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本帖最后由 419595891 于 2018-3-16 13:51 编辑
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* k& z( P# z* y2 ^6 m( b这是一篇文献里给出的script,但我放到S-plus里跑不出来,有错误,有可能是我不会用S-plus,所以请教大神下面的script是否有错,没有错的话怎么解决错误?能不能改动换到R里运行?
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; O, y! e- o/ c' \library(Design) - K" s2 }0 H& q$ Q% X( {
reri <- function(datsam)
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, o, `' n4 N; zfitlr <- glm(outcome ~ determinant A * determinant B, family=binomial, data=datsam) 2 ~5 I' A' g |! [" o8 E
reri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1
% B0 e+ g; t+ w* D- [5 T8 B} 8 W f, O1 w, ]7 }, Z
summary.bootstrap(bootstrap(dataset, reri(dataset), B=10000), probs=c(0.025,0.5, 0.975))
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————————————我是分割线——————————————————————————————
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+ V$ U3 c/ N" `4 f library(Design)+ C' A0 @. \ p( X- N5 t% Q
> reri <- function(datsam)* t- a) M% ?* @, r0 S* }
+ {fitlr <- glm(DM.0813 ~ smoking * rs163184, famliy=binomial, data=datsam)
8 e! C8 b* k, B6 @0 ]% {8 J+ reri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) -exp(fitlr$coef[3]) + 1}+ h/ A' P6 ^( U3 v; |2 G' Y
Last expression in function is an assignment b3 c6 ]4 [8 @. ]& _
(You probably wanted to return the left-hand side)$ Y# s7 T/ a& o
in: reri <- exp(fitlr$coef[2] + fitlr$coef[3] + fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1
; y6 Y+ H3 \2 M8 U `" h I> summary.bootstrap(bootstrap(smoking, reri(smoking), B=10000), probs=c(0.025,0.5, 0.975)) 2 B3 p0 h$ }2 u- \- s
Warning messages:
! U6 i Q+ M. a/ U% R& s Numerical expression has 7028 elements: only the first used in: model.frame(formula, data, na.action, dots)8 y6 ]9 Q& K, y' e
Problem: Object "DM.0813" not found 4 D9 I' K" P. E6 P* w$ A& l7 r( k
Use traceback() to see the call stack6 v3 i3 H$ t2 t5 W$ Y; u
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不明白这个错误什么意思,小白求指导……. ?. `# P( ?, B2 k
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