我有以下对象M
,我需要从中提取fstatistic
。它是由summaryC
生成的模型的函数aovp
生成的模型,这两个函数都来自包lmPerm
。我尝试过从正常线性模型和attr
,extract
和getElement
中的函数中提取值的提示,但没有成功。
任何人都可以给我一个暗示吗?
> str(M)
List of 2
$ Error: vegetation: NULL
$ Error: Within :List of 11
..$ NA : NULL
..$ terms :Classes 'terms', 'formula' length 3 Temp ~ depth
.. .. ..- attr(*, "variables")= language list(Temp, depth)
.. .. ..- attr(*, "factors")= int [1:2, 1] 0 1
.. .. .. ..- attr(*, "dimnames")=List of 2
.. .. .. .. ..$ : chr [1:2] "Temp" "depth"
.. .. .. .. ..$ : chr "depth"
.. .. ..- attr(*, "term.labels")= chr "depth"
.. .. ..- attr(*, "order")= int 1
.. .. ..- attr(*, "intercept")= int 1
.. .. ..- attr(*, "response")= int 1
.. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
..$ residuals : Named num [1:498] -46.9 -43.9 -46.9 -38.9 -41.9 ...
.. ..- attr(*, "names")= chr [1:498] "3" "4" "5" "6" ...
..$ coefficients : num [1:4, 1:4] -2.00 -1.00 -1.35e-14 1.00 2.59 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:4] "depth1" "depth2" "depth3" "depth4"
.. .. ..$ : chr [1:4] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
..$ aliased : Named logi [1:4] FALSE FALSE FALSE FALSE
.. ..- attr(*, "names")= chr [1:4] "depth1" "depth2" "depth3" "depth4"
..$ sigma : num 29
..$ df : int [1:3] 4 494 4
..$ r.squared : num 0.00239
..$ adj.r.squared: num -0.00367
..$ **fstatistic** : Named num [1:3] 0.395 3 494
.. ..- attr(*, "names")= chr [1:3] "value" "numdf" "dendf"
..$ cov.unscaled : num [1:4, 1:4] 0.008 -0.002 -0.002 -0.002 -0.002 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:4] "depth1" "depth2" "depth3" "depth4"
.. .. ..$ : chr [1:4] "depth1" "depth2" "depth3" "depth4"
..- attr(*, "class")= chr "summary.lmp"
- attr(*, "class")= chr "listof"
它是一个可重现的例子:
Temp=1:100
depth<- rep( c("1","2","3","4","5"), 100)
vegetation=rep( c("1","2"), 50)
df=data.frame(Temp,depth,vegetation)
M=summaryC(aovp(Temp~depth+Error(vegetation),df, perm=""))
答案 0 :(得分:0)
作为示例显示的str
输出,M
是两个列表的列表,第二个包含您想要的内容。因此,通过[[
进行列表提取可以解决问题:
> M[[2]][["fstatistic"]]
value numdf dendf
0.3946 3.0000 494.0000
如果这不是您想要的,请发表评论。