[R] AFT model time-dependent with weibull distribution
Rui Barradas
ruipbarradas at sapo.pt
Sat Aug 4 21:20:06 CEST 2012
Hello,
Vous êtes française?
It shows, in english it would be 'exponential', with an 'a'.
Worked with me, after reading the manual.
dataexp <- read.table(text="
vi Ti
1 26 5.79
2 26 1579.52
3 26 2323.70
4 28 68.85
[...]
73 38 0.39
74 38 1.13
75 38 0.09
76 38 2.38
", header=TRUE)
# Better in a post to R-Help is the output of dput()
# It looks like the following line, without the assignment.
dput(dataexp)
dataexp <-
structure(list(vi = c(26L, 26L, 26L, 28L, 28L, 28L, 28L, 28L,
30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 32L, 32L,
32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L,
34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L,
34L, 34L, 34L, 34L, 34L, 34L, 36L, 36L, 36L, 36L, 36L, 36L, 36L,
36L, 36L, 36L, 36L, 36L, 36L, 36L, 36L, 38L, 38L, 38L, 38L, 38L,
38L, 38L, 38L), Ti = c(5.79, 1579.52, 2323.7, 68.85, 426.07,
110.29, 108.29, 1067.6, 17.05, 22.66, 21.02, 175.88, 139.07,
144.12, 20.46, 43.4, 194.9, 47.3, 7.74, 0.4, 82.85, 9.88, 89.29,
215.1, 1.75, 0.79, 15.93, 3.91, 0.27, 0.69, 100.58, 27.8, 13.95,
53.24, 0.96, 4.15, 0.19, 0.78, 8.01, 31.75, 7.35, 6.5, 8.27,
33.91, 32.52, 3.16, 4.85, 2.78, 4.67, 1.31, 12.06, 36.71, 72.89,
1.97, 0.59, 2.58, 1.69, 2.71, 25.5, 0.35, 0.99, 3.99, 3.67, 2.07,
0.96, 5.35, 2.9, 13.77, 0.47, 0.73, 1.4, 0.74, 0.39, 1.13, 0.09,
2.38), status = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c("vi",
"Ti", "status"), row.names = c("1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17",
"18", "19", "20", "21", "22", "23", "24", "25", "26", "27", "28",
"29", "30", "31", "32", "33", "34", "35", "36", "37", "38", "39",
"40", "41", "42", "43", "44", "45", "46", "47", "48", "49", "50",
"51", "52", "53", "54", "55", "56", "57", "58", "59", "60", "61",
"62", "63", "64", "65", "66", "67", "68", "69", "70", "71", "72",
"73", "74", "75", "76"), class = "data.frame")
##Not run
#install.packages('eha')
library(eha)
library(survival)
dataexp$status <- rep(1, 76)
aftexp <- aftreg(Surv(Ti,status) ~ vi, data=dataexp,
dist="weibull", shape=1)
str(aftexp)
summary(aftexp)
coef(aftexp)
The exponential can be seen as a special case of the weibull
distribution with shape = 1.
And the help page of aftreg() says precisely to use the weibull with
shape = 1 if we want an exponential.
Em 04-08-2012 18:45, hafida escreveu:
> Dear R-community,
>> I have tried to estimate an EXPONENTIEL accelerated failure time(AFT)
>> power rule model with time-independent . For that purpose, I have used
>> the eha package.
