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
答案 0 :(得分：46)
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)
答案 1 :(得分：39)
使用data.table
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
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！
rbindlist 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。
答案 2 :(得分：5)
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]])
))
答案 3 :(得分：0)
addind tidyverse解决方案：
library(tidyverse)
bind_rows(sample.list)