如何按键拆分字典列表答案

【问题标题】：How to split a list of dicts by key如何按键拆分字典列表
【发布时间】：2021-10-26 22:30:49
【问题描述】：

我有以下字典列表：

list_of_dicts = [
    {"type": "X", "hour": 22},
    {"type": "A", "measure": "1"},
    {"type": "B", "measure": "2"},
    {"type": "X", "hour": 23},
    {"type": "A", "measure": "3"},
    {"type": "X", "hour": 24},
    {"type": "A", "measure": "4"},
    {"type": "B", "measure": "5"},
    {"type": "C", "measure": "6"}
]

如何将其拆分为一个 dict，其中键是 'type' = 'X' dicts 中的 'hour' 值，而值是两个 'type' = 'X' dicts 之间的其他 dicts？这就是我想使用这个例子获得的，但是两个 'type' = 'X' dicts 之间的间隔可以是可变的。

dict_of_dicts = {
    22: [
        {"type": "A", "measure": "1"},
        {"type": "B", "measure": "2"},
    ],
    23:[
        {"type": "A", "measure": "3"}
    ],
    24:[
        {"type": "A", "measure": "4"},
        {"type": "B", "measure": "5"},
        {"type": "C", "measure": "6"},
    ]
}

提前致谢！

【问题讨论】：

如果您想手动完成，我建议您尝试用文字描述进行此转换所需的步骤。

标签： python dictionary

【解决方案1】：

这段代码应该可以解决问题。
每次遇到包含“小时”键的字典时，它都会在结果字典中创建新元素。

res = {}
cur_key = None

for d in list_of_dicts:
    if "hour" in d:
        cur_key = d["hour"]
        res[cur_key] = []
    elif cur_key is not None:
        res[cur_key].append(d)

【讨论】：

【解决方案2】：

这是另一种使用列表理解和一个巧妙的小技巧（借用自 here）的方法来使 while 循环在其中工作：

list_of_dicts = [
    {"type": "X", "hour": 22},
    {"type": "A", "measure": "1"},
    {"type": "B", "measure": "2"},
    {"type": "X", "hour": 23},
    {"type": "A", "measure": "3"},
    {"type": "X", "hour": 24},
    {"type": "A", "measure": "4"},
    {"type": "B", "measure": "5"},
    {"type": "C", "measure": "6"}
]

def while_generator(lst):
    i = 0
    while i < len(lst) and lst[i]['type'] != 'X':
        yield lst[i]
        i += 1

dict_of_dicts = {
    d['hour']: [e for e in while_generator(list_of_dicts[i+1:])]
    for i, d in enumerate(list_of_dicts) if d['type'] == 'X'
}

print(dict_of_dicts)

打印：

{
    22: [
        {'type': 'A', 'measure': '1'}, 
        {'type': 'B', 'measure': '2'}
    ], 
    23: [
        {'type': 'A', 'measure': '3'}
    ], 
    24: [
        {'type': 'A', 'measure': '4'}, 
        {'type': 'B', 'measure': '5'}, 
        {'type': 'C', 'measure': '6'}
    ]
}

【讨论】：

【解决方案3】：

跟踪当前的“type X”列表以向其中添加其他字典。

list_of_dicts = [
    {"type": "X", "hour": 22},
    {"type": "A", "measure": "1"},
    {"type": "B", "measure": "2"},
    {"type": "X", "hour": 23},
    {"type": "A", "measure": "3"},
    {"type": "X", "hour": 24},
    {"type": "A", "measure": "4"},
    {"type": "B", "measure": "5"},
    {"type": "C", "measure": "6"}
]

dict_of_dicts = dict()
for d in list_of_dicts:
    if 'hour' in d: dict_of_dicts[d['hour']] = subList = []
    else:           subList.append(d)

print(dict_of_dicts)
{ 22: [{'type': 'A', 'measure': '1'},
       {'type': 'B', 'measure': '2'}],
  23: [{'type': 'A', 'measure': '3'}],
  24: [{'type': 'A', 'measure': '4'},
       {'type': 'B', 'measure': '5'},
       {'type': 'C', 'measure': '6'}]}

它可以在这样的理解中做到这一点，但它有点令人费解：

dict_of_dicts = { d['hour']:sl.append([]) or sl[-1]
                  for sl in [[]] for d in list_of_dicts
                  if 'hour' in d or sl[-1].append(d) }

【讨论】：