>> Please, consider this example:
> vi Ti
> 1 26 5.79
> 2 26 1579.52
> 3 26 2323.70
> 4 28 68.85
> 5 28 426.07
> 6 28 110.29
> 7 28 108.29
> 8 28 1067.60
> 9 30 17.05
> 10 30 22.66
> 11 30 21.02
> 12 30 175.88
> 13 30 139.07
> 14 30 144.12
> 15 30 20.46
> 16 30 43.40
> 17 30 194.90
> 18 30 47.30
> 19 30 7.74
> 20 32 0.40
> 21 32 82.85
> 22 32 9.88
> 23 32 89.29
> 24 32 215.10
> 25 32 1.75
> 26 32 0.79
> 27 32 15.93
> 28 32 3.91
> 29 32 0.27
> 30 32 0.69
> 31 32 100.58
> 32 32 27.80
> 33 32 13.95
> 34 32 53.24
> 35 34 0.96
> 36 34 4.15
> 37 34 0.19
> 38 34 0.78
> 39 34 8.01
> 40 34 31.75
> 41 34 7.35
> 42 34 6.50
> 43 34 8.27
> 44 34 33.91
> 45 34 32.52
> 46 34 3.16
> 47 34 4.85
> 48 34 2.78
> 49 34 4.67
> 50 34 1.31
> 51 34 12.06
> 52 34 36.71
> 53 34 72.89
> 54 36 1.97
> 55 36 0.59
> 56 36 2.58
> 57 36 1.69
> 58 36 2.71
> 59 36 25.50
> 60 36 0.35
> 61 36 0.99
> 62 36 3.99
> 63 36 3.67
> 64 36 2.07
> 65 36 0.96
> 66 36 5.35
> 67 36 2.90
> 68 36 13.77
> 69 38 0.47
> 70 38 0.73
> 71 38 1.40
> 72 38 0.74
> 73 38 0.39
> 74 38 1.13
> 75 38 0.09
> 76 38 2.38
>> aftexp<-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp),
>> dist="exponentiel")
> Error in Surv(time, status) : Time variable is not numeric
>
>> aftexp<-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp),
>> dist="exponentiel")
> Error in Surv(Ti, status) : object 'status' not found
>> status<- rep(1, 76)
>> status
> [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1
> [39] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1
>
>> cbind(dataexp, status)
> vi Ti status
> 1 26 5.79 1
> 2 26 1579.52 1
> 3 26 2323.70 1
> 4 28 68.85 1
> 5 28 426.07 1
> 6 28 110.29 1
> 7 28 108.29 1
> 8 28 1067.60 1
> 9 30 17.05 1
> 10 30 22.66 1
> 11 30 21.02 1
> 12 30 175.88 1
> 13 30 139.07 1
> 14 30 144.12 1
> 15 30 20.46 1
> 16 30 43.40 1
> 17 30 194.90 1
> 18 30 47.30 1
> 19 30 7.74 1
> 20 32 0.40 1
> 21 32 82.85 1
> 22 32 9.88 1
> 23 32 89.29 1
> 24 32 215.10 1
> 25 32 1.75 1
> 26 32 0.79 1
> 27 32 15.93 1
> 28 32 3.91 1
> 29 32 0.27 1
> 30 32 0.69 1
> 31 32 100.58 1
> 32 32 27.80 1
> 33 32 13.95 1
> 34 32 53.24 1
> 35 34 0.96 1
> 36 34 4.15 1
> 37 34 0.19 1
> 38 34 0.78 1
> 39 34 8.01 1
> 40 34 31.75 1
> 41 34 7.35 1
> 42 34 6.50 1
> 43 34 8.27 1
> 44 34 33.91 1
> 45 34 32.52 1
> 46 34 3.16 1
> 47 34 4.85 1
> 48 34 2.78 1
> 49 34 4.67 1
> 50 34 1.31 1
> 51 34 12.06 1
> 52 34 36.71 1
> 53 34 72.89 1
> 54 36 1.97 1
> 55 36 0.59 1
> 56 36 2.58 1
> 57 36 1.69 1
> 58 36 2.71 1
> 59 36 25.50 1
> 60 36 0.35 1
> 61 36 0.99 1
> 62 36 3.99 1
> 63 36 3.67 1
> 64 36 2.07 1
> 65 36 0.96 1
> 66 36 5.35 1
> 67 36 2.90 1
> 68 36 13.77 1
> 69 38 0.47 1
> 70 38 0.73 1
> 71 38 1.40 1
> 72 38 0.74 1
> 73 38 0.39 1
> 74 38 1.13 1
> 75 38 0.09 1
> 76 38 2.38 1
>
>> aftexp<-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp),
>> dist="exponentiel")
> Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control, :
> exponentiel is not an implemented distribution
>
>> aftexp<-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp),
>> dist="exponentiel")
> Error in Surv(time, status) : Time variable is not numeric
>> pleas help me to find a solution to my problem
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p4639174.html
> Sent from the R help mailing list archive at Nabble.com.
>
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