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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______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
