按行列出data.frames的快速矢量化合并

Question

关于在SO列表中合并data.frame的大多数问题与我试图在这里得到的内容并不完全相关，但可以随意证明我的错误。

我有一个data.frames列表。我想逐行“rbind”行到另一个data.frame。实质上，所有第一行形成一个data.frame，第二行形成第二个data.frame等等。 结果将是与原始data.frame中的行数相同的长度列表。到目前为止，data.frames的尺寸相同。

以下是一些数据。

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)))

这就是我提出的好的'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]])), ]
}

使用矢量化可以改善这种情况吗？当然，正确答案将包含一段代码。 “是”作为答案并不重要。

修改的

#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)

BENCHMARKING

我的列表越大，每个data.frame的行数越多。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)

Answer 2

使用data.table

可以更快地解决这个问题

编辑 - 更大的数据集显示data.table更加精彩。

# here are some sample data 
sample.list <- replicate(10000, data.frame(x = sample(1:100, 10), 
  y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)), simplify = F)

Gabor的快速解决方案：

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

##    user  system elapsed 
##   25.87    0.01   25.92 

即使使用data.frames，data.table函数rbindlist也会使甚至更快。

library(data.table)
fastbind.ith.rows <- function(i) rbindlist(lapply(sample.list, "[", i, TRUE))
system.time(fastbound <- lapply(1:nr, fastbind.ith.rows))

##    user  system elapsed 
##   13.89    0.00   13.89 

A data.table解决方案

这是一个使用data.tables的解决方案 - 它是类固醇的split解决方案。

# data.table solution
system.time({
    # change each element of sample.list to a data.table (and data.frame) this
    # is done instaneously by reference
    invisible(lapply(sample.list, setattr, name = "class", 
               value = c("data.table",  "data.frame")))
    # combine into a big data set
    bigdata <- rbindlist(sample.list)
    # add a row index column (by refere3nce)
    index <- as.character(seq_len(nr))
    bigdata[, `:=`(rowid, index)]
    # set the key for binary searches
    setkey(bigdata, rowid)
    # split on this -
    dt_list <- lapply(index, function(i, j, x) x[i = J(i)], x = bigdata)
    # if you want to drop the `row id` column
    invisible(lapply(dt_list, function(x) set(x, j = "rowid", value = NULL)))
    # if you really don't want them to be data.tables run this line
    # invisible(lapply(dt_list, setattr,name = 'class', value =
    # c('data.frame')))
})
################################
##    user  system elapsed    ##
##    0.08    0.00    0.08    ##
################################

data.table真棒！

使用rbindlist

的警告用户

rbindlist很快，因为它不会执行do.call(rbind,....)的检查。例如，它假定任何因子列与列表的第一个元素具有相同的级别。

Answer 3

这是我与plyr的尝试，但我喜欢G.格洛腾迪克的方法：

library(plyr)
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]])
      ))

Answer 4

addind tidyverse解决方案：

library(tidyverse)
bind_rows(sample.list)