[R] how to calculate multiple meta p values

Ana Marija @okov|c@@n@m@r|j@ @end|ng |rom gm@||@com
Wed Oct 30 23:35:43 CET 2019 

I also tried to do it this way:

d$META <- sapply(seq_len(nrow(d)), function(rn) {
  unlist(sumz(as.matrix(d[,.(LCL,Retina)])[rn,], weights =
as.vector(d[,.(wl,wr)])[rn,],
              na.action=na.fail)["p"])
})

but again I am getting error:
Error in sumz(as.matrix(d[, .(LCL, Retina)])[rn, ], weights = as.vector(d[,  :
  Must have at least two valid p values

for this reference these are details about my data frame:
> head(d)
    chr    pos         gene_id                     LCL          Retina
           wl           wr
1: chr1 775930 ENSG00000237094 0.3559520 9.72251e-05 31.62278 21.2838
2: chr1 815963 ENSG00000237094 0.2648080 3.85837e-06 31.62278 21.2838
3: chr1 816376 ENSG00000237094 0.3313120 3.85824e-06 31.62278 21.2838
4: chr1 817186 ENSG00000237094 0.0912854 3.75134e-06 31.62278 21.2838
5: chr1 817341 ENSG00000237094 0.1020520 3.75134e-06 31.62278 21.2838
6: chr1 817514 ENSG00000237094 0.0831412 3.82866e-06 31.62278 21.2838
> sapply(d,class)
        chr         pos     gene_id         LCL      Retina          wl
"character" "character" "character"   "numeric"   "numeric"   "numeric"
         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 <sokovic.anamarija using gmail.com> wrote:
>
> Hi Michael,
>
> this still doesn't work, by data frame has a few less columns now, but
> the principle is still the same:
>
> > head(d)
>     chr    pos         gene_id                     LCL
> Retina       wl           wr
> 1: chr1 775930 ENSG00000237094 0.3559520 9.72251e-05 31.62278 21.2838
> 2: chr1 815963 ENSG00000237094 0.2648080 3.85837e-06 31.62278 21.2838
> 3: chr1 816376 ENSG00000237094 0.3313120 3.85824e-06 31.62278 21.2838
> 4: chr1 817186 ENSG00000237094 0.0912854 3.75134e-06 31.62278 21.2838
> 5: chr1 817341 ENSG00000237094 0.1020520 3.75134e-06 31.62278 21.2838
> 6: chr1 817514 ENSG00000237094 0.0831412 3.82866e-06 31.62278 21.2838
>
> so solution for the first row should be:
> > sumz(c(0.3559520,9.72251e-05), weights = c(31.62278,21.2838), na.action = na.fail)
> sumz =  2.386896 p =  0.008495647
>
> when I run what you proposed in the last email:
>
> helper <- function(x) {
>   p <- sumz(as.numeric(x[4:5]), weights = as.numeric(x[6:7]))$p
>   p
> }
>
> d$META <- apply(d, MARGIN = 1, helper)
>
> I am getting:
>
> Error in sumz(as.numeric(x[4:5]), weights = as.numeric(x[6:7])) :
>   Must have at least two valid p values
>
> Please advise,
> Ana
>
> On Wed, Oct 30, 2019 at 5:02 AM Michael Dewey <lists using dewey.myzen.co.uk> wrote:
> >
> > Dear Ana
> >
> > Yes, when apply coerces q to a matrix it does so as a character matrix
> > because of the values in the first column. So you need to wrap the
> > references to x in helper in as.numeric() tat is to day like
> > as.numeric(x[2:4]) and similarly for the other one. Sorry 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 you proposed with my data frame q:
> > >
> > >> head(q)
> > >             ID                P             G              E
> > >   wb          wg           we
> > > 1:  rs1029830 0.0979931 0.0054060 0.39160 580.6436 40.6325 35.39774
> > > 2:  rs1029832 0.1501820 0.0028140 0.39320 580.6436 40.6325 35.39774
> > > 3: rs11078374 0.1701250 0.0009805 0.49730 580.6436 40.6325 35.39774
> > > 4:  rs1124961 0.1710150 0.7252000 0.05737 580.6436 40.6325 35.39774
> > > 5:  rs1135237 0.1493650 0.6851000 0.06354 580.6436 40.6325 35.39774
