社会科学问题研究的计算实践——1、社会网络基础（聚集系数与嵌入性、友谊悖论的验证）

学习资源来自，一个哲学学生的计算机作业 (karenlyu21.github.io)

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1、背景问题

“网络”由节点组成，节点之间可能有边相连。网络常常是对社会的一种有效抽象，节点代表社会中的行动者，边代表行动者之间的联系。我们可以用一个矩阵（称为邻接矩阵）来表示网络，在程序中一般就是对应一个二维数组。邻接矩阵在(i,j)的值，表示i与j之间边的权重。一种基本的情形是权重为0或1，0表示有边，1表示没有边。此时的网络也称“图”。例如：

这个矩阵就表示如下一个网络（或图），节点符合A,B,C,D,分别对应矩阵的1,2,3,4列和行。

一个节点与自己总是没有边的，因此，矩阵从左上到右下的对角线上都是0。对于一个无向图，邻接矩阵A(i,j)=A(j,i)，即这一矩阵是对称的，以从左上到右下的对角线为轴。

基于图的邻接矩阵，我们可以通过计算得到对应社会网络的一些结构信息。例如，把一行内的值相加，就可以得到，该行对应的人有多少朋友。另一个例子是矩阵的乘幂，例如A²(i,j)表示节点i和j之间的长度为2的路径一共有多少条，也可以解释为i和j的共同朋友的个数。

利用这些计算操作，我们可以对网络的性质做出分析，这可以用以检验和社会网络相关的理论假设。作业“聚集系数和嵌入性”就是网络分析的一个例子。另一个作业则检验了社会网络学说的一个理论假设，即“友谊悖论”。

2、计算实践

2.1、聚集系数与嵌入性

2.1.1、作业描述与算法思路

本次作业的主要任务是计算某一社会网络中，每个节点的聚集系数，以及每条边的嵌入性。

在一个网络中，不同人的社交密度或强度（intensity）不一样，有的人把自己的朋友聚集起来的能力很强——ta的朋友往往互相之间也认识。我们用“聚集系数”来刻画这一现象。节点i的聚集系数指的是，i的朋友们中，两两之间也是朋友的概率。

计算聚集系数的基本思路如下：

先分析邻接矩阵，用一个字典变量friend_dict存储“谁是谁的朋友”，字典的键（key）i指的是节点i∈[0,n)，节点i的朋友j,k,…∈[0,n)被存储在该键对应的值（value）里（值是一个列表）。接着，我再两两分析节点i的朋友（我把他们俩称为以节点i为中介的“间接朋友”），用friend_dict存储的信息验证，他们俩是否为“直接朋友”。节点i的聚集系数就是“直接朋友”和“间接朋友”的比值。

两个人之间的关系也有性质上的差别，有的关系深深嵌入了周围的社交圈中——不仅他们俩是朋友关系，而且他们俩的共同好友很多。我们用“嵌入性”来刻画这一现象。边(i,j)的嵌入性，指的是节点i与j的共同朋友数。

计算嵌入性的基本思路如下：

分析网络中的边。对于边(i,j)，检查，节点i的朋友有多少也是节点j的朋友，这一过程也是用friend_dict完成的。

2.1.2、编程实现与要点说明

提前说明：笔者在原文中未给出net3.txt，故可自行给出，这里本人直接照抄背景问题中的矩阵。。。

首先，我们读取样本数据文件，把文件里的矩阵数据存储在一个numpy 2d-array A里。

import numpy as np

def arrayGen(filename):
    f = open(filename, 'r')
    r_list = f.readlines()  # 返回包含size行的列表, size未指定则返回全部行
    f.close()
    A = []
    for line in r_list:
        if line == '\n':
            continue
        line = line.strip('\n')
        line = line.strip()  # 用于移除字符串头尾指定的字符（默认为空格或换行符）或字符序列，不能删除中间字符
        row_list = line.split()  # 通过指定分隔符对字符串进行切片
        for k in range(len(row_list)):
            row_list[k] = row_list[k].strip()
            row_list[k] = int(row_list[k])
        A.append(row_list)  # 用于在列表末尾添加新的对象
    n = len(A[0])
    A = np.array(A)
    return A, n

filename = input('请输入邻接矩阵文件名：')  # 将邻居矩阵net3.txt放在同一文件夹下即可
A, n = arrayGen(filename)

