r - 在 R 中组合列表的行

Question

我有一个格式列表:

[[1]]
 [1] "10"  "719" "99"  

[[2]]
 [1] "10"  "624" "85"  "888" "624" 

[[3]]
 [1] "1"   "894" "110" "344" "634"  

我想通过列表中第一个元素的唯一值进行合并，即。

[[1]]
 [1] "10"  "719" "99" "624" "85"  "888" "624" 

[[2]]
 [1] "1"   "894" "110" "344" "634"

有没有办法以最少的内存使用来做到这一点？

Answer 1

我会这样处理:

x <- list(c("10",  "719", "99"),
          c("10",  "624", "85",  "888", "624"),
          c("1",   "894", "110", "344", "634"))
first_elems <- sapply(x, "[", 1) # get 1st elem of each vector
(first_elems <- as.factor(first_elems)) # factorize (i.a. find all unique elems)
## [1] 10 10 1 
## Levels: 1 10
(group <- split(x, first_elems)) # split by 1st elem (divide into groups)
## $`1`
## $`1`[[1]]
## [1] "1"   "894" "110" "344" "634"
## 
## 
## $`10`
## $`10`[[1]]
## [1] "10"  "719" "99" 
## 
## $`10`[[2]]
## [1] "10"  "624" "85"  "888" "624"
## 
(result <- lapply(group, unlist)) # combine vectors in each group (list of vectors -> an atomic vector)
## $`1`
## [1] "1"   "894" "110" "344" "634"
## 
## $`10`
## [1] "10"  "719" "99"  "10"  "624" "85"  "888" "624"

编辑 :对于非重复 key ，请使用:

(result <- lapply(group, function(x) {
      c(x[[1]][1], unlist(lapply(x, "[", -1)))
   }))
## $`1`
## [1] "1"   "894" "110" "344" "634"
## 
## $`10`
## [1] "10"  "719" "99"  "624" "85"  "888" "624"

不需要太多额外的内存。除了结果列表，我们需要存储 as.factor 的结果(类数 + 元素数 x )。 split需要很少的额外内存 - x 中的向量没有深度复制。

至于性能，对于相当大的列表:

set.seed(1L)
n <- 100000
x <- vector('list', n)
for (i in 1:n)
   x[[i]] <- as.character(sample(1:1000, ceiling(runif(1, 1, 1000)), replace=TRUE))
object.size(x) # 2GB
## 2175165880 bytes

我在我的旧 Linux 笔记本电脑上获得了以下运行时间:

system.time(local({
   first_elems <- as.factor(sapply(x, "[", 1))
   group <- split(x, first_elems)
   result <- lapply(group, function(x) {
     c(x[[1]][1], unlist(lapply(x, "[", -1)))
   })
}))

##    user  system elapsed 
##   4.119   0.001   4.149 

看起来很有道理，我想。