累积粘贴(连接)由另一个变量分组的值

Question

我在处理R中的数据帧时遇到问题.我想根据另一列中单元格的值将单元格的内容粘贴到不同的行中.我的问题是我希望逐步(累积)打印输出.输出向量必须与输入向量的长度相同.这是一个类似于我正在处理的样本表:

id <- c("a", "a", "a", "b", "b", "b")
content <- c("A", "B", "A", "B", "C", "B")
(testdf <- data.frame(id, content, stringsAsFactors=FALSE))
#  id content
#1  a       A
#2  a       B
#3  a       A
#4  b       B
#5  b       C
#6  b       B

这是希望我希望结果看起来像:

result <- c("A", "A B", "A B A", "B", "B C", "B C B") 
result

#[1] "A"     "A B"   "A B A" "B"     "B C"   "B C B"

我不需要这样的东西:

ddply(testdf, .(id), summarize, content_concatenated = paste(content, collapse = " "))

#  id content_concatenated
#1  a                A B A
#2  b                B C B

Answer 1

您可以使用以下命令定义"累积粘贴"功能Reduce:

cumpaste = function(x, .sep = " ") 
          Reduce(function(x1, x2) paste(x1, x2, sep = .sep), x, accumulate = TRUE)

cumpaste(letters[1:3], "; ")
#[1] "a"       "a; b"    "a; b; c"

Reduce循环避免了从开始重新连接元素,因为它延长了下一个元素的先前串联.

按组应用:

ave(as.character(testdf$content), testdf$id, FUN = cumpaste)
#[1] "A"     "A B"   "A B A" "B"     "B C"   "B C B"

另一个想法是,可以在开始时连接整个向量,然后逐步substring:

cumpaste2 = function(x, .sep = " ")
{
    concat = paste(x, collapse = .sep)
    substring(concat, 1L, cumsum(c(nchar(x[[1L]]), nchar(x[-1L]) + nchar(.sep))))
}
cumpaste2(letters[1:3], " ;@-")
#[1] "a"           "a ;@-b"      "a ;@-b ;@-c"

这似乎也有点快:

set.seed(077)
X = replicate(1e3, paste(sample(letters, sample(0:5, 1), TRUE), collapse = ""))
identical(cumpaste(X, " --- "), cumpaste2(X, " --- "))
#[1] TRUE
microbenchmark::microbenchmark(cumpaste(X, " --- "), cumpaste2(X, " --- "), times = 30)
#Unit: milliseconds
#                  expr      min       lq     mean   median       uq      max neval cld
#  cumpaste(X, " --- ") 21.19967 21.82295 26.47899 24.83196 30.34068 39.86275    30   b
# cumpaste2(X, " --- ") 14.41291 14.92378 16.87865 16.03339 18.56703 23.22958    30  a

......这就是它cumpaste_faster.

Answer 2

data.table解决方案

library(data.table)
setDT(testdf)[, content2 := sapply(seq_len(.N), function(x) paste(content[seq_len(x)], collapse = " ")), by = id]
testdf

##    id content content2
## 1:  a       A        A
## 2:  a       B      A B
## 3:  a       A    A B A
## 4:  b       B        B
## 5:  b       C      B C
## 6:  b       B    B C B

Answer 3

这是一种ddply使用子集sapply以增量方式粘贴在一起的方法：

library(plyr)
ddply(testdf, .(id), mutate, content_concatenated = sapply(seq_along(content), function(x) paste(content[seq(x)], collapse = " ")))
  id content content_concatenated
1  a       A                    A
2  a       B                  A B
3  a       A                A B A
4  b       B                    B
5  b       C                  B C
6  b       B                B C B