The goal of this vignette is to take the data gathered from the expert elicitation, turn it into individual priors for each expert, and then sample from it with the algorithm written by Professor Jim Clark, at Duke University. The methods for splitting the individual beta distributions into a Dirichlet are detailed in the vignette: `Split-Beta.Rmd.`

The methods for sampling from the mixture distribution are given in: `test.Rmd.`

To set up the sampler, we need to assemble the priors and the data. First, the data. We take these from a previously completed run of the Gibbs sampler. What I did was simply load in the data, run `updateStates()`

once, and save it out to an rdata file:

femEx <- moveFem[, , k] maleEx <- moveMale[, , k] save(femEx, maleEx, file = 'data/moveData.rda')

In this section we will explain the sampling algorithm we are using within the Gibbs sampler, highlight how we can extend it to work with the mixture of Dirichlet's, and then test it with just the data.

Jim wrote the algorithm as part of the Schick et al. 2013a PLoS-ONE paper. We used Gibbs sampling to estimate the parameters as opposed to drawing directly from the posterior because we impute the missing data for location. Therefore each time through the Gibbs loop, missing states can change. So the algorithm (based on Gelman et al. 2014) looks like this:

- Draw from $x \sim Gamma(a, 1)$
- Set $\theta = \frac{x}{\sum_{i=1}^n x}$

This is the implementation of that in `R`

:

di <- matrix(rgamma(nreg * nreg, shape = malePrior[, , k] + maleDat[, , k], scale = 1), nreg, nreg) malePost[, , k] <- di / matrix(colSums(di), nreg, nreg, byrow = T)

where:

`nreg`

equals the number of regions (9)`malePrior[, , k]`

is the prior for the $k^{th}$ month`maleDat[, , k]`

is the data for the $k^{th}$ month`malePost[, , k]`

is the Dirichlet posterior

These components look like this. n.b. that in our original code, we chose different names for these data frames. Note also that the data is reinitialised and imputed each time through the loop, so we don't have a glimpse of it.

load('../data/eg_202_ng_25000_BIG_2500.rdata') k <- 12 round(maleMove[, , k], 2) # prior round(maleMoveProb[, , k], 2) # posterior

Ok, with the Dirichlet algorithm outlined, we need to use the sampling details from `test.Rmd`

to simulate the posterior. Specifically, we need to set up a structure where by each time we go through the Gibbs loop, we sample from the mixture distribution. Before we had just one prior (as shown above); now the prior is a mixture of all experts' answers to the elicitation exercise.

The process flow will be as in `test.Rmd`

. This is all within the gibbs loop (because the data update each time):

- Split the betas and assemble one Dirichlet matrix containing all priors (Prior to the Gibbs loop)
- Calculate
`c`

for all experts using the estimate data and the individual prior - Use that to assemble the $K$ probability vector
- At each iteration draw from a Multinomial distribution with probability $K$.
- Use the $K^{th}$ expert's Dirichlet distribution
- Merge it into the existing prior
- Sample from the Posterior

First thing that would be useful is to summarise the text of **each question** so we have it for reference. The reason I think this is necessary, is that we're going to have split this differently based on the direction of movement we ask about. I'll explain each question and data object that they refer to:

**Question 9** - Consider a population of 100 adult **female** right whales that are present in the northern regions during December and move *south* to the mid-Atlantic during January.

In this scenario, there are two possible transitions. The animals are in one of 7 possible Northern regions, and can either move south to MIDA or remain in their Northern region. The answers are in `dfList[[1]]`

(n.b. all the answers are different elements in this list):

library(eliciteg) dfList[[1]]

The moves to the south are represented with the `moveNortherntoMIDA`

column. The animals remaining in the Northern regions are in the `moveNortherntoNorthern`

. In this case *all* of these transitions need to be split, because the first one represents 7 transitions from North to MIDA, and the second part of the question refers to animals remaining in 7 different Northern regions. So we'll have a matrix with 7 transitions from North to MIDA populated, as well as 7 `North[i]`

to `North[i]`

transitions.

**Question 10** - Consider a population of 100 adult **female** right whales that are present in the mid-Atlantic during December and *remain* in the mid-Atlantic for any duration during January.

In this scenario, the animals are in the MIDA, and can make three transitions: 1) they can remain in MIDA; 2) move south to MIDA; or 3) move North. The answers are in `dfList[[2]]`

:

```
dfList[[2]]
```

This one is a bit more straightforward. We have the MIDA to MIDA transition - `moveMIDAtoMIDA`

- that does not need to be split. Then we have the MIDA to SEUS - `moveMIDAtoSEUS`

- which also does not need to be split. Finally, we have the MIDA to Northern regions - `moveMIDAtoNorthern`

- which **does** need to be split.

**Question 11** - Consider a population of 100 adult **female** right whales that are present in the southeastern United States during December and move *north* to the mid-Atlantic during January.

In this scenario, the animals are in the SEUS and can make one of two transitions. The answers are in `dfList[[3]]`

.

```
dfList[[3]]
```

So far this is the most straightforward, as neither of the questions need to be split. We have animals starting in the SEUS, and can either remain - `moveSEUStoSEUS`

- or move north to MIDA - `moveSEUStoMIDA`

Now we move on to the males. **Question 12** - Consider a population of 100 adult **male** right whales that are present in the northern regions during December and move *south* to the mid-Atlantic during January.

As before with females, in this scenario, there are two possible transitions. The animals are in one of 7 possible Northern regions, and can either move south to MIDA or remain in their Northern region. The answers are in `dfList[[4]]`

:

```
dfList[[4]]
```

The moves to the south are represented with the `moveNortherntoMIDA`

column. The animals remaining in the Northern regions are in the `moveNortherntoNorthern`

. In this case *all* of these transitions need to be split, because the first one represents 7 transitions from North to MIDA, and the second part of the question refers to animals remaining in 7 different Northern regions. So we'll have a matrix with 7 transitions from North to MIDA populated, as well as 7 `North[i]`

to `North[i]`

transitions.

**Question 13** - Consider a population of 100 adult **male** right whales that are present in the mid-Atlantic during December and *remain* in the mid-Atlantic for any duration during January.

In this scenario, the animals are in the MIDA, and can make three transitions: 1) they can remain in MIDA; 2) move south to MIDA; or 3) move North. The answers are in `dfList[[5]]`

:

```
dfList[[5]]
```

This one is a bit more straightforward. We have the MIDA to MIDA transition - `moveMIDAtoMIDA`

- that does not need to be split. Then we have the MIDA to SEUS - `moveMIDAtoSEUS`

- which also does not need to be split. Finally, we have the MIDA to Northern regions - `moveMIDAtoNorthern`

- which does need to be split.

**Question 14** - Consider a population of 100 adult **male** right whales that are present in the southeastern United States during December and move *north* to the mid-Atlantic during January.

In this scenario, the animals are in the SEUS and can make one of two transitions. The answers are in `dfList[[6]]`

.

```
dfList[[6]]
```

So far this is the most straightforward, as neither of the questions need to be split. We have animals starting in the SEUS, and can either remain - `moveSEUStoSEUS`

- or move north to MIDA - `moveSEUStoMIDA`

**Summary** ok then, what all do we have for each of the 6 questions? For **questions 9 and 12,** all the answers need to be split. For example, we have for Question 9, expert 6: `Beta(35, 4)`

and that will need to be split into `Beta(5, 0.5714)`

for each of the 7 *southern* transitions. Similarly, we have `Beta(65, 4)`

, which will split to `Beta(9.285714, 0.5714)`

for each of the 7 *remaining* transitions.

For **questions 10 and 13,** we need to split just one of the transitions - the `moveMIDAtoNorthern`

answer. For question 10, expert 6, we'd go from `Beta(20, 4)`

to `Beta(2.8571429, 0.5714286)`

for transition from MIDA to each of the 7 Northern regions.

For **questions 11 and 14,** we do not need to split any transitions. They can be assembled as is.

Now that we've gone through that, let's start in on assembling the data into a form that we can use within the sampler. That means taking these individual Beta distributions, and assembling one Dirichlet.

