我认为这将是一件微不足道的事情，但我仍然在调整编写代码而不是指向和点击电子表格时遇到一些麻烦。
month = as.integer(c(1,2,3,4,5,6,7,8,9,10,11,12))
remaining = c(1000,925,852,790,711,658,601,567,530,501,485,466)
left = c(75, 73, 62, 79, 53, 57, 34, 37, 29, 16, 19, 0)
KPdata = data.frame(month, remaining, left)
> KPdata
month remaining left
1 1 1000 75
2 2 925 73
3 3 852 62
4 4 790 79
5 5 711 53
6 6 658 57
7 7 601 34
8 8 567 37
9 9 530 29
10 10 501 16
11 11 485 19
12 12 466 12
如何计算每个月的Kaplan-Meier生存函数？请注意，我想手动执行此操作，我知道有些软件包可以帮我完成。
答案 0 :(得分：3)
我认为这就是你要做的。我们使用lag和cumprod来获取手动KM估算工具：
KPdata$KM_init <- lag((KPdata$remaining - KPdata$left) / KPdata$remaining)
KPdata[1,ncol(KPdata)] <- 1
KPdata$KM_final <- cumprod(KPdata$KM_init)
KPdata
month remaining left KM_init KM_final
1 1 1000 75 1.0000000 1.000
2 2 925 73 0.9250000 0.925
3 3 852 62 0.9210811 0.852
4 4 790 79 0.9272300 0.790
5 5 711 53 0.9000000 0.711
6 6 658 57 0.9254571 0.658
7 7 601 34 0.9133739 0.601
8 8 567 37 0.9434276 0.567
9 9 530 29 0.9347443 0.530
10 10 501 16 0.9452830 0.501
11 11 485 19 0.9680639 0.485
12 12 466 0 0.9608247 0.466
或者，我认为有一种不同形式的KM估算器看起来像这样（注意我添加了一行对应month = 0）：
month = as.integer(c(0,1,2,3,4,5,6,7,8,9,10,11,12))
remaining = c(1000,1000,925,852,790,711,658,601,567,530,501,485,466)
left = c(0,75, 73, 62, 79, 53, 57, 34, 37, 29, 16, 19, 0)
KPdata2 = data.frame(month, remaining, left)
KPdata2$KM_init <- (KPdata2$remaining - KPdata2$left) / KPdata2$remaining
KPdata2$KM_final <- cumprod(KPdata2$KM_init)
KPdata2
month remaining left KM_init KM_final
1 0 1000 0 1.0000000 1.000
2 1 1000 75 0.9250000 0.925
3 2 925 73 0.9210811 0.852
4 3 852 62 0.9272300 0.790
5 4 790 79 0.9000000 0.711
6 5 711 53 0.9254571 0.658
7 6 658 57 0.9133739 0.601
8 7 601 34 0.9434276 0.567
9 8 567 37 0.9347443 0.530
10 9 530 29 0.9452830 0.501
11 10 501 16 0.9680639 0.485
12 11 485 19 0.9608247 0.466
13 12 466 0 1.0000000 0.466
答案 1 :(得分：0)
我对这个问题和@bouncyball的鼓舞人心的答案非常关注，我想我会加上我的ha'penny值得尝试处理审查。这是出于原始问题的精神 - “手工”做事来制定关键见解。
## rename remaining -> survived; left -> died
month = as.integer(c(1,2,3,4,5,6,7,8,9,10,11,12))
survived = c(1000,925,852,790,711,658,601,567,530,501,485,466)
died = c(75, 73, 62, 79, 53, 57, 34, 37, 29, 16, 19, 0)
## arbitrary censoring @ 10 per time period
censored <- c(0, rep(10,11))
KPdata3 = data.frame(month, at.risk, censored, died, survived)
## define those at risk <= those who survived
## awful bit of R fiddling for (something simple like) offsetting the index in base R
len <- length(month)
at.risk <- c(survived[1],
survived[-len] - died[-len] - cumsum(censored[-len]) )
## note use of cumsum()
## censoring uses at risk, rather than survived/remained
KPdata3$KM_increment <- (KPdata3$at.risk - KPdata3$died)/ KPdata3$at.risk
## code credit to @bouncyball
KPdata3$KM_cumulative <- cumprod(KPdata3$KM_increment)
KPdata3
给这个......
month at.risk censored died survived KM_increment KM_cumulative
1 1 1000 0 75 1000 0.9250000 0.9250000
2 2 925 10 73 925 0.9210811 0.8520000
3 3 842 10 62 852 0.9263658 0.7892637
4 4 770 10 79 790 0.8974026 0.7082873
5 5 681 10 53 711 0.9221733 0.6531636
6 6 618 10 57 658 0.9077670 0.5929203
7 7 551 10 34 601 0.9382940 0.5563336
8 8 507 10 37 567 0.9270217 0.5157333
9 9 460 10 29 530 0.9369565 0.4832197
10 10 421 10 16 501 0.9619952 0.4648551
11 11 395 10 19 485 0.9518987 0.4424949
12 12 366 10 0 466 1.0000000 0.4424949
设置rep(0,11)会给出与@ bouncyball相同的答案。