分析这一网络，我们把“谁是谁的朋友”存储在一个字典变量friend_dict里，字典的键（key）N∈[0,n)是节点，节点i∈[0,n)的朋友被存储在该键对应的值（value）里。

friend_dict = {}
for i in range(n):
    friend_dict[i] = []
    for j in range(n):
        if A[i][j] ==1:
            friend_dict[i].append(j)

接着，我再两两分析节点i的朋友（他们俩是“间接朋友”，以节点i为中介），用friend_dict存储的信息验证，他们俩是否为“直接朋友”。以节点i为中介的间接朋友数存储在变量PossTriCl里，这些人中的直接朋友数存储在变量TriCl里。聚集系数ClCo=TriClPoss/TriCl。每一节点的聚集系数被存储在一个字典变量ClCo_dict里。

ClCo_dict = {}  # ClCo --> clustering coefficient
for(person, friend_list) in friend_dict.items():  # 以列表返回视图对象，是一个可遍历的key/value 对
    TriCl = 0  # 直接朋友
    for x in range(len(friend_list)):
        friend1 = friend_list[x]
        for y in range(x+1, len(friend_list)):
            friend2 = friend_list[y]
            if A[friend1][friend2] == 1:
                TriCl += 1;
    PossTriCl = len(friend_list)*(len(friend_list)-1)/2  # 间接朋友
    if PossTriCl == 0:
        ClCo = 0
    else:
        ClCo = TriCl / PossTriCl  # 聚集系数
    ClCo_dict[person] = ClCo

最后，我们在控制台里打印出各节点聚集系数的情况。

print()
print('clustering coefficient as following:')
ClCo_list = list(ClCo_dict.items())
ClCo_list.sort(key=lambda x: x[1], reverse=True)
if len(ClCo_dict) < 10:
    for item in ClCo_list:
        print('node', str(item[0]) + ':', '%.2f' % item[1])
else:
    for item in ClCo_list[0:10]:
        print('node', str(item[0]) + ':', '%.2f' % item[1])

控制台输出如下：

请输入邻接矩阵文件名：net3.txt

clustering coefficient as following:
node 1: 1.00
node 3: 1.00
node 0: 0.67
node 2: 0.67

接着计算网络中各边的嵌入性。利用friend_dict遍历所有边、计算两个人的共同朋友数。

embd_dict = {}
for(person, friend_list) in friend_dict.items():
    for friend_a in friend_list:
        if friend_a > person:
            embd_dict[person, friend_a] = 0
            for friend_b in friend_dict[friend_a]:
                if friend_b in friend_list:
                    embd_dict[person, friend_a] += 1

输出结果到控制台。

print()
print('embeddedness as following:')
embd_list = list(embd_dict.items())
embd_list.sort(key=lambda x: x[1], reverse=True)
if len(embd_list) < 10:
    for item in embd_list:
        print('edge', end=' ')
        print(item[0], end='')
        print(':', str(item[1]))
else:
    for item in embd_list[0:10]:
        print('edge', end=' ')
        print(item[0], end='')
        print(':', str(item[1]))

输出结果如下：

请输入邻接矩阵文件名：net3.txt

embeddedness as following:
edge (0, 2): 2
edge (0, 3): 2
edge (2, 3): 2
edge (0, 1): 1
edge (1, 2): 1

2.2、友谊悖论的验证

2.2.1、作业描述与算法思路

本次作业的主要任务是验证友谊悖论。这一悖论说的是，现实中多数人一般都感到朋友的朋友多于自己的朋友；尽管，直观上我们会觉得，既然有人觉得朋友的朋友多于自己的朋友，另一些人就会觉得朋友的朋友少于自己的朋友。

验证这一悖论，要在某个社会网络中进行。为了得到亟待验证的社会网络，一个方法当然是以现实生活的数据为基础、进行抽象和转化，我们在这次作业中采取的是另一个方法：基于“小世界”现象，生成一系列的网络。社会科学以“小世界”为主题的研究颇多，这些成果保证，我们生成的社会网络是对真实世界的合理模拟。

