您可以使用Map字典：

dmap = dict(zip(ordered_elements, index))
out = [[dmap[key] for key in lst] for lst in elements]

输出：

>>> out
[[308, 302], [302, 303], [405, 506, 609]]

>>> dmap
{'A': 308, 'B': 302, 'C': 303, 'D': 405, 'E': 506, 'F': 609}

• 准备Map（字母到索引）
• 遍历元素并赋值给输出（使用嵌套的for/list解析），例如

ordered_elements = ['A','B','C', 'D', 'E', 'F']

index= [308,302,303,405,506,609]

element_index_mapping = {letter:idx for e, letter in zip(ordered_elements, index)}

out = []
for letters in elements:
    indices_for_letters = []
    for letter in letters:
        indices_for_letters.append(element_index_mapping.get(letter))
    out.append(indices_for_letters)

我不推荐这样做，但如果你想剪短，也许你可以试试这个：

result = [[index[ordered_elements.index(el_l1)] for el_l1 in el] for el in elements]

虽然此处没有安全检查，但您可以按以下方式操作：

elements = [['A', 'B'], ['B', 'C'], ['D', 'E', 'F']]
ordered_elements = ['A', 'B', 'C', 'D', 'E', 'F']
index = [308, 302, 303, 405, 506, 609]

out = [[index[ordered_elements.index(c)] for c in sl] for sl in elements]

print(out)

• • 输出：**

[[308, 302], [302, 303], [405, 506, 609]]