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", "#0088
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 1 -resethway",
"gmx mdrun -ntmpi 8 -ntomp 1 -s benchPEP.tpr -nsteps 1 -resethway",
"gmx mdrun -ntmpi 8 -s benchPEP.tpr -nsteps 4000 -resetstep 3000",
"gmx m
categorical data, but not continuous, numeric data which are better
> handled with box plots or strip 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.
>
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. 10 3.7
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 Thu, Oct 14, 2021 at 10:10 AM Ana Marija
> wrote:
>
>> Hi All,
>>
>> I hav
Hi All,
I have a data frame like this:
> head(b)
LRET02LRET04LRET06LRET08LRET10LRET12LRET14
1 0 0.6931472 . 1.0986123 1.0986123 1.0986123 0.6931472
2 2.1972246 2.4849066 2.4849066 . 2.5649494 2.6390573 2.6390573
3 1.6094379 1.7917595 1.6094379
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Fri, Sep 17, 2021 at 12:22 PM Ana Marija
> wrote:
> >
> > Hi All,
> >
> > I plan to identify metabolite levels that differ between individuals
> > with various retinopa
Hi All,
I plan to identify metabolite levels that differ between individuals
with various retinopathy outcomes (DR or noDR). I plan to model
metabolite levels using linear mixed models ref as implemented in
lmm2met software. The model covariates will include: age, sex, SV1,
SV, and disease_conditi
)
> #[1] 145092258
>
> df[14509227,] # beyond nrow(df) by 2
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 15:12 de 16/09/21, Ana Marija escreveu:
> > Hi All,
> >
> > I have lines in file that look like this:
> >
> >> df[1450922
Hi All,
I have lines in file that look like this:
> df[14509227,]
SNP A1 A2 freq b se p N
1: NA NA NA NA NA
data looks like this:
> head(df)
SNP A1 A2 freq b se p N
1: rs74337086 G A 0.0024460 0.1627 0.1231 0.1865 218792
2: rs76388980 G
Hello,
I have this code, and when I run it:
> kbpowerf()
Error in n * rvec : non-numeric argument to binary operator
this is the code:
function (){
#USER SPECIFICATION PORTION
alpha=0.05 #DESIGNATED ALPHA
g=3 #NUMBER OF GROUPS
nvec=c(25,10,15) #GROUP SIZES
beta1vec=c(789.93,122.87,1871
Hello,
I am in process of writing a grant where I am explaining my planned
methylation analysis using R software "minfi". In the text of the
grant I am mentioning looking for samples containing outliers in the
multi-dimensional scaling (MDS) plot
https://rdrr.io/bioc/minfi/man/mdsPlot.html . My qu
rsplit(names, ":")[-2]]
out <- data[, .(rsid, ref_allele, eff_allele)][,
WGT := files[i]][]
}
return(out)
rm(data)
gc()
}
parallel::stopCluster(cl)
big_data <- rbindlist(lst_out, fill = TRUE)
On Wed, Dec 16, 2020 at 9:31 AM Ana Marija wrote:
>
assuming that the filename
> # is stored in files[i]
> files<-"retina.ENSG0135776.wgt.RDat"
> i<-1
> WGT<-rep(files[i],length(rsid))
> data<-data.frame(rsid=rsid,weight=a$top1,
> ref_allele=ref_allele,eff_allele,WGT=WGT)
> data
>
> Note that the
e and see if it contains a
> column named "blup" or just the values that were extracted from
> a$blup. Also, I assume that weight=blup looks for an object named
> "blup", which may not be there.
>
> Jim
>
> On Wed, Dec 16, 2020 at 1:20 PM Ana Marija
>
much experience with data tables I may be
> wrong, but I suspect that the column name "blup" may not be visible or
> even present in "data". I don't see it in "dd" above this code
> fragment.
>
> Jim
>
> On Wed, Dec 16, 2020 at 11:12 AM Ana Mar
Hello,
I made a terribly inefficient code which runs forever but it does run.
library(dplyr)
library(splitstackshape)
datalist = list()
files <- list.files("/WEIGHTS1/Retina", pattern=".RDat", ignore.case=T)
for(i in files)
{
a<-get(load(i))
names <- rownames(a)
data <- as.data.frame(cbind(name
lp page, ?corrplot .