“小世界”现象指的是，邻接矩阵中任意两个点之间，都有一条很短的路径可以通达。之所以“小世界”现象存在，是因为社会网络有如下两个性质：

• 同质性（homophily）：一个圈子里的人往往互相都认识，网络中应当有很多三角形（参见第二讲三元闭包）。我们用初始规则度数r模拟同质性，表示每一节点和编码相邻的节点之间的边数。这样，编码相近的节点之间会形成很多三角关系。

• 随机性（randomness）：除了小圈子以外，网络中还存在一些跨圈的边，这些边连接的两个节点没有共同朋友（没有形成三元闭包）。这些边的形成不是由于同处一个圈子，或多或少体现了随机性。是这些边保证了任意两点之间有一条很短的路径；否则，要连接不在同一圈子的节点时（例如，要把澳洲的一个庄园主和一个叙利亚难民连接起来），必须借助相邻节点之间的一条条规则边，甚至如此“翻山越岭”也不一定能连接起两个节点。我们用随机调整边占比p来模拟随机性：按照初始规则度数生成的边中，比例为p的边被删除， 同样数量的边随机地在任意两点间生成。

改变初始规则度数r、随机调整边占比p、总节点数n，我们就可以得到一系列不同的社会网络。它们都有“小世界”现象，只是同质性、随机性和规模有所不同。对于每一网络，我们都来检验友谊悖论是否成立。

为了检验某一网络是否存在友谊悖论，我一一检查网络中的每一节点。对于节点i，如果他的朋友数（度数degree）比他朋友的平均度数小不少，那么节点i就存在友谊悖论。如果节点i度数刚好小于邻居的平均度数，节点i很难明显感觉到朋友比自己更受欢迎。我具体设定的悖论觉知门槛是，节点i度数的1.1倍小于邻居的平均度数。对于整个网络，如果超过一半的节点都觉知到了友谊悖论，这一网络就存在友谊悖论。

2.2.2、编程实现与要点说明

提前说明：笔者在写文件做了时间记录，但是我这里有报错，故删了，代码略微变动，不影响结果处理。

首先，我们需要一个生成社会网络的函数neighbor_generator：输入总节点数n、初始规则度数r、随机调整边占比p，这个函数会输出一个表示社会网络的矩阵matr（numpy 2d-array）。

def neighbor_generator(n, r, p):
    file = open('./output/generator_logs_{}', 'w')
    matr = np.zeros((n, n))

初始规则边的建立：每一节点与相邻r个节点建立规则边。当n是偶数时，使节点i与编码∈[i+1, i+|r/2+1+(r mod 2)|]的节点建立边即可；当n是奇数时，执行同样的操作后，还需要删除(n−1)/2条边。这样操作后，总边数为(n×r)/2。

	file.write('Generating an orderly matrix with homophilous links...\n')
    edge_list = []
    for i in range(0, n):
        j_list = [(i + x) % n for x in range(1, int(r/2 + 1 + r % 2))]
        for j in j_list:
            matr[i][j] = 1
            matr[j][i] = 1
            if j > i:
                edge_list.append([i, j])
            else:
                edge_list.append([j, i])
    file.write('Writing the list of node pairs which do not have an edge...\n')
    empt_list = []
    for i in range(0, n):
        if i % 100 == 0:
            file.write('%i nodes processed...' % i)
        for j in range(i+1, n):
            if matr[i][j] == 0:
                empt_list.append([i, j])
    file.write('For now, number of edges in Matrix A: %i\n' % np.sum(matr))
    if r % 2 != 0:
        for i in range(0, n-1, 2):
            matr[i][i+1] = 0
            matr[i+1][i] = 0
            edge_list.remove([i, i+1])
            edge_list.remove([i, i+1])
    file.write('Number of edges in Matrix A: %i\n' % np.sum(matr))

