如何将字典键转换为值

Question

我一直试图用户的美丽汤解析一个网站，并获得必要的数据。现在我有了数据，我想用熊猫来组织数据，但为此，我想制作一本结构良好的字典。

我清理过的数据在字典data_dict中。这里，键是海龟的名字，值是有关海龟的一些信息。

data_dict = {'Aesop': ['AGE: 7 Years Old', 'WEIGHT: 6 lbs', 'SEX: Female', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie'], 'Caesar': ['AGE: 2 Years Old', 'WEIGHT: 4 lbs', 'SEX: Male', 'BREED: Greek Tortoise', 'SOURCE: hatched in house'], 'Sulla': ['AGE: 1 Year Old', 'WEIGHT: 1 lb', 'SEX: Male', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie']}

现在，我希望这本字典的键类似于"turtle_1“和下一个"turtle_2”等等。所以我写了代码：

name_num = 1
for name in list(data_dict.keys()):
    data_dict["turtle_"+str(name_num)] = data_dict.pop(name)
    name_num += 1

这改变了"Aesop"的钥匙..。(等等)到"turtle_1"..。诸若此类。现在，我希望海龟的名字(以前是键)是相应键的值。我希望data_dict看起来像这样：

data_dict = {'turtle_1': ['AGE: 7 Years Old', 'WEIGHT: 6 lbs', 'SEX: Female', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie', 'NAME: Aesop'], 'turtle_2': ['AGE: 2 Years Old', 'WEIGHT: 4 lbs', 'SEX: Male', 'BREED: Greek Tortoise', 'SOURCE: hatched in house', 'NAME: Caesar'], 'turtle_3': ['AGE: 1 Year Old', 'WEIGHT: 1 lb', 'SEX: Male', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie', 'NAME: Sulla']}

因此，键得到一个新的值，这是每个海龟的名字。谢谢，我非常感谢你的帮助。

Answer 1

附加姓名：

data_dict = {'Aesop': ['AGE: 7 Years Old', 'WEIGHT: 6 lbs', 'SEX: Female', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie'], 'Caesar': ['AGE: 2 Years Old', 'WEIGHT: 4 lbs',
                                                                                                                                                               'SEX: Male', 'BREED: Greek Tortoise', 'SOURCE: hatched in house'], 'Sulla': ['AGE: 1 Year Old', 'WEIGHT: 1 lb', 'SEX: Male', 'BREED: African Aquatic Sideneck Turtle', 'SOURCE: found in Lake Erie']}
name_num = 1
for name in list(data_dict.keys()):
    data_dict["turtle_"+str(name_num)] = data_dict.pop(name)
    data_dict["turtle_"+str(name_num)].append('NAME: '+name)
    name_num += 1

print(data_dict)

Answer 2

key_map = {}

for i,k in enumerate(data_dict.keys()):
    key_map['turtle_'+str(i+1)] = k # store reverse map from turtle_1 -> name1
    data_dict['turtle_'+str(i+1)] = data_dict.pop(k)

• 现在，您可以使用turtle_1访问哪个海龟对应于key_map。

>>> key_map
{'turtle_1': 'Aesop', 'turtle_2': 'Caesar', 'turtle_3': 'Sulla'}

Answer 3

就地:只是将另一个元素添加到与键(或名称)对应的列表中。

key_count = 1
for name in list(data_dict.keys()):
    newlist = data_dict.pop(name)
    newlist.append('NAME : ' + name)
    data_dict['turtle_' + str(key_count)] = newlist
    key_count = key_count +1