---
title: "Apportionment scenarios"
date: 2024-02-22
author: "Flavio Poletti"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Apportionment scenarios}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
```

## 1) Introduction
This vignette explores the implementation of different seat apportionment 
scenarios. There are generally two ways to assign seats of a parliament: Either
you distribute seats within each district independently or you distribute seats
according to the vote share for the whole election region (e.g. on a national 
level). We will focus on the difference in seat shares compared to vote shares
these two approaches can produce.

<br/>

## 2) Setup data 
We'll mainly use the data set for the 2019 Finnish parliamentary elections. 
The `finland2019` data set contains two data.frames: One with the number of votes
for each party, the other data.frame tells us how many seats (`election_mandates`) 
each district has.

```{r library}
library(proporz)
str(finland2019)
```

To make comparisons easier, we'll use matrices instead of data.frames. We create
a `votes_matrix` from the given data.frame.

```{r prepare_votes_matrix}
votes_matrix = pivot_to_matrix(finland2019$votes_df)
dim(votes_matrix)

# Let's look at all parties with at least 10k votes
knitr::kable(votes_matrix[rowSums(votes_matrix) > 10000,])
```

We also create a named vector that defines how many seats each district has 
(`district_seats`). Note that the order of district names in `district_seats` 
and columns in `votes_matrix` differ. However, this does not affect the analysis 
as we access districts by name (as does `biproporz()` by the way).

```{r prepare_district_seats}
district_seats = finland2019$district_seats_df$seats
names(district_seats) <- finland2019$district_seats_df$district_name
district_seats
```
<br/>

## 3) Distribute seats in every district independently

### 3.1) Function to distribute seats in districts

In Finland and many other jurisdictions, seats are assigned for each district 
independently. There is no biproportional apportionment among all districts. The 
following function, `apply_proporz`, calculates the seat distribution for each 
district and returns the number of seats per party and district as a matrix. 
The user can specify the apportionment method and a quorum threshold.

```{r apply_proporz}
apply_proporz = function(votes_matrix, district_seats, method, quorum = 0) {
    seats_matrix = votes_matrix
    seats_matrix[] <- NA
    
    # calculate proportional apportionment for each district (matrix column)
    for(district in names(district_seats)) {
        seats_matrix[,district] <- proporz(votes_matrix[,district],
                                           district_seats[district],
                                           quorum = quorum,
                                           method = method)
    }
    return(seats_matrix)
}
```


### 3.2) Actual distribution method (D'Hondt)

The Finnish election system uses the _D'Hondt_ method. We'll calculate the seat
distribution as a baseline to compare it with other methods.

```{r}
bydistrict_v0 = apply_proporz(votes_matrix, district_seats, "d'hondt")

bydistrict_v0[rowSums(bydistrict_v0) > 0,]
```

### 3.3) Alternative methods

We'll now look at alternative apportionment methods. The _Sainte-Laguë_ method 
(standard rounding) for example is impartial to party size. On the other hand, 
the _Huntington–Hill_ method favors small parties. Generally, Huntington-Hill 
is used with a quorum, otherwise all parties with more than zero votes get at 
least one seat. In this example, the quorum is 3% of votes in a district.

```{r}
bydistrict_v1 = apply_proporz(votes_matrix, district_seats,
                                      method = "sainte-lague")

bydistrict_v2 = apply_proporz(votes_matrix, district_seats,
                                      method = "huntington-hill", 
                                      quorum = 0.03)
```

Let's compare the seat distributions for these three methods. We get the number 
of party seats on the national level with `rowSums`.

```{r compare}
df_bydistrict = data.frame(
    D.Hondt = rowSums(bydistrict_v0),
    Sainte.Lague = rowSums(bydistrict_v1),
    Huntington.Hill = rowSums(bydistrict_v2)
)

# sort table by D'Hondt seats
df_bydistrict <- df_bydistrict[order(df_bydistrict[[1]], decreasing = TRUE),] 

# print parties with at least one seat
knitr::kable(df_bydistrict[rowSums(df_bydistrict) > 0,])
```

The actual political analysis is better left for people familiar with the party
system. However, it is fair to say that the election system within districts has
a significant impact on the number of seats in parliament. For example, the 
party with the most seats changes with Sainte-Laguë or Huntington-Hill.

### 3.4) Compare seat share with vote share

Let's now compare the vote shares on the national level with the seat share in 
parliament. Since every entry in `votes_matrix` is only one voter, we can simply 
use the row sums to get the vote shares. Otherwise, we'd have to weigh votes in 
each district according to the number of seats. Since disproportionality 
analysis is not the focus of this package, we'll simply look at the difference 
in shares for the actual distribution method.

