265 lines
10 KiB
Python
265 lines
10 KiB
Python
import numpy as np
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import pandas as pd
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from TOPSIS import TOPSIS
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import random
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links = pd.read_csv('links_processed.csv')
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links = links.to_numpy()
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Initial_partitions=60
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# for chuu in range(1, 17):
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# for dic in range(1):
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# tic()
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# chu = 10
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# zhong = 2
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# 给道路起点和终点标注序列,eg从1到500,
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# 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
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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;
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# 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
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# 这里是给道路起点和终点标注序列,也就是路口表注序列,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
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noi = 1
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node = np.hstack((node, np.zeros((len(node), 1))))
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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
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# 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
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# 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
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a = np.where(node[:i, 0] == node[i, 0])[0]
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b = np.where(node[:i, 1] == node[i, 1])[0]
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c = np.intersect1d(a, b) # intersect1d 返回两个数组的交集
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if c.size > 0:
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x = c.shape[0]
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y = 1
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else:
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x, y = 0, 1
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# 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
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if x > 0 and y > 0:
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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
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else:
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node[i, 2] = noi
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noi += 1
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node = np.concatenate((node[:int(len(node) / 2), 2].reshape(-1, 1), node[int(len(node) / 2):, 2].reshape(-1, 1)),
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axis=1)
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np.save('node.npy', node)
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# 这里的links多加了一行才能yanlinks,但这样yanlinks就不对了
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links = np.hstack((links, np.zeros((len(links), 1))))
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links = np.hstack((links, np.zeros((len(links), 1))))
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links = np.hstack((links, np.zeros((len(links), 1))))
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yanlinks = np.concatenate((node, links[:, [5, 6, 7, 4, 0, 1, 2, 3]], np.zeros((len(links), 4))), axis=1)
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yanlinks[:, 4] = np.arange(1, len(yanlinks) + 1)
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road = np.arange(1, node.shape[0] + 1)
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adjacency = np.zeros((len(road), len(road)))
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# 初始化分区
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for i in range(len(road)):
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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
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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
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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
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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
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temp = np.unique(np.intersect1d(np.arange(i + 1, node.shape[0]), np.concatenate((temp1, temp2, temp3, temp4))))
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if len(temp) > 0:
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adjacency[i, temp] = 1
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adjacency[temp, i] = 1
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row_sums = np.sum(adjacency, axis=1)
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# 找到全零行的索引
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zero_row_indices = np.where(row_sums == 0)[0]
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from sklearn.cluster import KMeans
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N = Initial_partitions # 设置聚类数目
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# 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,
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# 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。
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# 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。
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# 在计算每个点到聚类中心的距离时,就需要用到idx的值。
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Cluster_Label, idx = KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).labels_, KMeans(n_clusters=N).fit(
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yanlinks[:, [6, 7]]).cluster_centers_
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dis = 111000 * np.sqrt(
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(yanlinks[:, 6] - idx[:, 0].reshape(N, 1)) ** 2 + (yanlinks[:, 7] - idx[:, 1].reshape(N, 1)) ** 2)
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# 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
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mm, nn = np.min(dis, axis=1, keepdims=True), np.argmin(dis, axis=1)
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data = links[:, 4] # links第五行是路的长度
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if data.size > 0:
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m = data.shape[0]
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n = 1
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else:
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m, n = 0, 1
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pattern = np.zeros((m, n)) # zeros(m,n+1)返回由零组成的m×(n+1)数组
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pattern[:, 0] = data # 前n列为data中的数据
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pattern = np.hstack((pattern, np.zeros((len(pattern), 1))))
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pattern[:, 1] = -1
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center = np.zeros((N, n)) # 初始化聚类中心
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pattern[:, :n] = data.reshape(-1, n)
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center = np.hstack((center, np.zeros((len(center), 1))))
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# 初始化聚类中心
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for x in range(0, N):
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center[x, 1] = nn[x]
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center[x, 0] = data[int(center[x, 1])]
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pattern[int(center[x, 1]), 1] = x
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# 初始化距离和计数
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distance = np.zeros(N)
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num = np.zeros(N)
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# 初始化新的聚类中心
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new_center = np.zeros((N, n))
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unassigned_links = 2
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while unassigned_links > 0:
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print(unassigned_links)
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for x in range(0, Initial_partitions):
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try:
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selected_links = adjacency[pattern[:, 1] == x, :]
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unassigned_roads = np.where(np.sum(selected_links, axis=0) > 0)[0]
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selected_links = np.where(pattern[:, 1] > -1)[0]
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unassigned_roads = np.setdiff1d(unassigned_roads, selected_links) # 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。
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selected_links = np.where(pattern[:, 1] == x)[0] # 这里的yisou表示已经被分配到的道路编号
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bus = []
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road_evaluation = np.zeros((len(unassigned_roads), 2))
