Question

id  bleed   episodes
J1  0   0
JI  0   1
J1  0   1
JI  yes 0
J2  0   0
J2  0   1
J2  0   1
J2  0   1
J2  yes 0
J2  0   0
J3  0   1
J3  0   1
J3  0   0
J3  0   1
J3  yes 0
J3  0   0

我想折叠这些数据来计算每个人看起来像这样的出血发生次数的剧集数

id  episodes
J1  2
J2  3
J3  3

观察是在不同时间进行的，我没有包括时间变量，而是每周

Answer 1

使用此示例输入

dd <- structure(list(id = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("J1", "J2", "J3"), class = "factor"), 
    bleed = structure(c(1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L, 1L, 1L, 1L, 2L, 1L), .Label = c("0", "yes"), class = "factor"), 
    episodes = c(0L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 1L, 
    1L, 0L, 1L, 0L, 0L)), .Names = c("id", "bleed", "episodes"
), class = "data.frame", row.names = c(NA, -16L))

您可以使用dplyr

完成此任务

library(dplyr)

dd %>% group_by(id) %>% 
    mutate(bleed_count=cumsum(bleed=="yes")) %>%
    filter(bleed_count==0) %>%
    summarize(episodes = sum(episodes))

对布尔值使用cumsum()来跟踪出血的时间。然后我们在第一次出血之前仅将这些值相加

Answer 2

我们也可以使用data.table

执行此操作

library(data.table)
setDT(df1)[,  .(episodes = sum(episodes[seq_len(which(bleed == "yes")[1])])), id]
#    id episodes
#1: J1        2
#2: J2        3
#3: J3        3

数据

df1 <- structure(list(id = c("J1", "J1", "J1", "J1", "J2", "J2", "J2", 
"J2", "J2", "J2", "J3", "J3", "J3", "J3", "J3", "J3"), bleed = c("0", 
"0", "0", "yes", "0", "0", "0", "0", "yes", "0", "0", "0", "0", 
"0", "yes", "0"), episodes = c(0L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 
0L, 0L, 1L, 1L, 0L, 1L, 0L, 0L)), .Names = c("id", "bleed", "episodes"
), class = "data.frame", row.names = c(NA, -16L))

折叠重复观察计数给定变量的总和到某一点

2 个答案:

数据