x*******i
发帖数: 1590 | 1 我 run proc genmod, 有一个contrast statement
CONTRAST a*b 0 0 0 0 1 -1;
but in sas log
"WARNING: The contrast will not be tested due to some rows being
nonestimable."
output is
Contrast Results for GEE Analysis
Chi-
Contrast DF Square Pr > ChiSq Type
0 . . Score
不知道这个some rows being nonestimable是什么？有牛人帮个忙吧。 |
l*********s
发帖数: 5409 | 2 It means you are asking a wrong question; could be due to rank deficiency,
or overspecification. |
x*******i
发帖数: 1590 | 3 CONTRAST a*b 0 0 0 0 1 -1;
the last one '-1' is a reference group. is this the reason? |
A*******s
发帖数: 3942 | 4 check out the chapter about estimability in any linear theory textbook.
【在 x*******i 的大作中提到】
: 我 run proc genmod, 有一个contrast statement
: CONTRAST a*b 0 0 0 0 1 -1;
: but in sas log
: "WARNING: The contrast will not be tested due to some rows being
: nonestimable."
: output is
: Contrast Results for GEE Analysis
: Chi-
: Contrast DF Square Pr > ChiSq Type
: 0 . . Score
|
e****t
发帖数: 766 | 5 how many category of a and b.
i am not sure whether genmod also have "e" option to show all estimable
function. |
x*******i
发帖数: 1590 | 6 a is a continuous variable.
b has 6 categories, 6th is the reference.
【在 e****t 的大作中提到】
: how many category of a and b.
: i am not sure whether genmod also have "e" option to show all estimable
: function.
|
e****t
发帖数: 766 | 7 i guess you forget label ?
label
identifies the contrast on the output. A label is required for every
contrast specified. Labels can be up to 20 characters and must be enclosed
in single quotes. |
j*****e
发帖数: 182 | 8 Proc genmod;
class A B;
model outcome=A*B/noint;
contrast .....;
run;
This should work. |
x*******i
发帖数: 1590 | 9 我有label.
发贴时没写上。
【在 e****t 的大作中提到】
: i guess you forget label ?
: label
: identifies the contrast on the output. A label is required for every
: contrast specified. Labels can be up to 20 characters and must be enclosed
: in single quotes.
|
x*******i
发帖数: 1590 | 10 谢谢，是work了。
可是model变了。
我的是
model outcome=a b a*b;
这结果能用的上吗？
【在 j*****e 的大作中提到】
: Proc genmod;
: class A B;
: model outcome=A*B/noint;
: contrast .....;
: run;
: This should work.
|
|
|
e****t
发帖数: 766 | 11 your A is continouse, right ? did you put it in the class statement ?
i think if with continouse variable in the model, "noint" option will effect
your contrast.
post your "proc genmod" code. let us check .... |
j*****e
发帖数: 182 | 12 It is the same model. Just different parameterization.
You use "a b a*b" to check interaction. You use "a*b" to check contrast,
since it is easier to write the contrast this way.
【在 x*******i 的大作中提到】
: 谢谢，是work了。
: 可是model变了。
: 我的是
: model outcome=a b a*b;
: 这结果能用的上吗？
|
j*****e
发帖数: 182 | 13 It is the same model. Just different parameterization.
You use "a b a*b" to check interaction. You use "a*b" to check contrast,
since it is easier to write the contrast this way.
【在 x*******i 的大作中提到】
: 谢谢，是work了。
: 可是model变了。
: 我的是
: model outcome=a b a*b;
: 这结果能用的上吗？
|
x*******i
发帖数: 1590 | 14 my sas codes:
PROC genmod data=xx;
class provfs b;
model outcome=a b a*b;
repeated subject=provfs;
weight c;
CONTRAST 'test equal slopes'
a*b 1 -1 0 0 0 0,
a*b 0 1 -1 0 0 0,
a*b 0 0 1 -1 0 0,
a*b 0 0 0 1 -1 0,
a*b 0 0 0 0 1 -1;
run;
a is continuous. b is categorical with 6 groups,
thanks,
effect
【在 e****t 的大作中提到】
: your A is continouse, right ? did you put it in the class statement ?
