How to get the result of lrm() respectively?
I use lrm() to bulid a logistic model, and get the result as follows:
n <- 1000 # define sample size
y <- rep(0:1, 500)
age <- rnorm(n, 50, 10)
sex <- factor(sample(c('female','male'), n,TRUE))
f <- lrm(y ~ age + sex, x=TRUE, y=TRUE)
f
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 1000 LR chi2 1.50 R2 0.002 C 0.520
0 500 d.f. 2 g 0.088 Dxy 0.040
1 500 Pr(> chi2) 0.4714 gr 1.092 gamma 0.040
max |deriv| 2e-13 gp 0.022 tau-a 0.020
Brier 0.250
Coef S.E. Wald Z Pr(>|Z|)
Intercept 0.2206 0.3370 0.65 0.5127
age -0.0030 0.0065 -0.46 0.6485
sex=male -0.1455 0.1266 -1.15 0.2504
How to get the result above as data.frame respectively? like:
mydf$df1
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 1000 LR chi2 1.50 R2 0.002 C 0.520
0 500 d.f. 2 g 0.088 Dxy 0.040
1 500 Pr(> chi2) 0.4714 gr 1.092 gamma 0.040
max |deriv| 2e-13 gp 0.022 tau-a 0.020
Brier 0.250
mydf$df2
Coef S.E. Wald Z Pr(>|Z|)
Intercept 0.2206 0.3370 0.65 0.5127
age -0.0030 0.0065 -0.46 0.6485
sex=male -0.1455 0.1266 -1.15 0.2504
Try,
res = capture.output(print(f), append = F, sep = " ")
lapply(res, function(x) write.table(data.frame(x), 'res.csv' , append= T, sep=',' ))
Related
I am trying to create a summary table for my Cox model. However, when I use modelsummary function, it gives me a table that shows coef. But I want to display exp(coef) on my summary table. How can I change coef to exp(coef)?
I use this script to create a summary table:
modelsummary(model.1,
statistic='({conf.low}, {conf.high})',
stars=TRUE,
vcov = 'classical',
coef_omit = "Intercept",
coef_rename=c('ln_reb_capacity'='Relative rebel strength',
'terrcont'='Rebel territorial control', 'gdp'='Economic strength',
'bdbest'='Conflict intensity', 'roughterrain'='Rough terrain',
'loot'='Lootable resources', 'in_tpop'='Population size',
'powersharing'='Sharing Leadership'),
title = 'Table I.',
output='gt'
)
This is the summary table:
Table I.
─────────────────────────────────────────────────────────
Model 1
─────────────────────────────────────────────────────────
Relative rebel strength 0.125*
(0.016, 0.235)
Rebel territorial control -0.295+
(-0.638, 0.048)
Economic strength 0.000
(0.000, 0.000)
Conflict intensity 0.000
(0.000, 0.000)
Rough terrain 0.098
(-0.210, 0.405)
Lootable resources 0.105
(-0.298, 0.507)
Population size -0.119+
(-0.249, 0.011)
Sharing Leadership 0.046
(-0.393, 0.486)
─────────────────────────────────────────────────────────
Num.Obs. 260
AIC 1678.5
BIC 1707.0
RMSE 0.83
Std.Errors Classical
─────────────────────────────────────────────────────────
+ p < 0.1, * p < 0.05, ** p < 0.01, *** p < 0.001
─────────────────────────────────────────────────────────
Column names: , Model 1
Here is my result for Cox model:
Call:
coxph(formula = Surv(month_duration, EndConflict) ~ ln_reb_capacity +
terrcont + gdp + bdbest + roughterrain + loot + in_tpop +
powersharing, data = df)
n= 260, number of events= 183
(108 observations deleted due to missingness)
coef exp(coef) se(coef) z Pr(>|z|)
ln_reb_capacity 0.125154562 1.133323609 0.055831926 2.242 0.0250 *
terrcont -0.295113621 0.744446997 0.174927860 -1.687 0.0916 .
gdp -0.000004416 0.999995584 0.000017623 -0.251 0.8021
bdbest -0.000010721 0.999989279 0.000016057 -0.668 0.5043
roughterrain 0.097602616 1.102524573 0.156809154 0.622 0.5337
loot 0.104686159 1.110362079 0.205406301 0.510 0.6103
in_tpop -0.119020975 0.887789179 0.066355450 -1.794 0.0729 .
powersharing 0.046026931 1.047102610 0.224229347 0.205 0.8374
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
exp(coef) exp(-coef) lower .95 upper .95
ln_reb_capacity 1.1333 0.8824 1.0159 1.264
terrcont 0.7444 1.3433 0.5284 1.049
gdp 1.0000 1.0000 1.0000 1.000
bdbest 1.0000 1.0000 1.0000 1.000
roughterrain 1.1025 0.9070 0.8108 1.499
loot 1.1104 0.9006 0.7424 1.661
in_tpop 0.8878 1.1264 0.7795 1.011
powersharing 1.0471 0.9550 0.6747 1.625
Concordance= 0.617 (se = 0.023 )
Likelihood ratio test= 18.96 on 8 df, p=0.02
Wald test = 18.2 on 8 df, p=0.02
Score (logrank) test = 18.36 on 8 df, p=0.02
Thanks.
You could adjust the argument exponentiate. If it's TRUE, the estimate, conf.low, and conf.high statistics are exponentiated, and the std.error is transformed to exp(estimate)*std.error
(by the delta method).
modelsummary(model.1,
...,
exponentiate = TRUE
)
Hi stackoverflow community,
I am a recent R starter and today I tried several hours to figure out how to get a scientific p-value (e.g. 3*e-1) from a competing risk analysis using the cmprsk package.
I used:
sumary_J1<-crr(ftime, fstatus, cov1, failcode=2)
summary(sumary_J1)
And got
Call:
crr(ftime = ftime, fstatus = fstatus, cov1 = cov1, failcode = 2)
coef exp(coef) se(coef) z p-value
group1 0.373 1.45 0.02684 13.90 0.00
age 0.122 1.13 0.00384 31.65 0.00
sex 0.604 1.83 0.04371 13.83 0.00
bmi 0.012 1.01 0.00611 1.96 0.05
exp(coef) exp(-coef) 2.5% 97.5%
group1 1.45 0.689 1.38 1.53
age 1.13 0.886 1.12 1.14
sex 1.83 0.546 1.68 1.99
bmi 1.01 0.988 1.00 1.02
Num. cases = 470690 (1900 cases omitted due to missing values)
Pseudo Log-likelihood = -28721
Pseudo likelihood ratio test = 2229 on 4 df,
I can see the p-value column,but I only get two decimal places. I would like to see as many decimal places as possible or print those p-values in the format e.g. 3.0*e-3.
I tried all of those, but nothing worked so far:
summary(sumary_J1, digits=max(options()$digits - 5,10))
print.crr(sumary_J1, digits = 20)
print.crr(sumary_J1, digits = 3, scipen = -2)
print.crr(sumary_J1, format = "e", digits = 3)
Maybe someone is able to help me! Thanks!
Best,
Carolin
The use of digits=2 limits the number of digits to the right of the decimal point when used as an argument to a .summary value. The digits parameter does affect how results are displayed for summary.crr.
summary(z, digits=3) # using first example in `?cmprsk::crr`
#----------------------
#Competing Risks Regression
Call:
crr(ftime = ftime, fstatus = fstatus, cov1 = cov)
coef exp(coef) se(coef) z p-value
x1 0.2668 1.306 0.421 0.633 0.526
x2 -0.0557 0.946 0.381 -0.146 0.884
x3 0.2805 1.324 0.381 0.736 0.462
exp(coef) exp(-coef) 2.5% 97.5%
x1 1.306 0.766 0.572 2.98
x2 0.946 1.057 0.448 2.00
x3 1.324 0.755 0.627 2.79
Num. cases = 200
Pseudo Log-likelihood = -320
Pseudo likelihood ratio test = 1.02 on 3 df,
You can use formatC to control format:
formatC( summary(z, digits=5)$coef , format="e")
#------------>
coef exp(coef) se(coef) z p-value
x1 "2.6676e-01" "1.3057e+00" "4.2115e-01" "6.3340e-01" "5.2647e-01"
x2 "-5.5684e-02" "9.4584e-01" "3.8124e-01" "-1.4606e-01" "8.8387e-01"
x3 "2.8049e-01" "1.3238e+00" "3.8098e-01" "7.3622e-01" "4.6159e-01"
You also might search on [r] very small p-value
Here's the first of over 100 hits on that topic, which despite not very much attention, still has very useful information and coding examples: Reading a very small p-value in R
By looking at the function that prints the output of crr() (cmprsk::print.crr) you can see what is done to create the p-values displayed in the summary. The code below is taken from that function.
x <- sumary_J1
v <- sqrt(diag(x$var))
signif(v, 4) # Gives you the one-sided p-values.
v <- 2 * (1 - pnorm(abs(x$coef)/v))
signif(v, 4) # Gibes you the two-sided p-values.
I am using semPaths (semPlot package) to draw my structural equation models. After some trial and error, I have a pretty good script to show what I want. Except, I haven’t been able to figure out how to include the p-value/significance levels of the estimates/regression coefficients in the figure.
Can/how can I include significance levels either as e.g. p-value in the edge labels below the estimate or as a broken line for insignificance or …?
I am also interested in including the R-square, but not as critically as the significance level.
This is the script I am using so far:
semPaths(fitmod.bac.class2,
what = "std",
whatLabels = "std",
style="ram",
edge.label.cex = 1.3,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7 )
Example of one of the SemPath outputs
In this example the following are not significant:
Ignavibacteria -> First_C_CO2_ugC_gC_day, p = 0.096
pH -> Ignavibacteria, p = 0.151
cand_class_MB_A2_108 <-> Bacilli correlation, p = 0.054
I am a R-user and not really a coder, so I might just be missing a crucial point in the arguments.
I am testing a lot of different models at the moment, and would really like not to have to draw them all up by hand.
update:
Using semPlotModel: Am I right in understanding that semPlotModel doesn’t include the significance levels from the sem function (see my script and output below)? I am specifically looking to include the P(>|z|) for regressions and covariance.
Is it just me that is missing that, or is it not included? If it is not included, my solution is simply just to custom the edge labels.
{model.NA.UP.bac.class2 <- '
#LATANT VARIABLES
#REGRESSIONS
#soil organic carbon quality
c_Negativicutes ~ CN
#microorganisms
First_C_CO2_ugC_gC_day ~ c_Bacilli
First_C_CO2_ugC_gC_day ~ c_Ignavibacteria
First_C_CO2_ugC_gC_day ~ c_cand_class_MB_A2_108
First_C_CO2_ugC_gC_day ~ c_Negativicutes
#pH
c_Bacilli ~pH
c_Ignavibacteria ~pH
c_cand_class_MB_A2_108~pH
c_Negativicutes ~pH
#COVARIANCE
initial_water ~~ CN
c_cand_class_MB_A2_108 ~~ c_Bacilli
'
fitmod.bac.class2 <- sem(model.NA.UP.bac.class2, data=datapNA.UP.log, missing="ml", meanstructure=TRUE, fixed.x=FALSE, std.lv=FALSE, std.ov=FALSE)
summary(fitmod.bac.class2, standardized=TRUE, fit.measures=TRUE, rsq=TRUE)
out <- capture.output(summary(fitmod.bac.class2, standardized=TRUE, fit.measures=TRUE, rsq=TRUE))
}
Output:
lavaan 0.6-5 ended normally after 188 iterations
Estimator ML
Optimization method NLMINB
Number of free parameters 28
Number of observations 30
Number of missing patterns 1
Model Test User Model:
Test statistic 17.816
Degrees of freedom 16
P-value (Chi-square) 0.335
Model Test Baseline Model:
Test statistic 101.570
Degrees of freedom 28
P-value 0.000
User Model versus Baseline Model:
Comparative Fit Index (CFI) 0.975
Tucker-Lewis Index (TLI) 0.957
Loglikelihood and Information Criteria:
Loglikelihood user model (H0) 472.465
Loglikelihood unrestricted model (H1) 481.373
Akaike (AIC) -888.930
Bayesian (BIC) -849.697
Sample-size adjusted Bayesian (BIC) -936.875
Root Mean Square Error of Approximation:
RMSEA 0.062
90 Percent confidence interval - lower 0.000
90 Percent confidence interval - upper 0.185
P-value RMSEA <= 0.05 0.414
Standardized Root Mean Square Residual:
SRMR 0.107
Parameter Estimates:
Information Observed
Observed information based on Hessian
Standard errors Standard
Regressions:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
c_Negativicutes ~
CN 0.419 0.143 2.939 0.003 0.419 0.416
c_cand_class_MB_A2_108 ~
CN -0.433 0.160 -2.707 0.007 -0.433 -0.394
First_C_CO2_ugC_gC_day ~
c_Bacilli 0.525 0.128 4.092 0.000 0.525 0.496
c_Ignavibacter 0.207 0.124 1.667 0.096 0.207 0.195
c_c__MB_A2_108 0.310 0.125 2.475 0.013 0.310 0.301
c_Negativicuts 0.304 0.137 2.220 0.026 0.304 0.271
c_Bacilli ~
pH 0.624 0.135 4.604 0.000 0.624 0.643
c_Ignavibacteria ~
pH 0.245 0.171 1.436 0.151 0.245 0.254
c_cand_class_MB_A2_108 ~
pH 0.393 0.151 2.597 0.009 0.393 0.394
c_Negativicutes ~
pH 0.435 0.129 3.361 0.001 0.435 0.476
Covariances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
CN ~~
initial_water 0.001 0.000 2.679 0.007 0.001 0.561
.c_cand_class_MB_A2_108 ~~
.c_Bacilli -0.000 0.000 -1.923 0.054 -0.000 -0.388
Intercepts:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.c_Negativicuts 0.145 0.198 0.734 0.463 0.145 3.826
.c_c__MB_A2_108 1.038 0.226 4.594 0.000 1.038 25.076
.Frs_C_CO2_C_C_ -0.346 0.233 -1.485 0.137 -0.346 -8.115
.c_Bacilli 0.376 0.135 2.778 0.005 0.376 9.340
.c_Ignavibacter 0.754 0.170 4.424 0.000 0.754 18.796
CN 0.998 0.007 145.158 0.000 0.998 26.502
pH 0.998 0.008 131.642 0.000 0.998 24.034
initial_water 0.998 0.008 125.994 0.000 0.998 23.003
Variances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.c_Negativicuts 0.001 0.000 3.873 0.000 0.001 0.600
.c_c__MB_A2_108 0.001 0.000 3.833 0.000 0.001 0.689
.Frs_C_CO2_C_C_ 0.001 0.000 3.873 0.000 0.001 0.408
.c_Bacilli 0.001 0.000 3.873 0.000 0.001 0.586
.c_Ignavibacter 0.002 0.000 3.873 0.000 0.002 0.936
CN 0.001 0.000 3.873 0.000 0.001 1.000
initial_water 0.002 0.000 3.873 0.000 0.002 1.000
pH 0.002 0.000 3.873 0.000 0.002 1.000
R-Square:
Estimate
c_Negativicuts 0.400
c_c__MB_A2_108 0.311
Frs_C_CO2_C_C_ 0.592
c_Bacilli 0.414
c_Ignavibacter 0.064
Warning message:
In lav_model_hessian(lavmodel = lavmodel, lavsamplestats = lavsamplestats, :
lavaan WARNING: Hessian is not fully symmetric. Max diff = 5.15131396241486e-05
This example is taken from ?semPaths since we don't have your object.
library('semPlot')
modFile <- tempfile(fileext = '.OUT')
download.file('http://sachaepskamp.com/files/mi1.OUT', modFile)
Use semPlotModel to get the object without plotting. There you can inspect what is to be plotted. I just dug around without reading the docs until I found what it seems to be using.
After you run semPlotModel, the object has an element x#Pars which contains the edges, nodes, and the std which is being used for the edge labels in your case. semPaths also has an argument that allows you to make custom edge labels, so you can take the data you need from x#Pars and add your p-values:
x <- semPlotModel(modFile)
x#Pars
# label lhs edge rhs est std group fixed par
# 1 lambda[11]^{(y)} perfIQ -> pc 1.000 0.6219648 Group 1 TRUE 0
# 2 lambda[21]^{(y)} perfIQ -> pa 0.923 0.5664888 Group 1 FALSE 1
# 3 lambda[31]^{(y)} perfIQ -> oa 1.098 0.6550159 Group 1 FALSE 2
# 4 lambda[41]^{(y)} perfIQ -> ma 0.784 0.4609990 Group 1 FALSE 3
# 5 theta[11]^{(epsilon)} pc <-> pc 5.088 0.6131598 Group 1 FALSE 5
# 10 theta[22]^{(epsilon)} pa <-> pa 5.787 0.6790905 Group 1 FALSE 6
# 15 theta[33]^{(epsilon)} oa <-> oa 5.150 0.5709541 Group 1 FALSE 7
# 20 theta[44]^{(epsilon)} ma <-> ma 7.311 0.7874800 Group 1 FALSE 8
# 21 psi[11] perfIQ <-> perfIQ 3.210 1.0000000 Group 1 FALSE 4
# 22 tau[1]^{(y)} int pc 10.500 NA Group 1 FALSE 9
# 23 tau[2]^{(y)} int pa 10.374 NA Group 1 FALSE 10
# 24 tau[3]^{(y)} int oa 10.663 NA Group 1 FALSE 11
# 25 tau[4]^{(y)} int ma 10.371 NA Group 1 FALSE 12
# 11 lambda[11]^{(y)} perfIQ -> pc 1.000 0.6515609 Group 2 TRUE 0
# 27 lambda[21]^{(y)} perfIQ -> pa 0.923 0.5876948 Group 2 FALSE 1
# 31 lambda[31]^{(y)} perfIQ -> oa 1.098 0.6981974 Group 2 FALSE 2
# 41 lambda[41]^{(y)} perfIQ -> ma 0.784 0.4621919 Group 2 FALSE 3
# 51 theta[11]^{(epsilon)} pc <-> pc 5.006 0.5754684 Group 2 FALSE 14
# 101 theta[22]^{(epsilon)} pa <-> pa 5.963 0.6546148 Group 2 FALSE 15
# 151 theta[33]^{(epsilon)} oa <-> oa 4.681 0.5125204 Group 2 FALSE 16
# 201 theta[44]^{(epsilon)} ma <-> ma 8.356 0.7863786 Group 2 FALSE 17
# 211 psi[11] perfIQ <-> perfIQ 3.693 1.0000000 Group 2 FALSE 13
# 221 tau[1]^{(y)} int pc 10.500 NA Group 2 FALSE 9
# 231 tau[2]^{(y)} int pa 10.374 NA Group 2 FALSE 10
# 241 tau[3]^{(y)} int oa 10.663 NA Group 2 FALSE 11
# 251 tau[4]^{(y)} int ma 10.371 NA Group 2 FALSE 12
# 26 alpha[1] int perfIQ -2.469 NA Group 2 FALSE 18
As you can see there are more edge labels than ones that are plotted, and I have no idea how it chooses which to use, so I am just taking the first four from each group (since there are four edges shown and the stds match those. Maybe there is an option to plot all of them or select which ones you need--I haven't read the docs.
