Question

我有一个如下列表。

list1 = [
    ('Ram','Laxman','Bharat','Sita'),
    ('Ram','Ravan','Bharat','Sita'),
    ('Ram','Luv','Dashrat','Sita'),
    ('Dasrath','Kekei','Bharat','Ram'),
    ('Laxman','Bharat','Ram','Hanuman'),
    ('Hanuman','Sita','Kekei','Ravan'),
    ('Ram','Sita','Hanuman','Ravan')
]

我要过滤具有至少3个元组值匹配的列表数据，如果2个或多个元组具有至少3个值匹配，则列表中仅第一个元组应与其余元素一起存在。例如，在上面的列表中，下面的元组具有3个值匹配。

result = [
    ('Ram','Laxman','Bharat','Sita'),
    ('Ram','Luv','Dashrat','Sita'),
    ('Dasrath','Kekei','Bharat','Ram'),
    ('Hanuman','Sita','Kekei','Ravan')
]

Answer 1

from operator import add
from functools import reduce

def solve(xss):
    
    mems = [xss[0]]
    
    for xs in xss[1:]:        
        if len(set(reduce(add,mems)).intersection(set(xs))) < 3:
            mems = mems + [xs]
    return mems

Answer 2

我认为您的问题很有趣。我使用一些矩阵运算来解决它。我认为，这种方式将比通用循环快得多。

from collections import OrderedDict
import numpy as np
from sklearn.preprocessing import MultiLabelBinarizer

class DuplicateFinder(object):
    def __init__(self, raw_list):
        self.raw_list = raw_list
        self.multi_label_encoder = MultiLabelBinarizer()
        self.final_results = OrderedDict()
        self.title_select = []
        self.content_select = []
        self.not_select = []

    @property
    def match_result(self):
        label_matrix = self.multi_label_encoder.fit_transform(self.raw_list)
        return np.dot(label_matrix, label_matrix.T)

    @property
    def raw_results(self):
        return np.array(np.where(self.match_result >= 3)).T

    def solve(self):
        for result in self.raw_results:
            if result[0] == result[1]:
                continue
            if result[0] in self.content_select:
                continue
            if result[0] not in self.final_results:
                self.final_results[result[0]] = []
                self.final_results[result[0]].append(result[1])
                self.title_select.append(result[0])
                self.content_select.append(result[1])
            elif result[1] not in self.content_select + self.title_select:
                self.final_results[result[0]].append(result[1])
                self.content_select.append(result[1])
            else:
                continue
        self.not_select = list(set(range(self.match_result.shape[0])) - set(
            self.title_select + self.content_select
        ))

    def print_result(self):
        print(f"This is more than one matched: {self.final_results}")
        for key in self.final_results:
            print(self.raw_list[key])
        print(f"This is just one: {self.not_select}")
        for key in self.not_select:
            print(self.raw_list[key])

list1 = [
    ("Ram", "Laxman", "Bharat", "Sita"),
    ("Ram", "Ravan", "Bharat", "Sita"),
    ("Ram", "Luv", "Dashrat", "Sita"),
    ("Dasrath", "Kekei", "Bharat", "Ram"),
    ("Laxman", "Bharat", "Ram", "Hanuman"),
    ("Hanuman", "Sita", "Kekei", "Ravan"),
    ("Ram", "Sita", "Hanuman", "Ravan"),
]
solver = DuplicateFinder(list1)
solver.solve()
solver.print_result()

这是结果：

This is more than one matched: OrderedDict([(0, [1, 4]), (5, [6])])
('Ram', 'Laxman', 'Bharat', 'Sita')
('Hanuman', 'Sita', 'Kekei', 'Ravan')
This is just one: [2, 3]
('Ram', 'Luv', 'Dashrat', 'Sita')
('Dasrath', 'Kekei', 'Bharat', 'Ram')

过滤元组数据列表

2 个答案: