Python 相同的代码但不同的结果，怎么可能呢？

我已经把matlab的代码写成了python代码，但是它给出了不同的最佳适配值，这怎么可能呢？我想把这个matlab代码写成python，是不是方法不对？我应该实现不同的方法吗？matlabs最佳适配值为340，pytons最佳适配值为17？这怎么可能？ 这是python版本

import numpy as np
coord=[]
coord = np.array([[11, 47, 62],       
       [31, 37, 69],
       [32, 38, 46],
       [33, 46, 10],
       [34, 61, 33],
       [35, 62, 63],
       [36, 63, 69],
       [37, 32, 22],
       [38, 45, 35],
       [39, 59, 15],
       [40, 5, 6],
       [41, 10, 17],
       [42, 21, 10],
       [43, 5, 64],
       [44, 30, 15],
       [45, 39, 10],
       [46, 32, 39],
       [47, 25, 32],
       [48, 25, 55],
       [49, 48, 28],
       [50, 56, 37],
       [61, 40, 50]],dtype=np.int)
city=len(coord)
best_solution=[]
best_fitness=100000000
pop_size=1000  
CR=0.8 # %YÜZDE 90 OLASILIKLA CAPRAZLANIYOR.
MR=0.5
MaxIter=5000



# distance = np.zeros((coord.shape[0], coord.shape[0]))
distance = np.zeros([city,city])

for i in range(city):
    for j in range(city):
        distance[i][j] = np.sqrt((coord[i][1] - coord[j][1]) ** 2 + (coord[i][2] - coord[j][2]) ** 2)

population=np.zeros([pop_size,city],dtype=np.int)     
for i in range(pop_size):
    population[i][:]=np.random.permutation(city)

fitness=np.zeros([1,pop_size])   

for i in range(pop_size):
             fitness[0][i]=0
             for j in range(city-1):
                 # fitness[0][i]=fitness[0][i]+distance[[population[i][j]][population[i][j+1]]]
                 fitness[0][i]=fitness[0][i]+ distance[population[i][j]][population[i][j+1]]
                 fitness[0][i]=fitness[0][i]+ distance[population[i][city-1]][population[i][1]]
                 if best_fitness > fitness[0][i]: 
                     best_solution=population[i][:]
                     best_fitness=fitness[0][i] 




b=np.min(fitness) # there is no lower than this value but it gives 13 or 20.

这是matlab版本

clc 
clear
coord=[11 47 62;
       31 37 69;
       32 38 46;
       33 46 10;
       34 61 33;
       35 62 63;
       36 63 69;
       37 32 22;
       38 45 35;
       39 59 15;
       40 5 6;
       41 10 17;
       42 21 10;
       43 5 64;
       44 30 15;
       45 39 10;
       46 32 39;
       47 25 32;
       48 25 55;
       49 48 28;
       50 56 37;
       61 40 50];

  [city,~]=size(coord);

     best_solution=[];

   best_fitness=100000000;
 pop_size=1000;  
  CR=0.8; %YÜZDE 90 OLASILIKLA CAPRAZLANIYOR.
 MR=0.5;

  MaxIter=5000;

 %51x51 matriste her şehirden şehire uzaklıgı bul  
   distance=[];
   for i=1:city
       for j=1:city
           distance(i,j)=sqrt((coord(i,2)-coord(j,2))^2+(coord(i,3)-coord(j,3))^2);
       end
   end
   %rastgele 100 tane cozum olusturdu.
   population=[];
for i=1:pop_size
    population(i,:)=randperm(city);
end
 %population


 for i=1:pop_size
    fitness(i)=0; %başta bos olusturuyor dikkat.
    for j=1:city-1
        %dk 32:48
        %mesela 14 ile 22. sutunun arasındaki mesafe 
        %j+1 var diye city -1 yazdık.
        %sırayla 2 sehir arasındaki mesafe hesaplanıyor.