## take first four stds from each group, generate some p-values
l <- sapply(split(x#Pars$std, x#Pars$group), function(x) head(x, 4))
set.seed(1)
l <- sprintf('%.3f, p=%s', l, format.pval(runif(length(l)), digits = 2))
l
# [1] "0.622, p=0.27" "0.566, p=0.37" "0.655, p=0.57" "0.461, p=0.91" "0.652, p=0.20" "0.588, p=0.90" "0.698, p=0.94" "0.462, p=0.66"
Then you can plot the object with your new labels, edgeLabels = l
layout(1:2)
semPaths(
x,
edgeLabels = l,
ask = FALSE, title = FALSE,
what = 'std',
whatLabels = 'std',
style = 'ram',
edge.label.cex = 1.3,
layout = 'tree',
intercepts = FALSE,
residuals = FALSE,
sizeMan = 7
)
With the help from #rawr, I have worked it out. If anybody else needs to include estimates and p-value from Lavaan in their semPaths, here is how it can be done.
#extracting the parameters from the sem model and selecting the interactions relevant for the semPaths (here, I need 12 estimates and p-values)
table2<-parameterEstimates(fitmod.bac.class2,standardized=TRUE) %>% head(12)
#turning the chosen parameters into text
b<-gettextf('%.3f \n p=%.3f', table2$std.all, digits=table2$pvalue)
I can honestly say that I do not understand how the last bit of script works. This is copied from rawr's answer before a lot of trial and error until it worked. There might (quite possibly) be a nicer way to write it, but it works :)
#putting that list into edgeLabels in sempaths
semPaths(fitmod.bac.class2,
what = "std",
edgeLabels = b,
style="ram",
edge.label.cex = 1,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7
)
Just a small, but relevant detail for an improvement for the above answer.
The above code requires an inspection of the parameter table to count how many lines to maintain to specify as in %>%head(4).
We can exclude from the extracted parameter table those lines which lhs and rhs are not equal.
#extracting the parameters from the sem model and selecting the interactions relevant for the semPaths
table2<-parameterEstimates(fitmod.bac.class2,standardized=TRUE)%>%as.dataframe()
table2<-table2[!table2$lhs==table2$rhs,]
If the formula comprised also extra lines as those with ':=' those also will comprise the parameter table, and should be removed.
The remaining keeps the same...
#turning the chosen parameters into text
b<-gettextf('%.3f \n p=%.3f', table2$std.all, digits=table2$pvalue)
#putting that list into edgeLabels in sempaths
semPaths(fitmod.bac.class2,
what = "std",
edgeLabels = b,
style="ram",
edge.label.cex = 1,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7
)
I am trying to predict and graph models with species presence as the response. However I've run into the following problem: the ggpredict outputs are wildly different for the same data in glmer and glmmTMB. However, the estimates and AIC are very similar. These are simplified models only including date (which has been centered and scaled), which seems to be the most problematic to predict.
yntest<- glmer(MYOSOD.P~ jdate.z + I(jdate.z^2) + I(jdate.z^3) +
(1|area/SiteID), family = binomial, data = sodpYN)
> summary(yntest)
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod']
Family: binomial ( logit )
Formula: MYOSOD.P ~ jdate.z + I(jdate.z^2) + I(jdate.z^3) + (1 | area/SiteID)
Data: sodpYN
AIC BIC logLik deviance df.resid
1260.8 1295.1 -624.4 1248.8 2246
Scaled residuals:
Min 1Q Median 3Q Max
-2.0997 -0.3218 -0.2013 -0.1238 9.4445
Random effects:
Groups Name Variance Std.Dev.
SiteID:area (Intercept) 1.6452 1.2827
area (Intercept) 0.6242 0.7901
Number of obs: 2252, groups: SiteID:area, 27; area, 9
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.96778 0.39190 -7.573 3.65e-14 ***
jdate.z -0.72258 0.17915 -4.033 5.50e-05 ***
I(jdate.z^2) 0.10091 0.08068 1.251 0.21102
I(jdate.z^3) 0.25025 0.08506 2.942 0.00326 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr) jdat.z I(.^2)
jdate.z 0.078
I(jdat.z^2) -0.222 -0.154
I(jdat.z^3) -0.071 -0.910 0.199
The glmmTMB model + summary:
Tyntest<- glmmTMB(MYOSOD.P ~ jdate.z + I(jdate.z^2) + I(jdate.z^3) +
(1|area/SiteID), family = binomial("logit"), data = sodpYN)
> summary(Tyntest)
Family: binomial ( logit )
Formula: MYOSOD.P ~ jdate.z + I(jdate.z^2) + I(jdate.z^3) + (1 | area/SiteID)
Data: sodpYN
AIC BIC logLik deviance df.resid
1260.8 1295.1 -624.4 1248.8 2246
Random effects:
Conditional model:
Groups Name Variance Std.Dev.