随机调整边：对于已建立的规则边（存储在exc_edge中），比例为p的边被删除，并随机在无关系的两个节点之间（存储在exc_empt中）生成新边。

    file.write('Introducing randomness...\n')
    exc_edge = random.sample(edge_list, int(n * r * p / 2))  # 随机获取n*r*p/2个元素作为一个片断返回
    exc_empt = random.sample(empt_list, int(n * r * p / 2))
    for item in exc_edge:
        matr[item[0]][item[1]] = 0
        matr[item[1]][item[0]] = 0
    for item in exc_empt:
        matr[item[0]][item[1]] = 1
        matr[item[1]][item[0]] = 1

输出矩阵

    np.savetxt('./output/neighbor_original_{}', matr, fmt='%i')
    return matr

除此之外，我们还需要一个验证友谊悖论的函数paradox：输入一个给定的矩阵，输出一个bool值，即它是否满足友谊悖论。

def paradox(matr, n):

我先把“谁是谁的朋友”存储在字典变量fr_dict中，再用这个字典变量计算：节点i的度数，存储在dg_dict[i]['self']；ta邻居的平均度数，存储在dg_dict[i]['nb']。

	fr_dict = {}
    dg_dict = {}
    # generate fr_dict to store who's whose friend
    for i in range(0, n):
        fr_dict[i] = []
        for j in range(0, n):
            if matr[i][j] == 1:
                fr_dict[i].append(j)

    # generate the degrees of individuals
    for i in range(0, n):
        dg_dict[i] = {}
        dg_dict[i]['self'] = len(fr_dict[i])

    # generate the average degrees of neighbors
    for i in range(0, n):
        dg_dict[i]['nb'] = 0
        for friend in fr_dict[i]:
            if fr_dict[i] == []:
                dg_dict[i]['nb'] = 0
            else:
                dg_dict[i]['nb'] += dg_dict[friend]['self'] / len(fr_dict[i])

如果节点i的度数的1.1倍小于邻居的平均度数，ta就存在友谊悖论。如果网络中所有节点超过55%存在友谊悖论（pdScale > 0.55），这一网络就存在友谊悖论。

    pdIndex = 0
    for i in range(0, n):
        if 1.1 * dg_dict[i]['self'] < dg_dict[i]['nb']:
            pdIndex +=1
    pdScale = pdIndex / n
    return [pdScale, pdScale > 0.55]

由于生成矩阵使用了随机数，随机过程的干扰可能导致单次生成的矩阵是一个特殊情况，对验证结论产生很大影响。因此，我们还需要一个计算友谊悖论期望值的函数pdE_calculator：输入一组特定的自变量（总节点数n、初始规则度数r、随机调整边占比p），运行矩阵生成函数一定次数loopCount，计算出现友谊悖论的节点比例pdScale的平均值，我们就得到该网络的友谊悖论期望值pdE。

def pdE_calculator(n, r, p, loopCount = 100):  # paradox expectation
    k = 0
    pdSum = 0
    while True:
        matr = neighbor_generator(n, r, p)
        pdScale = paradox(matr, n)[0]
        k += 1
        pdSum += pdScale
        if k == int(loopCount):
            print('given n = %i, r = %i, p = %.2f, the paradox scale has been successfully calculated' % (n, r, p), end='\n')
            break
        if n >= 500 or loopCount >= 300:
            if k == 1 :
                print('wait a few seconds for the computer to process...', end ='\r')
        if n >= 500 and k % int(5000 / n) == 0 and k > 10:
            print('%i out of %i rounds have been successfully runned........' % (k, loopCount), end='\r')
        if loopCount >= 300 and k % 100 == 0 and k > 300:
            print('%i out of %i rounds have been successfully runned........' % (k, loopCount), end='\r')
    pdE = pdSum / loopCount
    return pdE

