我将通过手动迭代来合并列表：

sgrades_flat=['Barrett', 'Edan', '70', '45', '59', 'Bradshaw', 'Reagan', '96', '97', '88', 'Charlton', 'Caius', '73', '94', '80', 'Mayo', 'Tyrese', '88', '61', '36', 'Stern', 'Brenda', '90', '86', '45']

s_grades=['F', 'A', 'B', 'D', 'C']

it1 = iter(sgrades_flat)
it2 = iter(s_grades)

result = []
try:
    while True:
        for _ in range(5):
            result.append(next(it1))
        result.append(next(it2))

except StopIteration:
    pass

print(result)

印刷品

['Barrett', 'Edan', '70', '45', '59', 'F', 
 'Bradshaw', 'Reagan', '96', '97', '88', 'A',
 'Charlton', 'Caius', '73', '94', '80', 'B',
 'Mayo', 'Tyrese', '88', '61', '36', 'D', 
 'Stern', 'Brenda', '90', '86', '45', 'C']

(this仍然看起来像是一个坏主意，因为平面列表对于这样的数据结构是次优的）
请注意，没有任何手动迭代的一行程序也执行相同的操作：

import itertools
grouped_sgrades = list(itertools.chain.from_iterable(
       sgrades_flat[i:i+5]+[s_grades[i//5]] 
       for i in range(0,len(sgrades_flat),5)))

然而，为什么要把清单扁平化呢？

grouped_sgrades = [sgrades_flat[i:i+5]+[s_grades[i//5]] for i in range(0,len(sgrades_flat),5)]

result是一个很好的列表列表，它接近于某种结构化数据：

[['Barrett', 'Edan', '70', '45', '59', 'F'],
 ['Bradshaw', 'Reagan', '96', '97', '88', 'A'],
 ['Charlton', 'Caius', '73', '94', '80', 'B'],
 ['Mayo', 'Tyrese', '88', '61', '36', 'D'],
 ['Stern', 'Brenda', '90', '86', '45', 'C']]

您可以使用enumerate，然后相应地对sgrades_flat进行切片：

sgrades_flat = [
    'Barrett', 'Edan', '70', '45', '59',
    'Bradshaw', 'Reagan', '96', '97', '88',
    'Charlton', 'Caius', '73', '94', '80',
    'Mayo', 'Tyrese', '88', '61', '36',
    'Stern', 'Brenda', '90', '86', '45']

s_grades = ['F', 'A', 'B', 'D', 'C']

z = []
for i, x in enumerate(s_grades):
    z += sgrades_flat[5*i:5*(i+1)] + [x]

print(z)

我们需要同时迭代s_grades和sgrades_flat，每次需要迭代5个元素。我们可以使用基于sgrades_flat长度1/5的x1m2 n1 a列表解析。然后，我们可以使用列表解析中的第二级迭代将j绑定到i * 5索引。

[
 [*sgrades_flat[j:j+5], s_grades[i]] 
 for i in range(len(sgrades_flat) // 5)
 for j in (i * 5,)
]
# [['Barrett',  'Edan',   '70', '45', '59', 'F'], 
#  ['Bradshaw', 'Reagan', '96', '97', '88', 'A'], 
#  ['Charlton', 'Caius',  '73', '94', '80', 'B'], 
#  ['Mayo',     'Tyrese', '88', '61', '36', 'D'], 
#  ['Stern',    'Brenda', '90', '86', '45', 'C']]

或者，使用enumerate在s_grades上迭代：

[
 [*sgrades_flat[j:j+5], g] 
 for i, g in enumerate(s_grades)
 for j in (i * 5,)
]

# [['Barrett',  'Edan',   '70', '45', '59', 'F'], 
#  ['Bradshaw', 'Reagan', '96', '97', '88', 'A'], 
#  ['Charlton', 'Caius',  '73', '94', '80', 'B'], 
#  ['Mayo',     'Tyrese', '88', '61', '36', 'D'], 
#  ['Stern',    'Brenda', '90', '86', '45', 'C']]