Q1 The social interactions among a group of 18 monks were recorded by S. Sampson. The data include information about the top three choices for the networks ‘liking at time 3’, ‘disliking’, ‘positive influence’ and ‘negative influence’. Calculate the indegree for each of these four networks, where the networks are dichotomized. Next calculate the heterogeneity of incoming tie types using the IQV measure.
#grab needed matrices
samp = Sampson_Monastery[8:12]
#dichotomize
dsamp = lapply(samp,xDichotomize)
#transpose so we are looking at incoming ties
tdsamp = lapply(dsamp,t)
#get indegree
indeg = sapply(tdsamp,rowSums)
#get heterogeneity of tie types
temp = xMultipleTieComposition(tdsamp,Measures="IQVType")
#display indegree side by side with IQV
cbind(indeg,temp[,6,drop=F])
LikeT1 LikeT2 LikeT3 NegativeInfluence PositiveInfluence IQVType
ROMUALD 1 1 0 1 1 0.9375000
BONAVENTURE 5 7 6 0 2 0.8937500
AMBROSE 2 1 4 0 3 0.8750000
BERTHOLD 1 2 2 5 1 0.8884298
PETER 3 5 4 8 5 0.9720000
LOUIS 2 2 2 2 4 0.9722222
VICTOR 4 2 2 2 0 0.9000000
WINFRID 1 4 6 3 4 0.9490741
JOHN_BOSCO 9 8 4 1 6 0.9343112
GREGORY 7 8 6 3 11 0.9653061
HUGH 6 2 2 1 1 0.8506944
BONIFACE 1 2 2 0 2 0.9183673
MARK 4 3 5 2 6 0.9687500
ALBERT 2 2 1 2 1 0.9765625
AMAND 2 2 2 4 1 0.9504132
BASIL 2 1 3 4 2 0.9548611
ELIAS 1 2 2 6 2 0.8875740
SIMPLICIUS 2 3 3 6 1 0.9222222
Q2 What would the IQV measure tell us if we were to apply it to Sampson’s liking network at three time points?
#grab the three liking matrices
samp = Sampson_Monastery[8:10]
#dichotomize
dsamp = lapply(samp,xDichotomize)
#transpose
tdsamp = lapply(dsamp,t)
#get heterogeneity of tie types
temp = xMultipleTieComposition(tdsamp,Measures="IQVType")
#display just the IQV measure
temp[,4,drop=F]
IQVType
ROMUALD 0.7500000
BONAVENTURE 0.9907407
AMBROSE 0.8571429
BERTHOLD 0.9600000
PETER 0.9791667
LOUIS 1.0000000
VICTOR 0.9375000
WINFRID 0.8429752
JOHN_BOSCO 0.9523810
GREGORY 0.9931973
HUGH 0.8400000
BONIFACE 0.9600000
MARK 0.9791667
ALBERT 0.9600000
AMAND 1.0000000
BASIL 0.9166667
ELIAS 0.9600000
SIMPLICIUS 0.9843750
We can see that Louis and Amand have equal numbers of incoming ties across time periods. Romuald has the most variety, because he doesn't have any ties in Time 3. But all of the monks have fairly equal numbers of ties across time periods.
Q3 Bernard and Killworth recorded the number of amateur ham radio calls made over a one-month period, as monitored by a voice-activated recording device. Using the valued ties in the Ham Radio dataset, calculate the average and standard deviation of tie strength using the calls data.
#grab the calls data
calls = Bernard_HamRadio$Calls
#calculate statistics
temp = xValuedTieComposition(calls,Measures=c("AvStrength","SDStrength"))
#display just the mean and SD of tie strength
temp[,c(2,3)]
AvStrength SDStrength
a01 0.04651163 0.3049971
a02 3.81395349 7.3945147
a03 0.02325581 0.1524986
a04 0.90697674 2.3483931
a05 0.02325581 0.1524986
a06 0.04651163 0.3049971
a07 4.60465116 8.9286578
a08 0.02325581 0.1524986
a09 0.04651163 0.2130826
a10 0.44186047 1.0533919
a11 0.04651163 0.2130826
a12 0.04651163 0.3049971
a13 0.13953488 0.9149914
a14 0.41860465 0.9815575
a15 0.04651163 0.3049971
a16 1.51162791 3.9902983
a17 0.02325581 0.1524986
a18 2.72093023 4.7775546
a19 0.11627907 0.3909270
a20 0.27906977 0.7965874
a21 0.06976744 0.2577696
a22 0.93023256 2.6131112
a23 0.00000000 0.0000000
a24 0.72093023 1.9064500
a25 0.04651163 0.2130826
a26 0.34883721 1.1102063
a27 0.44186047 1.0757572
a28 0.72093023 1.5169767
a29 0.04651163 0.2130826
a30 0.02325581 0.1524986
a31 3.09302326 6.2519100
a32 0.20930233 0.5588312
a33 2.83720930 5.3715752
a34 0.00000000 0.0000000
a35 0.11627907 0.4980582
a36 0.00000000 0.0000000
a37 0.41860465 1.1387662
a38 0.65116279 1.6018400
a39 0.51162791 1.3517553
a40 0.04651163 0.2130826
a41 0.02325581 0.1524986
a42 0.30232558 0.6375132
a43 2.25581395 7.9195148
a44 0.39534884 1.1980050
>
Q4 Lin Freeman collected friendship networks of 32 network scientists at two time points, as well as the number of citations of each of the scientists’ work in the Social Sciences Citation Index. Dichotomize the network data at time 1, so that a tie exists when the colleague is considered a friend or close friend. Calculate the average number of citations of each person's alters.
