Question

我有以下列表：

values = [
    ['registrationController', 'regBean', 'firstName'],
    ['registrationController', 'regBean', 'surname'],
    ['registrationController', 'regBean', 'userName'],
    ['registrationController', 'regBean', 'password'],
    ['registrationController', 'regBean', 'confirmPassword'],
    ['registrationController', 'regBean', 'emailAddress'],
    ['registrationController', 'regBean', 'confirmEmail'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerOne', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerOne', 'answer'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerTwo', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerTwo', 'answer'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerThree', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerThree', 'answer'],
    ['registrationController', 'regBean', 'tAndCAccepted']
]

我正在尝试弄清楚如何删除此列表中所有列表中预设的值，并将这些值保留在当他们成为唯一的某人时得到这样的东西：

unique_values = [
    ['firstName'],
    ['surname'],
    ['userName'],
    ['password'],
    ['confirmPassword'],
    ['emailAddress'],
    ['confirmEmail'],
    ['securityQuestionAndAnswerOne', 'question'],
    ['securityQuestionAndAnswerOne', 'answer'],
    ['securityQuestionAndAnswerTwo', 'question'],
    ['securityQuestionAndAnswerTwo', 'answer'],
    ['securityQuestionAndAnswerThree', 'question'],
    ['securityQuestionAndAnswerThree', 'answer'],
    ['tAndCAccepted']
]

有什么想法可以实现吗？我尝试了各种方法，但无法真正找到可行的解决方案。

Answer 1

使用集合交集获取所有公共元素，并使用嵌套列表理解来构建清除列表：

common = set(values[0])
for lst in values[1:]:
    common = common.intersection(lst)

unique_values = [[v for v in lst if v not in common] for lst in values]

Answer 2

这是一个易于理解和阅读的版本。它将保留顺序。

common = set.intersection(*values)
reduced_values = [[value for value in l if value not in common] for l in values]

Answer 3

具有列表理解：

>>> [
    [elt for elt in line if not all(elt in subline for subline in values)]
    for line in values
]

[['firstName'], ['surname'], ['userName'], ['password'], ['confirmPassword'], ['emailAddress'], ['confirmEmail'], ['securityQuestionAndAnswerOne', 'question'], ['securityQuestionAndAnswerOne', 'answer'], ['securityQuestionAndAnswerTwo', 'question'], ['securityQuestionAndAnswerTwo', 'answer'], ['securityQuestionAndAnswerThree', 'question'], ['securityQuestionAndAnswerThree', 'answer'], ['tAndCAccepted']]

Answer 4

假设每个值在列表中都是唯一的，则可以执行以下操作：

from collections import Counter
from itertools import chain

values = [
    ['registrationController', 'regBean', 'firstName'],
    ['registrationController', 'regBean', 'surname'],
    ['registrationController', 'regBean', 'userName'],
    ['registrationController', 'regBean', 'password'],
    ['registrationController', 'regBean', 'confirmPassword'],
    ['registrationController', 'regBean', 'emailAddress'],
    ['registrationController', 'regBean', 'confirmEmail'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerOne', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerOne', 'answer'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerTwo', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerTwo', 'answer'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerThree', 'question'],
    ['registrationController', 'regBean', 'securityQuestionAndAnswerThree', 'answer'],
    ['registrationController', 'regBean', 'tAndCAccepted']
]

counts = Counter(chain.from_iterable(values))
result = [[e for e in value if counts[e] != len(values)] for value in values]

print(result)

输出

[['firstName'], ['surname'], ['userName'], ['password'], ['confirmPassword'], ['emailAddress'], ['confirmEmail'], ['securityQuestionAndAnswerOne', 'question'], ['securityQuestionAndAnswerOne', 'answer'], ['securityQuestionAndAnswerTwo', 'question'], ['securityQuestionAndAnswerTwo', 'answer'], ['securityQuestionAndAnswerThree', 'question'], ['securityQuestionAndAnswerThree', 'answer'], ['tAndCAccepted']]

想法是对值中的每个元素进行计数，并获得未出现在值的所有元素中的元素：

Answer 5

trunc_lists = [tuple(sublist[2:]) for sublist in values]
ans = [list(i) for i in set(trunc_lists)]

请注意，输出是无序的

删除多个嵌套列表中的重复列表项

5 个答案: