(ID = 1:8, Type = c("gmx mdrun -ntmpi 8 -ntomp 1 -s benchPEP.tpr -nsteps 10000 -resethway", "gmx mdrun -ntmpi 8 -ntomp 1 -s benchPEP.tpr -nsteps 10000 -resethway", "gmx mdrun -ntmpi 8 -s benchPEP.tpr -nsteps 4000 -resetstep 3000", "gmx mdrun -ntmpi 8 -s benchPEP.tpr -nsteps 4000 -resetstep