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
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
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,
.
>
>
> 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
% 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
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:
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 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$RTNPT
, 2020 at 11:43 AM Ana Marija wrote:
>
> Hello,
>
> I have a data frame as shown bellow.
> I want to create a new column PHENO which will be defined as follows:
> if CURRELIG==1 -> PHENO==1
> in the above subset those that have:
> PLASER==2 -> PHENO==2
> and
>
t RHS position 1 taken as TRUE when
assigning to type 'logical' (column 6 named 'PHENO')
Please advise,
Ana
On Wed, Sep 23, 2020 at 2:48 PM Jeremie Juste wrote:
>
>
> Hello Ana Marija,
>
> I cannot reproduce your error,
>
> with a$PHENO=ifelse(a$PLASER==2 |a$
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$PLATE))
[1] 34
I am trying to make a ne
0117 G117 4RWG569 2 1 1
> 5 fam0118 G118 5XAV049 1 1 2
> 6 fam0119 G119 cherry 1 2 1
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
> -
4:ncol(mt)) mt[,i] <- 1 + (names(mt)[i]== mt$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
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.
NP[SNP$density>400,]
and plot it again:
p <- ggplot(a, mapping = aes(x = mean, y = var))
p <- p + geom_density_2d() + geom_point() + my.theme + ggtitle("SNPS_red")
On Thu, Oct 8, 2020 at 3:52 PM Ana Marija wrote:
>
> Hello,
>
> I have a data frame like this:
>
>
Hi Abby,
thank you for getting back to me and for this useful information.
I'm trying to detect the outliers in my distribution based of mean and
variance. Can I see that from the plot I provided? Would outliers be
outside of ellipses? If so how do I extract those from my data frame,
based on whi
an open mind is that people keep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Fri, Oct 9, 2020 at 8:25 AM Ana Marija wrote:
>>
>> Hi Abby,
>>
>> thank you for gettin
a bad idea.
> And if it's a good idea, then how much to trim.
>
>
> On Sat, Oct 10, 2020 at 5:47 AM Ana Marija
> wrote:
> >
> > Hi Bert,
> >
> > Another confrontational response from you...
> >
> > You might have noticed that I use the wor
Hello,
I have a data frame with one column:
> remove
V1
1 ABAFT_g_4RWG569_BI_SNP_A10_35096
2 ABAFT_g_4RWG569_BI_SNP_B12_35130
3 ABAFT_g_4RWG569_BI_SNP_E09_35088
4 ABAFT_g_4RWG569_BI_SNP_E12_35136
5 ABAFT_g_4RWG569_BI_SNP_F11_35122
6 ABAFT_g_4RWG569_BI_SNP_F12_351
lename %in% as.character(remove$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
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
> >
> >
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
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 -
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
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
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
e and see if it contains a
> column named "blup" or just the values that were extracted from
> a$blup. Also, I assume that weight=blup looks for an object named
> "blup", which may not be there.
>
> Jim
>
> On Wed, Dec 16, 2020 at 1:20 PM Ana Marija
>
assuming that the filename
> # is stored in files[i]
> files<-"retina.ENSG0135776.wgt.RDat"
> i<-1
> WGT<-rep(files[i],length(rsid))
> data<-data.frame(rsid=rsid,weight=a$top1,
> ref_allele=ref_allele,eff_allele,WGT=WGT)
> data
>
> Note that the
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:
>
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
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
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
)
> #[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 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
-- 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 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
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
Hello,
I am trying to run this program FUSION.assoc_test.R from :
http://gusevlab.org/projects/fusion/#typical-analysis-and-output
[t.cri.asokovic@cri16in002 fusion_twas-master]$ Rscript FUSION.assoc_test.R
\
> --sumstats /gpfs/data/stranger-lab/anamaria/meta_gwas/META_CHR22_1.txt \
> --weights .