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for unassigned_road_index in range(len(unassigned_roads)):
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selected_links_lengths_float = (pattern[selected_links, 0]).tolist()
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unassigned_road_length_array = (pattern[unassigned_roads[unassigned_road_index], 0])
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unassigned_road_length_array = [unassigned_road_length_array]
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abrr = selected_links_lengths_float + unassigned_road_length_array
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road_evaluation[unassigned_road_index, 0] = np.var(abrr, ddof=1)
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aas = yanlinks[yanlinks[:, 4] == unassigned_roads[unassigned_road_index] + 1, 6:8]
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road_evaluation[unassigned_road_index, 1] = 111000 * np.sqrt(np.sum(
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(yanlinks[yanlinks[:, 4] == unassigned_roads[unassigned_road_index] + 1, 6:8] - idx[x, :]) ** 2))
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if road_evaluation.shape[0] > 1:
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m, n = TOPSIS(road_evaluation) # bestxuhao最优方案的序号,bestgoal最优得分
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else:
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n = 0
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# pattern[unassigned_roads[n - 1], 1] = x
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pattern[unassigned_roads[n], 1] = x
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except:
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continue
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unassigned_links = np.sum(pattern[:, 1] == -1)
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# 因为我的pattern是从0到39的编号,所以要变成1到40
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pattern[:, 1] = pattern[:, 1] + 1
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yanlinks[:, 3] = links[:, 9]
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yanlinks[:, 10] = pattern[:, 1]
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data_path = r''
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df2 = pd.read_csv(data_path + 'links_processed.csv')
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zero_rows = yanlinks[:, 10] == 0
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# 获取已删除行的索引
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deleted_rows_indices = np.where(zero_rows)[0]
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# 从 links 中删除 deleted_rows_indices 中指定的行
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df2 = df2.drop(deleted_rows_indices, errors='ignore')
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df2.to_csv(data_path + 'links_test1.csv', index=False)
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yanlinks = yanlinks[yanlinks[:, 10] != 0]
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yanlinks = yanlinks[yanlinks[:, 10] != -1, :]
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road = np.unique(np.concatenate((yanlinks[:, 1], yanlinks[:, 0]), axis=0))
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adjacency = np.zeros((len(road), len(road)))
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adregion = np.zeros((int(np.max(yanlinks[:, 4])), int(np.max(yanlinks[:, 4]))))
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for i in range(len(yanlinks[:, 0])):
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temp1 = np.where(node[:, 0] == node[i, 0])[0]
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temp2 = np.where(node[:, 1] == node[i, 0])[0]
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temp3 = np.where(node[:, 0] == node[i, 1])[0]
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temp4 = np.where(node[:, 1] == node[i, 1])[0]
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temp = np.unique(np.intersect1d(np.arange(i + 1, node.shape[0]), np.concatenate((temp1, temp2, temp3, temp4))))
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if len(temp) > 0:
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adregion[i, temp] = 1
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adregion[temp, i] = 1
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# 给adregion矩阵乘上权重(道路的分组编号)
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for i in range(len(yanlinks[:, 1])):
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# print(adregion[:, int(yanlinks[i, 4])])
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# print(int(yanlinks[i, 10]))
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adregion[:, int(yanlinks[i, 4]) - 1] = adregion[:, int(yanlinks[i, 4]) - 1] * int(yanlinks[i, 10])
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subregion_adj = np.zeros((Initial_partitions, Initial_partitions))
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# 计算adregion中的每个元素出现的频率(判断是强相关还是弱相关)
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for i in range(len(adregion[:, 1])):
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a = adregion[i, :]
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a = np.unique(a)
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a = a[a != 0]
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if a.size > 0:
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x = 1
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y = a.shape[0]
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else:
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x, y = 0, 1
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if y > 1:
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for j in range(len(a)):
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for u in range(len(a)):
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if j != u:
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# subregion_adj表示子区域的邻接关系,其中数值的大小表示区域之间的相关程度
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subregion_adj[int(a[j]) - 1, int(a[u]) - 1] += 1
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subregion_adj[int(a[u]) - 1, int(a[j]) - 1] += 1
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# 计算后存到directed_adjacency_matrix里
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directed_adjacency_matrix = subregion_adj.copy()
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# 对于子区域相关程度处于弱相关的邻接关系进行忽略
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min_value = np.min(np.max(subregion_adj, axis=0)) - 2
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subregion_adj[subregion_adj < min_value] = 0
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subregion_adj[subregion_adj > 1] = 1
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directed_adjacency_matrix[directed_adjacency_matrix > 1] = 1
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unique_values, unique_indices = np.unique(yanlinks[:, 10], return_index=True)
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Asb = 0 # 计算平均相似性
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for i in unique_values:
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wu = np.where(subregion_adj[int(i) - 1, :] == 1) # wu是元组
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smrjj_divide_smrjj_ = 0
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# 0726
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wu_1 = wu[0]
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for j in wu_1:
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selected_values_list = [yanlinks[yanlinks[:, 10] == j + 1][:, 5]]
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# 主区域邻接的一个区域速度均值与方差
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selected_values = np.array(selected_values_list)
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average = np.mean(selected_values)
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variance = np.var(selected_values)
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# 计算主区域的速度均值与方差
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selected_values1 = yanlinks[yanlinks[:, 10] == i][:, 5]
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average1 = np.mean(selected_values1)
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variance1 = np.var(selected_values1)
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smrjj = 2 * variance1 # jj情况下的smrjj
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smrjj_ = variance + variance1 + (average - average1) ** 2
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smrjj_divide_smrjj_one = smrjj / smrjj_
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smrjj_divide_smrjj_ += smrjj_divide_smrjj_one
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num_elements = len(wu[0]) # 计算分母NE
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Asb_one = smrjj_divide_smrjj_ / num_elements
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Asb += Asb_one
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Asb=Asb/Initial_partitions
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print('Asb=', Asb)
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Tvb = 0
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for i in unique_values:
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selected_values = yanlinks[yanlinks[:, 10] == i][:, 5]
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variance = np.var(selected_values)
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Tvb += variance
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print('Tvb=', Tvb)
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