Intro
I'm trying to construct a GLM that models the quantity (mass) of eggs the specimens of a fish population lays depending on its size and age.
Thus, the variables are:
eggW: the total mass of layed eggs, a continuous and positive variable ranging between 300 and 30000.
fishW: mass of the fish, continuous and positive, ranging between 3 and 55.
age: either 1 or 2 years.
No 0's, no NA's.
After checking and realising assuming a normal distribution was probably not appropriate, I decided to use a Gamma distribution. I chose Gamma basically because the variable was positive and continuous, with increasing variance with higher values and appeared to be skewed, as you can see in the image below.
Frequency distribution of eggW values:
fishW vs eggW:
The code
myglm <- glm(eggW ~ fishW * age, family=Gamma(link=identity),
start=c(mean(data$eggW),1,1,1),
maxit=100)
I added the maxit factor after seeing it suggested on a post of this page as a solution to glm.fit: algorithm did not converge error, and it worked.
I chose to work with link=identity because of the more obvious and straightforward interpretation of the results in biological terms rather than using an inverse or log link.
So, the code above results in the next message:
Warning messages: 1: In log(ifelse(y == 0, 1, y/mu)) : NaNs
produced 2: step size truncated due to divergence
Importantly, no error warnings are shown if the variable fishW is dropped and only age is kept. No errors are reported if a log link is used.
Questions
If the rationale behind the design of my model is acceptable, I would like to understand why these errors are reported and how to solve or avoid them. In any case, I would appreciate any criticism or suggestions.
You are looking to determine the weight of the eggs based upon age and weight of the fish correct? I think you need to use:
glm(eggW ~ fishW + age, family=Gamma(link=identity)
Instead of
glm(eggW ~ fishW * age, family=Gamma(link=identity)
Does your dataset have missing values?
Are your variables highly correlated?
Turn fishW * age into a seperate column and just pass that to the algo
Related
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
Background
I am trying to test for differences in wind speed data among different groups. For the purpose of my question, I am looking only on side wind (wind direction that is 90 deg from the individual), and I only care about the strength of the wind. Thus, I use absolute values. The range of wind speeds is 0.0004-6.8 m/sec and because I use absolute values, Gamma distribution describes it much better than normal distribution.
My data contains 734 samples from 68 individuals, with each individual having between 1-30 repeats. However, even if I reduce my samples to only include individuals with at least 10 repeats (which leaves me with 26 individuals and a total of 466 samples), I still get the problematic error.
The model
The full model is Wind ~ a*b + c*d + (1|individual), but for the purpose of this question, the simple model of Wind ~ 1 + (1|individual) gives the same singularity error, so I do not think that the explanatory variables are the problem.
The complete code line is glmer(Wind ~ 1 + (1|individual), data = X, family = Gamma(log))
The problem and the strange part
When running the model, I get the boundary (singular) fit: see ?isSingular error, although, as you can see, I use a very simple model and random structure. The strange part is that I can solve this by adding 0.1 to the Wind variable (i.e. glmer(Wind+0.1 ~ 1 + (1|Tag), data = X, family = Gamma(log)) does not give any error). I honestly do not remember why I added 0.1 the first time I did it, but I was surprised to see that it solved the error.
The question
Is this a problem with lme4? Am I missing something? Any ideas what might cause this and why does me adding 0.1 to the variable solve this problem?
Edit following questions
I am not sure what's the best way to add data, so here is a link to a csv file in Google drive
using glmmTMB does not produce any warnings with the basic formula glmmTMB(Wind ~ 1 + (1|Tag), data = X, family = Gamma(log)), but gives convergence problems warnings ('non-positive-definite Hessian matrix') when using the full model (i.e., Wind ~ a*b + c*d + (1|individual)), which are then solved if I scale the continuous variables
I'm trying to follow this paper: Using a data science approach to predict cocaine use frequency from depressive symptoms where they use glm, gam with the beck inventory depression. So I did found a similiar dataset to test those models. However I'm having a hard time with both models. For example I have two variables d64a and d64b, and they're coded with 1,2,3,4 meaning that they're ordinal. Also, in the paper y2 is only the value of 1 but i have also a variable extra (that can be dependent, the proportion of consume)
For the GAM model I have:
b<-gam(y2~s(d64a)+s(d64b),data=DATOS2)
but I have the following error:
Error in smooth.construct.tp.smooth.spec(object, dk$data, dk$knots) :
A term has fewer unique covariate combinations than specified maximum degrees of freedom
Meanwhile for the glm, I have the following:
d<-glm(y2~d64a+d64b,data=DATOS2)
I don't know since d64a and d64b are ordinal I have to use factor()?
The error message tells you that one or both of d64a and d64b do not have 9 (nine) unique values.
By default s(...) will create a basis with nine functions. You get this error if there are fewer than nine unique values in the covariate.
Check which covariates are affected using:
length(unique(d64a))
length(unique(d64b))
and see what the number of unique values is for each of the covariates you wish to include. Then set the k argument to the number returned above if it is less than nine. FOr example, assume that the above checks returned 5 and 7 unique covariates, then you would indicate this by setting k as follows:
b <- gam(y2 ~ s(d64a, k = 5) + s(d64b, k = 7), data = DATOS2)
I'm computing omega for several different scales; and get different warning messages for different scales with different omega functions in R. My questions are regarding how to interpret these warnings and if it is safe to report the retrieved omega statistics.
When I'm using the following function from the article "From alpha to omega: A practical solution to the pervasive problem of internal consistency estimation"
ci.reliability(subscale1, interval.type="bca", B=1000)
I get these warnings:
1: In lav_object_post_check(lavobject) :
lavaan WARNING: some estimated variances are negative
2: In lav_object_post_check(lavobject) :
lavaan WARNING: observed variable error term matrix (theta) is not positive definite; use inspect(fit,"theta") to investigate.
And it can be many of them!
What do they mean?
I still receive omega statistics; can they be interpreted or not?
When I use the function:
psych::omega(subscale1)
I get this warning:
Warning message:
In GPFoblq(L, Tmat = Tmat, normalize = normalize, eps = eps, maxit = maxit, :
convergence not obtained in GPFoblq. 1000 iterations used.
Again,
What does it mean; and can I use the omega-statistics that I get?
Note that these warnings appear on different subscales; so one subscale can be computed using one of the function but not the other and vice versa.
EDIT: If it helps: Subscale1 encompasses 4 items; the sample includes N>300. Also, I can run a CFA analysis on these 4 items in lavaan (Chi2=11.8, p<.001; CFI=0.98; RMSEA=0.123).
That particular article to which you are referring seems to be the British Journal of Psychology (2014), 105, 399–412© 2013 by Dunn, Baguley and Brunsden. The omega coefficient they discuss is actually what Rick Zinbarg and I refer to as omega_total. (McDonald developed two omega coefficients which has led to this confusion.)
You are having problems using omega in my psych package. The omega function in psych is meant to find omega_hiearchical as well as omega_total. Thus, it tries (by default) to extract three lower level factors and then, in turn, factor the resulting correlations of those factors. However, with only 4 variables in your sub scale, it can not find a meaningful 3 factor solution. You can specify that you want to find two factors:
omega(subscale1,2)
and it will work. However, omega_h is not particularly meaningful for 4 items.
Contrary to the suggestion of sample size, it is actually due to the number of items.
I think you might find the tutorial for finding omega_h using psych helpful:
[http://personality-project.org/r/psych/HowTo/R_for_omega.pdf]
I am doing an lm()regression with R where I use stock quotations. I used exponential weights for the regression : the older the data, the less weight. My weights formula is like this : alpha^(seq(685,1,by=-1))) (the data length is 685), and to find alpha I tried every value between 0.9 and 1.1 with a step of 0.0001 and I chose the alpha which minimizes the difference between the predicted values and the real values. This alpha is equal to 0.9992 so I would like to know if it is statistically different from 1.
In other words I would like to know if the weights are different from 1. Is it possible to achieve that and if so, how could I do this ?
I don't really know whether this question should be asked on stats.stackexchange but it involves Rso I hope it is not misplaced.