```{r seat_vote_share}
vote_shares = rowSums(votes_matrix)/sum(votes_matrix)

shares = data.frame(
    seats = rowSums(bydistrict_v0)/sum(district_seats),
    votes = vote_shares 
)
shares$difference <- shares$seats-shares$votes
shares <- round(shares, 4)

# Only look at parties with at least 0.5 % of votes
shares <- shares[shares$votes > 0.005,]
shares <- shares[order(shares$difference),]

shares
```

The difference between seat and vote share varies among parties from -1.5 to
+2.3 percentage points.

<br/>

## 4) Biproportional method

### 4.1) Biproportional party seats
We'll now look at biproportional apportionment and whether it better matches the
seat share with the national vote share. Keep in mind that simply using existing 
data sets with `biproporz` is not really suitable from a modeling perspective. 
Vote distributions are likely to be different if, for example, people know their 
vote also counts on a national level or if smaller parties choose to run in more 
districts.

With that being said, let's assume our data set is properly modeled already. We 
need to consider that voters in Finland can only vote for _one_ person in each 
district. They can't distribute as many votes as there are seats in their 
district among candidates/parties. Thus we need to use `use_list_votes=FALSE` 
as a parameter in `biproporz`.

```{r biprop, results="hide"}
seats_biproportional = biproporz(votes_matrix, 
                                 district_seats, 
                                 use_list_votes = FALSE)

# show only parties with seats
seats_biproportional[rowSums(seats_biproportional) > 0,]
```
```{r biprop.knit, echo = F}
knitr::kable(seats_biproportional[rowSums(seats_biproportional) > 0,])
```

Let's now look at the difference between vote and seat share.

```{r compare_vote_seat_shares}
vote_shares = rowSums(votes_matrix)/sum(votes_matrix)
seat_shares = rowSums(seats_biproportional)/sum(seats_biproportional)

range(vote_shares - seat_shares)
```

As we can see, the difference between vote and seat share ranges from -0.5 to
+0.2 percentage point. Biproportional apportionment matches the national vote 
share better than apportionment by district. This is expected however, since
biproportional apportionment actually considers the national vote share. 
Discussing the pros and cons of a regional representation compared to a 
priority on national vote shares is not within the scope of this vignette. The
following chunk shows the seat changes.

```{r compare_matrices}
seat_changes = seats_biproportional-bydistrict_v0

knitr::kable(seat_changes[rowSums(abs(seat_changes)) > 0,colSums(abs(seat_changes))>0])
```
### 4.2) Distribute seats among districts as well
Normally the number of seats per district is defined before an election, based
on census data. However, you *could* also assign the number of seats per 
district, based on the actual votes cast in every district. This way, each 
districts seat share is proportional to its vote share. To do this, we use the
total number of seats for the `district_seats` parameter.

```{r biprop_seats}
full_biproportional = biproporz(votes_matrix, 
                                district_seats = sum(district_seats),
                                use_list_votes = FALSE)

# party seat distribution has not changed
rowSums(full_biproportional) - rowSums(seats_biproportional)

# district seat distribution is different
colSums(full_biproportional) - colSums(seats_biproportional)
```

As we can see, the number of party seats does not change. However, districts 
where more people voted get a higher share of the seats. 2 districts gain seats,
4 districts lose a seat.

<br/>

## 5) Effect of a system change on seat distribution

As a last example, we'll look at the changes in seat distribution biproportional
apportionment had in the 2020 parliament election in the Swiss canton of Uri.
In Uri, 2020 was the first year the cantonal parliament was elected with 
biproportional apportionment for some municipalities. Previously, these four 
municipalities used the _Hagenbach-Bischoff_ method. We'll ignore 
all other municipalities which use a majoritarian electoral system.

```{r uri}
seats_old_system = apply_proporz(uri2020$votes_matrix, uri2020$seats_vector, "hagenbach-bischoff")

seats_new_system = biproporz(uri2020$votes_matrix, uri2020$seats_vector)

seats_new_system-seats_old_system
```

Compared to the previous election system there was a change of one seat (out of
37).