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I created lmm using lmer. I want then look at the contrast using emmeans (just an overview below)

$contrasts
 contrast                  estimate   SE  df lower.CL upper.CL t.ratio p.value
 A - B    14.3927 1.40 102     9.91    18.87  10.310 <.0001 
 A - C    -0.9537 1.45 154    -5.55     3.64  -0.659 0.5957 
 A - D    14.2376 1.45 154     9.64    18.84   9.845 <.0001 
 A - E    11.2341 1.51 155     6.42    16.05   7.420 <.0001 
 A - F    12.9106 1.60 154     7.84    17.98   8.089 <.0001 
 A - H    12.5250 1.45 154     7.93    17.12   8.661 <.0001 
 A - I    14.2004 1.45 154     9.60    18.80   9.819 <.0001 
 B - L    -15.3463 1.45 154   -19.94   -10.75 -10.612 <.0001 
 B - J    -0.1550 1.45 154    -4.75     4.44  -0.107 0.9487 
 B - K    -3.1586 1.51 155    -7.97     1.65  -2.086 0.0832

However, I only want to look at specific contrasts for example A - B and D - E

I know I can use the

emmeans(l_glm_l1_cage[[1]], pairwise ~ timegroups, adjust = "fdr", at=list(timegroups=c("A", "B", "D", "E")))

but this of course gives me the pairwise comparisons. Any suggestions?

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  • $\begingroup$ try reading ? contrast and vignette("contrasts", "emmeans") for starters. Note that in the method argument, you may specify a list of contrast coefficients. $\endgroup$
    – Russ Lenth
    Commented Oct 7, 2020 at 2:18
  • $\begingroup$ I recommend using a one-sided formula and calling contrast on the result. $\endgroup$
    – Russ Lenth
    Commented Oct 7, 2020 at 2:22
  • $\begingroup$ Thank you. I think I kind of got the solution. First I define the interested factors e.g. A=c(0,0,0,0,1,0) and B=c(0,1,0,0,0,0,0). Then I call contrast(emm1, method = list("A-B" = A - B)). This seemed to work $\endgroup$
    – efrem
    Commented Oct 8, 2020 at 15:47
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    $\begingroup$ I'd have thought that it should be A = c(1,0,0,0,0,0) but what you have is cirrect if A is the next-to-last level $\endgroup$
    – Russ Lenth
    Commented Oct 8, 2020 at 21:14
  • $\begingroup$ Yes, you are right, what I wrote does not match the data! Sorry about that $\endgroup$
    – efrem
    Commented Oct 9, 2020 at 17:00

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