>
> --
>
> David.
>
> On 11/6/20 6:08 AM, Ana Marija wrote:
>
> sorry forgot to attach the plot.
>
> On Fri, Nov 6, 2020 at 8:07 AM Ana Marija wrote:
>
> Hello
>
> I have data like this:
>
> head(my_data)
>
> subjects DIABDUR HB
sorry forgot to attach the plot.
On Fri, Nov 6, 2020 at 8:07 AM Ana Marija wrote:
>
> Hello
>
> I have data like this:
>
> > head(my_data)
> subjects DIABDUR HBA1C ESRD SEX AGE PHENO C1 C2
> 1 fam0110_G110 38 9.41 2 51 2 -
Hello
I have data like this:
> head(my_data)
subjects DIABDUR HBA1C ESRD SEX AGE PHENO C1 C2
1 fam0110_G110 38 9.41 2 51 2 -0.01144980 0.002661140
2 fam0113_G113 30 12.51 2 40 2 -0.00502052 -0.000929061
3 fam0114_G114 23 8.42
Makes sense, thank you!
On Wed, 21 Oct 2020 at 17:46, Rolf Turner wrote:
>
> On Wed, 21 Oct 2020 16:15:22 -0500
> Ana Marija wrote:
>
> > Hello,
> >
> > I have a data frame with one column:
> >
> > > remove
> >
> >
lename %in% as.character(remove$V1)]
> >
> >
> > Hope this helps,
> >
> > Rui Barradas
> >
> > Às 22:15 de 21/10/20, Ana Marija escreveu:
> >> Hello,
> >>
> >> I have a data frame with one column:
> >&g
Hello,
I have a data frame with one column:
> remove
V1
1 ABAFT_g_4RWG569_BI_SNP_A10_35096
2 ABAFT_g_4RWG569_BI_SNP_B12_35130
3 ABAFT_g_4RWG569_BI_SNP_E09_35088
4 ABAFT_g_4RWG569_BI_SNP_E12_35136
5 ABAFT_g_4RWG569_BI_SNP_F11_35122
6 ABAFT_g_4RWG569_BI_SNP_F12_351
a bad idea.
> And if it's a good idea, then how much to trim.
>
>
> On Sat, Oct 10, 2020 at 5:47 AM Ana Marija
> wrote:
> >
> > Hi Bert,
> >
> > Another confrontational response from you...
> >
> > You might have noticed that I use the wor
an open mind is that people keep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Fri, Oct 9, 2020 at 8:25 AM Ana Marija wrote:
>>
>> Hi Abby,
>>
>> thank you for gettin
Hi Abby,
thank you for getting back to me and for this useful information.
I'm trying to detect the outliers in my distribution based of mean and
variance. Can I see that from the plot I provided? Would outliers be
outside of ellipses? If so how do I extract those from my data frame,
based on whi
NP[SNP$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 wrote:
>
> Hello,
>
> I have a data frame like this:
>
>
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.
4:ncol(mt)) mt[,i] <- 1 + (names(mt)[i]== mt$PLATE)
Thanks!
On Tue, Sep 29, 2020 at 12:08 PM Ana Marija wrote:
>
> HI Bert,
>
> thank you for getting back to me.
> I tried this:
>
> > dat <- cbind(mc, matrix(0,ncol = 34))
> > head(dat)
> FID IID PLATE 1
0117 G117 4RWG569 2 1 1
> 5 fam0118 G118 5XAV049 1 1 2
> 6 fam0119 G119 cherry 1 2 1
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
> -
Hello,
I have a data frame like this:
> head(mc)
FID IID PLATE
1 fam0110 G110 4RWG569
2 fam0113 G113 cherry
3 fam0114 G114 cherry
4 fam0117 G117 4RWG569
5 fam0118 G118 5XAV049
6 fam0119 G119 cherry
...
> dim(mc)
[1] 16254
> length(unique(mc$PLATE))
[1] 34
I am trying to make a ne
t RHS position 1 taken as TRUE when
assigning to type 'logical' (column 6 named 'PHENO')
Please advise,
Ana
On Wed, Sep 23, 2020 at 2:48 PM Jeremie Juste wrote:
>
>
> Hello Ana Marija,
>
> I cannot reproduce your error,
>
> with a$PHENO=ifelse(a$PLASER==2 |a$
, 2020 at 11:43 AM Ana Marija wrote:
>
> Hello,
>
> I have a data frame as shown bellow.
> I want to create a new column PHENO which will be defined as follows:
> if CURRELIG==1 -> PHENO==1
> in the above subset those that have:
> PLASER==2 -> PHENO==2
> and
>
Hello,
I have a data frame as shown bellow.
I want to create a new column PHENO which will be defined as follows:
if CURRELIG==1 -> PHENO==1
in the above subset those that have:
PLASER==2 -> PHENO==2
and
those where RTNPTHY==1 -> PHENO==1
I tried doing this:
a$PHENO=ifelse(a$CURRELIG==1 | a$RTNPT
SE)
> barpos<-barplot(counts~name+CHR,data=d,beside=TRUE,names.arg=rep("",22))
> legend(40,22,c("new","old"),fill=c("gray20","gray80"))
> library(plotrix)
> staxlab(1,at=colMeans(barpos),labels=1:22)
>
> Jim
>
> On Fri,
Hello,
I am trying to overlay two histograms with this:
p <- ggplot(d, aes(CHR, counts, fill = name)) + geom_bar(position = "dodge")
p
but I am getting this error:
Error: stat_count() can only have an x or y aesthetic.
Run `rlang::last_error()` to see where the error occurred.
my data is this:
% had Type 1. (my TD covariate is reference for
the type of diabetes) In the attach is the description of the data.
Cheers,
Ana
On Tue, Sep 15, 2020 at 7:59 PM David Winsemius wrote:
>
>
> On 9/15/20 8:57 AM, Ana Marija wrote:
> > Hi Abby and David,
> >
> > Thanks
.
>
>
> https://groups.google.com/g/plink2-users?pli=1
>
>
> --
>
> David.
>
> On 9/14/20 6:29 AM, Ana Marija wrote:
> > Hello,
> >
> > I was running association analysis using --glm genotypic from:
> > https://www.cog-genomics.org/plink/2.0/assoc w
sorry not replace with NA but with empty string for a name, for example
for example this:
> geneSymbol["Ku8QhfS0n_hIOABXuE"]
Ku8QhfS0n_hIOABXuE
"MACC1"
would go when I subject it to
> geneSymbol["Ku8QhfS0n_hIOABXuE"]
Ku8QhfS0n_hIOABXuE
On Mon,
Hello,
I have a vector like this:
> head(geneSymbol)
Ku8QhfS0n_hIOABXuE Bx496XsFXiAlj.Eaeo W38p0ogk.wIBVRXllY
QIBkqIS9LR5DfTlTS8 BZKiEvS0eQ305U0v34 6TheVd.HiE1UF3lX6g
"MACC1""GGACT" "A4GALT"
"NPSR1-AS1""NPSR1-AS1" "AAAS"
it has around 15000 en
Hello,
I was running association analysis using --glm genotypic from:
https://www.cog-genomics.org/plink/2.0/assoc with these covariates:
sex,age,PC1,PC2,PC3,PC4,PC5,PC6,PC7,PC8,PC9,PC10,TD,array,HBA1C. The
result looks like this:
#CHROMPOSIDREFALTA1TESTOBS_CTB
ed for tutorials also. This list cannot
> substitute for such homework on your own.
>
> 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&qu
It seems that "treatment" and "patient" are just vectors.
> treatment
[1] "treat" "treat" "treat" "treat" "treat" "control" "control"
[8] "control" "control" "cont
1] "treatment"
> str(treatment)
chr [1:10] "treat" "treat" "treat" "treat" "treat" "control" "control" ...
but this is not the format I need.
On Fri, Jul 31, 2020 at 9:18 PM Ana Marija wrote:
>
> Hello,
>
Hello,
I have this file:
> a=load("paired_example.Rdata")
> a
[1] "rawdata" "treatment" "patient"
I can extract "rawdata" with:
dat<-local(get(load("paired_example.Rdata")))
Can you please advise how would I extract in data frame "treatment"
and "patient"?
Thanks
Ana
___
Thank you so much as.character(r) indeed resolved the issue!