Hello,
I am trying to use this program:
https://github.com/kenhanscombe/ukbtools
> my_ukb_data[1:3,1:3]
eid sex_f31_0_0 year_of_birth_f34_0_0
1 117 Female 1938
2 125 Female 1951
3 138Male 1961
> ukb_icd_diagno
so, always reply to list.
>
> On Wed, Jul 10, 2019, 6:08 PM Ana Marija
> wrote:
>
> > Hi Patrick,
> >
> > thanks for getting back to me, I tried that:
> >
> > > ukb_icd_diagnosis(my_ukb_data, eid = "117", icd.version = 10)
> > Error in u
;
> > Sent from my iPhone
> >
> > > On Jul 10, 2019, at 1:09 PM, Patrick (Malone Quantitative) <
> mal...@malonequantitative.com> wrote:
> > >
> > > First response: The ID column in your data is labeled "eid" but your
> > > function call refers
I have data frame which looks like this:
df=data.frame(
eye_problemsdisorders_f6148_0_1=c(A,C,D,NA,D,A,C,NA,B,A),
eye_problemsdisorders_f6148_0_2=c(B,C,NA,A,C,B,NA,NA,A,D),
eye_problemsdisorders_f6148_0_3=c(C,A,D,D,B,A,NA,NA,A,B),
eye_problemsdisorders_f6148_0_4=c(D,D,NA,B,A,C,NA,C,A,B),
which would be named "case" and
values inside would be: 1,1,0,1,1,1,0,0,1,1
so "case" column is where value "A" can be found in any column.
On Mon, Jul 29, 2019 at 12:53 PM Eric Berger wrote:
> You may have a typo/misstatement in your question.
> You define a
Thank you so much! Just to confirm here MARGIN=1 indicates that "A" should
appear at least once per row?
On Mon, Jul 29, 2019 at 1:53 PM Eric Berger wrote:
> df$case <- apply(df,MARGIN = 1,function(v) { as.integer("A" %in% v) })
>
>
> On Mon, Jul 29,
Hello,
I am filtering my data frame "tot" via:
controls=tot %>% filter_all(any_vars(. %in% c("E109", "E119","E149"))) %>%
filter_all(any_vars(. %in% c("Caucasian"))) %>% filter_all(any_vars(. %in%
c("No kinship found","Ten or more third-degree relatives identified")))
> dim(controls)
[1] 15381
some kind of ID variable? If so, it should be
> straightforward with the appropriate joining function.
>
> What have you tried?
>
> Also, please post in plain text.
>
>
> On Tue, Aug 6, 2019 at 2:16 PM Ana Marija
> wrote:
> >
> > Hello,
> >
> > I am
guide.
>
>
> On Tue, Aug 6, 2019 at 2:38 PM Ana Marija
> wrote:
> >
> > Hi Patrick,
> >
> > yes both controls and tot have "eid" column, please see attached
> >
> > Can you please tell em what means to post in "plain text" ?
&g
tor of 1s to controls before joining the datasets.
