python - Python索引错误: list assignment index out of range ,

Question

我正在尝试实现功能。它的工作应该是这样的:

1. 需要两个列表。
2. 标记一些索引，最好是居中的几个。
3. parent 双方都交换了标记的索引。
4. 其他索引按顺序转到其父元素。
5. 如果相同的元素已存在于该父元素中，它会映射并检查其他父元素相同的元素所在的位置并转到那里。

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

Answer 1

不确定您的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(与大多数其他语言一样)中，列表(更精确地说是序列)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 和 indB 列表各有 6 个元素 ([1..6])，它们的索引为:[ 0..5]
2. 您的 c1 和 c2 列表也有 6 个元素(索引也是 [0..5])
3. 但是，您使用 #1. 中的值作为 #2. 列表中的索引，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，您应该使用有效的索引值。要么:

• 在 indA 和 indB 中设置正确的值(这是我选择的值):

  a, b = pmxCrossover([0, 3, 1, 2, 5, 4], [4, 0, 2, 3, 5, 1])
  

• 无论何时遇到indA或indB值时减去1 em> 用作索引:

  c1[indA[i] - 1] = i
  

作为一般建议:每当遇到错误时，请在错误行之前添加打印语句(从中打印(部分)内容) ，这可能会给您提供一些线索，帮助您自己解决问题。

@EDIT0

发布原始代码(稍加修改的版本)，并进行索引转换:

• 在算法之前:(从每个元素中)减去 1 以获得有效索引
• 算法结束后:加 1 回到基于 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(可能性之一(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.