嗨，晚上来自德国：）
我对R很陌生，但我真的在理解极限。
基本上我有n个矩阵，它们在列表中。它们看起来像这样：
$edu
cue op split pred
1 edu < 1 TRUE
2 edu > 1 TRUE
3 edu == 1 TRUE
4 edu < 2 TRUE
5 edu > 2 TRUE
6 edu == 2 TRUE
7 edu < 3 TRUE
8 edu > 3 TRUE
9 edu == 3 TRUE
$religion
cue op split pred
1 religion == 0 TRUE
2 religion == 1 TRUE
3 religion == 0 FALSE
4 religion == 1 FALSE
$med_exp
cue op split pred
1 med_exp == 0 TRUE
2 med_exp == 1 TRUE
3 med_exp == 0 FALSE
4 med_exp == 1 FALSE
我需要的是与“expand.grid（）”类似的东西。我需要在所有可能的排列中混合在一起的所有项目（我已经检查了'combinat'包），但是在它们的原始列顺序中（med_exp例如不应该'拆分超过1且只有“==”作为运算符！） 。
如果我在表格中多次使用相同的“cuetest”（一个列表中的一行），我会适得其反。我最好在矩阵中需要数据，因为我想使用“parRapply”。这是表格应该是什么样子
cue op split pred cue op split pred cue op split pred
edu < 1 TRUE religion == 0 TRUE med_exp == 0 TRUE
edu < 1 TRUE religion == 0 TRUE med_exp == 1 TRUE
edu < 1 TRUE religion == 0 TRUE med_exp == 0 FALSE
edu < 1 TRUE religion == 0 TRUE med_exp == 1 FALSE
[..]
med_exp == 0 TRUE edu < 1 TRUE religion == 0 TRUE
问题是，expand.grid甚至会混淆子列，这只是给我带来内存麻烦而且没有必要。
以下是要试验的数据：
structure(list(edu = structure(list(cue = structure(c(1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L), .Label = "edu", class = "factor"), op = structure(c(1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("<", ">", "=="), class = "factor"),
split = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L,
1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L), pred = c(TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE)), .Names = c("cue", "op", "split",
"pred"), out.attrs = structure(list(dim = c(1L, 3L, 4L, 2L),
dimnames = structure(list(Var1 = "Var1=edu", Var2 = c("Var2=<",
"Var2=>", "Var2==="), Var3 = c("Var3=1", "Var3=2", "Var3=3",
"Var3=4"), Var4 = c("Var4=TRUE", "Var4=FALSE")), .Names = c("Var1",
"Var2", "Var3", "Var4"))), .Names = c("dim", "dimnames")), class = "data.frame", row.names = c(NA,
-24L)), edu_hus = structure(list(cue = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L), .Label = "edu_hus", class = "factor"), op = structure(c(1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("<", ">", "=="), class = "factor"),
split = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L,
1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L), pred = c(TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE)), .Names = c("cue", "op", "split",
"pred"), out.attrs = structure(list(dim = c(1L, 3L, 4L, 2L),
dimnames = structure(list(Var1 = "Var1=edu_hus", Var2 = c("Var2=<",
"Var2=>", "Var2==="), Var3 = c("Var3=1", "Var3=2", "Var3=3",
"Var3=4"), Var4 = c("Var4=TRUE", "Var4=FALSE")), .Names = c("Var1",
"Var2", "Var3", "Var4"))), .Names = c("dim", "dimnames")), class = "data.frame", row.names = c(NA,
-24L)), religion = structure(list(cue = structure(c(1L, 1L, 1L,
1L), .Label = "religion", class = "factor"), op = structure(c(1L,
1L, 1L, 1L), .Label = "==", class = "factor"), split = c(0L,
1L, 0L, 1L), pred = c(TRUE, TRUE, FALSE, FALSE)), .Names = c("cue",
"op", "split", "pred"), out.attrs = structure(list(dim = c(1L,
1L, 2L, 2L), dimnames = structure(list(Var1 = "Var1=religion",
Var2 = "Var2===", Var3 = c("Var3=0", "Var3=1"), Var4 = c("Var4=TRUE",
"Var4=FALSE")), .Names = c("Var1", "Var2", "Var3", "Var4"
))), .Names = c("dim", "dimnames")), class = "data.frame", row.names = c(NA,
-4L))), .Names = c("edu", "edu_hus", "religion"))
非常感谢， 马克
答案 0 :(得分：3)
请注意，dput提供的数据与之前显示的不同。此外，列表的元素是data.frames而不是矩阵。前者是有道理的，因为你有不同类型（因素，整数，逻辑）的混合，矩阵只能容纳一种类型。因此，我在这里给出的代码也返回一个data.frame。您可以随时使用as.matrix，但我建议您在大多数情况下不要这样做。
#create combinations of row indices
ind <- expand.grid(seq_len(nrow(dat[[3]])),
seq_len(nrow(dat[[2]])),
seq_len(nrow(dat[[1]])))
#use subsetting and cbind
res <- cbind(dat[[1]][ind[,3],],
dat[[2]][ind[,2],],
dat[[3]][ind[,1],])
head(res)
# cue op split pred cue op split pred cue op split pred
# 1 edu < 1 TRUE edu_hus < 1 TRUE religion == 0 TRUE
# 1.1 edu < 1 TRUE edu_hus < 1 TRUE religion == 1 TRUE
# 1.2 edu < 1 TRUE edu_hus < 1 TRUE religion == 0 FALSE
# 1.3 edu < 1 TRUE edu_hus < 1 TRUE religion == 1 FALSE
# 1.4 edu < 1 TRUE edu_hus > 1 TRUE religion == 0 TRUE
# 1.5 edu < 1 TRUE edu_hus > 1 TRUE religion == 1 TRUE
如果您有更多列，则可以使用lapply和do.call来概括该方法：
sl <- lapply(dat, function(df) seq_len(nrow(df)))
sl <- sl[rev(seq_along(sl))]
ind <- do.call(expand.grid, sl)
ind <- ind[,rev(seq_along(dat))]
res <- do.call(cbind, lapply(seq_along(dat), function(i) dat[[i]][ind[,i],]))