我有这个数据集：
ID Set Type Count
1 1 1 A NA
2 2 1 R NA
3 3 1 R NA
4 4 1 U NA
5 5 1 U NA
6 6 1 U NA
7 7 2 A NA
8 8 3 R NA
9 9 3 R NA
作为dputs：
mystart <- structure(list(ID = 1:9, Set = c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
3L, 3L), Type = structure(c(1L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L
), .Label = c("A", "R", "U"), class = "factor"), Count = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("ID", "Set", "Type",
"Count"), class = "data.frame", row.names = c(NA, -9L))
使用dplyr包我该如何获得：
ID Set Type Count
1 1 1 A A1
2 2 1 R A1R1
3 3 1 R A1R2
4 4 1 U A1R2U1
5 5 1 U A1R2U2
6 6 1 U A1R2U3
7 7 2 A A1
8 8 3 R R1
9 9 3 R R2
再次dputs：
myend <- structure(list(ID = 1:9, Set = c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
3L, 3L), Type = structure(c(1L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L
), .Label = c("A", "R", "U"), class = "factor"), Count = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 1L, 7L, 8L), .Label = c("A1", "A1R1", "A1R2",
"A1R2U1", "A1R2U2", "A1R2U3", "R1", "R2"), class = "factor")), .Names = c("ID",
"Set", "Type", "Count"), class = "data.frame", row.names = c(NA,
-9L))
<小时/> 简而言之，我想计算列
"type"中列"set"的观察结果，并累计打印此count(text)。
检查类似的帖子，我接近这个：
myend <- structure(list(ID = 1:9, Set = c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
3L, 3L), Type = structure(c(1L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L
), .Label = c("A", "R", "U"), class = "factor"), Count = c(1L,
1L, 2L, 1L, 2L, 3L, 1L, 1L, 2L)), .Names = c("ID", "Set", "Type",
"Count"), class = "data.frame", row.names = c(NA, -9L))
使用代码：
library(dplyr)
myend <- read.table("mydata.txt", header=TRUE, fill=TRUE)
myend %>%
group_by(Set, Type) %>%
mutate(Count = seq(n())) %>%
ungroup(myend)
非常感谢你的帮助，
答案 0 :(得分：7)
Base R版本：
aggregateGroup <- function(x){
vecs <- Reduce(f=function(a,b){ a[b] <- sum(a[b],1L,na.rm=TRUE); a },
init=integer(0),
as.character(x),
accumulate = TRUE)
# vecs is a list with something like this :
# [[1]]
# integer(0)
# [[2]]
# A
# 1
# [[3]]
# A R
# 1 1
# ...
# so we simply turn those vectors into characters using vapply and paste
# (excluding the first)
vapply(vecs,function(y) paste0(names(y),y,collapse=''),FUN.VALUE='')[-1]
}
split(mystart$Count,mystart$Set) <- lapply(split(mystart$Type,mystart$Set), aggregateGroup)
> mystart
ID Set Type Count
1 1 1 A A1
2 2 1 R A1R1
3 3 1 R A1R2
4 4 1 U A1R2U1
5 5 1 U A1R2U2
6 6 1 U A1R2U3
7 7 2 A A1
8 8 3 R R1
9 9 3 R R2
答案 1 :(得分：6)
dplyr版本：
mystart %>%
group_by(Set) %>%
mutate(Count = paste0('A', cumsum(Type == 'A'),
'R', cumsum(Type == 'R'),
'U', cumsum(Type == 'U'))) %>%
ungroup()
哪个收益
# A tibble: 9 x 4
ID Set Type Count
<int> <int> <chr> <chr>
1 1 1 A A1R0U0
2 2 1 R A1R1U0
3 3 1 R A1R2U0
4 4 1 U A1R2U1
5 5 1 U A1R2U2
6 6 1 U A1R2U3
7 7 2 A A1R0U0
8 8 3 R A0R1U0
9 9 3 R A0R2U0
<小时/> 如果你想省略计数为零的变量，你需要像它一样包围一个函数
mygroup <- function(lst) {
name <- names(lst)
vectors <- lapply(seq_along(lst), function(i) {
x <- lst[[i]]
char <- name[i]
x <- ifelse(x == 0, "", paste0(char, x))
return(x)
})
