262 lines
10 KiB
Python
262 lines
10 KiB
Python
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import numpy as np
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import pandas as pd
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import sklearn
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import Code_bus_correlation
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# import adregionf
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import fun
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import xfun
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import mutationnew
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import test
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import TOPSIS
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import crossnew
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import traceback
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import statistics
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from variancediy import variancediy
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from TOPSIS import TOPSIS
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from variancediy2 import variancediy2
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zhong = 8
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for zhong in [zhong]:
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for chu in [zhong * 5]:
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pass
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df = pd.read_excel('links.xls')
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links = df.to_numpy()
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df = pd.read_excel('stationid.xls')
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stationid = df.to_numpy()
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df = pd.read_excel('stationidwan.xls')
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stationid = df.to_numpy()
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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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noi = 1
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node = np.hstack((node, np.zeros((len(node), 1))))
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print(node.shape[0])
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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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print(i)
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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)), axis=1)
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np.savetxt('node.txt', 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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print(range(len(road)))
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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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from sklearn.cluster import KMeans
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N = chu # 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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cur, idx = KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).labels_, KMeans(n_clusters=N).fit(yanlinks[:, [6, 7]]).cluster_centers_
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# 计算每个点到聚类中心的距离
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dis = 111000 * np.sqrt((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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# 初始化聚类中心
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for x in range(0,N):
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center = np.hstack((center, np.zeros((len(center), 1))))
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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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mb = 10
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while mb > 1:
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print(mb)
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for x in range(0, N): # x表示当前聚类的编号
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try:
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yisou = adjacency[pattern[:, 1] == x, :]
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bound = np.where(np.sum(yisou, axis=0) > 0)[0]
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yisou = np.where(pattern[:, 1] > -1)[0]
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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。
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yisou = 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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# yisou = np.array([1778], dtype=int)
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# bound = np.array([1722, 1776, 1779, 1782])
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bus = []
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# for y=1:length(yisou)
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for y in range(len(yisou)):
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yicifangwen = yisou[y]
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yicifangwendesuoying = np.where(stationid[:, 5] == yicifangwen)[0]
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for dengyuyisourow in yicifangwendesuoying:
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bus.append(stationid[dengyuyisourow, 6])
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bouvar = np.zeros((len(bound), 2))
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for adad in range(len(bound)):
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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:
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# 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
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# bus 和stationid[stationid[:, 6] == bound[adad]不能直接和零比较，因为他们不是数
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# pattern[yisou, 0]和pattern[bound[adad], 0]不是一个数据类型
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yaozhuanchenfloat = (pattern[yisou-1, 0]).tolist()
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yaozhuanchenarray = (pattern[bound[adad]-1, 0]).tolist()
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bouvar[adad, 0] = variancediy(yaozhuanchenfloat,yaozhuanchenarray)* variancediy((np.concatenate((bus, stationid[stationid[:, 5] == bound[adad], 6]))).tolist())
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else:
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if variancediy(bus)> 0: # if var(bus)>0%已分配道路的速度方差
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yaozhuanchenfloat1 = (pattern[yisou-1, 0]).tolist() # 这里没问题！！！！
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yaozhuanchenarray1 = (pattern[bound[adad]-1, 0]).tolist()
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bouvar[adad, 0] = variancediy(yaozhuanchenfloat1, yaozhuanchenarray1) + variancediy(bus)
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else:
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yaozhuanchenfloat2 = (pattern[yisou-1, 0]).tolist()
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yaozhuanchenarray2 = (pattern[bound[adad]-1, 0]).tolist()
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bouvar[adad, 0] = variancediy(yaozhuanchenfloat2, yaozhuanchenarray2)
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bouvar[adad, 1] = 111000 * np.sqrt(np.sum((yanlinks[yanlinks[:, 4] == bound[adad], 6:8] - idx[x-1 , :]) ** 2))
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if bouvar.shape[0] > 1:
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m, n = TOPSIS(bouvar) # bestxuhao最优方案的序号，bestgoal最优得分
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else:
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n = 1
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pattern[bound[n-1], 1] = x
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except:
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continue
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mb = np.sum(pattern[:, 1] == -1)
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np.savetxt('pattern.txt', pattern)
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yanlinks[:, 10] = pattern[:, 1]
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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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print(range(len(yanlinks[:, 0])))
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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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np.save('adregion.npy', adregion)
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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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# 这里前面都没有问题！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！我说没有就没有！！！！！！！！！！！！！
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adr = np.zeros((chu, chu))
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# 计算adregion中的每个元素出现的频率
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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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print(a)
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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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adr[int(a[j])-1, int(a[u])-1] += 1
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adr[int(a[u]), int(a[j])] += 1
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# 计算后存到dadr里
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dadr = adr
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print(np.max(adr,axis=0))
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print(np.min(np.max(adr,axis=0)) - 2)
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# 假设 adr 和 dadr 分别是两个 NumPy 数组
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is_symmetric = np.allclose(adr, adr.T)
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print(is_symmetric)
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min_value = np.min(np.max(adr,axis=0)) - 2
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adr[adr < min_value] = 0
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adr[adr > 1] = 1
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dadr[dadr > 1] = 1
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np.savetxt('adr.txt', adr)
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np.savetxt('dadr.txt', dadr)
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np.savetxt('yanlinks.txt', yanlinks)
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