Compute a consensus from a TALE msa

Pick the most frequent element in each column of the alignment matrix.

Usage

tales_consensus(align)

Arguments

align

A TALE alignment as a character matrix, one row per array and one column per alignment position, with NA for gaps. as.matrix() on a tales_msa returns one (see as.matrix.tales_msa), filled with RVDs or domain codes.

Value

A vector of consensus elements in each column of align.

Details

A column has a consensus only when one element is strictly more common than every other. Where two or more are tied for most frequent – as happens whenever each array carries a different repeat at that position – the result is NA, because there is no agreement to report. NA is likewise returned when the most common thing at a position is a gap.

See also

Other TALE alignment: as.matrix.tales_msa(), is_tales_msa(), tales_align(), tales_consensus_match(), tales_msa(), tales_msa_params(), tales_msa_width(), validate_tales_msa()

Examples

# column 1: HD is a clear majority. column 2: a three-way tie, no consensus.
aln <- matrix(c("HD", "HD", "NI",
               "NG", "NI", "HD"),
             nrow = 3, dimnames = list(c("A1", "A2", "A3"), NULL))
aln
#>    [,1] [,2]
#> A1 "HD" "NG"
#> A2 "HD" "NI"
#> A3 "NI" "HD"
tales_consensus(aln)
#> [1] "HD" NA  

# From a tales_msa, such as tales_align() returns
msa <- tales_msa(data.frame(
  array_id = c("A1", "A1", "A1", "A2", "A2", "A3", "A3", "A3"),
  position_in_array = c(1L, 2L, 3L, 1L, 2L, 1L, 2L, 3L),
  alignment_position = c(1L, 2L, 3L, 1L, 3L, 1L, 2L, 3L),
  rvd = c("NTERM", "HD", "CTERM", "NTERM", "CTERM", "NTERM", "HD", "CTERM")
))
tales_consensus(as.matrix(msa))
#> [1] "NTERM" "HD"    "CTERM"