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fix: compare_lists_of_dicts的bug

This commit is contained in:
caoqianming 2023-12-18 13:01:19 +08:00
parent bf5d62e99d
commit a1e2f12df7
1 changed files with 6 additions and 3 deletions
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9
apps/utils/tools.py
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@ -220,9 +220,11 @@ def compare_lists_of_dicts(list1, list2, ignore_order=False):
"""比较两个列表，这里的列表包含字典（对象）"""
if ignore_order:
# 转换列表中的字典为元组列表，然后排序进行比较
sorted_list1 = sorted((tuple(sorted(d.items())) for d in list1))
sorted_list2 = sorted((tuple(sorted(d.items())) for d in list2))
return sorted_list1 == sorted_list2
if list1 and isinstance(list1[0], dict):
sorted_list1 = sorted((tuple(sorted(d.items())) for d in list1))
sorted_list2 = sorted((tuple(sorted(d.items())) for d in list2))
return sorted_list1 == sorted_list2
return list1 == list2
else:
# 按顺序比较列表中的字典
return all(compare_dicts(obj1, obj2) for obj1, obj2 in zip(list1, list2))
@ -230,6 +232,7 @@ def compare_lists_of_dicts(list1, list2, ignore_order=False):
def compare_values(val1, val2, ignore_order=False):
"""通用比较函数，也可以处理字典和列表。"""
print(val1, val2)
if isinstance(val1, list) and isinstance(val2, list):
# 假设这里我们关心列表中对象的顺序
return compare_lists_of_dicts(val1, val2, ignore_order)