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本帖最后由 419595891 于 2018-3-16 13:51 编辑 + @) Z) B7 R8 g% z- y; R8 [, n V
1 O+ B1 o* K$ h- J" u% W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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library(Design) ; [7 ?- L, [' y7 T' E* [* z+ J" N( P
reri <- function(datsam)
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$ {( h5 X. ^ I2 N: c7 ffitlr <- glm(outcome ~ determinant A * determinant B, family=binomial, data=datsam)
2 |$ y# U% q0 d: |reri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1
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summary.bootstrap(bootstrap(dataset, reri(dataset), B=10000), probs=c(0.025,0.5, 0.975))
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; k. x% b9 [# a% R& O4 I. k————————————我是分割线—————————————————————————————— 0 ~ S) d- |8 |/ y# D0 h& `
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8 d1 U$ v$ T; A library(Design)& S8 n( w* n0 d ~: j
> reri <- function(datsam)
4 T, U$ B* B, x( W/ L8 N+ {fitlr <- glm(DM.0813 ~ smoking * rs163184, famliy=binomial, data=datsam)
! \$ R: e7 ?, _; b+ reri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) -exp(fitlr$coef[3]) + 1}9 |* ?5 l" x( T; z' z* @
Last expression in function is an assignment
; G; J5 W; G. V* |$ o' O (You probably wanted to return the left-hand side)
- R) c9 j. }0 p" Z in: reri <- exp(fitlr$coef[2] + fitlr$coef[3] + fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1
( ]$ w" E0 Q, G/ D8 [! {> summary.bootstrap(bootstrap(smoking, reri(smoking), B=10000), probs=c(0.025,0.5, 0.975))
0 R- z$ q p! ^6 u1 n2 kWarning messages:- s) ^2 o& i( ]9 J2 ^
Numerical expression has 7028 elements: only the first used in: model.frame(formula, data, na.action, dots): c; h. C8 ^+ R( g
Problem: Object "DM.0813" not found ' s0 O. @! W8 {' G- b; }! Q
Use traceback() to see the call stack
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" Y7 k3 D' ~2 P. p, D$ K% |, ^' p* m$ O不明白这个错误什么意思,小白求指导……: P9 s1 j& J7 U" e1 d, t# p
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