On Sat, Jun 20, 2020 at 3:47 AM Ivan Krylov wrote:
>
> On Fri, 19 Jun 2020 19:36:41 -0500
> Ana Marija wrote:
>
> > Error in cat(x, file = file, sep = c(rep.int(sep, ncolumns - 1),
> > "\n"), :
s a foreach package but I try to avoid extras
>
> make your for statements:
>
> for ( a in rownames(f1) ) {
>
> # a will now be a row number rather than the value, so replace ' a ' in
> the paste0 with: f1[ a, 1]
>
> so
>
> ext <- paste0( "/ld/human
SNPs at the time and I need to do that for 300 pairs
On Fri, Jun 19, 2020 at 6:42 PM Rasmus Liland wrote:
>
> On 2020-06-19 14:34 -0500, Ana Marija wrote:
> >
> > I have two files (each has 300 lines)like this:
>
> The example looks quite similar to the R example in
&g
) You may have something that adds it but better to use
> something that works. So try using:
>
> library(readr)
> f1 <- read_tsv("1g.txt", col.names=F)
>
> This will give you a tibble with f1$X1 with the file in it
>
> then loop it with (a in as.list(f1[,1])
&g
not found
Am I am doing here something wrong?
Do I need any other libraries loaded?
Thanks
Ana
On Fri, Jun 19, 2020 at 3:49 PM Rasmus Liland wrote:
>
> On 2020-06-19 14:34 -0500, Ana Marija wrote:
> >
> > server <- "http://rest.ensembl.org";
> > ext &
uot;)
>
> r <- GET(paste(server, ext, sep = ""),
> content_type("application/json"))
>
> # You presumably need to do something with 'r' at the
> moment its over written by the next loop.. were
> #
Hello,
I have two files (each has 300 lines)like this:
head 1g.txt
rs6792369
rs1414517
rs16857712
rs16857703
rs12239392
...
head 1n.txt
rs1042779
rs2360630
rs10753597
rs7549096
rs2343491
...
For each pair of rs# from those two files I can run this command in R
library(httr)
library(jsonlite)
l
te in the presence of ties
as I understand high p-values here say I cannot claim statistical
support for a difference, but low p-values are not evidence of
sameness?
D should be the maximum difference between the two probability distributions?
On Wed, Jun 17, 2020 at 3:06 PM Ana Marija wrote:
>
&g
Hello,
I have p values from two distributions, Pold and Pnew
> head(m)
CHR POS MARKER Pnew Pold
1: 1 785989 rs2980300 0.1419 0.9521
2: 1 1130727 rs10907175 0.1022 0.4750
3: 1 1156131 rs2887286 0.3698 0.5289
4: 1 1158631 rs6603781 0.1929 0.2554
5: 1 1211292 rs6685064 0
t; ypos<--log10(x[,p])
> plot(x[,BP],ypos,ylim=c(0,max(ypos,na.rm=TRUE)*1.05),
> main=main,xlab=xlab,ylab=ylab,
> pch=pch,col=chrcol[x[,CHR]],cex=4,xaxt="n")
> abline(h=-log10(siglevel),lty=2)
> staxlab(1,at=(1:nchr) + 0.5,labels=chrlab)
> sigindx<-which(ypos
Hello,
Is there is a way to set colors in this plot to look like this one in
attach (different color for each CHR-there is 22 of them)?
library(qqman)
results_log <- read.table("meta_p_pos_chr.F", head=TRUE,stringsAsFactors=FALSE)
png("META.png")
manhattan(results_log,chr="CHR",bp="POS",p="META_
a(b$pheno),]$FLASER,NA)
> # use the valid PLASER values when FLASER if NA
> b[is.na(b$pheno),]$pheno<-ifelse(!is.na(b[is.na(b$pheno),]$PLASER),
> b[is.na(b$pheno),]$PLASER,NA)
> b
>
> I could write that mess in one straitjacket of conditional statements
> but my brain hurts eno
iate logic to get rid of only the
> ones you don't want.
>
> Jim
>
> On Sat, Jun 13, 2020 at 12:50 PM Ana Marija
> wrote:
> >
> > Hi Rasmus,
> >
> > thank you for getting back to be, the command your provided seems to
> > add all 11 NAs to 2s
&
; > > On Sat, Jun 13, 2020 at 10:46 AM Ana Marija wrote:
> > > >
> > > > I am trying to make a new column
> > > > "pheno" so that I reduce the number
> > > > of NAs
> > >
> > > it looks like those two NA values in
>
$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |
> is.na(b$PLASER) & b$FLASER == 2,2,1)
>
> and if I have it the wrong way round, swap FLASER and PLASER in the
> bit I have added.
