将包含一项的列表转换为字典值中的项本身

Question

例如，有一个包含键值对的字典，其中的值是具有不同“内容”的列表。有些列表只有一个元素。这些元素可以是不同类型的数据。

问题：将具有一个元素的列表类型键值转换为dict值中的元素本身的最有效方法是什么？

输入数据

{
  "key1": ["text"],
  "key2": ["text", "text"],
  "key3": [{"key0": 0}],
  "key4": [{"key0": 0, "key1": 1}],
  "key5": [],
  "key6": [[]],
  "key7": [["text"]],
  "key8": [[{"key0": 0}]]
}

预期输出数据

{
  "key1": "text",
  "key2": ["text", "text"],
  "key3": {"key0": 0},
  "key4": {"key0": 0, "key1": 1},
  "key5": "",
  "key6": "",
  "key7": "text",
  "key8": {"key0": 0}
}

我试过的

{k: v[0] for k, v in dct.items()}

带有问题和评论的我的代码输出

{
 'key1': 'text',
 'key2': 'text', <- ISSUE 1 should be ["text", "text"]
 'key3': {'key0': 0},
 'key4': {'key0': 0, 'key1': 1},
 'key5': <- ISSUE 2 Raises ERROR "IndexError: list index out of range" which is quite expected, should be ""
 'key6': [], <- ISSUE 3 should be ""
 'key7': ['text'], <- ISSUE 4 should be "text"
 'key8': [{'key0': 0}] <- ISSUE 5 should be {'key0': 0}
}

将 Input data 中的字典转换为 Expected output data 中的字典的最有效方法是什么？

Answer 1

为了解决这个问题，我专注于如何转换值：我创建了一个名为 delist 的函数来删除包含 1 个元素的列表：

def delist(value):
    while isinstance(value, list) and len(value) == 1:
        value = value[0]
    if value == []:
        value = ""
    return value

希望逻辑很容易理解。使用它：

data = {
  "key1": ["text"],
  "key2": ["text", "text"],
  "key3": [{"key0": 0}],
  "key4": [{"key0": 0, "key1": 1}],
  "key5": [],
  "key6": [[]],
  "key7": [["text"]],
  "key8": [[{"key0": 0}]]
}

new_data = {k: delist(v) for k, v in data.items()}

而new_data是

{'key1': 'text',
 'key2': ['text', 'text'],
 'key3': {'key0': 0},
 'key4': {'key0': 0, 'key1': 1},
 'key5': '',
 'key6': '',
 'key7': 'text',
 'key8': {'key0': 0}}

Answer 2

也许不是最优雅的（实际上很丑陋）但可以解决问题，以防我不会收到更智能的东西。

# input data
dct = {
  "key1": ["text"],
  "key2": ["text", "text"],
  "key3": [{"key0": 0}],
  "key4": [{"key0": 0, "key1": 1}],
  "key5": [],
  "key6": [[]],
  "key7": [["text"]],
  "key8": [[{"key0": 0}]]
}

# define function
def delist(dct):
    new_dct = {}
    for k, v in dct.items():
        if v:
            if len(v) > 1:
                new_dct[k] = v
            else:
                if v[0]:
                    if type(v[0])==list:
                        new_dct[k] = v[0][0]
                    else:
                        new_dct[k] = v[0]
                else:
                    new_dct[k] = ""         
        else:
            new_dct[k] = ""
    return new_dct

# apply function to input data
delist(dct)

Returns

{
 'key1': 'text',
 'key2': ['text', 'text'],
 'key3': {'key0': 0},
 'key4': {'key0': 0, 'key1': 1},
 'key5': '',
 'key6': '',
 'key7': 'text',
 'key8': {'key0': 0}
}

Answer 3

这里有一个 one-liner 使用 np.ravel()

d = {
  "key1": ["text"],
  "key2": ["text", "text"],
  "key3": [{"key0": 0}],
  "key4": [{"key0": 0, "key1": 1}],
  "key5": [],
  "key6": [[]],
  "key7": [["text"]],
  "key8": [[{"key0": 0}]]
}

out = {k: list(np.ravel(v))  if np.ravel(v).size>1 else (np.ravel(v)[0] if np.ravel(v).size==1 else '')  for k,v in d.items()}

输出：

{'key1': 'text',
 'key2': ['text', 'text'],
 'key3': {'key0': 0},
 'key4': {'key0': 0, 'key1': 1},
 'key5': '',
 'key6': '',
 'key7': 'text',
 'key8': {'key0': 0}}

将包含一项的列表转换为字典值中的项本身

Convert list with one item to item itself in dict value

python

dictionary

list

python-3.x

one-liner