基本上有 15 个问题从Question 1到Question 15作为父词典中的嵌套词典Questions。但现在我想用这个列表更新这本词典，其中也有 15 个问题：new_list = ['Question 6', 'Question 7', 'Question 8', 'Question 9', 'Question 10', 'Question 11', 'Question 12', 'Question 13', 'Question 14', 'Question 15', 'Question 16', 'Question 17', 'Question 18', 'Question 19', 'Question 20']这只是Question n + 5人类语言。但我一整天都在挣扎。我观察到的奇怪行为是：方法1-for ind, val in enumerate(list(q_dict['Questions'].keys())):
q_dict['Questions'][new_list[ind]] = q_dict['Questions'].pop(val)结果是：{'Paper Name': 'Paper 8.jpg', 'Category': 1, 'Questions': {'Question 16': {'Question Type': 1, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 17': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 18': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 19': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 20': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}}}方法2-for ind, val in enumerate(list(q_dict['Questions'].keys())):
q_dict['Questions'][new_list[ind]] = q_dict['Questions'][val] del q_dict['Questions'][val]结果：{'Paper Name': 'Paper 8.jpg', 'Category': 1, 'Questions': {'Question 6': {'Question Type': 1, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 7': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 8': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 9': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}, 'Question 10': {'Question Type': 3, 'Question Point': [], 'Expected Answers': [], 'Student Answers': []}}}
2 回答
慕的地8271018
TA贡献1796条经验 获得超4个赞
从您正在迭代的列表中添加或删除元素通常是一个坏主意，因为行为可能不是您所期望的。相反，完全替换问题字典：
q_dict['Questions'] = {
name: entry for name, entry in zip(new_list, q_dict['Questions'].values())}请注意，在 Python 3.6 之前，您可能需要对值进行排序，因为无法保证迭代顺序。
一只名叫tom的猫
TA贡献1906条经验 获得超2个赞
我想分享另一种解决此问题的方法，该方法不涉及额外的键列表（只需在每个问题键上添加 5）：
import re
def change_name(m_str):
a = re.match('\D+(\d+)', m_str)
return m_str.replace(str(a.groups()[0]), str(int(a.groups()[0])+5))
q_dict["Questions"] = {change_name(inner_key):inner_val for inner_key, inner_val in q_dict["Questions"].items()}
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