如何删除行匹配标准和行与之相邻

Question

我有下面的示例数据：

data <- data.table(ID = c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4), 
       date = c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6), 
       score = c(4,3,2,2,4,1,5,5,5,2,1,4,2,1,5,5,5,3,5,5,5,2,4,5)) 

    ID date score 
1: 1 1  4 
2: 1 2  3 
3: 1 3  2 
4: 1 4  2 
5: 1 5  4 
6: 1 6  1 
7: 2 1  5 
8: 2 2  5 
9: 2 3  5 
10: 2 4  2 
11: 2 5  1 
12: 2 6  4 
13: 3 1  2 
14: 3 2  1 
15: 3 3  5 
16: 3 4  5 
17: 3 5  5 
18: 3 6  3 
19: 4 1  5 
20: 4 2  5 
21: 4 3  5 
22: 4 4  2 
23: 4 5  4 
24: 4 6  5 
    ID date score 

我想消除某些行，并改变他人，基于其在表中的位置的部分。我有两个标准，每个ID：

1. 如果行有date == 1和score == 5，我想删除该行和所有后续行有score==5是该行后面紧跟着，直到score是不是5。 （因此，例如，对于I == 4，我想保留日期4,5,6的数据）。

2. 对于score == 5的所有其他日期，我想用其先前两个分数的平均值（或只是他们以前的分数，如果他们只有一个先前分数）替换他们的分数。

所以，我想结束了该表：

ID date score 
1: 1 1 4.0 
2: 1 2 3.0 
3: 1 3 2.0 
4: 1 4 2.0 
5: 1 5 4.0 
6: 1 6 1.0 
7: 2 4 2.0 
8: 2 5 1.0 
9: 2 6 4.0 
10: 3 1 2.0 
11: 3 2 1.0 
12: 3 3 1.5 
13: 3 4 1.5 
14: 3 5 1.5 
15: 3 6 3.0 
16: 4 4 2.0 
17: 4 5 4.0 
18: 4 6 3.0 

什么是去了解这一点的最好方法是什么？我想这是shift和.I的一些组合，但我还没有把它放在一起。

Answer 1

# find rows satisfying 1st condition 
torm = data[, if(score[1] == 5 & date[1] == 1) .I 
      , by = .(ID, rleid(score), cumsum(date == 1))]$V1 

library(caTools) # for running mean 

data[-torm # remove the extra rows 
    # add a running mean 
    ][, mn := runmean(score, 2, endrule = 'keep', align = 'right'), by = ID 
    # compute the new score - a little care needed here in case we only have 5's in a group 
    ][, new.score := ifelse(score == 5, mn[which(score != 5)[1]], score) 
    , by = .(ID, cumsum(score != 5))][] 
# ID date score mn new.score 
# 1: 1 1  4 4.0  4.0 
# 2: 1 2  3 3.5  3.0 
# 3: 1 3  2 2.5  2.0 
# 4: 1 4  2 2.0  2.0 
# 5: 1 5  4 3.0  4.0 
# 6: 1 6  1 2.5  1.0 
# 7: 2 4  2 2.0  2.0 
# 8: 2 5  1 1.5  1.0 
# 9: 2 6  4 2.5  4.0 
#10: 3 1  2 2.0  2.0 
#11: 3 2  1 1.5  1.0 
#12: 3 3  5 3.0  1.5 
#13: 3 4  5 5.0  1.5 
#14: 3 5  5 5.0  1.5 
#15: 3 6  3 4.0  3.0 
#16: 4 4  2 2.0  2.0 
#17: 4 5  4 3.0  4.0 
#18: 4 6  5 4.5  3.0 

Answer 2

na.locf随着从zoo包：

library(zoo) 

DF <- data.frame(ID = c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4), 
       date = c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6), 
       score = c(4,3,2,2,4,1,5,5,5,2,1,4,2,1,5,5,5,3,5,5,5,2,4,5)) 



#mark rows for deletion 

DF$markForDel=NA 

DF$markForDel[DF$date==1 & DF$score==5]=1 

DF$markForDel[DF$score!=5]=0 

DF$markForDel = zoo::na.locf(DF$markForDel) 


newDF = DF[DF$markForDel!=1,] 
rownames(newDF)=NULL 


#impute mean of previous score where score == 5 
newDF$score[newDF$score==5]=NA 

newDF$imputedScore = sapply(1:nrow(newDF),function(x) { 
ifelse(x>3 & is.na(newDF$score[x]),mean(c(newDF$score[x-1],newDF$score[x-2])),newDF$score[x]) })    


newDF$imputedScore = zoo::na.locf(newDF$imputedScore) 

输出：

newDF 
# ID date score markForDel imputedScore 
#1 1 1  4   0   4.0 
#2 1 2  3   0   3.0 
#3 1 3  2   0   2.0 
#4 1 4  2   0   2.0 
#5 1 5  4   0   4.0 
#6 1 6  1   0   1.0 
#7 2 4  2   0   2.0 
#8 2 5  1   0   1.0 
#9 2 6  4   0   4.0 
#10 3 1  2   0   2.0 
#11 3 2  1   0   1.0 
#12 3 3 NA   0   1.5 
#13 3 4 NA   0   1.5 
#14 3 5 NA   0   1.5 
#15 3 6  3   0   3.0 
#16 4 4  2   0   2.0 
#17 4 5  4   0   4.0 
#18 4 6 NA   0   3.0