>
> Jim
>
> On Sat, Jun 13, 2020 at 10:46 AM Ana Marija
> wrote:
> >
> > Hell
Hello
I have a data frame like this:
> head(b)
FID IID FLASER PLASER
1: fam1000 G1000 1 1
2: fam1001 G1001 1 1
3: fam1003 G1003 1 2
4: fam1005 G1005 1 1
5: fam1009 G1009 1 1
6: fam1052 G1052 1 1
...
> table(b$PLASER,b$FLASER, e
e shown on the x-axis?
On Thu, Jun 11, 2020 at 4:39 PM wrote:
>
> Your dots are too big!
>
> Add
>
> geom_points(... , size = 1
>
> May need to play... 0.5 or 0.1?
>
> On 11 Jun 2020 22:26, Ana Marija wrote:
>
> I tried it,
> ggplot( data = tmp.tidy) +geom
I tried it,
ggplot( data = tmp.tidy) +geom_point( aes(y = BP,x = CHR,color=key)
,position = "jitter" )
I got the attached
On Thu, Jun 11, 2020 at 4:18 PM wrote:
>
> Try adding
> position = "jitter" to the geom_point(...
>
>
>
> On 11 Jun 2020 21:41, Ana M
Hello,
I tried your code and this is what I got
I really need two groups side by side shown per chromosome as it is here:
https://imgur.com/a/pj40c
on the image there are 4 groups I do have only two
On Thu, Jun 11, 2020 at 11:52 AM wrote:
>
> On 2020-06-11 15:59, Ana Marija wrote:
>
me, and CHR would go from 1 to 22
On Thu, Jun 11, 2020 at 9:26 AM wrote:
>
> On 2020-06-11 14:54, Ana Marija wrote:
> > Hello,
> >
> > I expected it to look like this:
> > https://imgur.com/a/pj40c
> >
>
> Ah - so all on the one plot? - so you don't w
value: num 0.952 0.475 0.529 0.255 0.295 ...
Unfortunately qqman doesn't do this kind of overlay of two plots
On Wed, Jun 10, 2020 at 11:24 PM John wrote:
>
> On Wed, 10 Jun 2020 15:36:11 -0500
> Ana Marija wrote:
>
> > Hello,
> >
> > I have a data frame lik
mon wrote:
>
> Hi Ana,
> The problem may be that the JPEG device doesn't handle transparency.
> Perhaps PNG?
>
> Jim
>
> On Thu, Jun 11, 2020 at 6:48 AM Ana Marija
> wrote:
> >
> > Hello,
> >
> > I have a data frame like this:
> >
>
Hello,
I do have a file like this:
head M3.assoc.logistic.C
CHR SNP BP P
1 1:785989:T:C 785989 0.4544
1 1:785989:T:C 785989 0.689
1 1:1130727:A:C 1130727 0.05068
1 1:1130727:A:C 1130727 0.07381
1 1:1156131:T:C 1156131 0.6008
1 1:1156131:T:C 1156131 0.8685
...
And I don't have any "NA" or "inf" va
ve all rows that have at least one of "E102" or "E112"
>
>
> unwanted <- c("E102", "E112")
> no <- sapply(dat, function(x){
>grepl(paste(unwanted, collapse = "|"), x)
> })
> no <- apply(no, 1, any)
> dat[!no, ]
&
t;
>
>
> 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 Wed, Jun 3, 2020 at 7:56 AM Ana Marija
>
Hello.
I am trying to filter only rows that have ANY of these variables:
E109, E119, E149
so I did:
controls=t %>% filter_all(any_vars(. %in% c("E109", "E119","E149")))
than I checked what I got:
> s0 <- sapply(controls, function(x) grep('^E10', x, value = TRUE))
> d0=unlist(s0)
> d10=unique(d0)
Hi Jim,
not in this case, but thanks for asking!
Ana
On Mon, Jun 1, 2020 at 10:04 PM Jim Lemon wrote:
>
> So recombination sticks out its foot before us. Do you want to account
> for gene linkage?
