问题描述
我有一本字典,它的键映射到一个值列表。我正在尝试创建一个函数,该函数输出一个 dict,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,应该被持久化,列表的其他元素应该映射到字典中的这个元素。但是如果之前的第一个元素链接到另一个值,它应该链接到那个
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(bc 是第一个元素)。但由于“apples”链接到“fruit1”和“fruit2”,“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”之前已链接到“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)
键属性: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
有人就如何获得正确的输出提出建议吗?我在我的代码中维护一个 dict 捕获已重新链接的 dict 值,因此我始终可以将当前值路由到该值。虽然,似乎是某个地方的错误
解决方法
所以我想这不是一个灵活的解决方案,但它可以解决您的问题:
d = {
'apples': ['fruit1','fruit2'],'orange': ['fruit3','round'],'grape':['fruit2','fruit5'],'mango': ['round']
}
new = {}
# invraping dict
for k in d:
v = d[k]
for i in range(1,len(v)):
new[v[i]] = v[0]
new[k] = v[0]
# appliing your special rules
for k in new:
v = new[k]
if v in new:
new[k] = new[v]
print(new)
,
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'}
我们将它存储在新的字典中,在存储之前我们会不断检查是否有任何链接已经存储为值,如果是,那么我们使用链接值而不是创建新链接,否则我们创建新链接。