The data are stored in the `dfList.rda`

data object. What we want, then, is a data structure that will allow us to easily extract from this and apply it to the matrix prior. Let's assume that we have once slice of a (9 * 9 * 12) array. If we ignore all of Philip Hamilton's priors, that will leave us with something like this:

regNorth <- regID[!regID %in% c('MIDA', 'SEUS')] # this vector contains the Northern regions nNorthreg <- length(regNorth) # This will be used to divide down the Betas movePrior <- matrix(1, nrow = length(regID), ncol = length(regID)) rownames(movePrior) <- colnames(movePrior) <- regID movePrior2 <- movePrior movePrior

**However** this raises a question - namely what do we want the flat prior values to be? I've started at 1, but perhaps it should be lower? One reason I ask this, is that as you'll see in the Northern to Northern transitions, you can end up with prior values << 1.

Back to our reality. Each time through the loop, we'll sample an expert, and populate `movePrior`

. Let's show this for one expert for each of these two questions. Note that in the loop we'll sample the `expert`

with the $c$ and $K$ machinery previously covered; here I'll just choose one for illustration purposes.

Recall that we used the following parameterisation for each Beta distribution: $\alpha = n \times m_1$ and $\beta = n - n \times m_1$, where $n$ = `inputConfRaw`

, and $m_1$ = `moveNortherntoMIDA / 100`

. Obviously, the column corresponding to $m_1$ changes for each question.

Here are two functions to return these values per each of the `nreg`

transitions:

betaA <- function(n, m, nreg) { expBeta1a <- (n * m) / nreg expBeta1a } betaB <- function(n, m, nreg) { expBeta1b <- (n - (n * m)) / nreg expBeta1b }

First we'll update elements of `movePrior`

with data for the transitions south (North to MIDA). As noted above, the idea is that animals could be in any of the 7 Northern regions and could make a transition to MIDA:

expert <- 4 dfList[[1]][which(dfList[[1]]$expert == expert),] n <- dfList[[1]]$inputConfRaw[which(dfList[[1]]$expert == expert)] m <- dfList[[1]]$moveNortherntoMIDA[which(dfList[[1]]$expert == expert)] / 100 expBeta1a <- betaA(n, m, nreg = nNorthreg) expBeta1b <- betaB(n, m, nreg = nNorthreg) movePrior[rownames(movePrior) == 'MIDA', colnames(movePrior) %in% regNorth] <- expBeta1a movePrior

And then adding in the North to North transitions:

n <- dfList[[1]]$inputConfRaw[which(dfList[[1]]$expert == expert)] m <- dfList[[1]]$moveNortherntoNorthern[which(dfList[[1]]$expert == expert)] / 100 expBeta1a <- betaA(n, m, nreg = nNorthreg) expBeta1b <- betaB(n, m, nreg = nNorthreg) diag(movePrior)[which(rownames(movePrior) %in% regNorth)] <- expBeta1a movePrior

OK, that was `dfList[[1]]`

, i.e. Question 9, what does the process and result look like for `dfList[[2]]`

, i.e. Question 10? Recall from above that we need to split the MIDA to Northern transitions `moveMIDAtoNorthern`

, because it could go from MIDA to each of 7 areas in the North.

First for the remaining in MIDA transition (i.e., MIDA to MIDA):

expert2 <- 1 dfList[[2]][which(dfList[[2]]$expert == expert2),] n <- dfList[[2]]$inputConfRaw[which(dfList[[2]]$expert == expert2)] m <- dfList[[2]]$moveMIDAtoMIDA[which(dfList[[2]]$expert == expert2)] / 100 expBeta1a <- betaA(n, m, nreg = 1) expBeta1b <- betaB(n, m, nreg = 1) movePrior2[rownames(movePrior2) == 'MIDA', rownames(movePrior2) == 'MIDA'] <- expBeta1a movePrior2

And then adding in the MIDA to SEUS transition:

n <- dfList[[2]]$inputConfRaw[which(dfList[[2]]$expert == expert2)] m <- dfList[[2]]$moveMIDAtoSEUS[which(dfList[[2]]$expert == expert2)] / 100 expBeta1a <-betaA(n, m, nreg = 1) expBeta1b <- betaB(n, m, nreg = 1) movePrior2[rownames(movePrior2) == 'SEUS', colnames(movePrior2) == 'MIDA'] <- expBeta1a movePrior2

