结果列表到元组（如描述中所述，使用 tuple 函数）并且结果元组需要在列表中

Question

以下是我从字典列表和元组中的排序数据中提取的代码

def compress(data):
    res = []
    for idx, sub in enumerate(data, start=0):
        if idx == 0:
            res.append(tuple(sub.keys()))
            res.append(list(sub.values()))
        else:
            res.append(list(sub.values()))
    return tuple(res)

data = [
    {"a": 1, "b": 2, "c": 3},
    {"a": 4, "c": 6, "b": 5}
]

print(compress(data))

预期结果是：

(('a', 'b', 'c'), [(1, 2, 3), (6, 5, 4)])

我的代码无法返回预期的结果。

Answer 1

应该是

def compress(data):
    res = []
    for idx, sub in enumerate(data, start=0):
        if idx == 0:
            res.append(tuple(sub.keys()))
            res.append([])
            res[-1].append(tuple(sub.values()))
        else:
            res[-1].append(tuple(sub.values()))
    return tuple(res)

Answer 2

使用列表推导：

def compress(data):
  keys = tuple(sorted(data[0].keys()))
  values = [tuple(d[k] for k in keys) for d in data]
  return (keys, values)

>>> compress([{"a": 1, "b": 2, "c": 3},{"a": 4, "c": 6, "b": 5}])
(('a', 'b', 'c'), [(1, 2, 3), (4, 5, 6)])

可选：缺少键

如果某些字典中可能缺少某些键，您可以使用所有字典中的所有键，然后使用set 删除重复的键，然后使用d.get(k, default_value) 而不是d[k]：

def compress(data, default_value=None):
  keys = tuple(sorted(set(k for d in data for k in d.keys())))
  values = [tuple(d.get(k, default_value) for k in keys) for d in data]
  return (keys, values)

>>> data = [{'a':1, 'b': 2, 'c': 3}, {'a':11, 'b':12, 'd':14}]
>>> compress(data, 0)
(('a', 'b', 'c', 'd'), [(1, 2, 3, 0), (11, 12, 0, 14)])

可选：存储此数据的另一种方式

您可以将此字典列表重构为列表字典：

def refactor(data):
  keys = data[0].keys()
  return { k: [d[k] for d in data] for k in keys }

>>> refactor([{"a": 1, "b": 2, "c": 3},{"a": 4, "c": 6, "b": 5}])
{'a': [1, 4], 'b': [2, 5], 'c': [3, 6]}

同样，您可以小心丢失键：

def refactor(data):
  keys = set(k for d in data for k in d.keys())
  return { k: [d[k] for d in data if k in d] for k in keys }

>>> refactor([{'a':1, 'b': 2, 'c': 3}, {'a':11, 'b':12, 'd':14}])
{'d': [14], 'a': [1, 11], 'c': [3], 'b': [2, 12]}

Answer 3

所以我创建了你想要的，你可以把它放在一个函数中，如果你愿意的话

data = [
    {"a": 1, "b": 2, "c": 3},
    {"a": 4, "c": 6, "b": 5}
]

keys = [[],[]]
for d in data:
    items = list(d.items())
    l = []
    for i in items:
        l.append(i[1])
        if i[0] not in keys[0]:
            keys[0].append(i[0])
        else:
            continue
    keys[1].append(tuple(l))

keys = tuple(keys)

Answer 4

def compress(data):
    result = []    
    keys = tuple(data[0].keys()) # the keys
    result.append(keys)
    result2 = []
    result2.append(tuple([data[0][key] for key in keys]))
    result2.append(tuple([data[1][key] for key in keys[::-1]]))
    result.append(result2)
    return tuple(result)

data = [
    {"a": 1, "b": 2, "c": 3},
    {"a": 4, "c": 6, "b": 5}
]

print(compress(data))

打印：

(('a', 'b', 'c'), [(1, 2, 3), (6, 5, 4)])