我有一本字典,其键映射到值列表。我正在尝试创建一个函数,该函数输出一个字典,其键仅映射到一个值。在字典中,如果键映射到元素列表,则列表的第一个元素是正确的值,并且应该被持久化,并且列表的其他元素应该映射到字典中的该元素。但是,如果第一个元素之前已链接到另一个值,则它应该链接到该值
示例
输入:d = {
'apples': ['fruit1', 'fruit2'],
'orange': ['fruit3', 'round'],
'grape':['fruit2', 'fruit5'],
'mango': ['round']
}
预期输出:
o = {'apples': 'fruit1', # since fruit1 was the first element
'fruit2': 'fruit1', # fruit2 should link to the first element (fruit1)
'orange': 'fruit3', # first element persisted
'round': 'fruit3', # second element, round, links to the first, fruit3
'grape': 'fruit1', # should keep first element fruit2, but since fruit2 linked to fruit1 earlier, link to fruit1
'fruit5': 'fruit1', # since fruit2 links to fruit1
'mango': 'fruit3' # since round links to fruit 3
}
在此示例中,“apples”链接到输入中的fruit1 和fruit2。 “fruit1”应该是持续存在的值(因为它是第一个元素)。但由于“apples”链接到“fruit1”和“fruit2”,因此“fruit2”也应该链接到“fruit1”。
然后,当“grape”映射到“fruit2”时,“grape”应该重新链接到“fruit1”,因为“fruit2”之前链接到“fruit1”。同样,输出中的“mango”映射到“fruit3”,因为“round”之前链接到“fruit3”(对于橙色)
Key 属性:键中不存在字典的任何值
我的代码:
new_d = {}
relinked_items = {}
for key, values in d.items():
if len(values) == 1:
value = values[0]
if key not in new_d:
# if value has been relinked before, link to that
if value in relinked_items:
new_d[key] = relinked_items[value]
# hasnt been relinked
else:
new_d[key] = value
continue
target_value = values[0]
# link key to target value
new_d[key] = target_value
for value in values[1:]:
if target_value in relinked_items:
new_d[value] = relinked_items[target_value]
# hasnt been relinked
else:
new_d[value] = target_value
我的输出
{'apples': 'fruit1', # correct
'fruit2': 'fruit1', # correct
'fruit5': 'fruit2', # wrong. fruit2 maps to fruit1
'grape': 'fruit2', # wrong fruit2 maps to fruit1
'mango': 'round', # wrong. round maps to fruit3
'orange': 'fruit3', # correct
'round': 'fruit3'} # correct
有人建议如何获得正确的输出吗?我在代码中维护一个字典,捕获已重新链接的字典的值,因此我始终可以将当前值路由到该字典。虽然,似乎是某个地方的错误
最佳答案
这是我的方法:
q = dict()
for k, values in d.items():
c = values[0]
q[k] = q.get(c, c)
for v in range(1, len(values)):
q[values[v]] = q.get(c, c)
Output :
{'apples': 'fruit1',
'fruit2': 'fruit1',
'fruit5': 'fruit1',
'grape': 'fruit1',
'mango': 'fruit3',
'orange': 'fruit3',
'round': 'fruit3'}
我们将其存储在新的字典中,在存储之前,我们不断检查是否有任何链接已存储值,如果是,则我们使用链接值而不是创建新链接,否则我们创建新链接。
关于python - 解析字典以具有一对一的键值映射,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/67402501/