>
> On Tue, Aug 6, 2019 at 3:01 PM Ana Marija
> wrote:
> >
> > I really don't know how I would implement this
> >
> > On Tue, Aug 6, 2019 at 1:42 PM Patrick (Malone Quantitative) <
> mal...@malonequan
Hello,
I have a data frame which looks like this:
> head(pt)
eidQ phenoQ phenoH
1 117 -9 -9
2 125 -9 -9
3 138 -9 1
4 142 -9 -9
5 156 -9 -9
6 174 -9 -9
7 138 -9 1
8 1000127 2 1
9 1000690 2
does this look ok:
pt$pheno=ifelse(pt$phenoQ==-9 & pt$phenoH==-9,-9,ifelse(pt$phenoH==2 |
pt$phenoQ==2,2,1))
On Wed, Aug 7, 2019 at 1:40 PM Ana Marija wrote:
>
> Hello,
>
> I have a data frame which looks like this:
>
> > head(pt)
> eidQ phenoQ phenoH
> 1 10
Hello,
I made plot in attach using:
boxplot(flcn_M~subject,data=dx,col =
c("royalblue1","palevioletred1"),xlab="subjects",ylab="Expression
estimate in delta (log2)",boxwex = 0.2,frame.plot = FALSE)
stripchart(flcn_M~subject, vertical = TRUE, data = dx,method =
"jitter", add = TRUE,pch = 20, col=r
Hello,
I tried using qvalue function:
library(qvalue)
qval_obj=qvalue(pvalR)
pi1=1-qval_obj$pi0
but after running:
qval_obj=qvalue(pvalR)
Error in smooth.spline(lambda, pi0, df = smooth.df) :
missing or infinite values in inputs are not allowed
or
qval_obj=qvalu
Hi,
I created a bar plot with this code:
library(ggplot2)
df <- data.frame("prop" = c(7.75,70.42), "Name" = c("All Genes","RG Genes"))
p<-ggplot(data=df, aes(x=Name, y=prop,fill=Name)) +
geom_bar(stat="identity")+ labs(x="", y = "Proportion of cis
EQTLs")+ scale_fill_brewer(palette="Greens") +
70.42-7.75 or fold change 70.42/7.75. If they are absolute value you can
> also scale them in log scale and do the same. Hope this helps. Good luck.
>
> Vivek
>
> On Fri, Sep 27, 2019 at 9:32 PM Ana Marija
> wrote:
>
>> Hi Vivek,
>>
>> Thanks for getting back
Hello,
I have a dataframe (t1) with many columns, but the one I care about it this:
> unique(t1$sex_chromosome_aneuploidy_f22019_0_0)
[1] NA"Yes"
it has these two values.
I would like to remove from my dataframe t1 all rows which have "Yes"
in t1$sex_chromosome_aneuploidy_f22019_0_0
I tried
Hello,
I have a data frame tot which has many columns and many rows.
I am trying to find all columns that have say a value in any of their
rows that STARTS WITH: "E94"
for example there are columns like this:
> unique(tot$diagnoses_icd9_f41271_0_44)
[1] NA "E9420"
I tried:
s=select(tot,st
tot$newcol <- -9
> tot$newcol[e10] <- 1
> tot$newcol[e11] <- 2
>
>
> On both cases the 2 lines sapply/rowSums can be made one with
>
> rowSums(sapply(...)) > 0
>
>
> Hope this helps,
>
> Rui Barradas
>
> Às 20:52 de 05/10/19, Ana Marija escreveu:
>
Hello,
I would like to calculate a p value from two distributions, one looks like this:
> head(b)
gene_id number_of_eqtles_per_gene
1: ENSG0237683.5 5
2: ENSG0225972.1 267
3: ENSG0225630.197
4: ENSG0
er
>
> "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 Tue, Oct 22, 2019 at 6:18 PM Ana Marija
> wrote:
>>
>
Hello,
I have two data frames like this:
> head(l4)
X1X2 X3 X4 X5 variant_id pval_nominal gene_id.LCL
1 chr1 13550 G A b38 1:13550:G:A 0.375614 ENSG0227232
2 chr1 14671 G C b38 1:14671:G:C 0.474708 ENSG0227232
3 chr1 14677 G A b38 1:14677:G:A 0.699887 ENSG0
I also tried left_join but I got: Error: std::bad_alloc
> df3 <- left_join(l4, asign, by = c("chr","pos"))
Error: std::bad_alloc
> dim(l4)
[1] 166941635 8
> dim(asign)
[1] 107371528 5
On Wed, Oct 23, 2019 at 5:32 PM Ana Marija wrote:
>
>
ently the best way to do it if you
> have a database manager is to read the two datasets into tables and do
> the join via SQL or whatever language is available.
>
> Jim
>
> On Thu, Oct 24, 2019 at 10:17 AM Ana Marija
> wrote:
> >
> > no can you please send m
X2 X3 X4 X5
> [6] variant_id pval_nominal gene_id.LCL gene chr_pos
> [11] p.val.Retina
> <0 rows> (or 0-length row.names)
>
> It works okay, but there are no matches in the join. So I can't even
> guess what the proble
no can you please send me an example how the command would look like in my case?
