如何比较python中的两个字典并相应地替换值？

debugcn 发表于 Dev

法罕帕特尔

这是第一本词典：

[{'ITEMNO': None}, {'ITEM_TYPE': None}, {'CASE_QYT': None}, {'MIN_MINMAX_QUANTITY': None}, {'MAX_MINMAX_QUANTITY': None}, {'ONHAND': None}, {'ONHAND_SUBINV': None}, {'LOCATOR': None}, {'ORGANIZATION_ID': None}, {'INVENTORY_ITEM_ID': None}, {'CREATION_DATA': None}, {'INVENTORY_ITEM_STATUS_CODE': None}]

该字典包含值为 None 的所有关键元素。我有第二个字典，它可能包含第一个字典中的所有关键元素或少于第一个字典但具有相同关键元素的元素。如果键相同，我想比较这两个字典并用第二个字典的值元素替换 None 的值。第二本词典如下：

[{'ITEM_TYPE': 'SPR_ITEM_CPR'}, {'MIN_MINMAX_QUANTITY': '2'}, {'MAX_MINMAX_QUANTITY': '3'}, {'ONHAND': '162'}, {'ONHAND_SUBINV': 'RSP-SPARES'}, {'LOCATOR': '.RJG005D'}, {'ORGANIZATION_ID': '300000002445681'}, {'INVENTORY_ITEM_ID': '100000001537040'}, {'CREATION_DATE': '01-08-17'}, {'INVENTORY_ITEM_FLAG': 'Y'}, {'INVENTORY_ITEM_STATUS_CODE': 'Active'}]

结果字典应如下所示：

[{'ITEMNO': None}, {'ITEM_TYPE': 'SPR_ITEM_CPR'}, {'CASE_QTY': None}, {'MIN_MINMAX_QUANTITY': '2'}, {'MAX_MINMAX_QUANTITY': '3'}, {'ONHAND': '162'}, {'ONHAND_SUBINV': 'RSP-SPARES'}, {'LOCATOR': '.RJG005D'}, {'ORGANIZATION_ID': '300000002445681'}, {'INVENTORY_ITEM_ID': '100000001537040'}, {'CREATION_DATE': '01-08-17'}, {'INVENTORY_ITEM_FLAG': 'Y'}, {'INVENTORY_ITEM_STATUS_CODE': 'Active'}]

拉克什

尝试将 l2 转换为字典，然后形成所需的输出。

前任：

l1 = [{'ITEMNO': None}, {'ITEM_TYPE': None}, {'CASE_QYT': None}, {'MIN_MINMAX_QUANTITY': None}, {'MAX_MINMAX_QUANTITY': None}, {'ONHAND': None}, {'ONHAND_SUBINV': None}, {'LOCATOR': None}, {'ORGANIZATION_ID': None}, {'INVENTORY_ITEM_ID': None}, {'CREATION_DATA': None}, {'INVENTORY_ITEM_STATUS_CODE': None}]
l2 = [{'ITEM_TYPE': 'SPR_ITEM_CPR'}, {'MIN_MINMAX_QUANTITY': '2'}, {'MAX_MINMAX_QUANTITY': '3'}, {'ONHAND': '162'}, {'ONHAND_SUBINV': 'RSP-SPARES'}, {'LOCATOR': '.RJG005D'}, {'ORGANIZATION_ID': '300000002445681'}, {'INVENTORY_ITEM_ID': '100000001537040'}, {'CREATION_DATE': '01-08-17'}, {'INVENTORY_ITEM_FLAG': 'Y'}, {'INVENTORY_ITEM_STATUS_CODE': 'Active'}]

check_val = dict((i.keys()[0], i.values()[0]) for i in l2)
result = [{i.keys()[0]: check_val.get(i.keys()[0])} for i in l1 ]
print(result)

蟒蛇3

check_val = dict((list(i.keys())[0], list(i.values())[0]) for i in l2)
result = [{list(i.keys())[0]: check_val.get(list(i.keys())[0])} for i in l1 ]

输出：

[{'ITEMNO': None},
 {'ITEM_TYPE': 'SPR_ITEM_CPR'},
 {'CASE_QYT': None},
 {'MIN_MINMAX_QUANTITY': '2'},
 {'MAX_MINMAX_QUANTITY': '3'},
 {'ONHAND': '162'},
 {'ONHAND_SUBINV': 'RSP-SPARES'},
 {'LOCATOR': '.RJG005D'},
 {'ORGANIZATION_ID': '300000002445681'},
 {'INVENTORY_ITEM_ID': '100000001537040'},
 {'CREATION_DATA': None},
 {'INVENTORY_ITEM_STATUS_CODE': 'Active'}]

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