I want to run a nested logistic regression in R, but the examples I found online didn't help much. I read over an example from this website (Step by step procedure on how to run nested logistic regression in R) which is similar to my problem, but I found that it seems not resolved in the end (The questioner reported errors and I didn't see more answers).
So I have 9 predictors (continuous scores), and 1 categorical dependent variable (DV). The DV is called "effect", and it can be divided into 2 general categories: "negative (0)" and "positive (1)". I know how to run a simple binary logit regression (using the general grouping way, i.e., negative (0) and positive (1)), but this is not enough. "positive" can be further grouped into two types: "physical (1)" and "mental (2)". So I want to run a nested model which includes these 3 categories (negative (0), physical (1), and mental (2)), and reflects the nature that "physical" and "mental" are nested in "positive". Maybe R can compare these two models (general vs. detailed) together? So I created two new columns, one is called "effect general", in which the individual scores are "negative (0)" and "positive (1)"; the other is called "effect detailed", which contains 3 values - negative (0), physical (1), and mental (2). I ran a simple binary logit regression only using "effect general", but I don't know how to run a nested logit model for "effect detailed".
From the example I searched and other materials, the R package "mlogit" seems right, but I'm stuck with how to make it work for my data. I don't quite understand the examples in R-help, and this part in the example from this website I mentioned earlier (...shape='long', alt.var='town.list', nests=list(town.list)...) makes me very confused: I can see that my data shape should be 'wide', but I have no idea what "alt.var" and "nests" are...
I also looked at page 19 of the mlogit manual for examples of nested logit model calls. But I still cannot decide what I need in terms of options. (http://cran.r-project.org/web/packages/mlogit/mlogit.pdf)
Could someone provide me with detailed steps and notes on how to do it? I'm sure this example (if well discussed and resolved) is also going to help me and others a lot!
Thanks for your help!!!
I can help you with understanding the mlogit structure. When using the mlogit.data() command, specify choice = yourchoicevariable (and id.var = respondentid if you have a panel dataset, i.e. you have multiple responses from the same individual), along with the shape='wide' argument. The new data.frame created will be in long format, with a line for each choice situation, negative, physical, mental. So you will have 3 rows for which you only had one in the wide data format. Whatever your MN choice var is, it will now be a column of logical values, with TRUE for the row that the respondent chose. The row names will now have be in the format of observation#.level(choice variable) So in your case, if the first row of your dataset the person had a response of negative, you would see:
row.name | choice
1.negative | TRUE
1.physical | FALSE
1.mental | FALSE
Also not that the actual factor level for each choice is stored in an index called alt of the mlogit.data.frame which you can see by index(your.data.frame) and the observation number (i.e. the row number from your wide format data.frame) is stored in chid. Which is in essence what the row.name is telling you, i.e. chid.alt. Also note you DO NOT have to specify alt.var if your data is in wide format, only long format. The mlogit.data function does that for you as I have just described. Essentially, it takes unique(choice) when you specify your choice variable and creates the alt.var for you, so it is redundant if your data is in wide format.
You then specify the nests by adding to the mlogit() command a named list of the nests like this, assuming your factor levels are just '0','1','2':
mlogit(..., nests = c(negative = c('0'), positive = c('1','2')
or if the factor levels were 'negative','physical','mental' it would be the like this:
mlogit(..., nests = c(negative = c('negative'), positive = c('physical','mental')
Also note a nest of one still MUST be specified with a c() argument per the package documentation. The resulting model will then have the iv estimate between nests if you specify the un.nest.el=T argument, or nest specific estimates if un.nest.el=F
You may find Kenneth Train's Examples useful
Related
Hi y'all I'm fairly new to R and I'm supposed to calculate F statistic for this table
The code I have inputted is as follows:
# F-test
res.ftest <- var.test(TotalLength ~ SwimSpeed , data = my_data)
res.ftest
I know I have more than two levels from the other posts I have read online, but I am not sure what to change to get the outcome I want.
FIRST AND FOREMOST...If you invoke
?var.test()
you will note that the S3 version you called assumes lhs is numeric and rhs is a 2-level factor.
As for the rest, while I don't know the words to your specific work/school assignment here, the words shouldn't be "calculate an F-test", exactly. They should be "analyze these data appropriately". While there are a number of routes you could take, this is normally seen as a regression problem, NOT a problem of trying to compare two variances/complete a 1-way ANOVA which is what var.test() is designed to do. (Reading the documentation at, for example, https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/var.test should make this clear and is something you should always do when invoking R procedures.)
Using a subset of your data (please do this yourself for stack helpers next time rather than make someone here do it for you)...
df <- data.frame(
ID = 1:4,
TL = c(27.1,29.0,33.0,29.3),
SS = c(86.6,62.4,63.8,62.3)
)
cor.test(df$TL,df$SS) # reports t statistic
# or
summary(lm(df$TL ~ df$SS)) # reports F statistic
Note that F is simply t^2 here in the 2 variable case.
Lastly, I should add it is remotely, vaguely possible the assignment is to check if the variances of the 2 distributions are equal even though I can see no reason why anyone would want to know considering they are 2 different measures on two different underlying scales measuring 2 different things. However,
var.test(df$TL, df$SS)
will return a "result" should you take the assignment to mean compare the observed variances.
I need to perform a correlation analysis on a data frame constructed as follows:
Rows: Features --> gene variants related with different levels of severity of the disease we are studying, in a format of a Boolean matrix
Columns: Observations --> List of patients;
The discriminant of my analysis is, thus, the severity marked as follows:
A: less severe than expected
B: equal to what expected
C: more severe than expected
Suppose I have a lot more features than observations and I want to use the PTM function with a three-level annotation (i.e. A,B,C) as a match template. The function requires you to set the annotation.level.set.high parameter, but it's not clear to me how it works. For example, if I set annotation.level.set.high='A', does that mean I'm making a comparison between A vs B&C? So I can only do a comparison between two groups/classes even if I have multiple levels? Because my goal is to compare all levels with each other (i.e. A vs B vs C), but it is not clear to me how to achieve this comparison, if it is possible.
Thanks
I'm trying to do the compositional analysis of habitat use with the compana() function in the adehabitatHS package (I use adehabitat because I can't install adehabitatHS).
Compana() needs two matrices: one of habitat use and one of avaiable habitat.
When I try to run the function it doesn't work (it never stops), so I have to abort the RStudio session.
I read that one problem could be the 0-values in some habitat types for some animals in the 'avaiable' matrix, whereas other animals have positive values for the same habitat. As done by other people, I replaced 0-values with small values (0,001), ran compana and it worked BUT the lambda values returned me NaN.
The problem is similar to the one found here
adehabitatHS compana test returns lambda = NaN?
They said they resolved using as 'used' habitat matrix the counts (integers) and not the proportions.
I tried also this approach, but never changed (it freezes when there are 0-values in the available matrix, or returns NaN value for Lambda if I replace 0- values wit small values).
I checked all matrices and they are ok, so I'm getting crazy.
I have 6 animals and 21 habitat types.
Can you resolve this BIG problem?
PARTIALLY SOLVED: Asking to some researchers, they told me that the number of habitats shouldn't be higher than the number of animals.
In fact I merged some habitats in order to have six animals per six habitats and now the function works when I replace 0-values in the 'avaiable' matrix with small values (e.d. 0.001).
Unfortunately this is not what I wanted, because I needed to find values (rankings, Log-ratios, etc..) for each habitat type (originally they were 21).
I am working with a large federal dataset with thousands of observations and thousands of variables. Replicate weights are provided. I am using the "survey" package in R to apply these weights:
els.weighted=svrepdesign(data=els, repweights = ~els$F3F1PNLWT,
combined.weights = TRUE).
I am interested in some categorical descriptive characteristics of a subset of the population, such as family living arrangements. I want to get these sorted out into a contingency table that shows frequency. I would like to sort people based on four variables (none of which are binary, but all of which are numeric) This is what I would like to get:
.
The blank boxes are where the cross-tabulation/frequency counts would show. (I only put in 3 columns beneath F1COMP for brevity's sake, but it has 9 outcomes – indexed 1-9)
My current code: svyby(~F1FCOMP, ~F1RTRCC +BYS33C +F1A10 +byurban, els.weighted, svytotal)
This code does sort the data, but it sorts every single combination, by default. I want them pared down to represent only specific subpopulations of each variable. I tried:
svyby(~F1FCOMP, ~F1RTRCC==2 |F1RTRCC==3 +BYS33C==1 +F1A10==2 | F1A10==3 +byurban==3, els.weighted, svytotal)
But got stopped:
Error: unexpected '==' in "svyby(~F1FCOMP, ~F1RTRCC==2 |F1RTRCC==3 +BYS33C=="
Additionally, my current version of the code tells me how many cases occur for each combination, This is a picture of what my current output looks like. There are hundreds more rows, 1 for each combination, when I keep scrolling down.
This is a picture of what my current output looks like. There are hundreds more rows, 1 for each combination, when I keep scrolling down
.
You can see in that picture that I only get one number for F1FCOMP per row – the number of cases who fit the specified combination – a specific subpopulation. I want to know more about that subpopulation. That is, F1COMP has nine different outcomes (indexed 1-9), and I want to see how many of each subpopulation fits into each of the 9 outcomes of F1COMP.
My data is 1,785,000 records with 271 features. I'm trying to reduce number of features used to build the model.
Q1. while exploring the data I found that some features are almost all missing data, like only 25 records has value for this feature and the others records has missing values, so I thought that is not informative enough and it's better to eleminate those features, am I right? and if I am right, for what level I can do that, I mean if 90%, 80%, etc.. of each feature are missing values, when I can decide to get rid of these features? (taking in consideration that it is the dependent variable is N/Y and only %1.157 of the whole data is belonging to Y).
Q2. for each individual in the dataset, there are 64 trait_type listed, where each one can take one of the values [1 or 3 or 5]. my question is: if some trait-type take only value [5] or missing dat for all the record, does it have any value or again we can eliminate that feature?
Q3. if the choice is to delete these features, how to delete column from data.frame in R?
Thank you
Update:
I'm trying to use caret package to do the variable selection.
I applied this:
ctrl<- rfeControl(functions = lmFuncs, method="cv", verbose = FALSE, returnResamp=
"final")
lmprofile<- rfe(x,y, sizes = subsets, rfeControl = ctrl)
where x is the data.frame that have 270 dependant variables and y is the factor of the independent variable which has value Y/N. I got this error:
Error in { :
task 1 failed - "contrasts can be applied only to factors with 2 or more levels"
enter code here
In addition: There were 11 warnings (use warnings() to see them)
any help please?
Just because much of your data in one column is missing doesn't mean that column will not be predictive, it's just the same as having many of the same value in that column.
Of course there is a cutoff, if that column can only help you distinguish between a few cases (of many) then it can be removed and could only affect overall model strength a little.
To help you decide whether to keep the column, you could build a univariate model with it - where the dataset just includes that column and the dependant variable, and look at the strength of that model. If it's not much better than random, then it's probably safe to drop the column.