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import pandas as pd
import numpy as np
from itertools import combinations, permutations
from collections import defaultdict
import matplotlib.pyplot as plt
from collections import Counter
import seaborn as sns
def return_affiliation_df(dd):
Returns an affiliation index data frame per voting round
input: dd - data frame of voting history
output: affiliation_dict_list - data frame of affiliation index per voting round
flatten = lambda l: [item for sublist in l for item in sublist]
contestants = dd.index
pairs = list(combinations(contestants,2))
affiliation_dict_list = []
# iterate over voting rounds
for col_index in dd.columns:
# get the clusters who voted for particular people
voting_dict = dd.groupby(col_index).groups
# if not a name, remove from the voting dict
for i in ['','']:
if i in voting_dict.keys():
# everyone who voted in tribal council
in_tribal = set(flatten(voting_dict.values()))
# get all the people who received votes at tribal
candidates = voting_dict.keys()
candidates = [name for name in candidates if name in contestants]
# get the vote counts of the people who received votes
vote_counts = []
for voted in candidates:
affiliation_dict = defaultdict(int)
# initialize all pairs of people in tribal to have affiliation value of 0
for i in in_tribal:
for j in in_tribal:
if (i,j) in pairs:
affiliation_dict[(i,j)] = 0
# in the case of tiebreakers, or when only 2 people are at risk, set affiliation value
# of a pair of contestants to 1 if they voted together
if 'None' in voting_dict.keys() or len(candidates) < 3:
for voted in candidates:
for i in voting_dict[voted]:
for j in voting_dict[voted]:
if (i,j) in pairs:
affiliation_dict[(i,j)] = 1
# in the case of split votes:
# there are usually 3 names thrown out during split votes:
# the 2 targets of the majority affiliation, and the target of the
# people in the minority.
# get who the contestants at risk voted for
candidates_voted_for = dict()
for voted in candidates:
for other in candidates:
if voted in voting_dict[other]:
# cross contains all possible 3 pairs of the contestants at risk
cross = list(combinations(candidates,3))
# cluster_i represents the contestants who voted for contestant i
for (voted1, voted2, voted3) in cross:
cluster1 = voting_dict[voted1]
cluster2 = voting_dict[voted2]
cluster3 = voting_dict[voted3]
# set affiliation value of a pair of contestants to 1 if the they
# belong to the same cluster
for i in cluster1:
for j in cluster1:
if (i,j) in pairs:
affiliation_dict[(i,j)] = 1
for i in cluster2:
for j in cluster2:
if (i,j) in pairs:
affiliation_dict[(i,j)] = 1
for i in cluster3:
for j in cluster3:
if (i,j) in pairs:
affiliation_dict[(i,j)] = 1
# target_i is the vote of the ith contestant at risk
target1 = candidates_voted_for[voted1]
target2 = candidates_voted_for[voted2]
target3 = candidates_voted_for[voted3]
voteds = [voted1,voted2,voted3]
clusters = [cluster1, cluster2, cluster3]
targets = [target1, target2, target3]
index_pairs = combinations([0,1,2],2)
for (ix1, ix2) in index_pairs:
for cluster in clusters:
# check if more than 1 person votes for contestant ix1 and more than
# 1 person votes for contestant ix2, then they are candidates for the
# split vote.
if (len(clusters[ix1]) > 1) and (len(clusters[ix2]) > 1):
# check if the targets of contenstant ix1 and ix2 are in the same cluster
# if yes, then the people who voted for ix1 and ix2 did a split
# vote and are in an alliance
if (targets[ix1] in cluster) and (targets[ix2] in cluster):
everyone = flatten([clusters[ix1],clusters[ix2]])
for i in everyone:
for j in everyone:
if (i,j) in pairs:
# assign affiliation = 1 to all who voted for
# ix1 and ix2
affiliation_dict[(i,j)] = 1
# since ix1 and ix2 voted together, they are in a
# minority affiliation
if (voted[ix1], voted[ix2]) in pairs:
affiliation_dict[(voted[ix1], voted[ix2])] = 1
elif (voted[ix2], voted[ix1]) in pairs:
affiliation_dict[(voted[ix2], voted[ix1])] = 1
return pd.DataFrame(affiliation_dict_list).T
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