Allenfenqu/srg_kmeans.py

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import numpy as np
import pandas as pd
from TOPSIS import TOPSIS
import random
links = pd.read_csv('links_processed.csv')
links = links.to_numpy()
Initial_partitions=60
# 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
# 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
noi = 1
node = np.hstack((node, np.zeros((len(node), 1))))
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
# 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.save('node.npy', 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
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
row_sums = np.sum(adjacency, axis=1)
# 找到全零行的索引
zero_row_indices = np.where(row_sums == 0)[0]
from sklearn.cluster import KMeans
N = Initial_partitions # 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。
Cluster_Label, idx = KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).labels_, KMeans(n_clusters=N).fit(
yanlinks[:, [6, 7]]).cluster_centers_
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)
center = np.hstack((center, np.zeros((len(center), 1))))
# 初始化聚类中心
for x in range(0, N):
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))
unassigned_links = 2
while unassigned_links > 0:
print(unassigned_links)
for x in range(0, Initial_partitions):
try:
selected_links = adjacency[pattern[:, 1] == x, :]
unassigned_roads = np.where(np.sum(selected_links, axis=0) > 0)[0]
selected_links = np.where(pattern[:, 1] > -1)[0]
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。
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
bus = []
road_evaluation = np.zeros((len(unassigned_roads), 2))
for unassigned_road_index in range(len(unassigned_roads)):
selected_links_lengths_float = (pattern[selected_links, 0]).tolist()
unassigned_road_length_array = (pattern[unassigned_roads[unassigned_road_index], 0])
unassigned_road_length_array = [unassigned_road_length_array]
abrr = selected_links_lengths_float + unassigned_road_length_array
road_evaluation[unassigned_road_index, 0] = np.var(abrr, ddof=1)
aas = yanlinks[yanlinks[:, 4] == unassigned_roads[unassigned_road_index] + 1, 6:8]
road_evaluation[unassigned_road_index, 1] = 111000 * np.sqrt(np.sum(
(yanlinks[yanlinks[:, 4] == unassigned_roads[unassigned_road_index] + 1, 6:8] - idx[x, :]) ** 2))
if road_evaluation.shape[0] > 1:
m, n = TOPSIS(road_evaluation) # bestxuhao最优方案的序号bestgoal最优得分
else:
n = 0
# pattern[unassigned_roads[n - 1], 1] = x
pattern[unassigned_roads[n], 1] = x
except:
continue
unassigned_links = np.sum(pattern[:, 1] == -1)
# 因为我的pattern是从0到39的编号所以要变成1到40
pattern[:, 1] = pattern[:, 1] + 1
yanlinks[:, 3] = links[:, 9]
yanlinks[:, 10] = pattern[:, 1]
data_path = r''
df2 = pd.read_csv(data_path + 'links_processed.csv')
zero_rows = yanlinks[:, 10] == 0
# 获取已删除行的索引
deleted_rows_indices = np.where(zero_rows)[0]
# 从 links 中删除 deleted_rows_indices 中指定的行
df2 = df2.drop(deleted_rows_indices, errors='ignore')
df2.to_csv(data_path + 'links_test1.csv', index=False)
yanlinks = yanlinks[yanlinks[:, 10] != 0]
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]))))
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
# 给adregion矩阵乘上权重道路的分组编号
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])
subregion_adj = np.zeros((Initial_partitions, Initial_partitions))
# 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)
for i in range(len(adregion[:, 1])):
a = adregion[i, :]
a = np.unique(a)
a = a[a != 0]
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:
# 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
subregion_adj[int(a[j]) - 1, int(a[u]) - 1] += 1
subregion_adj[int(a[u]) - 1, int(a[j]) - 1] += 1
# 计算后存到directed_adjacency_matrix里
directed_adjacency_matrix = subregion_adj.copy()
# 对于子区域相关程度处于弱相关的邻接关系进行忽略
min_value = np.min(np.max(subregion_adj, axis=0)) - 2
subregion_adj[subregion_adj < min_value] = 0
subregion_adj[subregion_adj > 1] = 1
directed_adjacency_matrix[directed_adjacency_matrix > 1] = 1
unique_values, unique_indices = np.unique(yanlinks[:, 10], return_index=True)
Asb = 0 # 计算平均相似性
for i in unique_values:
wu = np.where(subregion_adj[int(i) - 1, :] == 1) # wu是元组
smrjj_divide_smrjj_ = 0
# 0726
wu_1 = wu[0]
for j in wu_1:
selected_values_list = [yanlinks[yanlinks[:, 10] == j + 1][:, 5]]
# 主区域邻接的一个区域速度均值与方差
selected_values = np.array(selected_values_list)
average = np.mean(selected_values)
variance = np.var(selected_values)
# 计算主区域的速度均值与方差
selected_values1 = yanlinks[yanlinks[:, 10] == i][:, 5]
average1 = np.mean(selected_values1)
variance1 = np.var(selected_values1)
smrjj = 2 * variance1 # jj情况下的smrjj
smrjj_ = variance + variance1 + (average - average1) ** 2
smrjj_divide_smrjj_one = smrjj / smrjj_
smrjj_divide_smrjj_ += smrjj_divide_smrjj_one
num_elements = len(wu[0]) # 计算分母NE
Asb_one = smrjj_divide_smrjj_ / num_elements
Asb += Asb_one
Asb=Asb/Initial_partitions
print('Asb=', Asb)
Tvb = 0
for i in unique_values:
selected_values = yanlinks[yanlinks[:, 10] == i][:, 5]
variance = np.var(selected_values)
Tvb += variance
print('Tvb=', Tvb)