匹配和计数R中的数据矩阵 [英] Match and Count the Data Matrix in R

问题描述

数据集如下:

 Gene SampleName
gene1    sample1
gene1    sample2
gene1    sample3
gene2    sample2
gene2    sample3
gene2    sample4
gene3    sample1
gene3    sample5

我的目标是制作一个像这样的数据矩阵:

My goal is to make a data matrix like this:

       gene1 gene2 gene3
gene1      -     2     1
gene2      -     -     0
gene3      -     -     -

gene1与gene2之所以是2，是因为它们共享相同的样本sample2和sample3. gene1 vs gene3为1，因为它们仅共享一个相同的样本-sample1.

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.

我的问题是我如何在R或Perl中实现这个目标?实际数据集要大得多.非常感谢您的帮助.

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.

这是R的dput(df)输出:

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")

推荐答案

您可以将crossprod(或tcrossprod)功能与table一起查看:

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

拖放对角线和下三角形以获得您显示的确切输出.

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                  

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