README.Rmd

---
output: github_document
---

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```{r setup, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>",
  fig.path = "man/figures/README-",
  out.width = "100%"
)
```

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# gnet

The goal of gnet is to ...

## Installation

~~You can install the released version of gnet from [CRAN](https://CRAN.R-project.org) with:~~

``` r
devtools::install_github("muriteams/gnet")
```

## Support

This material is based upon work support by, or in part by, the U.S.
Army Research Laboratory and the U.S. Army Research Office under
grant number W911NF-15-1-0577

Computation for the work described in this paper was supported by
the University of Southern California's Center for High-Performance
Computing (hpcc.usc.edu).

## Example

This is a basic example which shows you how to solve a common problem:

```{r example1, cache=FALSE}
library(ergmito)
library(gnet)


# A random sample of networks
set.seed(12)
nets <- rbernoulli(c(3,3,4,4,4))

# Are the edgecounts random??
fun <- function(g, y) mean(count_stats(g, "edges") - y, na.rm = TRUE)

mytest01 <- struct_test(nets ~ edges + balance, y = runif(5), R=1000, stat=fun)
```

```{r printing}
mytest01
```


```{r hist}
# Looking at the distribution
hist(mytest01$t, breaks=10, col="gray", border="transparent")
abline(v = mytest01$t0, col="steelblue", lwd=2, lty="dashed")
```

Using the fivenets dataset included in `ergmito`. In this example, we show how the model can be use to discover an association between 

```{r fivenets}

# Loading the fivents dataset. We actually know that data generating process,
# so we use these paramaters for the model
data(fivenets, package="ergmito")

# We will generate a group level variable that is related to the proportion of
# females in the group
set.seed(52)
y <- count_stats(fivenets ~ nodeocov("female"))
y <- y + rnorm(nnets(fivenets))

# Performing the struct test
f02 <- function(g, y) cor(count_stats(g ~ nodeocov("female")), y, use = "complete.obs")[1] 
mytest02 <- struct_test(
  fivenets ~ edges + nodematch("female"), y = y, R=3000,
  stat = f02
  )
mytest02
```

What would have we got if we use a linear regression only?

```{r ols}
# Li
x <- count_stats(fivenets ~ nodeocov("female"))
summary(lm(y ~ x))
```

```{r plot-comparison}
op <- par(mfrow=c(1, 2))
hist(mytest02$t, breaks=50, col="gray", border="transparent", main = "Null distribution of t",
     xlab = "Values of t")
abline(v = mytest02$t0, col="steelblue", lwd=2, lty="dashed")
plot(y ~ x, main = "Linear regression", xlab="nodeocov(\"female\")", ylab = "y")
abline(lm(y~x), lty="dashed", lwd=2)
par(op)
```


In this case, the test has higher power than the other 2.