Python - 删除列表之间的非唯一元素

Question

我有一个图像列表的字典,属于一类图像,如狗和猫.一些图像包含图像中的狗和猫,我想删除这些图像.

可以说我有

{'cat':[1,2,3], 'dog':[2,3,4]}

我们可以看到id为2和3的图像都有猫和狗.我想要排除这些图像以获得以下内容:

[[1],[4]]

到目前为止我试过这个:

from collections import Counter
img_ids = {'cat':[1,2,3], 'dog':[2,3,4]}
flattened = [item for sublist in img_ids.values() for item in sublist]
flattened_unique = [k for k, v in dict(Counter(flattened)).items() if v < 2]
filtered_ids_dfs = []
for key, val in img_ids.items():
  filtered = [x for x in val if x in flattened_unique]
  filtered_ids_dfs.append(filtered)
print(filtered_ids_dfs)

对此有更好或更优雅的解决方案吗？也可能有任意数量的类,所以我们的字典可能有猫,狗,鸡等.

Answer 1

首先,计算每个图像有多少个对象(例如猫,狗).然后找到只有一个对象的图像(唯一图像).最后,使用字典理解来查找唯一图像列表中的图像.

from collections import Counter

d = {'cat':[1,2,3], 'dog':[2,3,4], 'chicken': [2, 4, 5, 6]}

c = Counter([item for items in d.values() for item in items])
unique_images = set(k for k, count in c.iteritems() if count == 1)  # .items() in Python3

>>> {k: [item for item in items if item in unique_images] for k, items in d.iteritems()}  # .items() in Python3
{'cat': [1], 'chicken': [5, 6], 'dog': []}

Answer 2

只需使用套装:

d = {'cat':[1,2,3], 'dog':[2,3,4]}
common = set(d['cat']) & set(d['dog'])
out = [list(set(d['cat']) - common), list(set(d['dog']) - common)]

将其扩展到两个以上的键:

common = set.intersection(*(set(v) for k,v in d.items()))
out = [list(set(v) - common) for k,v in d.items()]