On Wed, Oct 23, 2019 at 6:16 PM Jim Lemon wrote:
>
> Yes. Have you tried the bigmemory package?
>
> Jim
>
> On Thu, Oct 24, 2019 at 10:08 AM Ana Marija
> wrote:
> >
> > Hi Jim,
&g
I am using R-3.6.1
and these libraries:
library(data.table)
library(dplyr)
On Wed, Oct 23, 2019 at 6:54 PM Duncan Murdoch wrote:
>
> On 23/10/2019 7:04 p.m., Ana Marija wrote:
> > I also tried left_join but I got: Error: std::bad_alloc
> >
> >> df3 <- left_joi
Hello,
I would like to use this package metap
to calculate multiple o values
I have my data frame with 3 p values
> head(tt)
RSG E B
1: rs2089177 0.9986 0.7153 0.604716
2: rs4360974 0.9738 0.7838 0.430228
3: rs6502526 0.9744 0.7839 0.4291
this is the function I was referring to:
https://www.rdocumentation.org/packages/metap/versions/1.1/topics/sumz
On Fri, Oct 25, 2019 at 6:31 PM Ana Marija wrote:
>
> Hello,
>
> I would like to use this package metap
> to calculate multiple o values
>
> I have my data
gt; There must be several ways of doing this but see below for an idea with
> comments in-line.
>
> On 26/10/2019 00:31, Ana Marija wrote:
> > Hello,
> >
> > I would like to use this package metap
> > to calculate multiple o values
> >
> > I have
Hello,
I am trying to calculate True Positive Rate, TPR with this procedure:
pvals=q$METAge
qval_obj=qvalue(pvals) #is false discovery rate
pi1=1-qval_obj$pi0 #TPR
pi1 #TPR
But I am getting this error:
Error in smooth.spline(lambda, pi0, df = smooth.df) :
missing or
1.053550e-01,4.812686e-01,1.404957e-01,9.835912e-02,4.373995e-01,
> 8.803856e-02)
> qval_obj=qvalue(pvals)
> qval_obj$pi0
> [1] 0.1981095
> pi1=1-qval_obj$pi0
> pi1
> [1] 0.8018905
>
> Jiim
>
> On Tue, Oct 29, 2019 at 8:45 AM Ana Marija
> wrote:
&g
:
missing or infinite values in inputs are not allowed
On Mon, Oct 28, 2019 at 6:02 PM Ana Marija wrote:
>
> can you please send me command you used to install it?
>
> On Mon, Oct 28, 2019 at 5:12 PM Jim Lemon wrote:
> >
> > Hi Ana,
> > Seems to work without er
not to install packages in my
> script but just to use `library` and manually get each of the packages that
> `library` complains about onto my machine. Once done, the script runs just
> fine after that.
>
> On October 28, 2019 4:08:03 PM PDT, Ana Marija
> wrote:
>
- (l4join$X1 %in% ajoin$chr) & (l4join$X2 %in% ajoin$pos)
> i2 <- (ajoin$chr %in% l4join$X1) & (ajoin$pos %in% l4join$X2)
>
> rm(l4join, ajoin) # don't need this any more, remove them
>
> # now the real fread's
> l4 <- data.table::fread(l4_file)
> asign <
6.024415e-01,
>> 2.459322e-02,2.873351e-01,8.477168e-01,1.351068e-02,
>> 1.053550e-01,4.812686e-01,1.404957e-01,9.835912e-02,4.373995e-01,
>> 8.803856e-02)
>> qval_obj=qvalue(pvals)
>> qval_obj$pi0
>> [1] 0.1981095
>> pi1=1-qval_obj$pi0
>> pi1
&
ry about that, I
> should have thought of it before.
>
> When I next update metap I will try to get it to degrade more gracefully
> when it finds an error.
