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Passing rate and network centrality #2

Open nepito opened 1 year ago

nepito commented 1 year ago

Soccermatics (David Sumpter):

Thomas settled on the following approach. He added up all the passes each player received and made, and compared these totals with the number of passes made by the player involved in the most passes. For example, in the Italy network above, Pirlo received the ball 78 times from the other seven players connected in the network. The second-most-passed player, Riccardo Montolivo, received it 39 times. So Montolivo received the ball on 78 − 39 = 39 fewer occasions than Pirlo. The average of these differences over all players divided by the total number of passes made gives a number between 0 and 1. If all the passes had been made to Pirlo, this number would be 1, while if all the players had received the ball exactly the same number of times then it would be 0.

Armed with these two measures, passing rate and network centrality, Thomas looked to see how well they predicted match outcome in the two Premier League seasons. He found that those teams that passed more when in possession of the ball scored more goals. A team that passed on average five times a minute scored around 20% more goals per season than a team that passed three times a minute. This difference is quite small, but it is significant, and it could make a major difference in the final league positions. Thomas also looked at centrality. Teams that focused their passing on only a few players scored fewer goals than teams that passed the ball more evenly between all members of the team. Here the difference is harder to quantify exactly, but roughly speaking, being decentralised gives a team an 8% advantage in terms of scoring rate.

Referencias

nepito commented 1 year ago
[30, 43, 60, 65, 37, 49, 40, 37, 24, 26, 13, None, None, 5, None, None, 17, 6, 5, None]
[21, 86, 63, 70, 42, 73, 77, 44, 25, 64, 11, None, None, None, None, None, None, 4, None, None]

Los máximos son 65 y 86. El total de pases fueron 580 y 457.

c=0.076 y c=0.065.

eq_1 = [player["statistics"][0]["passes"]["total"] for player in data["response"][0]["players"]]

c=0.087 y c=0.048

nepito commented 1 year ago

Aquí están los resultados con el Indice de Shannon: image

nepito commented 1 year ago

Aquí hay unos ejemplos: image

y los datos son los siguientes:

eq = [30, 43, 60, 65, 37, 49, 40, 37, 24, 26, 13, 5, 17, 6, 5]
eq_0 = [30, 43, 60, 79, 37, 49, 40, 37, 24, 26, 13, 5, 3, 6, 5]
eq_1 = [30, 43, 46, 79, 37, 49, 40, 37, 24, 26, 13, 5, 17, 6, 5]
eq_2 = [29, 42, 59, 79, 36, 48, 39, 36, 23, 25, 12, 4, 16, 5, 4]
eq_3 = np.ones(15)

Por último el código es este.

nepito commented 2 weeks ago

Así se obtiene los pases en un partido:

https://searchapi.wyscout.com/api/v1/team_stats/teams/1461/pass_distributions?match_id=5581940&type=passes&lang=en&token=$TOKEN

Aquí hay otro ejemplo:

https://searchapi.wyscout.com/api/v1/team_stats/teams/219/pass_distributions?match_id=5467635&type=passes&lang=en&token=$TOKEN