除此之外，我还定义了一个输出结果的函数pdOutput，便于反复调用。输入一组自变量值parameterList和对应的友谊悖论验证结果pdEList，它能在控制台打印结果、在csv文件中写入结果、使用matplotlib绘图。

def pdOutput(parameter, parameterList, pdEList, f):
    print('Paradox scales of different %s values as following...' % parameter)
    f.write('%s,' % parameter)
    for i in parameterList:
        print(i, end='/t')
        f.write(str(i)+',')
    print()
    f.write('\nParadox Expectation,')
    for pd in pdEList:
        print('%.3f' % pd, end='\t')
        f.write('%.3f' % pd + ',')
    print()
    f.write('\n')
    xPoints = np.array(parameterList)
    yPoints = np.array(pdEList)
    plotNumDict = {'n': 1, 'r': 2, 'p': 3}
    plt.subplot(2, 2, plotNumDict[parameter])
    plt.plot(xPoints, yPoints, marker='*', color='orange')
    plt.xlabel('%s values' % parameter)
    plt.ylabel('average paradox scales')
    plt.show()

先准备好记录输出的csv文件。

resultDir = './output'
if os.path.isdir(resultDir):
    pass
else:
    os.makedirs(resultDir)
fname = './output/paradox.csv'
f = open(fname, 'w')

我们先来验证，不同的网络规模n对友谊悖论发生期望的影响。给定,r=5,p=0.25，n∈{50,100,150,200,300,500,600,800,1000}。对于每一个n值，我们调用pdMean函数，设定loopCount为100，计算发生友谊悖论的比例，结果存储在n_pdList里。

n_start = time.time()
print('*** Calculating paradox scales given different n values ***')
n_list = [50, 100, 150, 200, 300, 500, 600, 800, 1000]  # more than 7 groups to make the tendency more explicit
n_pdList = []
for n in n_list:
    n_pdE = pdE_calculator(n=n, r=5, p=0.25)
    n_pdList.append(n_pdE)
pdOutput('n', n_list, n_pdList, f)
n_end = time.time()
n_span = n_end - n_start
n_time = '%i min %i s' % (n_span // 60, n_span % 60)
print('Paradox scales calculating of different n values cost', n_time)
print()

输出结果如下：

*** Calculating paradox scales given different n values ***
given n = 50, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 150, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 200, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 300, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 500, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 600, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 800, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 1000, r = 5, p = 0.25, the paradox scale has been successfully calculated
Paradox scales of different n values as following...
50/t100/t150/t200/t300/t500/t600/t800/t1000/t
0.466	0.464	0.466	0.465	0.469	0.470	0.466	0.467	0.466	
Paradox scales calculating of different n values cost 5 min 56 s

同理，我们再来验证，初始规则度数不同时，友谊悖论发生期望有何不同。给定n=100,p=0.25，r∈{3,4,5,6,7,8,9,10}，设定loopCount为100。

print('*** Calculating paradox scales given different r values ***')
r_list = [3, 4, 5, 6, 7, 8, 9, 10]
r_pdList = []
for r in r_list:
    r_pdE = pdE_calculator(n=100, r=r, p=0.25)
    r_pdList.append(r_pdE)
pdOutput('r', r_list, r_pdList, f)
r_end = time.time()
r_span = r_end - n_end
r_time = '%i min %i s' % (r_span // 60, r_span % 60)
print('Paradox scales calculating of different r values cost', r_time)
print()

输出结果如下：

*** Calculating paradox scales given different r values ***
given n = 100, r = 3, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 4, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 6, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 7, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 8, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 9, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 10, p = 0.25, the paradox scale has been successfully calculated
Paradox scales of different r values as following...
3/t4/t5/t6/t7/t8/t9/t10/t
0.531	0.505	0.468	0.468	0.443	0.428	0.404	0.400	
Paradox scales calculating of different r values cost 0 min 14 s

同理，我们最后来验证，随机调整边占比p不同时，友谊悖论发生期望有何不同。给定n=100,r=5,p∈{0.15,0.25,0.35,0.45,0.55,0.65,0.75,0.85,0.95}。