>
> Michael
>
> On 28/10/2019 19:06, Ana Marija wrote:
> > Hi Michael,
> >
> > I tried what y
wr
"numeric"
> sum(is.na(d$LCL))
[1] 0
> sum(is.na(d$Retina))
[1] 0
> sum(is.na(d$wl))
[1] 0
> sum(is.na(d$wr))
[1] 0
> dim(d)
[1] 1668837 7
On Wed, Oct 30, 2019 at 4:52 PM Ana Marija wrote:
>
> Hi Michael,
>
> this still doesn't work, by dat
Can you please get back to me about this, I need this meta p values
for manuscript I have to submit next week
On Wed, Oct 30, 2019 at 5:35 PM Ana Marija wrote:
>
> I also tried to do it this way:
>
> d$META <- sapply(seq_len(nrow(d)), function(rn) {
> unlist(sumz(as.matr
Hello,
I have 3 groups,let's call them g1, g2, g3. Each of them is a result
of analysis in between groups of conditions, and g1 looks like this
geneSymbol logFC t P.Value
adj.P.Val Beta
EXykpF1BRREdXnv9Xk MKI67 -0.3115880 -5.521186 5.77213
Hi,
I have 3 data frames like this:
> head(s11)
B_NoD
Ebfrl.7uOZfnjp_E7k 7.583709
ueQUrXd5FH554RlhZc 5.177791
0Uu3XrB6Bd14qoNeuc 4.680306
0t7nhVLii6tSAxtLhc 4.565023
fSUyR.vR7Xu0iR4nUU 2.885992
0Tm7hdRJxd9zoevPlA 2.866847
> head(s22)
B_DwoC
Ebfrl.7uOZfnj
Hi,
I am running this function:
library(psych)
corr.test.col.1to3 <- corr.test(allF[1:3], method = "spearman", use =
"complete.obs")
names(corr.test.col.1to3)
corr.test.col.1to3$p
and my result looks like this:
> corr.test.col.1to3$p
B_NoDB_DwoC B_DwC
B_NoD 0.000 0.000
Hello,
I did Mendelian randomization using this software:
https://cran.r-project.org/web/packages/MendelianRandomization/vignettes/Vignette_MR.pdf
library(MendelianRandomization)
f=read.table("246LDout272Biobank_Retina.txt", header=T)
> head(f)
rs exposure.beta exposure.se outcome.beta
Hello,
I have a data frame like this:
> head(dt,20)
chrpos gene_id pval_nominal pval_ret wl wr
1: chr1 54490 ENSG02272320.6084950 0.7837780 31.62278 21.2838
2: chr1 58814 ENSG02272320.2952110 0.8975820 31.62278 21.2838
3: chr1 60351 ENSG02272
-
> Dr. Gerrit Eichner Mathematical Institute, Room 212
> gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
> Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
> http://www.uni-giessen.de/eichner
>
---------
>
> Am 08.11.2019 um 16:02 schrieb Ana Marija:
> > I tried it but I got this error:
> >> udt <- unique(dt[c("chr", "pos", "gene_id")])
> > Error in `[.data.table`(dt, c("chr", "pos", "gene_id")) :
&g
: chr1 64931 ENSG0227232 0.276679 0.907037 31.62278 21.2838 0.5974800
On Fri, Nov 8, 2019 at 9:30 AM Ana Marija wrote:
>
> Thank you so much! Converting it to data frame resolved the issue!
>
> On Fri, Nov 8, 2019 at 9:19 AM Gerrit Eichner
> wrote:
> >
> > It seem
them.
About duplicated() function I know as well as about unique
On Fri, 8 Nov 2019 at 10:08, Boris Steipe wrote:
> Are you trying to eliminate duplicated rows from your dataframe? Because
> that would be better achieved with duplicated().
>
>
> B.
>
>
>
>
> > On
the - sign
>>
>> If you prefer, the "Tidyverse" world has what are purported to be more
>> user-friendly versions of such data handling functionality that you can use
>> instead.
