此数据表显示学生每年上学的月份。
DT = data.table(
Student = c(1, 1, 1, 1, 1, 1, 1, 1, 1,
2, 2, 2, 2, 2, 2, 2, 2,
3, 3, 3, 3, 3, 3, 3, 3),
Month = c(1, 2, 3, 5, 6, 7, 8, 11, 12,
2, 3, 4, 5, 7, 8, 9, 10,
1, 2, 3, 5, 6, 7, 8, 9))
DT
Student Month
1: 1 1
2: 1 2
3: 1 3
4: 1 5
5: 1 6
6: 1 7
7: 1 8
8: 1 11
9: 1 12
10: 2 2
11: 2 3
12: 2 4
13: 2 5
14: 2 7
15: 2 8
16: 2 9
17: 2 10
18: 3 1
19: 3 2
20: 3 3
21: 3 5
22: 3 6
23: 3 7
24: 3 8
25: 3 9
我想确定连续三个月的期间（由该期间的第一个月确定）。这是数据表和合格期间的可视化。
1 2 3 4 5 6 7 8 9 10 11 12
1 * * * * * * * * *
[-------] [-------]
[-------]
2 * * * * * * * *
[-------] [-------]
[-------] [-------]
3 * * * * * * * *
[-------] [-------]
[-------]
[-------]
所需的输出：
id First_month_in_the_period
1 1
1 5
1 6
2 2
2 3
2 7
2 8
3 1
3 5
3 6
3 7
正在寻找data.table（或dplyr）解决方案。
答案 0 :(得分：7)
使用standard method（cumsum...diff...condition）识别连续值的运行，然后将其与“学生”一起用作分组变量。在每个组中，根据每次运行的时间创建序列，然后添加到第一个月。
DT[ , .(start = if(.N >= 3) Month[1] + 0:(.N - 3)),
by = .(Student, r = cumsum(c(1L, diff(Month) > 1)))]
# Student r start
# 1: 1 1 1
# 2: 1 2 5
# 3: 1 2 6
# 4: 2 3 2
# 5: 2 3 3
# 6: 2 4 7
# 7: 2 4 8
# 8: 3 4 1
# 9: 3 5 5
# 10: 3 5 6
# 11: 3 5 7
等效的dplyr替代项：
DT %>%
group_by(Student, r = cumsum(c(1L, diff(Month) > 1))) %>%
summarise(list(data.frame(start = if(n() >= 3) Month[1] + 0:(n() - 3)))) %>%
tidyr::unnest()
# # A tibble: 11 x 3
# # Groups: Student [3]
# Student r start
# <dbl> <int> <dbl>
# 1 1 1 1
# 2 1 2 5
# 3 1 2 6
# 4 2 3 2
# 5 2 3 3
# 6 2 4 7
# 7 2 4 8
# 8 3 4 1
# 9 3 5 5
# 10 3 5 6
# 11 3 5 7
答案 1 :(得分：6)
使用tidyverse的解决方案。
library(tidyverse)
library(data.table)
DT2 <- DT %>%
arrange(Student, Month) %>%
group_by(Student) %>%
# Create sequence of 3
mutate(Seq = map(Month, ~seq.int(.x, .x + 2L))) %>%
# Create a flag to show if the sequence match completely with the Month column
mutate(Flag = map_lgl(Seq, ~all(.x %in% Month))) %>%
# Filter the Flag for TRUE
filter(Flag) %>%
# Remove columns
select(-Seq, -Flag) %>%
ungroup()
DT2
# # A tibble: 11 x 2
# Student Month
# <dbl> <dbl>
# 1 1 1
# 2 1 5
# 3 1 6
# 4 2 2
# 5 2 3
# 6 2 7
# 7 2 8
# 8 3 1
# 9 3 5
# 10 3 6
# 11 3 7
答案 2 :(得分：3)
这是一个解决方案，它使用data.table提供的分组，
seqfun <- function(month) {
n <- length(month)
tmp <- data.table(a=month[1:(n-2)],b=month[2:(n-1)],c=month[3:n])
month[which(apply(tmp,1,function(x){all(c(1,1)==diff(x))}))]}
Result <- DT[,seqfun(Month), by=Student]
names(Result) <- c("Student","Month")
> Result