注意到，p越大，随机数对矩阵生成的扰动就越大，我们需要增加循环数，亦即增加矩阵生成次数loopCount，保证友谊悖论节点比例的平均值能达致稳定。因此，我将循环数设为75/(1−p)。

print('*** Calculating paradox scales given different p values ***')
p_list = [0.15, 0.25, 0.35, 0.45, 0.55, 0.65, 0.75, 0.85, 0.95]
p_pdList = []
for p in p_list:
    p_loop = 75 / (1 - p)
    p_pdE = pdE_calculator(100, 5, p, p_loop)
    p_pdList.append(p_pdE)
pdOutput('p', p_list, p_pdList, f)
p_end = time.time()
p_span = p_end - r_end
p_time = '%i min %i s' % (p_span // 60, p_span % 60)
print('Paradox scales calculating of different p values cost', p_time)
print()

输出结果如下：

*** Calculating paradox scales given different p values ***
given n = 100, r = 5, p = 0.15, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.25, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.35, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.45, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.55, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.65, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.75, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.85, the paradox scale has been successfully calculated
given n = 100, r = 5, p = 0.95, the paradox scale has been successfully calculated
Paradox scales of different p values as following...
0.15/t0.25/t0.35/t0.45/t0.55/t0.65/t0.75/t0.85/t0.95/t
0.404	0.469	0.503	0.516	0.536	0.545	0.555	0.556	0.560	
Paradox scales calculating of different p values cost 0 min 59 s

2.2.3、结果分析

n	50	100	150	200	300	500	600	800	1000
Paradox Expectation	0.466	0.464	0.466	0.465	0.469	0.470	0.466	0.467	0.466
r	3	4	5	6	7	8	9	10
Paradox Expectation	0.531	0.505	0.468	0.468	0.443	0.428	0.404	0.400
p	0.15	0.25	0.35	0.45	0.55	0.65	0.75	0.85	0.95
Paradox Expectation	0.404	0.469	0.503	0.516	0.536	0.545	0.555	0.556	0.560

可以得出：

• 社会网络规模n对友谊悖论节点比例平均值的影响不明朗。
• 随着初始规则度数r变大，友谊悖论在全网络中的占比变小，友谊悖论更加弱。当r≥5时，友谊悖论在全网络中的比例低于0.5，不再是一个社会普遍的现象。
• 随着随机调整边占比p变大，友谊悖论在全网络中的占比变大，友谊悖论更加明显。当p≥0.35时，友谊悖论在全网络中的比例高于0.5，可以说这时友谊悖论是一个普遍现象。

进一步的验证，还需要优化模型、增加取值、增加pdE_calculator()循环数loopCount后，进行严谨的统计学分析。我们的初步结论是，当较r小、较p大时，友谊悖论在全网络中的比例高于0.5，可以说这时友谊悖论是一个普遍现象。

3、完整代码

附上可运行的完整代码，以供学习和交流！

聚集系数与嵌入性

import numpy as np


def arrayGen(filename):
    f = open(filename, 'r')
    r_list = f.readlines()  # 返回包含size行的列表, size 未指定则返回全部行
    f.close()
    A = []
    for line in r_list:
        if line == '\n':
            continue
        line = line.strip('\n')
        line = line.strip()  # 用于移除字符串头尾指定的字符（默认为空格或换行符）或字符序列，不能删除中间字符
        row_list = line.split()  # 通过指定分隔符对字符串进行切片
        for k in range(len(row_list)):
            row_list[k] = row_list[k].strip()
            row_list[k] = int(row_list[k])
        A.append(row_list)
    n = len(A[0])
    A = np.array(A)
    return A, n


filename = input('请输入邻接矩阵文件名：')
A, n = arrayGen(filename)

friend_dict = {}
for i in range(n):
    friend_dict[i] = []
    for j in range(n):
        if A[i][j] ==1:
            friend_dict[i].append(j)

# 计算聚集系数
ClCo_dict = {}  # ClCo表示clustering coefficient
for(person, friend_list) in friend_dict.items():  # 以列表返回视图对象，是一个可遍历的key/value 对
    TriCl = 0  # 直接朋友
    for x in range(len(friend_list)):
        friend1 = friend_list[x]
        for y in range(x+1, len(friend_list)):
            friend2 = friend_list[y]
            if A[friend1][friend2] == 1:
                TriCl += 1;
    PossTriCl = len(friend_list)*(len(friend_list)-1)/2  # 间接朋友
    if PossTriCl == 0:
        ClCo = 0
    else:
        ClCo = TriCl / PossTriCl  # 聚集系数
    ClCo_dict[person] = ClCo

print()
print('clustering coefficient as following:')
ClCo_list = list(ClCo_dict.items())
ClCo_list.sort(key=lambda x: x[1], reverse=True)
if len(ClCo_dict) < 10:
    for item in ClCo_list:
        print('node', str(item[0]) + ':', '%.2f' % item[1])
else:
    for item in ClCo_list[0:10]:
        print('node', str(item[0]) + ':', '%.2f' % item[1])

