1

我有以下代码用于从矩阵返回数组，但它不起作用。我想知道你是否可以帮我纠正它。谢谢一堆。

import numpy as np

class city:
    def  __init__(self,A,route):
        self.A=A
        self.route=route

    def  distance(self):
        A = np.array([[ 0,  10,    20,  30],[10,   0,    25,  20],[20,  25,     0,  15],[30,  20,    15,   0]])
        return A

    def route(self,A):
        route = random.sample(A, len(A[:,0]))
        return route

ob=city(route)
print(ob.route)

预期输出：

[(0,1),(1,2),(2,3)]
4

2 回答 2

0

import numpy as np import random class city: def init (self): self.distance()

def  distance(self):
    self.A = np.array([[ 0,  10,    20,  30],[10,   0,    25,  20],[20,  25,     0,  15],[30,  20,    15,   0]])
    self.B = [[random.randint(0,1) for j in range(self.A.shape[0])] for i in range(self.A.shape[1])]

def route(self):
    n=self.B.shape[0]
    nodes=list(range(n))
    route = [nodes.pop(random.sample(range(n-i),1)[0]) for i in range(n)]
    return [(route[i],route[i+1]) for i in range(n-1)]

ob=city() 打印(ob.route())

于 2018-11-28T19:05:10.623 回答
0

您可以选择接下来要访问的随机节点，并从要访问的节点列表中删除选择的节点（nodes）。

最后根据当前和下一个节点对值创建单元地址。

它确保每个节点只被选择一次（每行和每列一个）并且不会返回到同一个节点（a_ii）

import numpy as np
import random 

class city:
    def  __init__(self):
        self.distance()

    def  distance(self):
        self.A = np.array([[ 0,  10,    20,  30],[10,   0,    25,  20],[20,  25,     0,  15],[30,  20,    15,   0]])
        self.B = np.array([[random.randint(0,1) for j in range(self.A.shape[0])] for i in range(self.A.shape[1])])

    def route(self):
        n=self.B.shape[0]
        nodes=list(range(n))
        route = [nodes.pop(random.sample(range(n-i),1)[0]) for i in range(n)]
        return [(route[i],route[i+1]) for i in range(n-1)]

ob=city() 
print(ob.route())

输出：

[(1, 0), (0, 3), (3, 2)]
于 2018-11-28T06:35:45.940 回答