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In Example 8.21 we described how to fit a latent class model to data from the HELP dataset using SAS and R. Subjects were classified based on their observed (manifest) status on the following variables (on street or in shelter in past 180 days [homeless], CESD scores above 20, received substance abuse treatment [satreat], or linked to primary care [linkstatus]). We arbitrarily specify a three class solution.

In this example, we fit the same model using the `randomLCA()` function within the package of the same name.

**R**

We begin by reading in the data.

ds = read.csv("http://www.math.smith.edu/r/data/help.csv")

attach(ds)

library(randomLCA)

We start by creating a dichotomous variable with high scores on the CESD, and put this together as part of a dataframe to be given as input.

cesdcut = ifelse(cesd>20, 1, 0)

smallds = na.omit(data.frame(homeless, cesdcut, satreat, linkstatus))

results = randomLCA(smallds, nclass=3, notrials=1000)

summary(results)

This generates the following output:

Classes AIC BIC logLik

3 2092.968 2149.893 -1032.484

Class probabilities

Class 1 Class 2 Class 3

0.07846 0.70534 0.21620

Outcome probabilities

homeless cesdcut satreat linkstatus

Class 1 9.465e-06 0.5786 1.000e+00 9.538e-06

Class 2 4.375e-01 0.8322 9.988e-06 4.145e-01

Class 3 7.297e-01 0.8846 1.000e+00 3.971e-01

The results are equivalent to the results from the prior example, though the scientific notation for the observed prevalences in each class are hard to read. Other objects are available from the returned value, though they are also not in an easily digestible form:

> names(results)

[1] "fit" "nclass" "classp" "outcomep"

[5] "se" "np" "nobs" "logLik"

[9] "observed" "fitted" "deviance" "classprob"

[13] "bics" "random" "level2" "byclass"

[17] "blocksize" "call" "probit" "quadpoints"

[21] "patterns" "notrials" "freq"

> results$patterns

homeless cesdcut satreat linkstatus

1 0 0 0 0

2 0 0 0 1

3 0 0 1 0

4 0 0 1 1

5 0 1 0 0

6 0 1 0 1

7 0 1 1 0

8 0 1 1 1

9 1 0 0 0

10 1 0 0 1

11 1 0 1 0

12 1 0 1 1

13 1 1 0 0

14 1 1 0 1

15 1 1 1 0

16 1 1 1 1

> results$freq

[1] 17 10 16 1 82 62 33 9 15 9 4 4 64 45 37 23

We’ll address workarounds for these shortcomings in a future entry.

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