# 计算两个人共同的朋友数
embd_dict = {}
for(person, friend_list) in friend_dict.items():
    for friend_a in friend_list:
        if friend_a > person:
            embd_dict[person, friend_a] = 0
            for friend_b in friend_dict[friend_a]:
                if friend_b in friend_list:
                    embd_dict[person, friend_a] += 1

print()
print('embeddedness as following:')
embd_list = list(embd_dict.items())
embd_list.sort(key=lambda x: x[1], reverse=True)
if len(embd_list) < 10:
    for item in embd_list:
        print('edge', end=' ')
        print(item[0], end='')
        print(':', str(item[1]))
else:
    for item in embd_list[0:10]:
        print('edge', end=' ')
        print(item[0], end='')
        print(':', str(item[1]))

net3.txt

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 
0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 
0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 
0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 
0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 
0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 
0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 
0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 
0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 
0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 
1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 
0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 
0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 
0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 

友谊悖论

import numpy as np
import random
import os
import matplotlib.pyplot as plt
import time


def neighbor_generator(n, r, p):
    file = open('./output/generator_logs_{}', 'w')
    matr = np.zeros((n, n))

    # 初始规则边建立
    file.write('Generating an orderly matrix with homophilous links...\n')
    edge_list = []
    for i in range(0, n):
        j_list = [(i + x) % n for x in range(1, int(r/2 + 1 + r % 2))]
        for j in j_list:
            matr[i][j] = 1
            matr[j][i] = 1
            if j > i:
                edge_list.append([i, j])
            else:
                edge_list.append([j, i])
    file.write('Writing the list of node pairs which do not have an edge...\n')
    empt_list = []
    for i in range(0, n):
        if i % 100 == 0:
            file.write('%i nodes processed...' % i)
        for j in range(i+1, n):
            if matr[i][j] == 0:
                empt_list.append([i, j])
    file.write('For now, number of edges in Matrix A: %i\n' % np.sum(matr))
    if r % 2 != 0:
        for i in range(0, n-1, 2):
            matr[i][i+1] = 0
            matr[i+1][i] = 0
            edge_list.remove([i, i+1])
            edge_list.append([i, i+1])
    file.write('Number of edges in Matrix A: %i\n' % np.sum(matr))

    # 随机调整边
    file.write('Introducing randomness...\n')
    exc_edge = random.sample(edge_list, int(n * r * p / 2))  # 随机获取n*r*p/2个元素作为一个片断返回
    exc_empt = random.sample(empt_list, int(n * r * p / 2))
    for item in exc_edge:
        matr[item[0]][item[1]] = 0
        matr[item[1]][item[0]] = 0
    for item in exc_empt:
        matr[item[0]][item[1]] = 1
        matr[item[1]][item[0]] = 1

    np.savetxt('./neighbor_original_{}', matr, fmt='%i')
    return matr

# 友谊悖论
def paradox(matr, n):
    fr_dict = {}
    dg_dict = {}
    # generate fr_dict to store who's whose friend
    for i in range(0, n):
        fr_dict[i] = []
        for j in range(0, n):
            if matr[i][j] == 1:
                fr_dict[i].append(j)

    # generate the degrees of individuals
    for i in range(0, n):
        dg_dict[i] = {}
        dg_dict[i]['self'] = len(fr_dict[i])