: i think if with continouse variable in the model, "noint" option will effect
: your contrast.
: post your "proc genmod" code. let us check ....
|
e****t
发帖数: 766 | 15 what do you want to test ?
you want to test each of linear combination = zero. or all five simutanious
= zero.
if first, try
CONTRAST 'test 1 ' a*b 1 -1 0 0 0 0 ;
CONTRAST 'test 2 ' a*b 0 1 -1 0 0 0 ;
CONTRAST 'test 3 ' a*b 0 0 1 -1 0 0 ;
CONTRAST 'test 4 ' a*b 0 0 0 1 -1 0 ;
CONTRAST 'test 5 ' a*b 0 0 0 0 1 -1 ;
there will be five p values ;
if second, i guess what you specify in contrast is right.
there will be one p value . i am not sure whether is about your data |
j*****e
发帖数: 182 | 16 Try this.
PROC genmod data=xx;
class provfs b/param=ref ref=last;
model outcome=a b a*b;
repeated subject=provfs;
weight c;
CONTRAST 'test equal slope'
a*b 1 -1 0 0 0 ;
run;
For simultanous testing, just use the p-vale for the a*b term(5 df).
By the way, it seems like you don't understand the model parameterization. |
x*******i
发帖数: 1590 | 17 各位，不好意思，确实不是统计专业出生却在做一些统计的活，不太懂一些名词。
有了问题，就到处google，然后依葫芦画瓢。
我是想test 6 interactions(b*a) coefficients 的不同。网上找到了这个：
http://www.ats.ucla.edu/stat/sas/faq/compreg3.htm
里面有一个sas codes
"PROC GLM DATA=htwt2 ;
CLASS age ;
MODEL weight = age height age*height / SOLUTION ;
CONTRAST 'test equal slopes' age*height 1 -1 0,
age*height 0 1 -1 ;
RUN;“
例子里是Ho: B1 = B2 = B3
我想test ho: B1 = B2 = B3=B4= B5 = B6,所以就模仿了他的contrast code.
这是一个contrast statement
CONTRAST 'test equa |
A*******s
发帖数: 3942 | 18 这应该是说B5-B6是non-estimable的
【在 x*******i 的大作中提到】
: 各位，不好意思，确实不是统计专业出生却在做一些统计的活，不太懂一些名词。
: 有了问题，就到处google，然后依葫芦画瓢。
: 我是想test 6 interactions(b*a) coefficients 的不同。网上找到了这个：
: http://www.ats.ucla.edu/stat/sas/faq/compreg3.htm
: 里面有一个sas codes
: "PROC GLM DATA=htwt2 ;
: CLASS age ;
: MODEL weight = age height age*height / SOLUTION ;
: CONTRAST 'test equal slopes' age*height 1 -1 0,
: age*height 0 1 -1 ;
|
j*****e
发帖数: 182 | 19 B5-B6 corresponds to the value of B5 in the B*A parameter estimation section
of the output. The slope estimate of B6 is given by A. This is a typical
ANCOVA parameterization.
【在 A*******s 的大作中提到】
: 这应该是说B5-B6是non-estimable的
|
x*******i
发帖数: 1590 | 20 楼上各位，那我这个non-estimable问题是不是data的问题呀？
另外，jsdagre (na), 用了你的code后（class provfs b/param=ref ref=last;），出
现了out of memory的问题。如果没有/param=ref ref=last，则还好。 |
j*****e
发帖数: 182 | 21 First of all, you did not tell the whole story.
What kind of outcome do you have? What is the modeling distribution and what
is the link function here? By default, it is normal with identity link. Is
this what you want? If so, what not use proc mixed?
Second, what does weight=c means?
Third, you can write two class statements, one with provfs and one with b
using the options that I gave.
You have to understand the parameterization of linear models before you use
SAS. Otherwise, it will be a disa
【在 x*******i 的大作中提到】
: 楼上各位，那我这个non-estimable问题是不是data的问题呀？
: 另外，jsdagre (na), 用了你的code后（class provfs b/param=ref ref=last;），出
: 现了out of memory的问题。如果没有/param=ref ref=last，则还好。
|
