r - data.frames 列表按行快速矢量化合并

Question

关于在 SO 上的列表中合并 data.frame 的大多数问题与我在这里想要表达的内容不太相关，但请随意证明我错了。

我有一个 data.frames 列表。我想将行“rbind”到另一个 data.frame 中。本质上，所有第一行形成一个 data.frame，第二行形成第二个 data.frame，依此类推。结果将是一个与原始 data.frame 中的行数相同长度的列表。到目前为止，data.frames 的尺寸是相同的。

这里有一些可以使用的数据。

sample.list <- list(data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),
        data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)))

这就是我想出的优秀的 ol' for 循环。

#solution 1
my.list <- vector("list", nrow(sample.list[[1]]))
for (i in 1:nrow(sample.list[[1]])) {
    for (j in 1:length(sample.list)) {
        my.list[[i]] <- rbind(my.list[[i]], sample.list[[j]][i, ])
    }
}

#solution 2 (so far my favorite)
sample.list2 <- do.call("rbind", sample.list)
my.list2 <- vector("list", nrow(sample.list[[1]]))

for (i in 1:nrow(sample.list[[1]])) {
    my.list2[[i]] <- sample.list2[seq(from = i, to = nrow(sample.list2), by = nrow(sample.list[[1]])), ]
}

可以使用矢量化来改进这一点而不需要太多脑损伤吗？当然，正确的答案将包含一段代码。 "is"作为答案不算数。

编辑

#solution 3 (a variant of solution 2 above)
ind <- rep(1:nrow(sample.list[[1]]), times = length(sample.list))
my.list3 <- split(x = sample.list2, f = ind)

基准测试

我已经使我的列表更大，每个数据帧有更多行。我对结果进行了基准测试，结果如下:

#solution 1
system.time(for (i in 1:nrow(sample.list[[1]])) {
    for (j in 1:length(sample.list)) {
        my.list[[i]] <- rbind(my.list[[i]], sample.list[[j]][i, ])
    }
})
   user  system elapsed 
 80.989   0.004  81.210 

# solution 2
system.time(for (i in 1:nrow(sample.list[[1]])) {
    my.list2[[i]] <- sample.list2[seq(from = i, to = nrow(sample.list2), by = nrow(sample.list[[1]])), ]
})
   user  system elapsed 
  0.957   0.160   1.126 

# solution 3
system.time(split(x = sample.list2, f = ind))
   user  system elapsed 
  1.104   0.204   1.332 

# solution Gabor
system.time(lapply(1:nr, bind.ith.rows))
   user  system elapsed 
  0.484   0.000   0.485 

# solution ncray
system.time(alply(do.call("cbind",sample.list), 1,
                .fun=matrix, ncol=ncol(sample.list[[1]]), byrow=TRUE,
                dimnames=list(1:length(sample.list),names(sample.list[[1]]))))
   user  system elapsed 
 11.296   0.016  11.365

Answer 1

试试这个:

bind.ith.rows <- function(i) do.call(rbind, lapply(sample.list, "[", i, TRUE))
nr <- nrow(sample.list[[1]])
lapply(1:nr, bind.ith.rows)