Student Month
1: 1 1
2: 1 5
3: 1 6
4: 2 2
5: 2 3
6: 2 7
7: 2 8
8: 3 1
9: 3 5
10: 3 6
11: 3 7
基本上，它使用组月份向量，创建3个向量来比较diff，并检查两个diff是否相距1。如果是，则返回原始月份向量的索引。 / p>
一点点细节。假设我们有
month <- c(1,2,3,5,6,7,8,11,12)
，我们计算出tmp data.table（注意：您还可以使用rollapply中的zoo函数来创建类似的表格，我将在非常底部）
a b c
1: 1 2 3
2: 2 3 5
3: 3 5 6
4: 5 6 7
5: 6 7 8
6: 7 8 11
7: 8 11 12
当我们将diff跨行使用时，我们得到
> apply(tmp,1,function(x){all(c(1,1)==diff(x))})
[1] TRUE FALSE FALSE TRUE TRUE FALSE FALSE
真实值是我们感兴趣的指标。
如上所述，使用zoo库的rollapply，我们可以拥有
> apply(c(1,1)==rollapply(month,width=3,FUN=diff),1,all)
[1] TRUE FALSE FALSE TRUE TRUE FALSE FALSE
获取特定学生感兴趣的索引的布尔向量。
答案 3 :(得分：2)
这是一个base R解决方案，它创建了可以应用于data.table的函数：
cons3fun<-function(x,n){
consec.list<-split(x,cumsum(c(1,diff(x)!=1))) #Splits into list based on consecutive numbers
min.len.seq<-consec.list[which(sapply(consec.list,length)>(n-1))] #Selects only the list elements >= to n
seq.start<-lapply(min.len.seq,function(i) i[1:(length(i)-(n-1))]) #Extracts the first number of each sequence of n
return(as.vector(unlist(seq.start))) #Returns result as a vector
}
请注意，此功能将使您可以相当容易地更改要查找的连续数字的数量。在这里，您将使用n=3。然后，您可以使用data.table或dplyr应用此功能。我将使用data.table，因为您使用过。
DT[,cons3fun(Month,3),by=.(Student)]
希望您觉得这很有用。祝你好运！
答案 4 :(得分：1)
这是我使用tidyverse的方法：
> as_tibble(DT) %>%
arrange(Student, Month) %>%
group_by(Student) %>%
# create an identifier for the start of the sequence
mutate(seq_id = ifelse(row_number() == 1 | Month - lag(Month) > 1,
letters[row_number()], NA)) %>%
fill(seq_id) %>%
# add another grouping level (sequence identifier)
group_by(Student, seq_id) %>%
# only keep data with attendance in 3 or more consecutive months
filter(length(seq_id) > 2) %>%
# n consecutive months => n - 2 periods
slice(1:(n() - 2)) %>%
# clean up
ungroup() %>%
select(Student, Month)
# A tibble: 11 x 2
# Student Month
# <dbl> <dbl>
#1 1 1
#2 1 5
#3 1 6
#4 2 2
#5 2 3
#6 2 7
#7 2 8
#8 3 1
#9 3 5
#10 3 6
#11 3 7
答案 5 :(得分：1)
另一种data.table方法...
#first, clculate the difference between months, by student.
ans <- DT[, diff := shift( Month, type = "lead" ) - Month ), by = .(Student)]
#then filter rows that are at the start of 2 consecutive differences of 1
#also, drop the temporary diff-column
ans[ diff == 1 & shift( diff, type = "lead" ) == 1,][, diff := NULL][]
瞧
# Student Month
# 1: 1 1
# 2: 1 5
# 3: 1 6
# 4: 2 2
# 5: 2 3
# 6: 2 7
# 7: 2 8
# 8: 3 1
# 9: 3 5
# 10: 3 6
# 11: 3 7