This vignette contains answers to questions received from users or posted on discussion boards like Cross Validated and Stack Overflow
Contents
- What are EMMs/lsmeans?
- What is the fastest way to obtain EMMs and pairwise comparisons?
- I wanted comparisons, but all I get is (nothing)
- The model I fitted is not supported by emmeans
- I have three (or two or four) factors that interact
- I have covariate(s) that interact(s) with factor(s)
- I have covariate(s) and am fitting a polynomial model
- Some “significant” comparisons have overlapping confidence intervals
- All my pairwise comparisons have the same P value
- emmeans() doesn’t work as expected
- All or some of the results are NA
- If I analyze subsets of the data separately, I get different results
- My lsmeans/EMMs are way off from what I expected
- Why do I get
Inf
for the degrees of freedom? - I get exactly the same comparisons for each “by” group
- My ANOVA F is significant, but no pairwise comparisons are
- I wanted differences, but instead I got ratios (or odds ratios)
- I asked for a Tukey adjustments, but that’s not what I got
emmeans()
completely ignores my P-value adjustmentsemmeans()
gives me pooled t tests, but I expected Welch’s t- I want to see more digits
What are EMMs/lsmeans?
Estimated marginal means (EMMs), a.k.a. least-squares means, are predictions on a reference grid of predictor settings, or marginal averages thereof. See details in the “basics” vignette.
What is the fastest way to obtain EMMs and pairwise comparisons?
There are two answers to this (i.e., be careful what you wish for):
- Don’t think; just fit the first model that comes to mind and run
emmeans(model, pairwise ~ treatment)
. This is the fastest way; however, the results have a good chance of being invalid. - Do think: Make sure you fit a model that really explains the responses. Do diagnostic residual plots, include appropriate interactions, account for heteroscadesticity if necessary, etc. This is the fastest way to obtain appropriate estimates and comparisons.
The point here is that emmeans()
summarizes the
model, not the data directly. If you use a bad model, you will
get bad results. And if you use a good model, you will get appropriate
results. It’s up to you: it’s your research—is it important?
I wanted comparisons, but all I get is (nothing)
This happens when you have only one estimate; and you can’t compare it with itself! This is turn can happen when you have a situation like this: you have fitted
mod <- lm(RT ~ treat, data = mydata)
and treat
is coded in your dataset with numbers 1, 2, 3,
… . Since treat
is a numeric predictor,
emmeans()
just reduces it to a single number, its mean,
rather than separate values for each treatment. Also, please note that
this is almost certainly NOT the model you want, because it forces an
assumption that the treatment effects all fall on a straight line. You
should fit a model like
mod <- lm(RT ~ factor(treat), data = mydata)
then you will have much better luck with comparisons.
The model I fitted is not supported by emmeans
You may still be able to get results using qdrg()
(quick
and dirty reference grid). See ?qdrg
for details and
examples.
I have three (or two or four) factors that interact
Perhaps your question has to do with interacting factors, and you want to do some kind of post hoc analysis comparing levels of one (or more) of the factors on the response. Some specific versions of this question…
- Perhaps you tried to do a simple comparison for one treatment and got a warning message you don’t understand
- You do pairwise comparisons of factor combinations and it’s just too much – want just some of them
- How do I even approach this?
My first answer is: plots almost always help. If you have factors A,
B, and C, try something like emmip(model, A ~ B | C)
, which
creates an interaction-style plot of the predictions against B, for each
A, with separate panels for each C. This will help visualize what
effects stand out in a practical way. This can guide you in what
post-hoc tests would make sense. See the “interactions” vignette for more discussion
and examples.
I have covariate(s) that interact(s) with factor(s)
This is a situation where it may well be appropriate to compare the
slopes of trend lines, rather than the EMMs. See the
help("emtrends"()
“)` and the discussion of this topic in the “interactions” vignette
I have covariate(s) and am fitting a polynomial model
You need to be careful to define the reference grid consistently. For
example, if you use covariates x
and xsq
(equal to x^2
) to fit a quadratic curve, the default
reference grid uses the mean of each covariate – and
mean(xsq)
is usually not the same as
mean(x)^2
. So you need to use at
to ensure
that the covariates are set consistently with respect to the model. See
this subsection of the “basics”
vignette for an example.