And finally, the ones that need to be split:

n <- dfList[[2]]$inputConfRaw[which(dfList[[2]]$expert == expert2)] m <- dfList[[2]]$moveMIDAtoNorthern[which(dfList[[2]]$expert == expert2)] / 100 expBeta1a <- betaA(n, m, nreg = nNorthreg) expBeta1b <- betaB(n, m, nreg = nNorthreg) movePrior2[rownames(movePrior2) %in% regNorth, colnames(movePrior2) == 'MIDA'] <- expBeta1a movePrior2

Note above that the splitting we assign the split value to all of the regions in `regNorth`

Let's examine them together to see if the differences make sense given the structure of the questions.

dfList[[1]][dfList[[1]]$expert == expert, ] movePrior dfList[[2]][dfList[[2]]$expert == expert2, ] movePrior2

So this seems to make sense to me. For expert 4 in the first answer, they had a majority of whales staying in the northern regions (30 going south, 70 remaining). So all of the diagonals are 0.7, and indeed higher than the transitions to MIDA. The matrix is set up as from columns to rows. So the MIDA row has been populated with 0.3 for the *from* Northern *to* MIDA transitions. Their confidence in their answers was low (7).

In contrast for expert 1 in the next question, they had much higher confidence - so we expect the values in the prior to be higher, and indeed they are. Also, the individual has equal probability of MIDA to MIDA as they do with MIDA to SEUS; those values are identical at 4.29. Since the number of whales must sum to 100, the value moving North is smaller (20), and that's reflected in the matrix.

Ok, so it at least passes the intuition test. Now we need to assemble all of these into one matrix/array/list. This will be static because these priors never change. However, they should be in the same form as the data, which is an array, because I'm going to have to add them together. I'm not 100% sure it matters all that much, because I'm just going to be sending them both to `calcC()`

.

If we use an array, which is fine, than what are the dimensions that we need? We have:

- one month's transition (December to January),
- 9 regions
- 8 experts
- 3 questions
- 2 sexes (M/F)

We need to add a 9x9 slice to a 9x9 slice for each question, where above each array slice is a month. So here we'll just have one slice for the December to January transition. So for each question (1-6) we'll have 8 slices, where each slice represents one experts' prior that has been split based on their answers. I could then just have a recursive list, where each question has a sub-list containing each expert's answers, or each questions' answers. If the latter, we'd have `list[[1]]`

correspond to question 9, and then within that list component, we'd have 8 sub-lists, each containing a 9*9 slice that corresponded to each expert.

What would that look like? Let's try it with fake-ish data:

nexp <- 8 nreg <- 9 nq <- 6 allqs <- list(q1 = vector('list', nexp), q2 = vector('list', nexp), q3 = vector('list', nexp), q4 = vector('list', nexp), q5 = vector('list', nexp), q6 = vector('list', nexp)) for(i in seq_along(1:nq)){ for(j in seq_along(1:nexp)){ allqs[[i]][[j]] <- matrix(data = round(runif(nreg * nreg), 0) + j, ncol = nreg, nrow = nreg) } } allqs[[2]][[2]]

Not the most elegant way of doing it, I don't think, but it works and is fast enough for this application. Ok, let's populate it with real data now:

# This chunk of code will iterate over the question/expert combinations and populate a list element with the new prior nexp <- 8 nreg <- 9 nq <- 6 flatpriorVal <- 0.001 allqs <- list(q1 = vector('list', nexp), q2 = vector('list', nexp), q3 = vector('list', nexp), q4 = vector('list', nexp), q5 = vector('list', nexp), q6 = vector('list', nexp)) for(i in seq_along(1:nq)){ for(j in seq_along(1:nexp)){ allqs[[i]][[j]] <- matrix(data = flatpriorVal, ncol = nreg, nrow = nreg) rownames(allqs[[i]][[j]]) <- colnames(allqs[[i]][[j]]) <- regID } }

Ok, the list now contains the flat prior, and we can populate it. The goal of this will be to iterate over each question, peel off the answers, and the split them up and put them into the correct list element. To be anal for now, I'm going to split them into two code blocks for females (q9-11) and males (q12-14).

