189 lines
7.1 KiB
Python
189 lines
7.1 KiB
Python
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import pandas as pd
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import matplotlib.pyplot as plt
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from sklearn.cluster import KMeans
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from sklearn.preprocessing import StandardScaler
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import numpy as np
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Initial_partitions=60
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# 加载数据
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df = pd.read_csv('links_processed.csv', usecols=[0, 1, 2, 3, 4])
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df.columns = ['start_lat', 'start_long', 'end_lat', 'end_long', 'speed']
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# 计算路段的中心点
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df['center_lat'] = ((df['start_lat'] + df['end_lat']) / 2).round(7)
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df['center_long'] = ((df['start_long'] + df['end_long']) / 2).round(7)
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# 提取用于聚类的特征
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features = df[['center_lat', 'center_long']]
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# 数据标准化
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scaler = StandardScaler()
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scaled_features = scaler.fit_transform(features)
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# 运行KMeans算法
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kmeans = KMeans(n_clusters=Initial_partitions, n_init=10) # 假设我们想要划分为40个区域
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kmeans.fit(scaled_features)
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# 将聚类结果添加到原始数据中
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df['cluster'] = kmeans.labels_
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df['cluster'] = df['cluster'] + 1
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df=df.to_numpy()
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links = pd.read_csv('links_processed.csv')
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links = links.to_numpy()
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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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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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# 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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yanlinks[:, 3] = links[:, 9]
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yanlinks[:, 10] = df[:, 7]
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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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# adregion矩阵表示路段之间的邻接关系
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np.save('adregion.npy', adregion)
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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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# np.save('subregion_adj.npy', subregion_adj)
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# np.save('yanlinks.npy', yanlinks)
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