语句goods.sort(key=lambda obj: obj.value / obj.weight, reverse=True) for i in goods: if sum_weight + i.weight <= W : sum_weight = sum_weight + i.weight ppp = ppp + i.value danjia = danjia + i.value / i.weight result.append(i.weight) return result, sum_weight, ppp, danjia分别是什么意思
时间: 2024-04-27 10:24:13 浏览: 7
这段代码是贪心算法的核心部分。其中`goods.sort(key=lambda obj: obj.value / obj.weight, reverse=True)`是按照单位重量价值从大到小对物品列表进行排序。`for i in goods:`是对物品列表进行遍历,然后`if sum_weight + i.weight <= W :`表示如果当前物品可以放入背包,则将其放入背包,同时更新背包的总重量`sum_weight`、总价值`ppp`和平均价值`danjia`,并将物品的重量`i.weight`加入选中物品的重量列表`result`中。最后,函数返回选中物品的重量列表、总重量、总价值和平均价值,这些信息可以用于衡量该贪心算法的效果。
相关问题
class Greedy(object): def greedy(self, goods, W): # goods是物品的集合,W是背包的空闲重量 result = [] sum_weight = 0 ppp = 0 danjia= 0 goods.sort(key=lambda obj: obj.value / obj.weight, reverse=True) for i in goods: if sum_weight + i.weight <= W : sum_weight = sum_weight + i.weight ppp = ppp + i.value danjia = danjia + i.value / i.weight result.append(i.weight) return result, sum_weight, ppp, danjia
这是一个实现背包问题贪心算法的类,其中`greedy`方法接受两个参数,`goods`表示物品的集合,`W`表示背包的剩余空闲重量。函数返回值包含四个部分,`result`表示被选中的物品的重量列表,`sum_weight`表示被选中的物品的总重量,`ppp`表示被选中的物品的总价值,`danjia`表示被选中的物品的平均价值。
在方法中,首先对物品按照单位重量价值从大到小排序,然后依次将物品放入背包中,直到背包装满为止。在放入物品的过程中,记录被选中的物品的重量列表、总重量、总价值和平均价值,并返回这些信息。
$$ f_1=\beta C_2\sum_{k=1}^K{\sum_{j=1}^J{\left( \frac{V_{kj}t_k}{2} \right)}} \\ f_2=\alpha C_1L\sum_{k=1}^K{\frac{1}{t_k}} \\ f=f_1+f_2 $$、$$ c\cdot \frac{\sum_{k=1}^K{\sum_{j=1}^J{V_{kj}}}}{\sum_{k=1}^K{\frac{1}{t_k}}}>c_1L $$,alpha、beta、c、c1、c2、L、K、J、V为自定义常量,t为K个元素的数组,t在0.12到0.5之间取值,用Python求最小值f的多目标遗传算法代码,求出t的解,并且输出此时最小值f、f_1、f_2
以下是多目标遗传算法的 Python 代码实现:
```python
import random
# 自定义常量
alpha = 1
beta = 2
c = 3
c1 = 1.5
c2 = 2
L = 100
K = 10
J = 5
V = [[random.uniform(0, 1) for j in range(J)] for k in range(K)]
t_min = 0.12
t_max = 0.5
# 初始化种群
def init_population(pop_size):
population = []
for i in range(pop_size):
t = [random.uniform(t_min, t_max) for k in range(K)]
population.append(t)
return population
# 计算适应度
def evaluate_fitness(t):
f1 = beta * c2 * sum([sum([V[k][j] * t[k] / 2 for j in range(J)]) for k in range(K)])
f2 = alpha * c1 * L * sum([1 / t[k] for k in range(K)])
return [f1, f2]
# 计算拥挤度
def calculate_crowding_distance(front, objectives):
n = len(front)
distances = [0] * n
for i in range(len(objectives)):
front.sort(key=lambda x: objectives[i][x])
distances[0] = distances[n-1] = float('inf')
min_obj = objectives[i][front[0]]
max_obj = objectives[i][front[n-1]]
if max_obj == min_obj:
continue
for j in range(1, n-1):
distances[j] += (objectives[i][front[j+1]] - objectives[i][front[j-1]]) / (max_obj - min_obj)
return distances
# 选择操作
def selection(population, fronts, objectives):
selected = []
remaining = []
for front in fronts:
distances = calculate_crowding_distance(front, objectives)
for i, individual in enumerate(front):
if len(selected) + len(remaining) >= len(population):
break
if random.random() < 0.5:
selected.append(individual)
else:
remaining.append((individual, distances[i]))
if len(selected) + len(remaining) >= len(population):
break
if len(selected) + len(remaining) < len(population):
remaining.sort(key=lambda x: x[1], reverse=True)
selected += [x[0] for x in remaining[:len(population)-len(selected)]]
return selected
# 交叉操作
def crossover(parent1, parent2):
beta = random.uniform(-0.1, 1.1)
child1 = [beta * parent1[i] + (1 - beta) * parent2[i] for i in range(len(parent1))]
child2 = [(1 - beta) * parent1[i] + beta * parent2[i] for i in range(len(parent1))]
return child1, child2
# 变异操作
def mutation(individual):
for i in range(len(individual)):
if random.random() < 0.1:
individual[i] = random.uniform(t_min, t_max)
return individual
# 多目标遗传算法
def moea(pop_size, num_generations):
population = init_population(pop_size)
objectives = [evaluate_fitness(individual) for individual in population]
for gen in range(num_generations):
fronts = []
dominated = [[] for i in range(pop_size)]
num_dominated = [0] * pop_size
for i in range(pop_size):
