import numpy as np import pandas as pd import sklearn import Code_bus_correlation # import adregionf import fun import xfun import mutationnew import test import TOPSIS import crossnew import traceback import statistics from variancediy import variancediy from TOPSIS import TOPSIS from variancediy2 import variancediy2 zhong = 8 for zhong in [zhong]: for chu in [zhong * 5]: pass df = pd.read_excel('links.xls') links = df.to_numpy() df = pd.read_excel('stationid.xls') stationid = df.to_numpy() df = pd.read_excel('stationidwan.xls') stationid = df.to_numpy() # for chuu in range(1, 17): # for dic in range(1): # tic() # chu = 10 # zhong = 2 # 给道路起点和终点标注序列，eg从1到500， # 因为一个路口可以是好几个道路的起点或终点，所以同一路口就会有同样的标记 node = np.concatenate((links[:, :2], links[:, 2:4]), axis=0) # np.concatenate 函数会将这两个子数组沿着轴 0 连接起来; # axis 是指在数组操作时沿着哪个轴进行操作。当axis=0时，表示在第一个维度上进行拼接操作。这里就是纵轴 noi = 1 node = np.hstack((node, np.zeros((len(node), 1)))) print(node.shape[0]) for i in range(node.shape[0]): # node.shape[0] 是指 node 数组的第一维大小，即 node 数组的行数 print(i) # node[:i, 0] 表示从 node 数组的第一行到第 i-1 行的所有行的第一列构成的数组 # np.where() 函数返回一个包含下标的元组，后面的[0]就代表返回第一个元素的下标 a = np.where(node[:i, 0] == node[i, 0])[0] b = np.where(node[:i, 1] == node[i, 1])[0] c = np.intersect1d(a, b) # intersect1d 返回两个数组的交集 if c.size > 0: x = c.shape[0] y = 1 else: x, y = 0, 1 # 在 node 数组的最后添加一列全为0的列，并将添加后的新数组重新赋值给 node if x > 0 and y > 0: node[i, 2] = node[min(c), 2] # 如果c是矩阵，则min(A)是包含每一列的最小值的行向量 else: node[i, 2] = noi noi += 1 node = np.concatenate((node[:int(len(node)/2), 2].reshape(-1, 1), node[int(len(node)/2):, 2].reshape(-1, 1)), axis=1) np.savetxt('node.txt', node) # 这里的links多加了一行才能yanlinks，但这样yanlinks就不对了 links = np.hstack((links, np.zeros((len(links), 1)))) links = np.hstack((links, np.zeros((len(links), 1)))) links = np.hstack((links, np.zeros((len(links), 1)))) yanlinks = np.concatenate((node, links[:, [5,6,7,4,0,1,2,3]], np.zeros((len(links), 4))), axis=1) yanlinks[:,4] = np.arange(1, len(yanlinks)+1) road = np.arange(1, node.shape[0] + 1) adjacency = np.zeros((len(road), len(road))) #初始化分区 print(range(len(road))) for i in range(len(road)): temp1 = np.where(node[:, 0] == node[i, 0])[0] # 找出第一列每个数字在第一列出现的位置 temp2 = np.where(node[:, 1] == node[i, 0])[0] # 找出第一列每个数字在第二列出现的位置 temp3 = np.where(node[:, 0] == node[i, 1])[0] # 找出第二列每个数字在第一列出现的位置 temp4 = np.where(node[:, 1] == node[i, 1])[0] # 找出第二列每个数字在第二列出现的位置 temp = np.unique(np.intersect1d(np.arange(i + 1, node.shape[0]), np.concatenate((temp1, temp2, temp3, temp4)))) if len(temp) > 0: adjacency[i, temp] = 1 adjacency[temp, i] = 1 from sklearn.cluster import KMeans N = chu # 设置聚类数目 # 利用 K-Means 算法对 yanlinks 矩阵的第 7 列和第 8 列（即经度和纬度）进行聚类， # 将样本分成 N 类，idx是一个N x 2的矩阵，其中N是聚类数目。 # idx的每一行就是一个聚类中心，其中第一列是该中心的经度，第二列是该中心的纬度。 # 在计算每个点到聚类中心的距离时，就需要用到idx的值。 cur, idx = KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).labels_, KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).cluster_centers_ # 计算每个点到聚类中心的距离 dis = 111000 * np.sqrt((yanlinks[:, 6] - idx[:, 0].reshape(N, 1)) ** 2 + (yanlinks[:, 7] - idx[:, 1].reshape(N, 1)) ** 2) # 找到每个点最近的聚类中心，mm是最小值，nn是最小值在向量的索引 mm, nn = np.min(dis, axis=1, keepdims=True), np.argmin(dis, axis=1) data = links[:, 4] # links第五行是路的长度 if data.size > 0: m = data.shape[0] n = 1 else: m, n = 0, 1 pattern = np.zeros((m, n)) # zeros(m,n+1)返回由零组成的m×(n+1)数组 pattern[:, 0] = data # 前n列为data中的数据 pattern = np.hstack((pattern, np.zeros((len(pattern), 1)))) pattern[:, 1] = -1 center = np.zeros((N, n)) # 初始化聚类中心 pattern[:, :n] = data.reshape(-1, n) # 初始化聚类中心 for x in range(0,N): center = np.hstack((center, np.zeros((len(center), 1)))) center[x, 1] = nn[x] center[x, 0] = data[int(center[x, 1])] pattern[int(center[x, 1]), 1] = x # 初始化距离和计数 distance = np.zeros(N) num = np.zeros(N) # 初始化新的聚类中心 new_center = np.zeros((N, n)) mb = 10 while mb > 1: print(mb) for x in range(0, N): # x表示当前聚类的编号 try: yisou = adjacency[pattern[:, 1] == x, :] bound = np.where(np.sum(yisou, axis=0) > 0)[0] yisou = np.where(pattern[:, 1] > -1)[0] bound = np.setdiff1d(bound, yisou) # bound 是一个向量，表示与聚类 x 相关的未被分配到聚类中的道路的编号。 