>>
>>
>> Bert
>>
>> On Fri, Nov 8, 2019 at 7:38 AM Ana M
Hi,
I was using this library, qqman
https://cran.r-project.org/web/packages/qqman/vignettes/qqman.html
to create QQ plot, attached. How would I change this default abline to
start from the beginning of my QQ line?
This is my code:
qq(dd$P, main = "Q-Q plot of GWAS p-values")
Thanks
Ana
details about my data if it is helpful:
> median(dd$P,na.rm = FALSE)
[1] 0.000444
> mean(dd$P,na.rm = FALSE)
[1] 0.000461
> min(dd$P,na.rm = FALSE)
[1] 9.89e-08
> max(dd$P,na.rm = FALSE)
[1] 0.001
On Tue, Nov 12, 2019 at 2:07 PM Ana Marija wrote:
>
> Hi,
>
> what I know
I agree with Abby. That would defeat the purpose of a QQ plot.
> >
> > On Mon, Nov 11, 2019, 9:54 PM Abby Spurdle wrote:
> >
> > > Hi
> > >
> > > I'm not familiar with the qqman package, or GWAS studies.
> > > However, my guess would be th
> wrote:
> >>
> >> I agree with Abby. That would defeat the purpose of a QQ plot.
> >>
> >> On Mon, Nov 11, 2019, 9:54 PM Abby Spurdle wrote:
> >>
> >>> Hi
> >>>
> >>> I'm not familiar with the qqman package
Just do I need to change the axis when I multiply with 1000 and what
should I put on my axis?
On Tue, Nov 12, 2019 at 3:07 PM Ana Marija wrote:
>
> Hi Duncan,
>
> yes I choose for QQ plot only P<1e-3 and multiplying everything with
> 1000 works great!
> This should no
the smallest p value in my dataset goes to 9.89e-08. How do I make
that known on the new QQ plot with multiplied with 1000 values
On Tue, Nov 12, 2019 at 3:37 PM Ana Marija wrote:
>
> Just do I need to change the axis when I multiply with 1000 and what
> should I put on my axis?
>
&
PM Ana Marija wrote:
>
> the smallest p value in my dataset goes to 9.89e-08. How do I make
> that known on the new QQ plot with multiplied with 1000 values
>
> On Tue, Nov 12, 2019 at 3:37 PM Ana Marija
> wrote:
> >
> > Just do I need to change the axis when I mult
ight grey so the overlap is more
> visible.
>
> Michael
>
> On 12/11/2019 22:04, Ana Marija wrote:
> > why I selected only those with P<0.003 to put on QQ plot is because
> > the original data set contains 5556249 points and when I extract only
> > P<0.001 I am getting
Hello,
I have a data frame like this (a matrix):
head(calc.rho)
rs9900318 rs8069906 rs9908521 rs9908336 rs9908870 rs9895995
rs56192520 0.903 0.268 0.327 0.327 0.327 0.582
rs3764410 0.928 0.276 0.336 0.336 0.336 0.598
rs145984817 0.
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, Nov 14, 2019 at 10:50 AM Ana Marija
> wrote:
>>
>> Hello,
>>
>>
what would be the approach to remove variable that has at least 2
correlation coefficients >0.8?
this is the whole output of the head()
> head(calc.rho)
rs56192520 rs3764410 rs145984817 rs1807401 rs1807402 rs35350506
rs56192520 1.000 0.976 0.927 0.927 0.927
s of the variables with at most one absolute value greater than
> 0.8 ignoring the diagonal values because I don't care about those". If
> so:
>
> colnames(calc.jim)[colSums(abs(calc.jim)>0.8)<3]
>
> Any more tricks?
>
> Jim
>
> On Fri, Nov 15, 2019 at 8:1
contain bugs, but something along
> these lines should get you where you want to be.
>
> Oh, and depending on how strict you want to be with the remaining
> correlations, you could use complete linkage clustering (will retain
> more variables, some correlations will be above 0.8)
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