IndexError:列表分配索引超出范围,Python

时间:2019-10-17 02:23:36

标签: python python-3.x list pycharm

我正在尝试实现功能。它的工作方式应该是这样的:

  1. 需要两个列表。
  2. 标记一些索引,最好居中。
  3. 父母双方都切换标记的索引。
  4. 其他索引顺序进入其父元素。
  5. 如果该父元素中已经存在相同的元素,它将映射并检查其他同父元素在何处并到达那里。
import random
def pm(indA, indB):
    size = min(len(indA), len(indB))
    c1, c2 = [0] * size, [0] * size

    # Initialize the position of each indices in the individuals
    for i in range(1,size):
        c1[indA[i]] = i
        c2[indB[i]] = i

    crosspoint1 = random.randint(0, size)
    crosspoint2 = random.randint(0, size - 1)
    if crosspoint2 >= crosspoint1:
        crosspoint2 += 1
    else:  # Swap the two cx points
        crosspoint1, crosspointt2 = crosspoint2, crosspoint1


    for i in range(crosspoint1, crosspoint2):
        # Keep track of the selected values
        temp1 = indA[i]
        temp2 = indB[i]
        # Swap the matched value
        indA[i], indA[c1[temp2]] = temp2, temp1
        indB[i], indB[c2[temp1]] = temp1, temp2
        # Position bookkeeping
        c1[temp1], c1[temp2] = c1[temp2], c1[temp1]
        c2[temp1], c2[temp2] = c2[temp2], c2[temp1]
        return indA, indB

a,b = pm([3, 4, 8, 2, 7, 1, 6, 5],[4, 2, 5, 1, 6, 8, 3, 7])

错误:

in pm
    c1[indA[i]] = i
IndexError: list assignment index out of range

2 个答案:

答案 0 :(得分:2)

不确定代码中是否还有其他错误(我没有运行它),但这是对此的解释。在 Python 中(与其他大多数语言一样),列表(序列更精确)索引基于 0

>>> l = [1, 2, 3, 4, 5, 6]
>>>
>>> for e in l:
...     print(e, l.index(e))
...
1 0
2 1
3 2
4 3
5 4
6 5
>>>
>>> l[0]
1
>>> l[5]
6
>>> l[6]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

总结您的问题:

  1. 您的 indA indB 列表各有6个元素( [1..6] ),以及它们的索引: [ 0..5]
  2. 您的 c1 c2 列表也包含6个元素(索引也为 [0..5]
  3. 但是,您将 #1。 中的值用作 #2。 中列表的索引,并且值 6 是一个问题,因为没有这样的索引

要解决您的问题,应使用有效的索引值。要么:

  • indA indB 中有适当的值(这是我选择的值):

    a, b = pmxCrossover([0, 3, 1, 2, 5, 4], [4, 0, 2, 3, 5, 1])
    
  • 减去 1 任何地方,您遇到的 indA indB 用作索引:

    c1[indA[i] - 1] = i
    

一般建议:每当遇到错误时,在错误行之前添加 print 语句(在之前)(从中打印(部分)内容) ,这可能会为您提供线索,从而可以自己解决问题。

@ EDIT0

发布原始代码(略有修改的版本),并进行索引转换:

  • 在算法之前:从每个元素中减去1以得到有效的索引
  • 算法之后:加1可返回基于1的索引

code00.py

#!/usr/bin/env python3

import sys
import random


def pmx_crossover(ind_a, ind_b):
    size = min(len(ind_a), len(ind_b))
    c1, c2 = [0] * size, [0] * size

    # Initialize the position of each indices in the individuals
    for i in range(1, size):
        c1[ind_a[i]] = i
        c2[ind_b[i]] = i
    # Choose crossover points
    crosspoint1 = random.randint(0, size)
    crosspoint2 = random.randint(0, size - 1)
    if crosspoint2 >= crosspoint1:
        crosspoint2 += 1
    else:  # Swap the two cx points
        crosspoint1, crosspointt2 = crosspoint2, crosspoint1

    # Apply crossover between cx points
    for i in range(crosspoint1, crosspoint2):
        # Keep track of the selected values
        temp1 = ind_a[i]
        temp2 = ind_b[i]
        # Swap the matched value
        ind_a[i], ind_a[c1[temp2]] = temp2, temp1
        ind_b[i], ind_b[c2[temp1]] = temp1, temp2
        # Position bookkeeping
        c1[temp1], c1[temp2] = c1[temp2], c1[temp1]
        c2[temp1], c2[temp2] = c2[temp2], c2[temp1]
    return ind_a, ind_b


def main():
    #initial_a, initial_b = [1, 2, 3, 4, 5, 6, 7, 8], [3, 7, 5, 1, 6, 8, 2, 4]
    initial_a, initial_b = [1, 4, 2, 3, 6, 5], [5, 1, 3, 4, 6, 2]
    index_offset = 1
    temp_a = [i - index_offset for i in initial_a]
    temp_b = [i - index_offset for i in initial_b]
    a, b = pmx_crossover(temp_a, temp_b)
    final_a = [i + index_offset for i in a]
    final_b = [i + index_offset for i in b]
    print("Initial: {0:}, {1:}".format(initial_a, initial_b))
    print("Final:   {0:}, {1:}".format(final_a, final_b))


if __name__ == "__main__":
    print("Python {0:s} {1:d}bit on {2:s}\n".format(" ".join(item.strip() for item in sys.version.split("\n")), 64 if sys.maxsize > 0x100000000 else 32, sys.platform))
    main()
    print("\nDone.")

输出(一种可能(由于random.randint)):

[cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q058424002]> "e:\Work\Dev\VEnvs\py_064_03.07.03_test0\Scripts\python.exe" code00.py
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 22:22:05) [MSC v.1916 64 bit (AMD64)] 64bit on win32

Initial: [1, 4, 2, 3, 6, 5], [5, 1, 3, 4, 6, 2]
Final:   [1, 3, 2, 4, 6, 5], [5, 1, 4, 3, 6, 2]

Done.

答案 1 :(得分:0)

c1超出范围,因为在您的for的第四个索引中,indA [4]的值为6。 而且c1索引的范围是0-5(长度是6)。

使用c1[indA[i]] = i 您尝试做c1[6] = 4