Some “significant” comparisons have overlapping confidence intervals
That can happen because it is just plain wrong to use
[non-]overlapping CIs for individual means to do comparisons. Look
at the printed results from something like
emmeans(mymodel, pairwise ~ treatment)
. In particular, note
that the SE
values are not the same*, and may even
have different degrees of freedom. Means are one thing statistically,
and differences of means are quite another thing. Don’t ever mix them
up, and don’t ever use a CI display for comparing means.
I’ll add that making hard-line decisions about “significant” and “non-significant” is in itself a poor practice. See the discussion in the “basics” vignette
All my pairwise comparisons have the same P value
This will happen if you fitted a model where the treatments you want
to compare were put in as a numeric predictor; for example
dose
, with values of 1, 2, and 3. If dose
is
modeled as numeric, you will be fitting a linear trend in those dose
values, rather than a model that allows those doses to differ in
arbitrary ways. Go back and fit a different model using
factor(dose)
instead; it will make all the difference. This
is closely related to the next topic.
emmeans() doesn’t work as expected
Equivalently, users ask how to get post hoc comparisons when we have covariates rather than factors. Yes, it does work, but you have to tell it the appropriate reference grid.
But before saying more, I have a question for you: Are you sure your model is meaningful?
- If your question concerns only two-level predictors such as
sex
(coded 1 for female, 2 for male), no problem. The model will produce the same predictions as you’d get if you’d used these as factors. - If any of the predictors has 3 or more levels, you may have
fitted a nonsense model, in which case you need to fit a different model
that does make sense before doing any kind of post hoc
analysis. For instance, the model contains a covariate
brand
(coded 1 for Acme, 2 for Ajax, and 3 for Al’s), this model is implying that the difference between Acme and Ajax is exactly equal to the difference between Ajax and Al’s, owing to the fact that a linear trend inbrand
has been fitted. If you had instead coded 1 for Ajax, 2 for Al’s, and 3 for Acme, the model would produce different fitted values. Ask yourself if it makes sense to havebrand = 2.319
. If not, you need to fit another model usingfactor(brand)
in place ofbrand
.
Assuming that the appropriateness of the model is settled, the
current version of emmeans automatically casts
two-value covariates as factors, but not covariates having higher
numbers of unique values. Suppose your model has a covariate
dose
which was experimentally varied over four levels, but
can sensibly be interpreted as a numerical predictor. If you want to
include the separate values of dose
rather than the mean
dose
, you can do that using something like
emmeans(model, "dose", at = list(dose = 1:4))
, or
emmeans(model, "dose", cov.keep = "dose")
, or
emmeans(model, "dose", cov.keep = "4")
. There are small
differences between these. The last one regards any covariate having 4
or fewer unique values as a factor.
See “altering the reference grid” in the “basics” vignette for more discussion.
All or some of the results are NA
The emmeans package uses tools in the estimability package to determine whether its results are uniquely estimable. For example, in a two-way model with interactions included, if there are no observations in a particular cell (factor combination), then we cannot estimate the mean of that cell.
When some of the EMMs are estimable and others are not, that is information about missing information in the data. If it’s possible to remove some terms from the model (particularly interactions), that may make more things estimable if you re-fit with those terms excluded; but don’t delete terms that are really needed for the model to fit well.
When all of the estimates are non-estimable, it could be symptomatic of something else. Some possibilities include:
- An overly ambitious model; for example, in a Latin square design, interaction effects are confounded with main effects; so if any interactions are included in the model, you will render main effects inestimable.
- Possibly you have a nested structure that needs to be included in
the model or specified via the
nesting
argument. Perhaps the levels that B can have depend on which level of A is in force. Then B is nested in A and the model should specifyA + A:B
, with no main effect forB
. - Modeling factors as numeric predictors (see also the related section on covariates). To illustrate,
suppose you have data on particular state legislatures, and the model
includes the predictors
state_name
as well asdem_gov
which is coded 1 if the governor is a Democrat and 0 otherwise. If the model was fitted withstate_name
as a factor or character variable, butdem_gov
as a numeric predictor, then, chances are,emmeans()
will return non-estimable results. If instead, you usefactor(dem_gov)
in the model, then the fact thatstate_name
is nested indem_gov
will be detected, causing EMMs to be computed separately for each party’s states, thus making things estimable. - Some other things may in fact be estimable. For illustration, it’s easy to construct an example where all the EMMs are non-estimable, but pairwise comparisons are estimable:
pg <- transform(pigs, x = rep(1:3, c(10, 10, 9)))
pg.lm <- lm(log(conc) ~ x + source + factor(percent), data = pg)
emmeans(pg.lm, consec ~ percent)
## $emmeans
## percent emmean SE df asymp.LCL asymp.UCL
## 9 nonEst NA NA NA NA
## 12 nonEst NA NA NA NA
## 15 nonEst NA NA NA NA
## 18 nonEst NA NA NA NA
##
## Results are averaged over the levels of: source
## Results are given on the log (not the response) scale.
## Confidence level used: 0.95
##
## $contrasts
## contrast estimate SE df t.ratio p.value
## percent12 - percent9 0.1796 0.0561 23 3.202 0.0114
## percent15 - percent12 0.0378 0.0582 23 0.650 0.8613
## percent18 - percent15 0.0825 0.0691 23 1.194 0.5201
##
## Results are averaged over the levels of: source
## Results are given on the log (not the response) scale.
## P value adjustment: mvt method for 3 tests
The “messy-data” vignette has more examples and discussion.
If I analyze subsets of the data separately, I get different results
Estimated marginal means summarize the model that you fitted
to the data – not the data themselves. Many of the most common models
rely on several simplifying assumptions – that certain effects are
linear, that the error variance is constant, etc. – and those
assumptions are passed forward into the emmeans()
results.
Doing separate analyses on subsets usually comprises departing from that
overall model, so of course the results are different.
My lsmeans/EMMs are way off from what I expected
First step: Carefully read the annotations below the output. Do they
say something like “results are on the log scale, not the response
scale”? If so, that explains it. A Poisson or logistic model involves a
link function, and by default, emmeans()
produces its
results on that same scale. You can add type = "response"
to the emmeans()
call and it will put the results of the
scale you expect. But that is not always the best approach. The “transformations” vignette has examples
and discussion.
Why do I get Inf
for the degrees of freedom?
This is simply the way that emmeans labels asymptotic results (that is, estimates that are tested against the standard normal distribution – z tests – rather than the t distribution). Note that obtaining quantiles or probabilities from the t distribution with infinite degrees of freedom is the same as obtaining the corresponding values from the standard normal. For example:
## [1] 1.281552 1.644854 1.959964
## [1] 1.281552 1.644854 1.959964
so when you see infinite d.f., that just means its a z test or a z confidence interval.
I get exactly the same comparisons for each “by” group
As mentioned elsewhere, EMMs summarize a model, not the
data. If your model does not include any interactions between the
by
variables and the factors for which you want EMMs, then
by definition, the effects for the latter will be exactly the same
regardless of the by
variable settings. So of course the
comparisons will all be the same. If you think they should be different,
then you are saying that your model should include interactions between
the factors of interest and the by
factors.
My ANOVA F is significant, but no pairwise comparisons are
First of all, you should not be making binary decisions of “significant” or “nonsignificant.” This is a simplistic view of P values that assigns an unmerited magical quality to the value 0.05. It is suggested that you just report the P values actually obtained, and let your readers decide how significant your findings are in the context of the scientific findings.
But to answer the question: This is a common misunderstanding of ANOVA. If F has a particular P value, this implies only that some contrast among the means (or effects) has the same P value, after applying the Scheffe adjustment. That contrast may be very much unlike a pairwise comparison, especially when there are several means being compared. Having an F statistic with a P value of, say, 0.06, does not imply that any pairwise comparison will have a P value of 0.06 or smaller. Again referring to the paragraph above, just report the P value for each pairwise comparison, and don’t try to relate them to the F statistic.
Another consideration is that by default, P values for pairwise comparisons are adjusted using the Tukey method, and the adjusted P values can be quite a bit larger than the unadjusted ones. (But I definitely do not advocate using no adjustment to “repair” this problem.)
I wanted differences, but instead I got ratios (or odds ratios)
When a transformation or link involves logs, then, unlike other transformations, comparisons can be back-transformed into ratios – and that is the default behavior. If you really want differences and not ratios, you can re-grid the means first. Re-gridding starts anew with everything on the response scale, and no memory of the transformation.
PS – A side effect is that this causes the tests to be done using SEs obtained by the delta method on the re-gridded scale, rather than on the link scale. Re-gridding can be used with any transformation, not just logs, and it has the same side effect.
I asked for Tukey adjustments, but that’s not what I got
There are two reasons this could happen:
- There is only one comparison in each
by
group (see next topic). - A Tukey adjustment is inappropriate. The Tukey adjustment is
appropriate for pairwise comparisons of means. When you have some other
set of contrasts, the Tukey method is deemed unsuitable and the Sidak
method is used instead. A suggestion is to use
"mvt"
adjustment (which is exact); we don’t default to this because it can require a lot of computing time for a large set of contrasts or comparisons.
emmeans()
completely ignores my P-value
adjustments
This happens when there are only two means (or only two in each
by
group). Thus there is only one comparison. When there is
only one thing to test, there is no multiplicity issue, and hence no
multiplicity adjustment to the P values.
If you wish to apply a P-value adjustment to all tests
across all groups, you need to null-out the by
variable and
summarize, as in the following:
EMM <- emmeans(model, ~ treat | group) # where treat has 2 levels
pairs(EMM, adjust = "sidak") # adjustment is ignored - only 1 test per group
summary(pairs(EMM), by = NULL, adjust = "sidak") # all are in one group now
Note that if you put by = NULL
inside the call
to pairs()
, then this causes all
treat
,group
combinations to be compared.
emmeans()
gives me pooled t tests, but I
expected Welch’s t
It is important to note that emmeans()
and its relatives
produce results based on the model object that you provide –
not the data. So if your sample SDs are wildly different, a model fitted
using lm()
or aov()
is not a good model,
because those R functions use a statistical model that presumes that the
errors have constant variance. That is, the problem isn’t in
emmeans()
, it’s in handing it an inadequate model
object.
Here is a simple illustrative example. Consider a simple one-way experiment and the following model:
mod1 <- aov(response ~ treat, data = mydata)
emmeans(mod1, pairwise ~ treat)
This code will estimate means and comparisons among treatments. All
standard errors, confidence intervals, and t statistics are
based on the pooled residual SD with N - k degrees of freedom
(assuming N observations and k treatments). These
results are useful only if the underlying assumptions of
mod1
are correct – including the assumption that the error
SD is the same for all treatments.
Alternatively, you could fit the following model using generalized least-squares:
mod2 = nlme::gls(response ~ treat, data = mydata,
weights = varIdent(form = ~1 | treat))
emmeans(mod2, pairwise ~ treat)
This model specifies that the error variance depends on the levels of
treat
. This would be a much better model to use when you
have wildly different sample SDs. The results of the
emmeans()
call will reflect this improvement in the
modeling. The standard errors of the EMMs will depend on the individual
sample variances, and the t tests of the comparisons will be in
essence the Welch t statistics with Satterthwaite degrees of
freedom.
To obtain appropriate post hoc estimates, contrasts, and
comparisons, one must first find a model that successfully explains the
peculiarities in the data. This point cannot be emphasized enough. If
you give emmeans()
a good model, you will obtain correct
results; if you give it a bad model, you will obtain incorrect results.
Get the model right first.
I want to see more digits
The summary display of emmeans()
and other functions
uses an optimal-digits routine that shows only relevant digits. If you
temporarily want to see more digits, add
|> as.data.frame()
to the code line. In addition, piping
|> data.frame()
will disable formatting of test
statistics and P values (but unfortunately will also suppress messages
below the tabular output).