def test_percentages(self): neutral_range = 0.001 self.G.node[0]['opinion'] = 0.001 # neutral self.G.node[1]['opinion'] = 0.01 # positive self.G.node[2]['opinion'] = -0.5 # negative percentages_exp = [1/3,1/3,1/3] np.testing.assert_array_equal(cp.percentages(self.G,neutral_range),percentages_exp) self.G.node[2]['opinion'] = 0.5 percentages_exp = [2/3,0/3,1/3] np.testing.assert_array_equal(cp.percentages(self.G,neutral_range),percentages_exp) self.G.node[1]['opinion'] = -0.001 self.G.node[2]['opinion'] = -0.01 percentages_exp = [0/3,1/3,2/3] np.testing.assert_array_equal(cp.percentages(self.G,neutral_range),percentages_exp)
def test_percentages(self):
neutral_range = 0.001
self.G.node[0]['opinion'] = 0.001 # neutral
self.G.node[1]['opinion'] = 0.01 # positive
self.G.node[2]['opinion'] = -0.5 # negative
percentages_exp = [1 / 3, 1 / 3, 1 / 3]
np.testing.assert_array_equal(cp.percentages(self.G, neutral_range),
percentages_exp)
self.G.node[2]['opinion'] = 0.5
percentages_exp = [2 / 3, 0 / 3, 1 / 3]
np.testing.assert_array_equal(cp.percentages(self.G, neutral_range),
percentages_exp)
self.G.node[1]['opinion'] = -0.001
self.G.node[2]['opinion'] = -0.01
percentages_exp = [0 / 3, 1 / 3, 2 / 3]
np.testing.assert_array_equal(cp.percentages(self.G, neutral_range),
percentages_exp)
# If strategy 1 is not a smart peer, the 2 forceful peers can be added simultaneously
if STRATEGY1 != 'smart':
gm.add_forceful(G, STRATEGY1, BUDGET1, STRATEGY2, BUDGET2)
# If strategy 1 is smart then ofrceful peer with strategy 2 is added first then smart peer
else:
#Add one forceul peer to the graph
gm.add_one_forceful(G, STRATEGY2, BUDGET2)
## Linear programming method to beat an existing peer with minimum budget
BUDGET1 = gm.add_smart(G, ALPHA, BUDGET2, NEUTRAL_RANGE)
# Calculate final opinion vector by equation, considering the presence of
# 2 forceful peers in the last 2 indices of the graph
#R_inf = cp.R_inf(G,ALPHA)
# Calculate final opinion vector by iterations
R_itr = gm.R_itr(G, ALPHA)
# Calculate the percentage of normal peers that followed either of the forceful peers
tmp = cp.percentages(G, NEUTRAL_RANGE)
# Update S1_wins_list for analytical reason
if (tmp[0] > 0.5): S1_wins_list.append(1)
else: S1_wins_list.append(0)
# Update percentages arrays
cp.update_percentages(tmp, S1_followers, S2_followers, neutral, i, wins)
i += 1
# Asserting that R_inf calculated by equation and iteration are equal to decimal places
#try:
# np.testing.assert_array_almost_equal(R_inf,R_itr,4)
#except AssertionError:
# sys.exit('ConvergenceError: convergence of R_inf is not correct to 4 decimal places\nProgram will terminate')
wins[0] = (wins[0] / SIMULATIONS) * 100
wins[1] = (wins[1] / SIMULATIONS) * 100
wins[2] = (wins[2] / SIMULATIONS) * 100
# If strategy 1 is not a smart peer, the 2 forceful peers can be added simultaneously
if STRATEGY1 != 'smart':
gm.add_forceful(G, STRATEGY1, BUDGET1, STRATEGY2, BUDGET2)
# If strategy 1 is smart then ofrceful peer with strategy 2 is added first then smart peer
else:
#Add one forceul peer to the graph
gm.add_one_forceful(G, STRATEGY2, BUDGET2)
## Linear programming method to beat an existing peer with minimum budget
BUDGET1 = gm.add_smart(G,ALPHA,BUDGET2, NEUTRAL_RANGE)
# Calculate final opinion vector by equation, considering the presence of
# 2 forceful peers in the last 2 indices of the graph
#R_inf = cp.R_inf(G,ALPHA)
# Calculate final opinion vector by iterations
R_itr = gm.R_itr(G,ALPHA)
# Calculate the percentage of normal peers that followed either of the forceful peers
tmp = cp.percentages(G,NEUTRAL_RANGE)
# Update S1_wins_list for analytical reason
if (tmp[0]>0.5): S1_wins_list.append(1)
else: S1_wins_list.append(0)
# Update percentages arrays
cp.update_percentages(tmp, S1_followers, S2_followers, neutral, i, wins)
i += 1
# Asserting that R_inf calculated by equation and iteration are equal to decimal places
#try:
# np.testing.assert_array_almost_equal(R_inf,R_itr,4)
#except AssertionError:
# sys.exit('ConvergenceError: convergence of R_inf is not correct to 4 decimal places\nProgram will terminate')
wins[0] = (wins[0]/SIMULATIONS)*100
wins[1] = (wins[1]/SIMULATIONS)*100
wins[2] = (wins[2]/SIMULATIONS)*100