R - 使用列表名称列出 data.frame

Question

所以我有这个 list :

structure(list(scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L, scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L, scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L), .Names = c("scaf", "pos", "scaf", "pos", "scaf", 
"pos"))

我想得到一个 data.frame 以便两列是 scaf和 pos
这是我得到的:

do.call(rbind, poor.loci)

想要的结果:

scaf     pos
HE638    8
HE638    8
HE638    8

Answer 1

以下是三个可供考虑的选项:

选项 1

直接选择所有其他元素并手动创建您的 data.frame .

setNames(
  data.frame(unlist(poor.loci[c(TRUE, FALSE)], use.names = FALSE),
             unlist(poor.loci[c(FALSE, TRUE)], use.names = FALSE)),
  unique(names(poor.loci)))
#    scaf pos
# 1 HE638   8
# 2 HE638   8
# 3 HE638   8

选项 2

转换您的 list成“长”data.frame和 reshape它变成你想要的形式。也可以使用 melt 使用“reshape2”包来完成和 dcast而不是 stack和 reshape .

X <- stack(lapply(poor.loci, as.character))
X$ID <- ave(X$values, X$values, FUN = seq_along)
reshape(X, direction = "wide", idvar="ID", timevar="ind")
#   ID values.scaf values.pos
# 1  1       HE638          8
# 3  2       HE638          8
# 5  3       HE638          8

选项 3

重新安排您的 list并转换重新排列的 list到 data.frame :

A <- unique(names(poor.loci))
data.frame(
  setNames(
    lapply(A, function(x) unlist(poor.loci[names(poor.loci) %in% x], 
                                 use.names = FALSE)), A))