> > > 6: rs11867934 0.0757972 0.0006140 0.00327 580.6436 40.6325 35.39774
> > >
> > > so the solution of the first row would be this:
> > >> sumz(c(0.0979931,0.0054060,0.39160), weights = c(580.6436,40.6325,35.39774), na.action = na.fail)
> > > sumz =  1.481833 p =  0.06919239
> > >
> > > I tried applying the function you wrote:
> > > helper <- function(x) {
> > >    p <- sumz(x[2:4], weights = x[5:7])$p
> > >    p
> > > }
> > >
> > > With:
> > >
> > > q$META <- apply(q, MARGIN = 1, helper)
> > >
> > > # I want to make a new column in q named META with results
> > > but I got this error:
> > >   Error in sumz(x[2:4], weights = x[5:7]) :
> > >    Must have at least two valid p values
> > >
> > > Please advise,
> > > Ana
> > >
> > > On Sun, Oct 27, 2019 at 9:49 AM Michael Dewey <lists using dewey.myzen.co.uk> wrote:
> > >>
> > >> Dear Ana
> > >>
> > >> 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 my data frame with 3 p values
> > >>>> head(tt)
> > >>>             RS            G           E          B
> > >>> 1: rs2089177   0.9986   0.7153   0.604716
> > >>> 2: rs4360974   0.9738   0.7838   0.430228
> > >>> 3: rs6502526   0.9744   0.7839   0.429160
> > >>> 4: rs8069906   0.7184   0.4918   0.521452
> > >>> 5: rs9905280   0.7205   0.4861   0.465758
> > >>> 6: rs4313843   0.9804   0.8522   0.474313
> > >>>
> > >>> and data frame with corresponding weights for each of the p values
> > >>> from the tt data frame
> > >>>
> > >>>> head(df)
> > >>>          wg       we             wb                RS
> > >>> 1 40.6325 35.39774 580.6436 rs2089177
> > >>> 2 40.6325 35.39774 580.6436 rs4360974
> > >>> 3 40.6325 35.39774 580.6436 rs6502526
> > >>> 4 40.6325 35.39774 580.6436 rs8069906
> > >>> 5 40.6325 35.39774 580.6436 rs9905280
> > >>> 6 40.6325 35.39774 580.6436 rs4313843
> > >>>
> > >>> RS column is the same in df and tt
> > >>>
> > >>
> > >> So you can create a new data-frame with merge()
> > >>
> > >> newdata <- merge(tt, df)
> > >>
> > >> which will use RS as the key to merge them on.
> > >>
> > >> The write a function of one argument, a seven element vector, which
> > >> picks out the p-values and the weights and feeds them to sumz().
> > >> Something like
> > >>
> > >> helper <- function(x) {
> > >>    p <- sumz(x[2:4], weights = x[5:7])$p
> > >>    p
> > >> }
> > >> Note you need to check that 2:4 and 5:7 are actually where they are in
> > >> the row of newdat.
> > >>
> > >> Then use apply() to apply that to the rows of newdat.
> > >>
> > >> I have not tested any of this but the general idea should be OK even if
> > >> the details are wrong.
> > >>
> > >> Michael
> > >>
> > >>
> > >>> How to use this sunz() function to create a new data frame which would
> > >>> look the same as tt only it would have additional column, say named
> > >>> "META" which has calculated meta p values for each row
> > >>>
> > >>> This i s example of how much would be p value in the first row:
> > >>>
> > >>>> sumz(c(0.9986,0.7153,0.604716), weights = c(40.6325,35.39774,580.6436), na.action = na.fail)
> > >>> p =  0.6940048
> > >>>
> > >>> Thanks
> > >>> Ana
> > >>>
> > >>> ______________________________________________
> > >>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > >>> 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.
> > >>>
> > >>
> > >> --
> > >> Michael
> > >> http://www.dewey.myzen.co.uk/home.html
> > >
> >
> > --
> > Michael
> > http://www.dewey.myzen.co.uk/home.html

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