我如何迭代字典列表并合并字典以形成新的更短的dict列表？

Question

我有一个航空公司的机票价格列表，它有一个'price'，一个'tickettype'，表示票价是否为“单程”（与往返相反，并且通过一个整数代码映射到另一个行程列表）。但是我收到的名单是重复的。在

[
{'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'等于，'tickettypecode'等于，而{}等于，则列表中有一个字典，因此结尾为：

^{pr2}$

我试过很多方法，但都被难住了。在

Answer 1

效率极低的解决方案，但起点是：

answer = []
for myDict in myList:
    for d in answer:
        if d['oneway']==myDict['oneway'] and d['price']==myDict['price'] and d['tickettype']==myDict['tickettype']:
            break
    else:
        answer.append(myDict)

希望这有帮助

Answer 2

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，如果以前没有看到过，则将其复制；如果有，则合并字段。在

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()

Answer 3

我会这样做：

import copy

def merge(iterable, keys, update):
    merged = {}
    for d in iterable:
        merge_key = tuple(d[k] for k in keys)
        m = merged.get(merge_key)
        if m:
            for u in update:
                m[u].extend(d[u])
        else:
            merged[merge_key] = copy.deepcopy(d)

    return list(merged.values())  # list(dict_view)

我在你的考试中测试过：

^{pr2}$

我得到了：

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