如何用数学公式简化Python分块遍历逻辑，移除loop变量？

需求：不借助额外库，将列表lst按8条为一组分块，为1-8、9-16、17-24等区间的记录匹配字典dct的对应值。现有代码通过loop变量实现逻辑，希望移除该变量及相关判断语句，用更简洁易读的数学公式替代record-loop，同时避免((record - 1) % 8) + 1这类不够直观的写法。

现有代码 #

dct = {1:'one', 2:'two', 3:'three', 4:'four', 5:'five', 6:'six', 7:'seven', 8:'eight'}

lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]

loop = 0
for record in lst:
    print(f'record id: {record:<10} dct id: {record-loop:<10} dct value: {dct.get(record - loop)}')


    if record % 8 == 0:
        loop += 8
        print('--- loop finished ---')

当前输出 #

record id: 1          dct id: 1          dct value: one
record id: 2          dct id: 2          dct value: two
record id: 3          dct id: 3          dct value: three
record id: 4          dct id: 4          dct value: four
record id: 5          dct id: 5          dct value: five
record id: 6          dct id: 6          dct value: six
record id: 7          dct id: 7          dct value: seven
record id: 8          dct id: 8          dct value: eight
--- loop finished ---
record id: 9          dct id: 1          dct value: one
record id: 10         dct id: 2          dct value: two
record id: 11         dct id: 3          dct value: three
record id: 12         dct id: 4          dct value: four
record id: 13         dct id: 5          dct value: five
record id: 14         dct id: 6          dct value: six
record id: 15         dct id: 7          dct value: seven
record id: 16         dct id: 8          dct value: eight
--- loop finished ---
record id: 17         dct id: 1          dct value: one
record id: 18         dct id: 2          dct value: two
record id: 19         dct id: 3          dct value: three
record id: 20         dct id: 4          dct value: four
record id: 21         dct id: 5          dct value: five
record id: 22         dct id: 6          dct value: six
record id: 23         dct id: 7          dct value: seven
record id: 24         dct id: 8          dct value: eight
--- loop finished ---

注意：直接使用record%8会得到0而非8，不符合需求，因此需要更合适的数学公式。

优化方案 #

方案1：整除计算组内位置 #

用record - 8 * ((record - 1) // 8)替代record-loop，逻辑是先算出当前记录所在的组（从0开始计数），再用记录值减去组的起始值（组号×8），得到组内1-8的位置。

优化后代码：

dct = {1:'one', 2:'two', 3:'three', 4:'four', 5:'five', 6:'six', 7:'seven', 8:'eight'}
lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]

for record in lst:
    dct_id = record - 8 * ((record - 1) // 8)
    print(f'record id: {record:<10} dct id: {dct_id:<10} dct value: {dct.get(dct_id)}')
    if dct_id == 8:
        print('--- loop finished ---')

方案2：模运算变种 #

利用(record % 8) or 8，当记录是8的倍数时，record%8结果为0，or运算会返回8；其他情况返回模运算的结果1-7，正好符合需求。

优化后代码：

dct = {1:'one', 2:'two', 3:'three', 4:'four', 5:'five', 6:'six', 7:'seven', 8:'eight'}
lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]

for record in lst:
    dct_id = (record % 8) or 8
    print(f'record id: {record:<10} dct id: {dct_id:<10} dct value: {dct.get(dct_id)}')
    if dct_id == 8:
        print('--- loop finished ---')

两个方案都移除了loop变量和相关判断，代码更简洁，逻辑清晰易读。

内容的提问来源于stack exchange，提问作者Roman Toasov