>
> JIm
>
> On Tue, Jun 2, 2020 at 11:55 AM Ana Marija
> wrote:
>
r","Chr"))
nn2<-merge(nn1,ret1,by=c("Marker","Chr"))
> Marker3<-nn2$Marker
> length(Marker3)
[1] 3742962
> Marker4<-nn1$Marker
> length(Marker4)
[1] 373
On Mon, Jun 1, 2020 at 8:50 PM Ana Marija wrote:
>
> Hi David,
>
> that is a
43494 9
How would I rewrite this code so that is merging by Chr and Marker
column? It seems that a Marker can be under a few Chr.
On Mon, Jun 1, 2020 at 8:41 PM David Winsemius wrote:
>
>
> On 6/1/20 5:40 PM, Ana Marija wrote:
> > Hi Jim,
> >
> > thank you so much
ctors=FALSE)
>
> # merge the three data frames on "Marker"
> nn1<-merge(neu1,nep1,by="Marker")
> nn2<-merge(nn1,ret1,by="Marker")
> # get the common "Marker" strings
> Marker3<-nn2$Marker
> # subset all three data frames on Marker3
&g
Hello,
I have 3 data frames which have about 3.4 mill lines (but they don't have
exactly the same number of lines)...they look like this:
> neu1=neu[order(neu$Marker),]
> head(neu1)
ChrBP Marker MAF A1 A2 Direction pValueN
209565 1 10012 1:10012:G:T 0.
4156 G A + 0.484813 1641
2 10 10645 10:10645:A:C 0.216027 C A + 0.73597 1641
Can you please tell me what colClasses=colClassvec suppose to do?
Thanks
Ana
On Mon, Jun 1, 2020 at 4:13 PM David Winsemius
wrote:
>
> On 6/1/20 1:37 PM, Ana Marija wrote:
&g
coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Jun 1, 2020 at 1:38 PM Ana Marija
> wrote:
>
>> Hello,
>>
>> I have a dataframe like this:
>>
>> Chr
Hello,
I have a dataframe like this:
ChrBP Marker MAF A1 A2 Direction pValueN
1 10 10625 10:10625:A:G 0.416562 G A - 0.558228 1594
2 10 10645 10:10645:A:C 0.215182 C A - 0.880622 1594
...
which I load with:
NEU <- read.table("g
ous(breaks
= unique(d))
ed
where:
> head(e)
ESRD pheno
11 1
21 1
31 2
41 1
51 1
> sapply(e,class)
ESRD pheno
"integer" "factor"
On Fri, May 22, 2020 at 1:52 PM Ana Marija wrote:
>
> Hello,
>
> I made t
Hello,
I made the plot in attach via:
ed<-ggplot(e) +
geom_bar(aes(x = ESRD, fill =
factor(pheno,labels=c("control","case"+scale_fill_manual(values=c("#56B4E9","#E7B800"))+labs(fill="pheno")
ed
How do I show only 1 and 2 on x axis?
Thanks
Ana
_
; space=0,ylab="Percentage",xaxt="n",ylim=c(0,25))
> > > text(mean(barpos),23,
> > > "Cases: n=848, nulls=26, median=7.3, mean=7.45, sd=1.96")
> > > box()
> > > par(mar=c(3,4,0,2))
> > > barplot(100*controlhist,names.arg=names
arpos=barplot(100*casehist,names.arg=names(casepct),col="orange",
> > space=0,ylab="Percentage",xaxt="n",ylim=c(0,25))
> > text(mean(barpos),23,
> > "Cases: n=848, nulls=26, median=7.3, mean=7.45, sd=1.96")
> > box()
> > par(
the result would basically look something like this on in attach or
the overlay of those two plots
On Thu, May 21, 2020 at 5:23 PM Ana Marija wrote:
>
> Hello,
>
> I have a data frame like this:
> > head(a)
> FID IID FLASER PLASER DIABDUR HBA1C ESRD pheno
Hello,
I have a data frame like this:
> head(a)
FID IID FLASER PLASER DIABDUR HBA1C ESRD pheno
1 fam1000-03 G1000 1 1 38 10.21 control
2 fam1001-03 G1001 1 1 15 7.31 control
3 fam1003-03 G1003 1 2 17 7.01case
4 fam1005-
ing a is where it finds the remaining parameters.
