d****z
发帖数: 301 | 1 一直折腾不出来.好几个月了.
牛细胞生长情况, 有两种细胞,3个时间点,每个有两个重复.
想找出差异表达的基因...
结果一直疯狂报错:
warning: Stopped because of infinite likelihood.
跪求建议意见!!!!
proc mixed data=long;
class rep genotype time;
model y=rep genotype time genotype*time /outp=check;
random rep*time;
lsmeans genotype*time;
ods output tests3=results;
ods output lsmeans=lsmeans;
by gene;
run; | w*****n
发帖数: 375 | 2 remove rep from model statement | d****z
发帖数: 301 | 3 多谢!!
我也觉得rep不是fixed变量.老板建议这样的,我再去找他讨论下.
而且去掉了也不成,还是一样报错.
很奇怪!
【在 w*****n 的大作中提到】
: remove rep from model statement
| w*****n
发帖数: 375 | 4 Your random statement might be wrong, and I suspect that you don't
have enough degrees of freedom to estimate so many parameters.What's
your sample size?
Generate variable "id", which has two reps, and try the following
simple model first:
proc mixed data=long;
class id genotype time;
model y = genotype time genotype*time /outp=check;
random intercept/sub = id;
lsmeans genotype*time;
ods output tests3=results;
ods output lsmeans=lsmeans;
by gene;
run;
【在 d****z 的大作中提到】
: 多谢!!
: 我也觉得rep不是fixed变量.老板建议这样的,我再去找他讨论下.
: 而且去掉了也不成,还是一样报错.
: 很奇怪!
| w******4
发帖数: 488 | 5 I think this is a split-plot design, where genotype is the whole-plot factor
, time is the sub-plot factor. You can try this:
proc mixed data=long;
class rep genotype time;
model y=genotype time genotype*time /outp=check;
random rep(genotype);
lsmeans genotype*time;
ods output tests3=results;
ods output lsmeans=lsmeans;
by gene;
run; |
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