    # generate the average degrees of neighbors
    for i in range(0, n):
        dg_dict[i]['nb'] = 0
        for friend in fr_dict[i]:
            if fr_dict[i] == []:
                dg_dict[i]['nb'] = 0
            else:
                dg_dict[i]['nb'] += dg_dict[friend]['self'] / len(fr_dict[i])
    pdIndex = 0
    for i in range(0, n):
        if 1.1 * dg_dict[i]['self'] < dg_dict[i]['nb']:
            pdIndex += 1
    pdScale = pdIndex / n
    return [pdScale, pdScale > 0.55]  # 这里？

# 期望计算
def pdE_calculator(n, r, p, loopCount = 100):  # paradox expectation
    k = 0
    pdSum = 0
    while True:
        matr = neighbor_generator(n, r, p)
        pdScale = paradox(matr, n)[0]
        k += 1
        pdSum += pdScale
        if k == int(loopCount):
            print('given n = %i, r = %i, p = %.2f, the paradox scale has been successfully calculated' % (n, r, p), end='\n')
            break
        if n >= 500 or loopCount >= 300:
            if k == 1:
                print('wait a few seconds for the computer to process...', end ='\r')
        if n >= 500 and k % int(5000 / n) == 0 and k > 10:
            print('%i out of %i rounds have been successfully runned........' % (k, loopCount), end='\r')
        if loopCount >= 300 and k % 100 == 0 and k > 300:
            print('%i out of %i rounds have been successfully runned........' % (k, loopCount), end='\r')
    pdE = pdSum / loopCount
    return pdE

# 输出
def pdOutput(parameter, parameterList, pdEList, f):
    print('Paradox scales of different %s values as following...' % parameter)
    f.write('%s,' % parameter)
    for i in parameterList:
        print(i, end='/t')
        f.write(str(i)+',')
    print()
    f.write('\nParadox Expectation,')
    for pd in pdEList:
        print('%.3f' % pd, end='\t')
        f.write('%.3f' % pd + ',')
    print()
    f.write('\n')
    xPoints = np.array(parameterList)
    yPoints = np.array(pdEList)
    plotNumDict = {'n': 1, 'r': 2, 'p': 3}
    plt.subplot(2, 2, plotNumDict[parameter])
    plt.plot(xPoints, yPoints, marker='o', color='orange')
    plt.xlabel('%s values' % parameter)
    plt.ylabel('average paradox scales')
    plt.show()

resultDir = './output'
if os.path.isdir(resultDir):
    pass
else:
    os.makedirs(resultDir)
fname = './output/paradox.csv'
f = open(fname, 'w')

# test

n_start = time.time()
print('*** Calculating paradox scales given different n values ***')
n_list = [50, 100, 150, 200, 300, 500, 600, 800, 1000]  # more than 7 groups to make the tendency more explicit
n_pdList = []
for n in n_list:
    n_pdE = pdE_calculator(n=n, r=5, p=0.25)
    n_pdList.append(n_pdE)
pdOutput('n', n_list, n_pdList, f)
n_end = time.time()
n_span = n_end - n_start
n_time = '%i min %i s' % (n_span // 60, n_span % 60)
print('Paradox scales calculating of different n values cost', n_time)
print()


print('*** Calculating paradox scales given different r values ***')
r_list = [3, 4, 5, 6, 7, 8, 9, 10]
r_pdList = []
for r in r_list:
    r_pdE = pdE_calculator(n=100, r=r, p=0.25)
    r_pdList.append(r_pdE)
pdOutput('r', r_list, r_pdList, f)
r_end = time.time()
r_span = r_end - n_end
r_time = '%i min %i s' % (r_span // 60, r_span % 60)
print('Paradox scales calculating of different r values cost', r_time)
print()

print('*** Calculating paradox scales given different p values ***')
p_list = [0.15, 0.25, 0.35, 0.45, 0.55, 0.65, 0.75, 0.85, 0.95]
p_pdList = []
for p in p_list:
    p_loop = 75 / (1 - p)
    p_pdE = pdE_calculator(100, 5, p, p_loop)
    p_pdList.append(p_pdE)
pdOutput('p', p_list, p_pdList, f)
p_end = time.time()
p_span = p_end - r_end
p_time = '%i min %i s' % (p_span // 60, p_span % 60)
print('Paradox scales calculating of different p values cost', p_time)
print()