Question

正如标题所说，我想将每个群集保存在pdf文件的单独页面上。

示例数据：

structure(list(P1 = c("ATCG00490", "AT5G17710", "AT2G42910", 
"AT4G23600", "AT3G61540", "AT2G05990"), P2 = c("AT5G38420", "AT5G20070", 
"AT5G04230", "AT1G08200", "AT4G30910", "AT5G52100"), clique = structure(list(
    `930` = integer(0), `2090` = integer(0), `3120` = c(2L, 3L, 
    231L), `3663` = integer(0), `3704` = integer(0), `4156` = c(19L, 
    27L)), .Names = c("930", "2090", "3120", "3663", "3704", 
"4156"), class = "AsIs")), .Names = c("P1", "P2", "clique"), row.names = c(930L, 
2090L, 3120L, 3663L, 3704L, 4156L), class = "data.frame")

有些行属于许多群集，有些行只属于单个群集。当然，必须考虑所有可能的变体。

如果可能，我只想保留至少有两个成员的集群。

如果每行都属于单个集群，那就是我使用的函数：

pdf("clusters.pdf", , width=12, height=18)
lapply(split(data_cluster, data_cluster$cluster), function(d) {
  grid::grid.newpage()
  gridExtra::grid.table(d)
}
)
dev.off()

也许它会帮助别人找到答案。

编辑：

我在准备示例数据时犯了一个错误...请查看我的原始数据，而不是你会发现那不是那么简单（至少在我看来）。

structure(list(P1 = c("ATCG00490", "AT5G17710", "AT2G42910", 
"AT4G23600", "AT3G61540", "AT2G05990"), P2 = c("AT5G38420", "AT5G20070", 
"AT5G04230", "AT1G08200", "AT4G30910", "AT5G52100"), clique = structure(list(
    `930` = integer(0), `2090` = integer(0), `3120` = c(2L, 3L, 
    231L), `3663` = integer(0), `3704` = integer(0), `4156` = c(19L, 
    27L)), .Names = c("930", "2090", "3120", "3663", "3704", 
"4156"), class = "AsIs")), .Names = c("P1", "P2", "clique"), row.names = c(930L, 
2090L, 3120L, 3663L, 3704L, 4156L), class = "data.frame")

集群在pdf的单独页面上。每行可以属于不同的集群

0 个答案: