I am getting different results from lmPerm based on the order in which I enter the variables in the function call.
For example, placing NCF.pf before TotalProperties yields the following:
pfit <- lmp(NetCashOps ~ NCF.pf + TotalProperties, data = sub.pm, subset = Presence == 1)
summary(pfit)
...
Coefficients:
Estimate Iter Pr(Prob)
NCF.pf 4.581e-01 51 1
TotalProperties 5.246e+04 5000 <2e-16 ***
but, when I switch the order of the coefficients in the formula and place TotalProperties before NCF.pf, the p-value on NCF.pf becomes significant
pfit2 <- lmp(NetCashOps ~ TotalProperties + NCF.pf, data = sub.pm, subset = Presence == 1)
summary(pfit2)
...
Coefficients:
Estimate Iter Pr(Prob)
TotalProperties 5.246e+04 5000 <2e-16 ***
NCF.pf 4.581e-01 5000 <2e-16 ***
Am I missing something? Why would the p-values be different just because I switched the order of the variables in the function call?
Update - Data Source and lm Output (11/11/2016)
The data can be found on GitHub at this link.
When calling the standard lm function twice (reversing the order of the variables on the second call), the p-values are identical (see below). Hence, unlike when using the lmPerm function, the order of the variables doesn't matter with lm.
fit1 <- lm(NetCashOps ~ NCF.pf + TotalProperties, data = sub.pm, subset = Presence == 1)
summary(fit1)
...
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 7.088e+05 2.258e+05 3.138 0.0019 **
NCF.pf 4.581e-01 1.112e-01 4.121 5.11e-05 ***
TotalProperties 5.246e+04 9.519e+03 5.511 8.76e-08 ***
fit2 <- lm(NetCashOps ~ TotalProperties + NCF.pf, data = sub.pm, subset = Presence == 1)
summary(fit2)
...
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 7.088e+05 2.258e+05 3.138 0.0019 **
TotalProperties 5.246e+04 9.519e+03 5.511 8.76e-08 ***
NCF.pf 4.581e-01 1.112e-01 4.121 5.11e-05 ***
Thanks!
I already saw 2 close votes to migrate this to Cross Validated, but in my humble opinion this should stay on Stack Overflow. It is true, that t-statistic and p-value are not invariant to the specification order of terms, under the non-pivoted QR factorization strategy used by lm and lmp, but as shown in the new edit, for OP's data, these statistic should be invariant. So there must be something sensitive at programming level.
My quick diagnose suggests, that if we set seqs = TRUE, rather than using the default FALSE, we would get consistent result:
## I have subsetted data with `Presence == 1` into a new dataset `dat`
## I have also renamed variable name for simplicity
coef(summary(lmp(y ~ x1 + x2, dat, seqs = TRUE)))
# Estimate Iter Pr(Prob)
#(Intercept) 2.019959e+06 5000 0
#x1 4.580840e-01 5000 0
#x2 5.245619e+04 5000 0
coef(summary(lmp(y ~ x2 + x1, dat, seqs = TRUE)))
# Estimate Iter Pr(Prob)
#(Intercept) 2.019959e+06 5000 0
#x2 5.245619e+04 5000 0
#x1 4.580840e-01 5000 0
Note, the Pr(Prob) should be "< 2e-16" when printed by summary, but when using coef to obtain a matrix, those tiny values are 0.
The documentation of ?lmp mentions a little on this part:
The SS will be calculated _sequentially_, just as ‘lm()’ does; or
they may be calculated _uniquely_, which means that the SS for
each source is calculated conditionally on all other sources.
I am at the moment not sure what SS is (as I am not a user of lmPerm), but this sounds like that for consistent result, we should set seqs = TRUE.
Related
I have the following model:
ModelPower <- lmer(DV ~ GroupAbstract * Condition_Cat_Abs + (1|Participant) + (1 + GroupAbstract|Stimulus), data = Dataset)
This model gives the following output:
Random effects:
Groups Name Variance Std.Dev. Corr
Participant (Intercept) 377.401 19.427
Stimulus (Intercept) 91.902 9.587
GroupAbstractOutgroup 2.003 1.415 -0.40
Residual 338.927 18.410
Number of obs: 16512, groups: Participant, 344; Stimulus, 32
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 65.8962 2.0239 59.6906 32.559 < 0.0000000000000002 ***
GroupAbstractOutgroup -0.9287 0.5561 129.9242 -1.670 0.0973 .
Condition_Cat_AbsSecondOrderIn -2.2584 0.4963 16103.9277 -4.550 0.00000539 ***
Condition_Cat_AbsSecondOrderOut -7.0821 0.4963 16103.9277 -14.270 < 0.0000000000000002 ***
GroupAbstractOutgroup:Condition_Cat_AbsSecondOrderIn -3.0229 0.7019 16103.9277 -4.307 0.00001665 ***
GroupAbstractOutgroup:Condition_Cat_AbsSecondOrderOut 7.8765 0.7019 16103.9277 11.222 < 0.0000000000000002 ***
I am interested in the interaction "GroupAbstractOutgroup:Condition_Cat_AbsSecondOrderIn" and I am trying to estimate the sample size to detect an effect size of at least -2 using the R package simr. the original slope is -3.02 so I specify the new one:
ModelPower#beta[names(fixef(ModelPower)) %in% "GroupAbstractOutgroup:Condition_Cat_AbsSecondOrderIn"] <- -2
However, regardless of how I specify the powerSim function both for the main effects and interactions (see some examples below), I get power of 0% and the following error when running lastResult()$errors 'object is not a matrix'. I know what the error should mean but even after converting the original data frame and the table of fixed effects to a matrix, the error is still there and I am not sure what it is referring to and how to get the actual output. Any help would be much appreciated!
Examples of the powerSim function:
powerSim(ModelPower, test=fixed("GroupAbstract", "anova"), nsim=10, seed=1)
powerSim(ModelPower, test=fixed("GroupAbstractOutgroup:Condition_Cat_AbsSecondOrderIn", "anova"), nsim=10, seed=1)
I have following model:
> summary(mymodel.gml6)
Call:
glm(formula = anumber ~ poly(coun, 2, raw = TRUE) + pharm+
pharm:patus, family = poisson, data = mydata)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.4805 -1.7070 -0.7171 0.4482 12.6264
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.2786022 0.0570305 -4.885 1.03e-06 ***
poly(coun, 2, raw = TRUE)1 0.2217527 0.0162746 13.626 < 2e-16 ***
poly(coun, 2, raw = TRUE)2 -0.0156164 0.0009538 -16.372 < 2e-16 ***
pharmyes 0.3945798 0.0343844 11.476 < 2e-16 ***
pharmno:patusyes 0.0178374 0.0352272 0.506 0.613
pharmyes:patusyes 0.2206909 0.0311678 7.081 1.43e-12 ***
pharm as well as patus are factors containing the values "yes" and "no". Now, I would like to remove the term pharmno:patusyes as it is not significant. I tried with the update method, but it did not work:
mymodel<-update(mymodel, .~.-pharmno:patusyes)
The questions here and here are slightly different as they focus on the removal of a complete interaction term.
R isn't letting you do it, because it shouldn't be done. The interaction is represented by two variables; you need to retain both or neither. It doesn't matter if the p-value reported on the line associated with one of those variables is significant, as you will get different patterns of significance with the same data with different ways of specifying the same model. To test the interaction, you need a simultaneous test of both variables. It seems to me there is something strange in your model but not enough information is presented to see, nonetheless, you should be able to do drop1(mymodel.glm6, test="LRT") to get that test in R.
I have a regression model for some time series data investigating drug utilisation. The purpose is to fit a spline to a time series and work out 95% CI etc. The model goes as follows:
id <- ts(1:length(drug$Date))
a1 <- ts(drug$Rate)
a2 <- lag(a1-1)
tg <- ts.union(a1,id,a2)
mg <-lm (a1~a2+bs(id,df=df1),data=tg)
The summary output of mg is:
Call:
lm(formula = a1 ~ a2 + bs(id, df = df1), data = tg)
Residuals:
Min 1Q Median 3Q Max
-0.31617 -0.11711 -0.02897 0.12330 0.40442
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.77443 0.09011 8.594 1.10e-11 ***
a2 0.13270 0.13593 0.976 0.33329
bs(id, df = df1)1 -0.16349 0.23431 -0.698 0.48832
bs(id, df = df1)2 0.63013 0.19362 3.254 0.00196 **
bs(id, df = df1)3 0.33859 0.14399 2.351 0.02238 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
I am using the Pr(>|t|) value of a2 to test if the data under investigation are autocorrelated.
Is it possible to extract this value of Pr(>|t|) (in this model 0.33329) and store it in a scalar to perform a logical test?
Alternatively, can it be worked out using another method?
A summary.lm object stores these values in a matrix called 'coefficients'. So the value you are after can be accessed with:
a2Pval <- summary(mg)$coefficients[2, 4]
Or, more generally/readably, coef(summary(mg))["a2","Pr(>|t|)"]. See here for why this method is preferred.
The package broom comes in handy here (it uses the "tidy" format).
tidy(mg) will give a nicely formated data.frame with coefficients, t statistics etc. Works also for other models (e.g. plm, ...).
Example from broom's github repo:
lmfit <- lm(mpg ~ wt, mtcars)
require(broom)
tidy(lmfit)
term estimate std.error statistic p.value
1 (Intercept) 37.285 1.8776 19.858 8.242e-19
2 wt -5.344 0.5591 -9.559 1.294e-10
is.data.frame(tidy(lmfit))
[1] TRUE
Just pass your regression model into the following function:
plot_coeffs <- function(mlr_model) {
coeffs <- coefficients(mlr_model)
mp <- barplot(coeffs, col="#3F97D0", xaxt='n', main="Regression Coefficients")
lablist <- names(coeffs)
text(mp, par("usr")[3], labels = lablist, srt = 45, adj = c(1.1,1.1), xpd = TRUE, cex=0.6)
}
Use as follows:
model <- lm(Petal.Width ~ ., data = iris)
plot_coeffs(model)
To answer your question, you can explore the contents of the model's output by saving the model as a variable and clicking on it in the environment window. You can then click around to see what it contains and what is stored where.
Another way is to type yourmodelname$ and select the components of the model one by one to see what each contains. When you get to yourmodelname$coefficients, you will see all of beta-, p, and t- values you desire.
We're trying to model a count variable with excessive zeros using a zero-inflated poisson (as implemented in pscl package). Here is a (simplified) output showing both categorical and continuous explanatory variables:
library(pscl)
> m1 <- zeroinfl(y ~ treatment + some_covar, data = d, dist =
"poisson")
> summary(m1)
Count model coefficients (poisson with log link):
Estimate Std. Error z value Pr(>|z|)
(Intercept) 3.189253 0.102256 31.189 < 2e-16 ***
treatmentB -0.282478 0.107965 -2.616 0.00889 **
treatmentC 0.227633 0.103605 2.197 0.02801 *
some_covar 0.002190 0.002329 0.940 0.34706
Zero-inflation model coefficients (binomial with logit link):
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.67251 0.74961 0.897 0.3696
treatmentB -1.72728 0.89931 -1.921 0.0548 .
treatmentC -0.31761 0.77668 -0.409 0.6826
some_covar -0.03736 0.02684 -1.392 0.1640
summary gave us some good answers but we are looking for a ANOVA-like table. So, the question is: is it ok to use car::Anova to obtain such table?
> Anova(m1)
Analysis of Deviance Table (Type II tests)
Response: y
Df Chisq Pr(>Chisq)
treatment 2 30.7830 2.068e-07 ***
some_covar 1 0.8842 0.3471
It seems to work fine but i'm not really sure whether is a valid approach since documentation is missing (seems like is only considering the 'count model' part?). Do you recommend to follow this approach or there is a better way?
I have checked out the defensive methods as was discussed in this post in order to prevent this error but it still doesn't go away.
model<-lmer(Proportion~Plot+Treatment+(1|Plot/Treatment),binomial,data=data)
Error in if (REML) p else 0L : argument is not interpretable as logical
tl;dr you should use glmer instead. Because you haven't named your arguments, R is interpreting them by position (order). lmer's third argument is REML, so R thinks you're specifying REML=binomial, which isn't a legitimate value. family is the third argument to glmer, so this would work (sort of: see below) if you used glmer, but it's usually safer to name the arguments explicitly if there's any possibility of getting confused.
A reproducible example would be nice, but:
model <- glmer(Proportion~Plot+Treatment+(1|Plot/Treatment),
family=binomial,data=data)
is a starting point. I foresee a few more problems though:
if your data are not Bernoulli (0/1) (which I'm guessing not since your response is called Proportion), then you need to include the total number sampled in each trial, e.g. by specifying a weights argument
you have Plot and Treatment as both fixed and as random-effect grouping variables in your model; that won't work. I see that Crawley really does suggest this in the R book (google books link).
Do not do it the way he suggests, it doesn't make any sense. Replicating:
library(RCurl)
url <- "https://raw.githubusercontent.com/jejoenje/Crawley/master/Data/insects.txt"
dd <- read.delim(text=getURL(url),header=TRUE)
## fix typo because I'm obsessive:
levels(dd$treatment) <- c("control","sprayed")
library(lme4)
model <- glmer(cbind(dead,alive)~block+treatment+(1|block/treatment),
data=dd,family=binomial)
If we look at the among-group standard deviation, we see that it's zero for both groups; it's exactly zero for block because block is already included in the fixed effects. It need not be for the treatment:block interaction (we have treatment, but not the interaction between block and treatment, in the fixed effects), but is because there's little among-treatment-within-block variation:
VarCorr(model)
## Groups Name Std.Dev.
## treatment:block (Intercept) 2.8736e-09
## block (Intercept) 0.0000e+00
Conceptually, it makes more sense to treat block as a random effect:
dd <- transform(dd,prop=dead/(alive+dead),ntot=alive+dead)
model1 <- glmer(prop~treatment+(1|block/treatment),
weights=ntot,
data=dd,family=binomial)
summary(model)
## ...
## Formula: prop ~ treatment + (1 | block/treatment)
## Random effects:
## Groups Name Variance Std.Dev.
## treatment:block (Intercept) 0.02421 0.1556
## block (Intercept) 0.18769 0.4332
## Number of obs: 48, groups: treatment:block, 12; block, 6
##
## Fixed effects:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.1640 0.2042 -5.701 1.19e-08 ***
## treatmentsprayed 3.2434 0.1528 21.230 < 2e-16 ***
Sometimes you might want to treat it as a fixed effect:
model2 <- update(model1,.~treatment+block+(1|block:treatment))
summary(model2)
## Random effects:
## Groups Name Variance Std.Dev.
## block:treatment (Intercept) 5.216e-18 2.284e-09
## Number of obs: 48, groups: block:treatment, 12
##
## Fixed effects:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.5076 0.0739 -6.868 6.50e-12 ***
## treatmentsprayed 3.2676 0.1182 27.642 < 2e-16 ***
Now the block-by-treatment interaction variance is effectively zero (because block soaks up more variability if treated as a fixed effect). However, the estimated effect of spraying is very nearly identical.
If you're worried about overdispersion you can add an individual-level random effect (or use MASS::glmmPQL; lme4 no longer fits quasi-likelihood models)
dd <- transform(dd,obs=factor(seq(1:nrow(dd))))
model3 <- update(model1,.~.+(1|obs))
## Random effects:
## Groups Name Variance Std.Dev.
## obs (Intercept) 4.647e-01 6.817e-01
## treatment:block (Intercept) 1.138e-09 3.373e-05
## block (Intercept) 1.813e-01 4.258e-01
## Number of obs: 48, groups: obs, 48; treatment:block, 12; block, 6
##
## Fixed effects:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.1807 0.2411 -4.897 9.74e-07 ***
## treatmentsprayed 3.3481 0.2457 13.626 < 2e-16 ***
The observation-level effect has effectively replaced the treatment-by-block interaction (which is now close to zero). Again, the estimated spraying effect has hardly changed (but its standard error is twice as large ...)