return(do.call("paste0", vectors))
}
mystart %>%
group_by(Set) %>%
mutate(Count = mygroup(list(A = cumsum(Type == 'A'),
R = cumsum(Type == 'R'),
U = cumsum(Type == 'U')))) %>%
ungroup()
这会产生
# A tibble: 9 x 4
ID Set Type Count
<int> <int> <chr> <chr>
1 1 1 A A1
2 2 1 R A1R1
3 3 1 R A1R2
4 4 1 U A1R2U1
5 5 1 U A1R2U2
6 6 1 U A1R2U3
7 7 2 A A1
8 8 3 R R1
9 9 3 R R2
答案 2 :(得分：4)
一行解决data.table
你必须先做
require(data.table)
mystart <- as.data.table(mystart)
然后只使用一行
mystart[, .(Type,
count = paste0(
'A',
cumsum(Type == 'A'),
'R',
countR = cumsum(Type == 'R'),
'U',
countU = cumsum(Type == 'U')
)),
by = c('Set')]
首先，您需要每个类型的cumsum并通过＆＃39; set＆＃39;
将它们粘贴在一起 cumsum(Type=='A')等于计数，因为当Type==A时，它是1，否则它是0。
您也希望将它们粘贴到一列中。因此，paste0()很好用。
您仍然需要Type列，因此我在行中添加了Type。
输出：
Set Type count
1: 1 A A1R0U0
2: 1 R A1R1U0
3: 1 R A1R2U0
4: 1 U A1R2U1
5: 1 U A1R2U2
6: 1 U A1R2U3
7: 2 A A1R0U0
8: 3 R A0R1U0
9: 3 R A0R2U0
希望这有帮助。
顺便说一下，如果你想忽略count 0，你必须自己设计一些if-esle条款。
基本上你想要这个：如果cumsum(something) ==0，NULL，esle paste0('something', cumsum(something))，那么你paste0()他们在一起。
它会变得讨厌，我不会写它。你明白了这个想法
答案 3 :(得分：3)
为此问题提供了许多优雅的解决方案。我仍在寻找dplyr方式（在固定类型上没有 - cumsum）。该函数足够通用，可以处理Type的其他值。
借助custom function作为解决方案：
library(dplyr)
mystart %>% group_by(Set, Type) %>%
mutate(type_count = row_number()) %>%
mutate(TypeMod = paste0(Type,type_count)) %>%
group_by(Set) %>%
mutate(Count = cumCat(TypeMod, type_count)) %>%
select(-type_count, -TypeMod)
cumCat <- function(x, y){
retVal <- character(length(x))
prevVal = ""
lastGrpVal = ""
for ( i in seq_along(x)){
if(y[i]==1){
lastGrpVal = prevVal
}
retVal[i] = paste0(lastGrpVal,x[i])
prevVal = retVal[i]
}
retVal
}
# # Groups: Set [3]
# ID Set Type Count
# <int> <int> <fctr> <chr>
# 1 1 1 A A1
# 2 2 1 R A1R1
# 3 3 1 R A1R2
# 4 4 1 U A1R2U1
# 5 5 1 U A1R2U2
# 6 6 1 U A1R2U3
# 7 7 2 A A1
# 8 8 3 R R1
# 9 9 3 R R2
答案 4 :(得分：3)
这是一个基本解决方案。
我们可以将原始字母粘贴到字母组的seq_along以获取最后2个字符，然后使用paste将结果Reduce粘贴到上一个结果的最后一个元素。
除此之外，我们使用ave按组计算。
fun <- function(x,y) paste0(x[length(x)],y,seq_along(y))
mystart$Count <- ave(as.character(mystart$Type),mystart$Set,
FUN = function(x) unlist(Reduce(fun,split(x,x),init=NULL,accumulate = TRUE)))
# ID Set Type Count
# 1 1 1 A A1
# 2 2 1 R A1R1
# 3 3 1 R A1R2
# 4 4 1 U A1R2U1
# 5 5 1 U A1R2U2
# 6 6 1 U A1R2U3
# 7 7 2 A A1
# 8 8 3 R R1
# 9 9 3 R R2
详情
split(x,x)分割字母，如第一组所示：
with(subset(mystart,Set==1),split(Type,Type))
# $A
# [1] "A"
#
# $R
# [1] "R" "R"
#
# $U
# [1] "U" "U" "U"
然后fun执行此类操作，由Reduce帮助：
fun(NULL,"A") # [1] "A1"
fun("A1",c("R","R")) # [1] "A1R1" "A1R2"
fun(c("A1R1","A1R2"),c("U","U","U")) # [1] "A1R2U1" "A1R2U2" "A1R2U3"
奖金解决方案
这个使用rle并避免split的其他基本解决方案为给定示例提供了相同的输出（每当Type值按集合分组时），但不是mystart2 <- rbind(mystart,mystart)。
fun2 <- function(x){
rle_ <- rle(x)
suffix <- paste0(x,sequence(rle_$length))
prefix <- unlist(mapply(rep,
lag(unlist(
Reduce(paste0,paste0(rle_$values,rle_$lengths),accumulate=TRUE)
),rle_$lengths[1]),
each=rle_$lengths))
prefix[is.na(prefix)] <- ""
paste0(prefix,suffix)
}
mystart$Count2 <-ave(as.character(mystart$Type), mystart$Set,FUN=fun2)