Reputation: 11657
I ran outlierTest from the car
package (outliers <- car::outlierTest(mod)
) and it returned the following list object
rstudent unadjusted p-value Bonferonni p
101812 -15.10404 0.0000000000000000000000000000000000000000000000000016115 0.00000000000000000000000000000000000000000000037466
137979 -14.65162 0.0000000000000000000000000000000000000000000000013829000 0.00000000000000000000000000000000000000000032152000
138209 -13.73731 0.0000000000000000000000000000000000000000006308000000000 0.00000000000000000000000000000000000014666000000000
124997 -13.59990 0.0000000000000000000000000000000000000000041618000000000 0.00000000000000000000000000000000000096759000000000
62602 -13.42890 0.0000000000000000000000000000000000000000424190000000000 0.00000000000000000000000000000000000986230000000000
78196 -13.42374 0.0000000000000000000000000000000000000000454740000000000 0.00000000000000000000000000000000001057200000000000
195658 -12.52146 0.0000000000000000000000000000000000058519000000000001030 0.00000000000000000000000000000136059999999999999007
124931 -12.07768 0.0000000000000000000000000000000014182999999999999254956 0.00000000000000000000000000032975000000000000085551
177689 -11.34566 0.0000000000000000000000000000079362999999999990162584378 0.00000000000000000000000184519999999999991835422130
82302 -10.66309 0.0000000000000000000000000153630000000000009271702167845 0.00000000000000000000357189999999999999516628564927
I want to extract the dataframe indices that it references on the left side. I assumed these would be accessible via row.names
, but was incorrect. How do I grab these?
Upvotes: 0
Views: 102
Reputation: 173527
By using str()
on the output of outlierTest
it seems the row names that end up being printed are just stored in a names attribute of several of the object list elements, like rstudent.
So you could just do something like attr(x = x$rstudent,"names")
.
Upvotes: 3