#get the network data
t1 = xDichotomize(Freeman_EIES$AcquaintanceT1)
#get the citations data
cit = Freeman_EIES$Attributes$Citations
#calculate composition measures
temp = xAlterCompositionCon(t1, cit)
#display each person's own citations along with their alters' citations
cbind(cit,temp[,"AvAlterV",drop=F])
#is there a correlation between a person's citations and their friends'?
cor(cit,temp[,"AvAlterV"])
cit AvAlterV
b01 19 23.032258
b02 3 19.709677
b03 170 17.322581
b06 23 21.709677
b08 16 16.096774
b10 6 5.387097
b11 1 9.096774
b13 9 18.225806
b14 6 4.483871
b18 40 21.838710
b19 15 20.935484
b20 54 15.387097
b21 4 20.096774
b22 46 21.903226
b23 17 20.709677
b24 32 16.161290
b25 23 18.677419
b26 1 12.774194
b27 34 20.483871
b32 64 20.838710
b33 11 22.903226
b35 11 18.870968
b36 31 19.709677
b37 18 22.580645
b38 4 21.580645
b39 0 22.290323
b40 4 17.225806
b41 56 11.516129
b42 12 21.322581
b43 2 17.419355
b44 0 18.387097
b45 1 9.290323
> cor(cit,temp[,"AvAlterV"])
[1] 0.08908977
>
Q5 Freeman also collected the discipline of each scientist (with codes 1 = sociology, 2 = anthropology, 3 = mathematics/statistics, 4 = other). Using the same network data as in the previous question, calculate the diversity with respect to discipline of each person's ego network.
#get the network data (as we did before)
t1 = xDichotomize(Freeman_EIES$AcquaintanceT1)
#get the discipline data
dis = Freeman_EIES$Attributes$Discipline
#calculate composition measures
temp = xAlterCompositionCat(t1, dis)
#display just the part we want
temp[,"IQV",drop=F]
IQV
b01 0.8657648
b02 0.9166667
b03 0.6452426
b06 0.7712665
b08 0.7327824
b10 0.8888889
b11 0.7916667
b13 0.8429752
b14 0.9256198
b18 0.7962963
b19 0.8216761
b20 0.7210884
b21 0.8661333
b22 0.8129252
b23 0.8115942
b24 0.7572016
b25 0.8051708
b26 0.7210884
b27 0.6893424
b32 0.7327824
b33 0.7722667
b35 0.7654321
b36 0.4270833
b37 0.7712665
b38 0.6648199
b39 0.7253333
b40 0.9135802
b41 0.6574331
b42 0.7712665
b43 0.8806584
b44 0.8836292
b45 0.9479167
>
Q6 Considering the discipline data for the same scientists, see if there is any evidence of homophily. In other words, calculate a measure of ego–alter similarity. We recommend Yule's Q.
#get the network data (as before)
t1 = xDichotomize(Freeman_EIES$AcquaintanceT1)
#get the discipline data (as before)
dis = Freeman_EIES$Attributes$Discipline
#calculate ego-alter similarity measures for categorical attributes
temp = xEgoAlterSimilarityCat(t1, dis)
#display just the part we want
temp[,"YulesQ",drop=F]
YulesQ
b01 NaN
b02 1.00000000
b03 -0.72602740
b06 0.64705882
b08 -0.66666667
b10 -0.28000000
b11 -0.19148936
b13 0.28000000
b14 0.83132530
b18 0.55555556
b19 0.36842105
b20 0.44000000
b21 1.00000000
b22 1.00000000
b23 0.36842105
b24 -0.17241379
b25 -0.03225806
b26 0.44000000
b27 0.77777778
b32 0.71929825
b33 1.00000000
b35 0.43089431
b36 0.89090909
b37 0.64705882
b38 0.73333333
b39 1.00000000
b40 1.00000000
b41 0.73333333
b42 0.64705882
b43 0.54838710
b44 1.00000000
b45 1.00000000
>
Q7 Consider Padgett’s marriage network of Florentine families. Characterize the wealth of the families that each family is connected to.
#get the network data
m = Padgett_FlorentineFamilies$Marriage
#get the wealth data
w = Padgett_FlorentineFamilies$Attributes$Wealth
#calculate measures characterizing wealth of families connected by marriage
temp = xAlterCompositionCon(m, w)
#display just the part we want
temp[,c("MinAlterV","MaxAlterV","WAvAlterV")]
MinAlterV MaxAlterV WAvAlterV
ACCIAIUOLI 103 103 103.00000
ALBIZZI 8 103 47.66667
BARBADORI 20 103 61.50000
BISCHERI 8 146 67.66667
CASTELLANI 49 146 83.33333
GINORI 36 36 36.00000
GUADAGNI 36 48 42.50000
LAMBERTESCHI 8 8 8.00000
MEDICI 10 55 31.00000
PAZZI 10 10 10.00000
PERUZZI 20 146 70.00000
PUCCI Inf -Inf NaN
RIDOLFI 48 146 99.00000
SALVIATI 48 103 75.50000
STROZZI 20 49 35.00000
TORNABUONI 8 103 46.00000
Q8 Consider Padgett’s marriage network again. Calculate the constraint measure of structural holes, as well as the density and fragmentation of each node’s ego network.
#get the network data (if you haven't already)
m = Padgett_FlorentineFamilies$Marriage
#calculate measures of structural holes
temp = xStructuralHoles(m, Measures="Constraint", Include="EgoNetwork")
#display just the part we want
temp[,c(2,3)]