rnewbie - 2 months ago 5x
R Question

# Manual calculation of the KM estimator

I thought this would be a trivial thing to do but I still have some trouble adjusting to writing code instead of pointing and clicking on a spreadsheet.

``````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
``````

How do I calculate the Kaplan Meier survival function at each month? Note that I want to do this manually, I am aware that there are packages which will do it for me.

I think this is what you're trying to do. We use `lag` and `cumprod` to get a manual KM estimator:

``````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
``````

Alternatively, I think there's a different form of a KM estimator that looks like this (note that I've added a row corresponding to `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
``````
Source (Stackoverflow)