Extract p-value from aov

问题

I am looking to extract the p-value generated from an anova in R.

Here is what I am running:

test <- aov(asq[,9] ~ asq[,187])
summary(test)

Yields:

              Df Sum Sq Mean Sq F value    Pr(>F)    
asq[, 187]     1   3.02 3.01951  12.333 0.0004599 ***
Residuals   1335 326.85 0.24483                      
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
12 observations deleted due to missingness

When I look a the structure, this is what I see. I usually can work through lists to get what I need, but I am having a hard time with this one. A Google searched also seemed to reveal much simpler structures than I am getting.

NOTE: ASQ is my data frame.

str(test)

List of 13
 $ coefficients : Named num [1:2] 0.2862 0.0973
  ..- attr(*, "names")= chr [1:2] "(Intercept)" "asq[, 187]"
 $ residuals    : Named num [1:1337] 0.519 0.519 -0.481 -0.481 -0.481 ...
  ..- attr(*, "names")= chr [1:1337] "1" "2" "3" "4" ...
 $ effects      : Named num [1:1337] -16.19 -1.738 -0.505 -0.505 -0.505 ...
  ..- attr(*, "names")= chr [1:1337] "(Intercept)" "asq[, 187]" "" "" ...
 $ rank         : int 2
 $ fitted.values: Named num [1:1337] 0.481 0.481 0.481 0.481 0.481 ...
  ..- attr(*, "names")= chr [1:1337] "1" "2" "3" "4" ...
 $ assign       : int [1:2] 0 1
 $ qr           :List of 5
  ..$ qr   : num [1:1337, 1:2] -36.565 0.0273 0.0273 0.0273 0.0273 ...
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : chr [1:1337] "1" "2" "3" "4" ...
  .. .. ..$ : chr [1:2] "(Intercept)" "asq[, 187]"
  .. ..- attr(*, "assign")= int [1:2] 0 1
  ..$ qraux: num [1:2] 1.03 1.02
  ..$ pivot: int [1:2] 1 2
  ..$ tol  : num 1e-07
  ..$ rank : int 2
  ..- attr(*, "class")= chr "qr"
 $ df.residual  : int 1335
 $ na.action    :Class 'omit'  Named int [1:12] 26 257 352 458 508 624 820 874 1046 1082 ...
  .. ..- attr(*, "names")= chr [1:12] "26" "257" "352" "458" ...
 $ xlevels      : list()
 $ call         : language aov(formula = asq[, 9] ~ asq[, 187])
 $ terms        :Classes 'terms', 'formula' length 3 asq[, 9] ~ asq[, 187]
  .. ..- attr(*, "variables")= language list(asq[, 9], asq[, 187])
  .. ..- attr(*, "factors")= int [1:2, 1] 0 1
  .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. ..$ : chr [1:2] "asq[, 9]" "asq[, 187]"
  .. .. .. ..$ : chr "asq[, 187]"
  .. ..- attr(*, "term.labels")= chr "asq[, 187]"
  .. ..- attr(*, "order")= int 1
  .. ..- attr(*, "intercept")= int 1
  .. ..- attr(*, "response")= int 1
  .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  .. ..- attr(*, "predvars")= language list(asq[, 9], asq[, 187])
  .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
  .. .. ..- attr(*, "names")= chr [1:2] "asq[, 9]" "asq[, 187]"
 $ model        :'data.frame':  1337 obs. of  2 variables:
  ..$ asq[, 9]  : int [1:1337] 1 1 0 0 0 1 1 1 0 0 ...
  ..$ asq[, 187]: int [1:1337] 2 2 2 2 2 2 2 2 2 2 ...
  ..- attr(*, "terms")=Classes 'terms', 'formula' length 3 asq[, 9] ~ asq[, 187]
  .. .. ..- attr(*, "variables")= language list(asq[, 9], asq[, 187])
  .. .. ..- attr(*, "factors")= int [1:2, 1] 0 1
  .. .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. .. ..$ : chr [1:2] "asq[, 9]" "asq[, 187]"
  .. .. .. .. ..$ : chr "asq[, 187]"
  .. .. ..- attr(*, "term.labels")= chr "asq[, 187]"
  .. .. ..- attr(*, "order")= int 1
  .. .. ..- attr(*, "intercept")= int 1
  .. .. ..- attr(*, "response")= int 1
  .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  .. .. ..- attr(*, "predvars")= language list(asq[, 9], asq[, 187])
  .. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
  .. .. .. ..- attr(*, "names")= chr [1:2] "asq[, 9]" "asq[, 187]"
  ..- attr(*, "na.action")=Class 'omit'  Named int [1:12] 26 257 352 458 508 624 820 874 1046 1082 ...
  .. .. ..- attr(*, "names")= chr [1:12] "26" "257" "352" "458" ...
 - attr(*, "class")= chr [1:2] "aov" "lm"

回答1:

summary(test)[[1]][["Pr(>F)"]]

回答2:

since the suggest above didn't work for me this is how i managed to solve it:

sum_test = unlist(summary(test))

then looking at the names with

names(sum_test)

i have"Pr(>F)1" and "Pr(>F)2", when the first it the requested value, so

sum_test["Pr(>F)1"]

will give the requested value

回答3:

I know this is old but I looked around online and didn't find an explanation or general solution and this thread is one of the first things that comes up in a Google search.

Aniko is right, the easiest way is looking in summary(test).

tests <- summary(test)
str(tests)

That gives you a list of 1 for an independent measures aov object but it could have multiple items with repeated measures. With the repeated measures each item in the list is defined by the error term for the item in the list. Where a lot of new people get confused is that if it's between measures the one lone list item isn't named. So, they don't really notice that and don't understand why using a typical selector doesn't work.

In the independent measures case something like the following works.

tests[[1]]$'Pr(>F)'

In repeated measures it's similar but you could also use named items like...

myModelSummary$'Error: subject:A'[[1]]$'Pr(>F)'

Note there I still had to do that list selection because each one of the list items in the repeated measures model is again a list of 1.

回答4:

Check out str(summary(test)) - that's where you see the p-value.

回答5:

Somewhat shorter, than in BurningLeo's advice:

summary(test)[[1]][[1,"Pr(>F)"]]

回答6:

summary(aov(y~factor(x)))[[1]][[5]][1]

回答7:

unlist(summary(myAOV)[[2]])[[9]]

2 and 9 are the positions of p-value in myAOV model

来源：https://stackoverflow.com/questions/3366506/extract-p-value-from-aov

标签

anova