I'm going to redo this now as a two element list, instead of a 6 element list. Let's start with the data as before. The logic will be that the priors all start at 0, and then we will fill these prior values up as we did before, one transition type at a time. Then we'll add the three slices together. That's the idea anyway.

# Question 9 - requires splitting: 1) northern to mida; 2) northern to northern listElement <- 1 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] dfList[[listElement]][which(dfList[[listElement]]$expert == expert[j]),] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveNortherntoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 # part 1 - Northern to MIDA m2 <- dfList[[listElement]]$moveNortherntoNorthern[which(dfList[[listElement]]$expert == expert[j])] / 100 # Part 2- Northern to Northern expBeta1a <- betaA(n, m, nreg = nNorthreg) expBeta1a2 <- betaA(n, m2, nreg = nNorthreg * nNorthreg) movePrior[rownames(movePrior) == 'MIDA', colnames(movePrior) %in% regNorth] <- expBeta1a movePrior[rownames(movePrior) %in% regNorth, colnames(movePrior) %in% regNorth] <- expBeta1a2 allqs[[listElement]][[j]] <- movePrior } # Question 10 - requires splitting MIDA to Northern transitions `moveMIDAtoNorthern` listElement <- 2 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveMIDAtoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a <- betaA(n, m, nreg = 1) movePrior[rownames(movePrior) == 'MIDA', rownames(movePrior) == 'MIDA'] <- expBeta1a m2 <- dfList[[listElement]]$moveMIDAtoSEUS[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a2 <-betaA(n, m2, nreg = 1) movePrior[rownames(movePrior) == 'SEUS', colnames(movePrior) == 'MIDA'] <- expBeta1a2 m3 <- dfList[[listElement]]$moveMIDAtoNorthern[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a3 <- betaA(n, m3, nreg = nNorthreg) movePrior[rownames(movePrior) %in% regNorth, colnames(movePrior) == 'MIDA'] <- expBeta1a3 allqs[[listElement]][[j]] <- movePrior } # Question 11 - doesn't require splitting listElement <- 3 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] dfList[[listElement]][which(dfList[[listElement]]$expert == expert[j]),] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveSEUStoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 # part 1 - SEUS to MIDA m2 <- dfList[[listElement]]$moveSEUStoSEUS[which(dfList[[listElement]]$expert == expert[j])] / 100 # Part 2- SEUS to SEUS expBeta1a <- betaA(n, m, nreg = 1) expBeta1a2 <- betaA(n, m2, nreg = 1) movePrior[rownames(movePrior) == 'MIDA', colnames(movePrior) == 'SEUS'] <- expBeta1a movePrior[rownames(movePrior) == 'SEUS', colnames(movePrior) == 'SEUS'] <- expBeta1a2 allqs[[listElement]][[j]] <- movePrior }