        fitness(i)=fitness(i)+distance(population(i,j),population(i,j+1));
    end
    %son sehirden bastaki sehire donme icin.
    fitness(i)=fitness(i)+distance(population(i,city),population(i,1));
    if  best_fitness>fitness(i)  %fitness(i)<best_fitness
        best_solution=population(i,:);
        best_fitness=fitness(i);
    end
 end  

---

由于随机生成涉及randperm和np.random.permutation，因此最佳适应度在每次迭代中都会发生变化。但是你是对的，Python脚本生成的数字大约是10-20，而Matlab则是数百

在python脚本中，您正在调用：

适合度[0][i]=适合度[0][i]+距离[population[i][j]][population[i][j+1]] 适合度[0][i]=适合度[0][i]+距离[population[i][city-1]][population[i][1]]

在相同的for循环中，Matlab脚本中不会出现这种情况，请删除python脚本中的缩进，如下所示：

import numpy as np
coord=[]
coord = np.array([[11, 47, 62],       
       [31, 37, 69],
       [32, 38, 46],
       [33, 46, 10],
       [34, 61, 33],
       [35, 62, 63],
       [36, 63, 69],
       [37, 32, 22],
       [38, 45, 35],
       [39, 59, 15],
       [40, 5, 6],
       [41, 10, 17],
       [42, 21, 10],
       [43, 5, 64],
       [44, 30, 15],
       [45, 39, 10],
       [46, 32, 39],
       [47, 25, 32],
       [48, 25, 55],
       [49, 48, 28],
       [50, 56, 37],
       [61, 40, 50]],dtype=np.int)
city=len(coord)
best_solution=[]
best_fitness=100000000
pop_size=1000  
CR=0.8 # %YÜZDE 90 OLASILIKLA CAPRAZLANIYOR.
MR=0.5
MaxIter=5000



# distance = np.zeros((coord.shape[0], coord.shape[0]))
distance = np.zeros([city,city])

for i in range(city):
    for j in range(city):
        distance[i][j] = np.sqrt((coord[i][1] - coord[j][1]) ** 2 + (coord[i][2] - coord[j][2]) ** 2)
#print(distance)
population=np.zeros([pop_size,city],dtype=np.int)     
for i in range(pop_size):
    population[i][:]=np.random.permutation(city)

fitness=np.zeros([1,pop_size])   

for i in range(pop_size):
    fitness[0][i]=0
    for j in range(city-1):
        # fitness[0][i]=fitness[0][i]+distance[[population[i][j]][population[i][j+1]]]
        fitness[0][i]=fitness[0][i]+ distance[population[i][j]][population[i][j+1]]
    fitness[0][i]=fitness[0][i]+ distance[population[i][city-1]][population[i][1]]
    if best_fitness > fitness[0][i]: 
        best_solution=population[i][:]
        best_fitness=fitness[0][i] 

b=np.min(fitness) # there is no lower than this value but it gives 13 or 20.

---

但这条线的适合度[0][i]=适合度[0][i]+距离[population[i][city-1]][population[i][1]]应该在i循环中。它在循环中吗？在spider程序中，它看起来像是脱离了i循环。我通过命令提示符在python解释器中运行脚本，在上面的示例中：fitness[0][i]=fitness[0][i]+distance[population[i][city-1]][population[i][1]]在i for循环中，fitness[0][i]=fitness[0][i]+distance[population[i][i][j]]我的第二个问题是：我知道matlabs和pythons，0-1索引。但我在python中没有使用-1，除了最后一个适应度方程。为什么我必须使用city-1？？？为什么在其他代码中没有任何-1。还有第一个方程，没有city-1，在i和j循环中。第二健身功能。在matlab中只有i循环。对于j for循环，在matlab和python脚本中都有city-1，这是因为您正在调用人口[j+1]，它比city的长度限制高一个。记住，如果有任何有用的评论或答案，请向上投票：