>
> You may also need to play with the sep =, and collapse = parameters to
> paste() to get the precise layout you want.
>
> Michael
>
> On 19/05/2020 17:21, Ana Marija wrote:
> > Hi Michael,
> >
> >
Hi Michael,
can you please send me code how that would be done?
Thanks
Ana
On Tue, May 19, 2020 at 11:18 AM Michael Dewey wrote:
>
> Dear Ana
>
> Perhaps paste together SNP and GENE using paste() and then supply that
> as the snp parameter.
>
> Michael
>
> On 19/05/2
Hello,
I am making manhattan plot with:
library(qqman)
manhattan(a, chr="CHR", bp="BP", snp="SNP", p="P",annotatePval = 0.0001)
and I would like to annotate these two SNPs which are above the
threshold so that they have GENE name beside them:
> a[a$SNP=="rs4081570",]
SNPP CHR
", unlist(tot), value = TRUE)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Às 20:24 de 15/05/20, Ana Marija escreveu:
> > Hello,
> >
> > this command was running for more than 2 hours
> > grep("E10",tot,value=T)
> > and no output
>
ch start with E10.
Thanks
Ana
On Fri, May 15, 2020 at 12:13 PM Jeff Newmiller
wrote:
>
> Read about regular expressions... they are extremely useful.
>
> df1 <- tot %>% filter_all(any_vars(grepl( '^E10', .)))
>
> It is bad form not to put spaces around the <
Hello,
I have a data frame:
> dim(tot)
[1] 502536 1093
How would I extract from it all strings that start with E10?
I know how to extract all rows that contain with E10
df0<-tot %>% filter_all(any_vars(. %in% c('E10')))
> dim(df0)
[1] 5105 1093
but I just need a vector of strings that start
Hello,
I got this error:
Error: Cannot use `+.gg()` with a single argument. Did you
accidentally put + on a new line?
After running this:
data(murders)
library(ggplot2)
library(dplyr)
library(ggplot2)
ggplot(data=murders)
#define the slope of the line
r<-murders %>% summarize(rate=sum(total)/sum
>
> t.test(x, mu = beta0)$statistic
>
>
> But in this case the estimator is the estimator for the mean value.
>
> Hope this helps,
>
> Rui Barradas
>
> Às 15:54 de 06/05/20, Ana Marija escreveu:
> > Thanks Patrick, so in conclusion this is fine?
> > z-sco
unit
> normal distribution. So like a t-test with infinite df.
>
>
> On Wed, May 6, 2020 at 10:41 AM Rui Barradas wrote:
> >
> > Hello,
> >
> > By z-scores do you mean function help('scale')?
> >
> > Hope this helps,
> >
&g
gt; Hope this helps,
>
> Rui Barradas
>
> Às 15:31 de 06/05/20, Ana Marija escreveu:
> > Hi Rui,
> >
> > Thank you for getting back to me. Is there is a better way to
> > calculate Z scores if I have p values, SE and Beta?
> >
> > Thanks
> &g
I guess I can have
z-score=Beta/StdErr
On Wed, May 6, 2020 at 9:37 AM Ana Marija wrote:
>
> thanks, can you please tell em what would be the way not to get the
> absolute (always positive values)
>
> On Wed, May 6, 2020 at 9:33 AM Rui Barradas wrote:
> >
> > Hel
If it's all you need/want, then the
> > answer is yes, you can.
> >
> > Hope this helps,
> >
> > Rui Barradas
> >
> > Às 14:28 de 06/05/20, Ana Marija escreveu:
> >> Hello,
> >>
> >> Can I
; answer is yes, you can.
>
> Hope this helps,
>
> Rui Barradas
>
> Às 14:28 de 06/05/20, Ana Marija escreveu:
> > Hello,
> >
> > Can I apply the quantile function qt() this way?
> > qt(pvals/2, 406-34, l
Hello,
Can I apply the quantile function qt() this way?
qt(pvals/2, 406-34, lower.tail = F)
to get the T-scores?
Thanks
Ama
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PLEASE do read
PDT, "Patrick (Malone Quantitative)" <
> mal...@malonequantitative.com> wrote:
> >"I tried this but I am not sure if this is correct:"
> >
> >Does it provide the expected result for all possible combinations of
> >1/2/NA
> >for both variables?
&g
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