yisou = np.where(pattern[:, 1] == x )[0] # 这里的yisou表示已经被分配到的道路编号 # yisou = np.array([1778], dtype=int) # bound = np.array([1722, 1776, 1779, 1782]) bus = [] # for y=1:length(yisou) for y in range(len(yisou)): yicifangwen = yisou[y] yicifangwendesuoying = np.where(stationid[:, 5] == yicifangwen)[0] for dengyuyisourow in yicifangwendesuoying: bus.append(stationid[dengyuyisourow, 6]) bouvar = np.zeros((len(bound), 2)) for adad in range(len(bound)): if len(np.concatenate([bus, stationid[stationid[:, 5] == bound[adad], 6]])) > 1 and variancediy((np.concatenate([bus, stationid[stationid[:, 5] == bound[adad], 6]])).tolist()) > 0: # np.var(np.concatenate((bus, stationid[stationid[:, 6] == bound[adad], 6]有可能为无穷大，当这里的两个变量为空集时 # bus 和stationid[stationid[:, 6] == bound[adad]不能直接和零比较，因为他们不是数 # pattern[yisou, 0]和pattern[bound[adad], 0]不是一个数据类型 yaozhuanchenfloat = (pattern[yisou-1, 0]).tolist() yaozhuanchenarray = (pattern[bound[adad]-1, 0]).tolist() bouvar[adad, 0] = variancediy(yaozhuanchenfloat,yaozhuanchenarray)* variancediy((np.concatenate((bus, stationid[stationid[:, 5] == bound[adad], 6]))).tolist()) else: if variancediy(bus)> 0: # if var(bus)>0%已分配道路的速度方差 yaozhuanchenfloat1 = (pattern[yisou-1, 0]).tolist() # 这里没问题！！！！ yaozhuanchenarray1 = (pattern[bound[adad]-1, 0]).tolist() bouvar[adad, 0] = variancediy(yaozhuanchenfloat1, yaozhuanchenarray1) + variancediy(bus) else: yaozhuanchenfloat2 = (pattern[yisou-1, 0]).tolist() yaozhuanchenarray2 = (pattern[bound[adad]-1, 0]).tolist() bouvar[adad, 0] = variancediy(yaozhuanchenfloat2, yaozhuanchenarray2) bouvar[adad, 1] = 111000 * np.sqrt(np.sum((yanlinks[yanlinks[:, 4] == bound[adad], 6:8] - idx[x-1 , :]) ** 2)) if bouvar.shape[0] > 1: m, n = TOPSIS(bouvar) # bestxuhao最优方案的序号，bestgoal最优得分 else: n = 1 pattern[bound[n-1], 1] = x except: continue mb = np.sum(pattern[:, 1] == -1) np.savetxt('pattern.txt', pattern) yanlinks[:, 10] = pattern[:, 1] yanlinks = yanlinks[yanlinks[:, 10] !=-1, :] road = np.unique(np.concatenate((yanlinks[:, 1], yanlinks[:, 0]), axis=0)) adjacency = np.zeros((len(road), len(road))) adregion = np.zeros((int(np.max(yanlinks[:, 4])),int(np.max(yanlinks[:, 4])))) print(range(len(yanlinks[:, 0]))) for i in range(len(yanlinks[:, 0])): temp1 = np.where(node[:, 0] == node[i, 0])[0] temp2 = np.where(node[:, 1] == node[i, 0])[0] temp3 = np.where(node[:, 0] == node[i, 1])[0] temp4 = np.where(node[:, 1] == node[i, 1])[0] temp = np.unique(np.intersect1d(np.arange(i + 1, node.shape[0]), np.concatenate((temp1, temp2, temp3, temp4)))) if len(temp) > 0: adregion[i, temp] = 1 adregion[temp, i] = 1 np.save('adregion.npy', adregion) for i in range(len(yanlinks[:, 1])): # print(adregion[:, int(yanlinks[i, 4])]) # print(int(yanlinks[i, 10])) adregion[:, int(yanlinks[i, 4])-1] = adregion[:, int(yanlinks[i, 4])-1] * int(yanlinks[i, 10]) # 这里前面都没有问题！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！我说没有就没有！！！！！！！！！！！！！ adr = np.zeros((chu, chu)) # 计算adregion中的每个元素出现的频率 for i in range(len(adregion[:, 1])): a = adregion[i, :] a = np.unique(a) a = a[a != 0] print(a) if a.size > 0: x =1 y =a.shape[0] else: x, y = 0, 1 if y > 1: for j in range(len(a)): for u in range(len(a)): if j != u: adr[int(a[j])-1, int(a[u])-1] += 1 adr[int(a[u]), int(a[j])] += 1 # 计算后存到dadr里 dadr = adr print(np.max(adr,axis=0)) print(np.min(np.max(adr,axis=0)) - 2) # 假设 adr 和 dadr 分别是两个 NumPy 数组 is_symmetric = np.allclose(adr, adr.T) print(is_symmetric) min_value = np.min(np.max(adr,axis=0)) - 2 adr[adr < min_value] = 0 adr[adr > 1] = 1 dadr[dadr > 1] = 1 np.savetxt('adr.txt', adr) np.savetxt('dadr.txt', dadr) np.savetxt('yanlinks.txt', yanlinks)