[R] Creating submatrices from a dataframe, depending on factors in sample names
David L Carlson
dcarlson at tamu.edu
Mon Dec 1 22:57:25 CET 2014
I may have misunderstood, but does this do what you want?
> df.mat <- as.matrix(df)
> same <- lapply(1:3, function(x) df.mat[grep(paste0("_", x),
+ rownames(df.mat)), grep(paste0("_", x), colnames(df.mat))])
> same
[[1]]
HQ673618_1 HQ674317_1 EU686630_1
HQ673618_1 NA 90.8 89.8
HQ674317_1 90.8 NA 98.6
EU686630_1 89.8 98.6 NA
[[2]]
EU686593_2 JN166322_2 EU491340_2
EU686593_2 NA 98.1 96.8
JN166322_2 98.1 NA 97.5
EU491340_2 96.8 97.5 NA
[[3]]
AB694259_3 AB694258_3 AB694462_3
AB694259_3 NA 98.3 95.9
AB694258_3 98.3 NA 95.8
AB694462_3 95.9 95.8 NA
> Diff <- as.matrix(expand.grid(1:3, 1:3))
> Diff <- Diff[Diff[,1]<Diff[,2],]
> different <- lapply(seq_len(nrow(Diff)), function(x)
+ df.mat[grep(paste0("_", Diff[x,1]), rownames(df.mat)),
+ grep(paste0("_", Diff[x,2]), colnames(df.mat))])
> different
[[1]]
EU686593_2 JN166322_2 EU491340_2
HQ673618_1 89.6 89.8 88.9
HQ674317_1 97.7 98.4 97.4
EU686630_1 98.4 98.9 97.7
[[2]]
AB694259_3 AB694258_3 AB694462_3
HQ673618_1 87.8 88.2 88.3
HQ674317_1 94.9 96.2 95.1
EU686630_1 95.4 96.4 95.8
[[3]]
AB694259_3 AB694258_3 AB694462_3
EU686593_2 94.4 95.6 94.8
JN166322_2 95.3 96.5 95.9
EU491340_2 96.5 97.7 96.0
-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Bert Gunter
Sent: Monday, December 1, 2014 11:46 AM
To: Tim Richter-Heitmann
Cc: r-help at r-project.org
Subject: Re: [R] Creating submatrices from a dataframe, depending on factors in sample names
I do not have the patience to study your request carefully, but does
the following help?
> a <- 1:3
> x <- outer(a,a,paste,sep=".")
> x
[,1] [,2] [,3]
[1,] "1.1" "1.2" "1.3"
[2,] "2.1" "2.2" "2.3"
[3,] "3.1" "3.2" "3.3"
> x[upper.tri(x)]
[1] "1.2" "1.3" "2.3"
> x[upper.tri(x,diag=TRUE)]
[1] "1.1" "1.2" "2.2" "1.3" "2.3" "3.3"
This gives you a vector all possible pairs (including identical pairs
or not) of values of a, which you could then loop over as an index to
do what you want, I think.
If this is not what you want, just ignore without replying.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Mon, Dec 1, 2014 at 8:47 AM, Tim Richter-Heitmann
<trichter at uni-bremen.de> wrote:
> Hello there,
>
> this is a cross-post of a stack-overflow question, which wasnt answered, but
> is very important for my work. Apologies for breaking any rules, but i do
> hope for some help from the list instead:
>
> I have a huge matrix of pairwise similarity percentages between different
> samples. The samples are belonging to groups. The groups are determined by
> the suffix "_n" in the row.names/header names.
> In the first step, i wanted to create submatrices consisting of all pairs
> within single groups (i.e. for all samples from "_1").
> However, I realized that i need to know all pairwise submatrices, between
> all combination of groups. So, i want to create (a list of) vectors that are
> named "_n1 vs _n2" (or similar) for all combinations of n, as illustrated by
> the colored rectangulars:
>
> http://i.stack.imgur.com/XMkxj.png
>
> Reproducible code, as provided by helpful Stack Overflow members, dealing
> with identical "_n"s.
>
>
> df <- structure(list(HQ673618_1 = c(NA, 90.8, 89.8, 89.6, 89.8,
> 88.9,
> 87.8, 88.2, 88.3), HQ674317_1 = c(90.8, NA, 98.6, 97.7, 98.4,
> 97.4, 94.9, 96.2, 95.1), EU686630_1 = c(89.8, 98.6, NA, 98.4,
> 98.9, 97.7, 95.4, 96.4, 95.8), EU686593_2 = c(89.6, 97.7, 98.4,
> NA, 98.1, 96.8, 94.4, 95.6, 94.8), JN166322_2 = c(89.8, 98.4,
> 98.9, 98.1, NA, 97.5, 95.3, 96.5, 95.9), EU491340_2 = c(88.9,
> 97.4, 97.7, 96.8, 97.5, NA, 96.5, 97.7, 96), AB694259_3 = c(87.8,
> 94.9, 95.4, 94.4, 95.3, 96.5, NA, 98.3, 95.9), AB694258_3 = c(88.2,
> 96.2, 96.4, 95.6, 96.5, 97.7, 98.3, NA, 95.8), AB694462_3 = c(88.3,
> 95.1, 95.8, 94.8, 95.9, 96, 95.9, 95.8, NA)), .Names =
> c("HQ673618_1",
> "HQ674317_1", "EU686630_1", "EU686593_2", "JN166322_2",
> "EU491340_2",
> "AB694259_3", "AB694258_3", "AB694462_3"), class = "data.frame",
> row.names = c("HQ673618_1",
> "HQ674317_1", "EU686630_1", "EU686593_2", "JN166322_2",
> "EU491340_2",
> "AB694259_3", "AB694258_3", "AB694462_3"))
>
>
> indx <- gsub(".*_", "", names(df))
> sub.matrices <- lapply(unique(indx), function(x) {
> temp <- which(indx %in% x)
> df[temp, temp]
> })
> unique_values <- lapply(sub.matrices, function(x) x[upper.tri(x)])
> names(unique_values) <- unique(indx)
>
> This code needs to be expanded to form sub.matrices for any combination of
> unique indices in temp.
>
>
> Thank you so much!
>
>
>
>
> --
> Tim Richter-Heitmann (M.Sc.)
> PhD Candidate
>
>
>
> International Max-Planck Research School for Marine Microbiology
> University of Bremen
> Microbial Ecophysiology Group (AG Friedrich)
> FB02 - Biologie/Chemie
> Leobener Straße (NW2 A2130)
> D-28359 Bremen
> Tel.: 0049(0)421 218-63062
> Fax: 0049(0)421 218-63069
>
> ______________________________________________
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