Reputation: 41
The dataset looks like this:
Gene SampleName
gene1 sample1
gene1 sample2
gene1 sample3
gene2 sample2
gene2 sample3
gene2 sample4
gene3 sample1
gene3 sample5
My goal is to make a data matrix like this:
gene1 gene2 gene3
gene1 - 2 1
gene2 - - 0
gene3 - - -
gene1
vs gene2
is 2
because they share the same samples sample2
and sample3
. gene1
vs gene3
is 1 because they only share one same sample - sample1
.
My question is how can I achieve this goal in R or Perl? The actual data set is much larger. I highly appreciate your help.
Here's the dput(df)
output for R:
df <- structure(list(Gene = c("gene1", "gene1", "gene1", "gene2", "gene2",
"gene2", "gene3", "gene3"), SampleName = c("sample1", "sample2",
"sample3", "sample2", "sample3", "sample4", "sample1", "sample5"
)), .Names = c("Gene", "SampleName"), row.names = c(NA, -8L), class = "data.frame")
Upvotes: 3
Views: 131
Reputation: 50647
perl -lane'
$s{$F[0]}++ or push @k, $F[0];
$h{$F[1]}{$F[0]} = 1;
END {
$, = "\t";
print "", @k;
for $c (@k) {
print $c, map {
$u = $_;
($c eq $u) ? "-" : scalar grep $_->{$c} && $_->{$u}, values %h;
} @k;
}
}
' file
output
gene1 gene2 gene3
gene1 - 2 1
gene2 2 - 0
gene3 1 0 -
Upvotes: 2
Reputation: 193597
You can look at the crossprod
(or tcrossprod
) function along with table
:
out <- tcrossprod(table(df))
out
# Gene
# Gene gene1 gene2 gene3
# gene1 3 2 1
# gene2 2 3 0
# gene3 1 0 2
Drop the diagonal and the lower-triangle to get the exact output you show.
diag(out) <- NA
out[lower.tri(out)] <- NA
print.table(out) ## print.table deals with NAs differently
# Gene
# Gene gene1 gene2 gene3
# gene1 2 1
# gene2 0
# gene3
Upvotes: 8