# Question 12 - requires splitting: 1) northern to mida; 2) northern to northern listElement <- 4 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] dfList[[listElement]][which(dfList[[listElement]]$expert == expert[j]),] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveNortherntoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 # part 1 - Northern to MIDA m2 <- dfList[[listElement]]$moveNortherntoNorthern[which(dfList[[listElement]]$expert == expert[j])] / 100 # Part 2- Northern to Northern expBeta1a <- betaA(n, m, nreg = nNorthreg) expBeta1a2 <- betaA(n, m2, nreg = nNorthreg * nNorthreg) movePrior[rownames(movePrior) == 'MIDA', colnames(movePrior) %in% regNorth] <- expBeta1a movePrior[rownames(movePrior) %in% regNorth, colnames(movePrior) %in% regNorth] <- expBeta1a2 allqs[[listElement]][[j]] <- movePrior } # Question 13 - requires splitting MIDA to Northern transitions `moveMIDAtoNorthern` listElement <- 5 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveMIDAtoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a <- betaA(n, m, nreg = 1) movePrior[rownames(movePrior) == 'MIDA', rownames(movePrior) == 'MIDA'] <- expBeta1a m2 <- dfList[[listElement]]$moveMIDAtoSEUS[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a2 <-betaA(n, m2, nreg = 1) movePrior[rownames(movePrior) == 'SEUS', colnames(movePrior) == 'MIDA'] <- expBeta1a2 m3 <- dfList[[listElement]]$moveMIDAtoNorthern[which(dfList[[listElement]]$expert == expert[j])] / 100 expBeta1a3 <- betaA(n, m3, nreg = nNorthreg) movePrior[rownames(movePrior) %in% regNorth, colnames(movePrior) == 'MIDA'] <- expBeta1a3 allqs[[listElement]][[j]] <- movePrior } # Question 14 - doesn't require splitting listElement <- 6 expert <- dfList[[listElement]]$expert nexp <- length(expert) for(j in seq_along(1:nexp)){ movePrior <- allqs[[listElement]][[j]] dfList[[listElement]][which(dfList[[listElement]]$expert == expert[j]),] n <- dfList[[listElement]]$inputConfRaw[which(dfList[[listElement]]$expert == expert[j])] m <- dfList[[listElement]]$moveSEUStoMIDA[which(dfList[[listElement]]$expert == expert[j])] / 100 # part 1 - SEUS to MIDA m2 <- dfList[[listElement]]$moveSEUStoSEUS[which(dfList[[listElement]]$expert == expert[j])] / 100 # Part 2- SEUS to SEUS expBeta1a <- betaA(n, m, nreg = 1) expBeta1a2 <- betaA(n, m2, nreg = 1) movePrior[rownames(movePrior) == 'MIDA', colnames(movePrior) == 'SEUS'] <- expBeta1a movePrior[rownames(movePrior) == 'SEUS', colnames(movePrior) == 'SEUS'] <- expBeta1a2 allqs[[listElement]][[j]] <- movePrior }

No we take these split-out elements and add them together to form a 2 element recursive list.

priorList <- list() priorList$females$exp1 <- allqs[[1]][[1]] + allqs[[2]][[1]] + allqs[[3]][[1]] priorList$females$exp2 <- allqs[[1]][[2]] + allqs[[2]][[2]] + allqs[[3]][[2]] priorList$females$exp3 <- allqs[[1]][[3]] + allqs[[2]][[3]] + allqs[[3]][[3]] priorList$females$exp4 <- allqs[[1]][[4]] + allqs[[2]][[4]] + allqs[[3]][[4]] priorList$females$exp5 <- allqs[[1]][[5]] + allqs[[2]][[5]] + allqs[[3]][[5]] priorList$females$exp6 <- allqs[[1]][[6]] + allqs[[2]][[6]] + allqs[[3]][[6]] priorList$females$exp7 <- allqs[[1]][[7]] + allqs[[2]][[7]] + allqs[[3]][[7]] priorList$females$exp8 <- allqs[[1]][[8]] + allqs[[2]][[8]] + allqs[[3]][[8]] priorList$males$exp1 <- allqs[[4]][[1]] + allqs[[5]][[1]] + allqs[[6]][[1]] priorList$males$exp2 <- allqs[[4]][[2]] + allqs[[5]][[2]] + allqs[[6]][[2]] priorList$males$exp3 <- allqs[[4]][[3]] + allqs[[5]][[3]] + allqs[[6]][[3]] priorList$males$exp4 <- allqs[[4]][[4]] + allqs[[5]][[4]] + allqs[[6]][[4]] priorList$males$exp5 <- allqs[[4]][[5]] + allqs[[5]][[5]] + allqs[[6]][[5]] priorList$males$exp6 <- allqs[[4]][[6]] + allqs[[5]][[6]] + allqs[[6]][[6]] priorList$males$exp7 <- allqs[[4]][[7]] + allqs[[5]][[7]] + allqs[[6]][[7]] priorList$males$exp8 <- allqs[[4]][[8]] + allqs[[5]][[8]] + allqs[[6]][[8]] devtools::use_data(priorList, overwrite = TRUE)

robschick/eliciteg documentation built on May 25, 2017, 7:38 a.m.

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