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本帖最后由 419595891 于 2018-3-16 13:51 编辑 ' E6 s0 j1 q" T6 x9 j
/ k/ }6 B* k2 M( k' Z5 [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) 1 j: o' I4 b5 ?0 N( ]
reri <- function(datsam)
% H" |) G3 ^5 ~1 m0 R! j+ N) F{ / K+ p& q/ g) {- S! O; z; Y
fitlr <- glm(outcome ~ determinant A * determinant B, family=binomial, data=datsam)
% b8 A; T* `* I; n( L# Xreri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1 . a$ _$ X* c/ m
}
% {9 C) P+ r. [/ i( X. n+ osummary.bootstrap(bootstrap(dataset, reri(dataset), B=10000), probs=c(0.025,0.5, 0.975)) 7 `1 G; m2 N* I0 ~8 Y6 Y/ P
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————————————我是分割线—————————————————————————————— 5 H3 P! C2 r" x# c1 L
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library(Design)
, S6 M. p" K: x9 l/ {> reri <- function(datsam)
: a, z P* r9 _/ v' e m+ {fitlr <- glm(DM.0813 ~ smoking * rs163184, famliy=binomial, data=datsam)6 f+ L H4 V8 y- V! _1 A
+ reri <- exp(fitlr$coef[2]+fitlr$coef[3]+fitlr$coef[4]) - exp(fitlr$coef[2]) -exp(fitlr$coef[3]) + 1}. Q5 ]5 H3 F2 A
Last expression in function is an assignment
- l J8 n3 W+ Y) K" z5 W' Q0 X (You probably wanted to return the left-hand side)
% f3 Z# X( i' l) n' p% D6 e in: reri <- exp(fitlr$coef[2] + fitlr$coef[3] + fitlr$coef[4]) - exp(fitlr$coef[2]) - exp(fitlr$coef[3]) + 1- p# o1 t+ \: q! ~$ }) ?5 O, x5 e1 X
> summary.bootstrap(bootstrap(smoking, reri(smoking), B=10000), probs=c(0.025,0.5, 0.975))
) @' _, ?* Y; dWarning messages:
% m j9 G* S6 V. U7 r Numerical expression has 7028 elements: only the first used in: model.frame(formula, data, na.action, dots)3 o: b% v/ i8 _$ U; H
Problem: Object "DM.0813" not found ( C0 n) @4 J! F7 U% U
Use traceback() to see the call stack
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+ s) w D/ T: E* @. Y不明白这个错误什么意思,小白求指导……$ K2 b) y; Y, L9 C! V a w
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