Localized list
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A Localized list or local list is a technique used under systems of party-list proportional representation to determine which party candidates are elected from the party list. Local lists differ from open lists or closed lists. As with open lists, local lists allow the electorate to vote for individual candidates, but that preference is expressed through local or district level election processes.
Voting in this system takes place at the district level, where each party is represented by a single candidate, just like in first-past-the-post or other single-winner systems. However, the candidate with the largest number of votes in a district is not necessarily the one that is elected. This is because of the proportionality requirement of the system.
To ensure that each party receives a proportional share of seats relative to its share of the popular vote, the first step in ballot counting is to count the votes going to each party either overall (at-large) or by multi-member constituencies. The results by party are then used to divide the number of seats proportionately among the different parties.
The party list is composed of the candidates running in each district. Once the number of seats won by a party is known, that party's candidates with the highest percentages of votes in their district are the ones elected until the number of seats corresponding to that party have been filled.
This system affords voters a way of voting for individual candidates. However, the system is designed to ensure proportionality, so the candidate with the highest popular vote in a single local constituency may not be elected, because his or her party-mates in other constituencies may have a higher voter share, and candidates with fewer votes can be elected because they are the best candidates in their party's list. Furthermore, it is possible for more than one candidates to be elected in a single district, or for no candidate to be elected.
Localized lists are used in Italy for provincial elections and were used, from 1948 to 2001, for senatorial elections.
Examples
How this mechanism operates can best be understood with some numerical examples. The examples below imagines a four-member constituency containing four equal-sized districts. The constituency is represented by four elected members but elected members are understood to represent the whole constituency, not just the individual districts. The two examples are virtually identical, except that in the second example, the number of votes secured by the Blue Party is slightly higher in District four.
| District 1 | District 2 | District 3 | District 4 |
|---|---|---|---|
| John (Reds), 600 | James (Yellows), 350 | Hughes (Reds), 470 | Anne (Reds), 390 |
| Carew (Yellows), 200 | Paul (Greens), 250 | Joshua (Blues), 290 | Mary (Blues), 280 |
| Andrew (Blues), 150 | Charles (Reds), 210 | Duncan (Yellows), 180 | Trevor (Greens), 170 |
| Millie (Greens), 50 | Ronnie (Blues), 190 | Patty (Greens), 60 | Michael (Yellows), 160 |
| 1000 voters | 1000 voters | 1000 voters | 1000 voters |
Four seats must be filled. The largest remainder method with Hare quota is used.
Reds received 1670 votes, Blues received 910 votes, Yellows received 890 votes, Greens received 530 votes. Red won two seats, while Blues and Yellows one seat each.
John, James, Hughes and Joshua are elected. District 3 elects two candidates, while District 4 none. However, this fact is not seen as a problem, because all winners represent their own party-list at-large, and not their single district.
| District 1 | District 2 | District 3 | District 4 |
|---|---|---|---|
| John (Reds), 600 | James (Yellows), 350 | Hughes (Reds), 470 | Anne (Reds), 370 |
| Carew (Yellows), 200 | Paul (Greens), 250 | Joshua (Blues), 290 | Mary (Blues), 300 |
| Andrew (Blues), 150 | Charles (Reds), 210 | Duncan (Yellows), 180 | Trevor (Greens), 170 |
| Millie (Greens), 50 | Ronnie (Blues), 190 | Patty (Greens), 60 | Michael (Yellows), 160 |
| 1000 voters | 1000 voters | 1000 voters | 1000 voters |
Four seats must be filled. The largest remainder method with Hare quota is used.
Reds received 1650 votes, Blues received 930 votes, Yellows received 890 votes, Greens received 530 votes. Red won two seats, while Blues and Yellows one seat each.
John, James, Hughes and Mary are elected. Anne lost her race, while her runner-up won. Again, this fact is not seen as a problem, because Mary represents the Blues at-large, and not simply District 4.