python - 排序方法 : The truth value of an array with more than one element is ambiguous. 使用 a.any() 或 a.all()

Question

我在使用 sorted() 方法时遇到问题。我在循环中使用此方法对我在循环的每一步中升级的列表进行排序。第一次迭代有效，但第二次迭代无效，并给出下一个错误:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() 

这是我的代码:

import numpy as np
import random as rd
import math

Poblacion = 10
pressure = int(0.3*Poblacion)
mutation_chance = 0.08

Modelo = np.array([[0.60,0.40,0.90],[0.26,0.20,0.02],[0.80,0.00,0.05]])

x = np.array([[1.0,2.0,3.0],[0.70,0.50,0.90],[0.10,0.40,0.20]])
y = np.array([[4.10,0.72,2.30],[1.43,0.30,1.01],[0.40,0.11,0.18]])

def crearIndividuo():
    return[np.random.random((3, 3))]

def crearPoblacion(): 
    return [crearIndividuo() for i in range(Poblacion)]

def calcularFitness(individual): 
    error = 0
    i=0
    for j in x:
        error += np.array(individual).dot(j)-y[i]
        i += 1
        error = np.linalg.norm(error,ord=1) 
        fitness = math.exp(-error)
    return fitness

def selection_and_reproduction(population):
    puntuados = [ (calcularFitness(i), i) for i in population] 
    puntuados = [i[1] for i in sorted(puntuados)] 
    population = puntuados

    selected =  puntuados[(len(puntuados)-pressure):] 

    j=0
    while (j < int(len(population)-pressure)):
        padre = rd.sample(selected, 2)

        population[j] = 0.5*(np.array(padre[0]) + np.array(padre[1]))
        j += 1
        population[j] = 1.5*np.array(padre[0]) - 0.5*np.array(padre[1])
        j += 1
        population[j] = -0.5*np.array(padre[0]) + 1.5*np.array(padre[1])
        j += 1
    return population

population = crearPoblacion()

for l in range(3):
    population = selection_and_reproduction(population)

print("final population: ", population)

错误发生在以下行:

puntuados = [i[1] for i in sorted(puntuados)] 

我不知道我做错了什么(我不是Python专家)。谁能帮我吗？

提前致谢。

Answer 1

问题出现在元组共享相同的第一个元素的地方，所以

sorted(puntuados)

必须比较两个元组的第二个元素以确定它们的相对顺序，此时您会遇到此异常。

你可以使用

sorted(graded, key=lambda x: x[0])

如果您只想根据元组的第一个元素进行排序，可以解决您的问题。