python - 我如何遍历字典列表并合并字典以形成新的更短的字典列表？

Question

我有一个包含'price'、'tickettype' 的航空公司航类票价列表，指示票价是否为“单程”(相对于往返并通过整数代码映射到另一个旅程列表。但我收到的列表是重复的。

[
{'price' : 1800, 'oneway' : 1, 'inboundJourneys' : [], "outboundJourneys": [3], 'tickettypecode' : 'SDS'},
{'price' : 1800, 'oneway' : 1, 'inboundJourneys' : [9,10,11], "outboundJourneys": [], 'tickettypecode' : 'SDS'},
{'price' : 1800, 'oneway' : 1, 'inboundJourneys' : [14,16], "outboundJourneys": [], 'tickettypecode' : 'SDS'},
{'price' : '2300', 'oneway' : 1, 'inboundJourneys' : [], "outboundJourneys": [6,8,9], 'tickettypecode' : 'TAR'},
{'price' : 2300, 'oneway' : 1, 'inboundJourneys' : [12,13,14], "outboundJourneys": [3], 'tickettypecode' : 'TAR'},
{'price' : 900, 'oneway' : 1, 'inboundJourneys' : [], "outboundJourneys": [18,19,20], 'tickettypecode' : 'GED'},
{'price' : 900, 'oneway' : 1, 'inboundJourneys' : [14,16,17], "outboundJourneys": [], 'tickettypecode' : 'GED'},
{'price' : 1200, 'oneway' : 1, 'inboundJourneys' : [], "outboundJourneys": [25], 'tickettypecode' : 'ABC'},
{'price' : 1200, 'oneway' : 1, 'inboundJourneys' : [32], "outboundJourneys": [], 'tickettypecode' : 'ABC'}
]

我需要的是:

其中 'price' 相等且 'tickettypecode' 相等且 'oneway' 相等，列表中有一个字典，因此结束于:

[
{'price' : 1800, 'oneway' : 1, 'inboundJourneys' : [9,10,11,14,16], "outboundJourneys": [3], 'tickettypecode' : 'SDS'},
{'price' : 2300, 'oneway' : 1, 'inboundJourneys' : [12,13,14], "outboundJourneys": ['6,8,9'], 'tickettypecode' : 'TAR'},
{'price' : 900, 'oneway' : 1, 'inboundJourneys' : [14,16,17], "outboundJourneys": [18,19,20], 'tickettypecode' : 'GED'},
{'price' : 1200, 'oneway' : 1, 'inboundJourneys' : [32], "outboundJourneys": [25], 'tickettypecode' : 'ABC'}
]

我已经尝试了很多方法，但我很难过。

Answer 1

假设合并列表中项目的顺序无关紧要，只需浏览列表中的每个项目，如果您以前没有看到它，就复制它，如果您有，则合并这些字段。

merged = {}

for item in original:
    key = (item['price'], item['tickettypecode'], item['oneway'])
    if key in merged:
        for mergekey in ['inboundJourneys','outboundJourneys']:
            # assign extended copy rather than using list.extend()
            merged[key][mergekey] = merged[key][mergekey] + item[mergekey]
    else:
        merged[key] = item.copy()

mergedlist = merged.values()