SiteID:area (Intercept) 1.6490 1.2841
area (Intercept) 0.6253 0.7908
Number of obs: 2252, groups: SiteID:area, 27; area, 9
Conditional model:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.96965 0.39638 -7.492 6.78e-14 ***
jdate.z -0.72285 0.18250 -3.961 7.47e-05 ***
I(jdate.z^2) 0.10096 0.08221 1.228 0.21941
I(jdate.z^3) 0.25034 0.08662 2.890 0.00385 **
---
ggpredict outputs
testg<-ggpredict(yntest, terms ="jdate.z[all]")
> testg
# Predicted probabilities of MYOSOD.P
# x = jdate.z
x predicted std.error conf.low conf.high
-1.95 0.046 0.532 0.017 0.120
-1.51 0.075 0.405 0.036 0.153
-1.03 0.084 0.391 0.041 0.165
-0.58 0.072 0.391 0.035 0.142
-0.14 0.054 0.390 0.026 0.109
0.35 0.039 0.399 0.018 0.082
0.79 0.034 0.404 0.016 0.072
1.72 0.067 0.471 0.028 0.152
Adjusted for:
* SiteID = 0 (population-level)
* area = 0 (population-level)
Standard errors are on link-scale (untransformed).
testgTMB<- ggpredict(Tyntest, "jdate.z[all]")
> testgTMB
# Predicted probabilities of MYOSOD.P
# x = jdate.z
x predicted std.error conf.low conf.high
-1.95 0.444 0.826 0.137 0.801
-1.51 0.254 0.612 0.093 0.531
-1.03 0.136 0.464 0.059 0.280
-0.58 0.081 0.404 0.038 0.163
-0.14 0.054 0.395 0.026 0.110
0.35 0.040 0.402 0.019 0.084
0.79 0.035 0.406 0.016 0.074
1.72 0.040 0.444 0.017 0.091
Adjusted for:
* SiteID = NA (population-level)
* area = NA (population-level)
Standard errors are on link-scale (untransformed).
The estimates are completely different and I have no idea why.
I did try to use both the ggeffects package from CRAN and the developer version in case that changed anything. It did not. I am using the most up to date version of glmmTMB.
This is my first time asking a question here so please let me know if I should provide more information to help explain the problem.
I checked and the issue is the same when using predict instead of ggpredict, which would imply that it is a glmmTMB issue?
GLMER:
dayplotg<-expand.grid(jdate.z=seq(min(sodp$jdate.z), max(sodp$jdate.z), length=92))
Dfitg<-predict(yntest, re.form=NA, newdata=dayplotg, type='response')
dayplotg<-data.frame(dayplotg, Dfitg)
head(dayplotg)
> head(dayplotg)
jdate.z Dfitg
1 -1.953206 0.04581691
2 -1.912873 0.04889584
3 -1.872540 0.05195598
4 -1.832207 0.05497553
5 -1.791875 0.05793307
6 -1.751542 0.06080781
glmmTMB:
dayplot<-expand.grid(jdate.z=seq(min(sodp$jdate.z), max(sodp$jdate.z), length=92),
SiteID=NA,
area=NA)
Dfit<-predict(Tyntest, newdata=dayplot, type='response')
head(Dfit)
dayplot<-data.frame(dayplot, Dfit)
head(dayplot)
> head(dayplot)
jdate.z SiteID area Dfit
1 -1.953206 NA NA 0.4458236
2 -1.912873 NA NA 0.4251926
3 -1.872540 NA NA 0.4050944
4 -1.832207 NA NA 0.3855801
5 -1.791875 NA NA 0.3666922
6 -1.751542 NA NA 0.3484646
I contacted the ggpredict developer and figured out that if I used poly(jdate.z,3) rather than jdate.z + I(jdate.z^2) + I(jdate.z^3) in the glmmTMB model, the glmer and glmmTMB predictions were the same.
I'll leave this post up even though I was able to answer my own question in case someone else has this question later.
I have a dataset that I am using to build generalised linear models. The response variable is binary (absence/presence) and the explanatory variables are categorical.
CODE
library(tidyverse)
library(AICcmodavg)
# Data
set.seed(123)
t <- tibble(ID = 1:100,
A = as.factor(sample(c(0, 1), 100, T)),
B = as.factor(sample(c("black", "white"), 100, T)),
C = as.factor(sample(c("pos", "neg", "either"), 100, T)))
# Candidate set of models - Binomial family because response variable
# is binary (0 for absent & 1 for present)
# Global model is A ~ B_black + C_either
m1 <- glm(A ~ 1, binomial, t)
m2 <- glm(A ~ B, binomial, t)
m3 <- glm(A ~ C, binomial, t)
m4 <- glm(A ~ B + C, binomial, t)
# List with all models
ms <- list(null = m1, m_B = m2, m_C = m3, m_BC = m4)
# Summary table
aic_tbl <- aictab(ms)
PROBLEM
I want to build a table like the one below that summarises the coefficients, standard errors, and Akaike weights of the models within my candidate set.
QUESTION
Can anyone suggest how to best build this table using my list of models and AIC table?
Just to point it out: broom gets you half-way to where you want to get by turning the model output into a dataframe, which you can then reshape.
library(broom)
bind_rows(lapply(ms, tidy), .id="key")
key term estimate std.error statistic p.value
1 null (Intercept) -0.12014431182649532 0.200 -0.59963969517107030 0.549
2 m_B (Intercept) 0.00000000000000123 0.283 0.00000000000000433 1.000
3 m_B Bwhite -0.24116205496397874 0.401 -0.60071814968372905 0.548
4 m_C (Intercept) -0.47957308026188367 0.353 -1.35892869678271544 0.174
5 m_C Cneg 0.80499548069651150 0.507 1.58784953814722285 0.112
6 m_C Cpos 0.30772282333522433 0.490 0.62856402205887851 0.530
7 m_BC (Intercept) -0.36339654526926718 0.399 -0.90984856337213305 0.363
8 m_BC Bwhite -0.25083209866475475 0.408 -0.61515191157571303 0.538
9 m_BC Cneg 0.81144822536950656 0.508 1.59682131202527056 0.110
10 m_BC Cpos 0.32706970242195277 0.492 0.66527127770403538 0.506
And if you must insist of the layout of your table, I came up with the following (arguably clumsy) way of rearranging everything:
out <- bind_rows(lapply(ms, tidy), .id="mod")
t1 <- out %>% select(mod, term, estimate) %>% spread(term, estimate) %>% base::t
t2 <- out %>% select(mod, term, std.error) %>% spread(term, std.error) %>% base::t
rownames(t2) <- paste0(rownames(t2), "_std_e")
tmp <- rbind(t1, t2[-1,])
new_t <- as.data.frame(tmp[-1,])
colnames(new_t) <- tmp[1,]
new_t
Alternatively, you may want to familiarise yourself with packages that are meant to display model output for publication, e.g. texreg or stargazer come to mind:
library(texreg)
screenreg(ms)
==================================================
null m_B m_C m_BC
--------------------------------------------------
(Intercept) -0.12 0.00 -0.48 -0.36
(0.20) (0.28) (0.35) (0.40)
Bwhite -0.24 -0.25
(0.40) (0.41)
Cneg 0.80 0.81
(0.51) (0.51)
Cpos 0.31 0.33
(0.49) (0.49)
--------------------------------------------------
AIC 140.27 141.91 141.66 143.28
BIC 142.87 147.12 149.48 153.70
Log Likelihood -69.13 -68.95 -67.83 -67.64
Deviance 138.27 137.91 135.66 135.28
Num. obs. 100 100 100 100
==================================================
*** p < 0.001, ** p < 0.01, * p < 0.05