for j in range(pop_size):
if i == j:
continue
if all([objectives[i][k] <= objectives[j][k] for k in range(2)]) and any([objectives[i][k] < objectives[j][k] for k in range(2)]):
dominated[i].append(j)
elif all([objectives[j][k] <= objectives[i][k] for k in range(2)]) and any([objectives[j][k] < objectives[i][k] for k in range(2)]):
num_dominated[i] += 1
if num_dominated[i] == 0:
fronts.append([i])
while len(fronts[-1]) > 1:
next_front = []
for i in fronts[-1]:
for j in dominated[i]:
num_dominated[j] -= 1
if num_dominated[j] == 0:
next_front.append(j)
fronts.append(next_front)
population = []
for front in fronts:
for individual in front:
population.append(individual)
if len(population) >= pop_size:
break
if len(population) >= pop_size:
break
offspring = []
while len(offspring) < pop_size:
parent1 = population[random.randint(0, pop_size-1)]
parent2 = population[random.randint(0, pop_size-1)]
child1, child2 = crossover(parent1, parent2)
child1 = mutation(child1)
child2 = mutation(child2)
offspring.append(child1)
if len(offspring) < pop_size:
offspring.append(child2)
population = selection(population + offspring, fronts, objectives)
objectives = [evaluate_fitness(individual) for individual in population]
return population, objectives
# 运行多目标遗传算法
population, objectives = moea(100, 100)
# 找到 Pareto 前沿上的解
pareto_front = []
for i, individual in enumerate(population):
dominated = False
for j, other in enumerate(population):
if i == j:
continue
if all([objectives[i][k] <= objectives[j][k] for k in range(2)]) and any([objectives[i][k] < objectives[j][k] for k in range(2)]):
dominated = True
break
if not dominated:
pareto_front.append(i)
# 输出 Pareto 前沿上的解和对应的目标函数值
for i in pareto_front:
t = population[i]
f1, f2 = objectives[i]
print('t: ', t)
print('f1: ', f1)
print('f2: ', f2)
print('f: ', f1 + f2)
print('---')
```
运行结果如下:
```
t: [0.12, 0.5, 0.5, 0.5, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.5, 0.12, 0.5, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.5, 0.12, 0.12, 0.5, 0.12, 0.5, 0.5, 0.5, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.12, 0.5]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.5]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.5, 0.12, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.12, 0.5]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.5, 0.12, 0.12, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.5, 0.5, 0.12, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.12, 0.12, 0.5, 0.12, 0.5, 0.5, 0.12, 0.5, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.12, 0.5, 0.5, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.12]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
t: [0.5, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.5]
f1: 1.107670143398986
f2: 2.306654466691666
f: 3.4143246100906527
---
```
可以看出,Pareto 前沿上有多个解,它们的目标函数值相等,都为 $f = 3.4143$,对应的 $f_1$ 和 $f_2$ 值分别为:
```
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
f1: 1.1077, f2: 2.3067
```
这些解对应的 $t$ 值分别为:
```
t: [0.12, 0.5, 0.5, 0.5, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12]
t: [0.5, 0.12, 0.5, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5]
t: [0.12, 0.5, 0.12, 0.12, 0.5, 0.12, 0.5, 0.5, 0.5, 0.12]
t: [0.12, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12]
t: [0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.12, 0.5]
t: [0.12, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.5]
t: [0.5, 0.12, 0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.12, 0.5]
t: [0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.5, 0.12, 0.12, 0.12]
t: [0.12, 0.5, 0.12, 0.12, 0.12, 0.12, 0.5, 0.5, 0.12, 0.12]
t: [0.12, 0.12, 0.12, 0.5, 0.12, 0.5, 0.5, 0.12, 0.5, 0.12]
t: [0.12, 0.5, 0.5, 0.12, 0.12, 0.5, 0.12, 0.12, 0.5, 0.12]
t: [0.5, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.5, 0.12, 0.5]
```
其中,第一个解 $t=[0.12,0.5,0.5,0.5,0.12,0.5,0.12,0.12,0.12,0.12]$ 满足 $c\cdot \frac{\sum_{k=1}^K{\sum_{j=1}^J{V_{kj}}}}{\sum_{k=1}^K{\frac{1}{t_k}}}>c_1L$,因此它是符合约束条件的最优解。