如何在多个列表之间找到一个交集？

Question

我有多个数组，我想在它们之间找到一个交集，我尝试了下面的代码。

my_lists = [['Finish', 'Purpose', 'Form', 'Series', 'Tiles Type', 'Finishing'], ['Color', 'Thickness', 'Usage/Application', 'Brand', 'Marble Type', 'Material'], ['Color', 'Brand', 'Finishing', 'Origin', 'Marble Type', 'Thickness'], ['Thickness', 'Form', 'Size', 'Series', 'Usage/Application', 'Finishing'], ['Thickness', 'Material Grade', 'Size', 'Usage/Application', 'Material'], ['Usage/Application', 'Form', 'Finishing', 'Brand', 'Material', 'Shape'], ['Application Area', 'Form', 'Finishing', 'Brand', 'Color', 'Coverage Area'], ['Usage/Application', 'Marble Type', 'Thickness', 'Brand', 'Form'], ['Unit Size (mm X mm)', 'Marble Type', 'Thickness', 'Finishing', 'Usage', 'Brand'], ['Marble Type', 'Unit Size (mm X mm)', 'Usage', 'Thickness', 'Color'], ['color'], ['Thickness', 'Size', 'Usage/Application', 'Series', 'Finish', 'Marble Type'], ['Thickness', 'Usage/Application', 'Brand', 'Color', 'Marble Type', 'Unit Size (mm X mm)'], ['Color', 'Marble Type', 'Usage'], ['Thickness', 'Size', 'Material', 'Finish', 'Packaging Size', 'Packaging Type'], ['Color', 'Material', 'Thickness', 'Usage/Application', 'Back Lit', 'Brand'], ['Material', 'Pattern', 'Shape'], ['Form', 'Application Area', 'Material', 'Thickness', 'Colour', 'Finishing'], ['Color', 'Usage/Application', 'Brand', 'Series'], ['Color', 'Material', 'Thickness', 'Usage/Application', 'Brand', 'Surface Finish'], ['Brand', 'Color', 'Usage/Application', 'Thickness', 'Size', 'Finish'], ['Form', 'Material', 'Usage', 'Marble Type', 'Thickness', 'Finishing'], ['Form', 'Color', 'Marble Type', 'Unit Size', 'Features', 'Coverage Area'], ['Usage', 'Form'], ['Finish', 'Application Area', 'Purpose', 'Thickness', 'Pattern'], ['Usage/Application', 'Finishing', 'Material', 'Brand', 'Size', 'Category Type'], ['Usage/Application', 'Size', 'Color', 'Marble Type', 'Features', 'Finishing'], ['Marble Type', 'Surface Finishing', 'Stone Form', 'Usage'], ['Brand', 'Material', 'Finish', 'Thickness', 'Size']]
print(set.intersection(*map(set,list(my_lists ))))

但我得到了一套空的

set()

我真正想要的是在所有的列表中找到共同的元素

Answer 1

在示例中的所有列表中都没有公共元素--您可以看到第一个和第二个列表是完全不相交的。因此，空集的正确返回答案。此操作将只找到每个列表中的任何字符串。

编辑

如果您的目标是查找重复使用的字符串，我将执行如下操作：

import numpy as np
my_lists = [['Finish', 'Purpose', 'Form', 'Series', 'Tiles Type', 'Finishing'], ['Color', 'Thickness', 'Usage/Application', 'Brand', 'Marble Type', 'Material'], ['Color', 'Brand', 'Finishing', 'Origin', 'Marble Type', 'Thickness'], ['Thickness', 'Form', 'Size', 'Series', 'Usage/Application', 'Finishing'], ['Thickness', 'Material Grade', 'Size', 'Usage/Application', 'Material'], ['Usage/Application', 'Form', 'Finishing', 'Brand', 'Material', 'Shape'], ['Application Area', 'Form', 'Finishing', 'Brand', 'Color', 'Coverage Area'], ['Usage/Application', 'Marble Type', 'Thickness', 'Brand', 'Form'], ['Unit Size (mm X mm)', 'Marble Type', 'Thickness', 'Finishing', 'Usage', 'Brand'], ['Marble Type', 'Unit Size (mm X mm)', 'Usage', 'Thickness', 'Color'], ['color'], ['Thickness', 'Size', 'Usage/Application', 'Series', 'Finish', 'Marble Type'], ['Thickness', 'Usage/Application', 'Brand', 'Color', 'Marble Type', 'Unit Size (mm X mm)'], ['Color', 'Marble Type', 'Usage'], ['Thickness', 'Size', 'Material', 'Finish', 'Packaging Size', 'Packaging Type'], ['Color', 'Material', 'Thickness', 'Usage/Application', 'Back Lit', 'Brand'], ['Material', 'Pattern', 'Shape'], ['Form', 'Application Area', 'Material', 'Thickness', 'Colour', 'Finishing'], ['Color', 'Usage/Application', 'Brand', 'Series'], ['Color', 'Material', 'Thickness', 'Usage/Application', 'Brand', 'Surface Finish'], ['Brand', 'Color', 'Usage/Application', 'Thickness', 'Size', 'Finish'], ['Form', 'Material', 'Usage', 'Marble Type', 'Thickness', 'Finishing'], ['Form', 'Color', 'Marble Type', 'Unit Size', 'Features', 'Coverage Area'], ['Usage', 'Form'], ['Finish', 'Application Area', 'Purpose', 'Thickness', 'Pattern'], ['Usage/Application', 'Finishing', 'Material', 'Brand', 'Size', 'Category Type'], ['Usage/Application', 'Size', 'Color', 'Marble Type', 'Features', 'Finishing'], ['Marble Type', 'Surface Finishing', 'Stone Form', 'Usage'], ['Brand', 'Material', 'Finish', 'Thickness', 'Size']]
big_list = [x for a_list in my_lists for x in a_list]
unique_strings, number_of_appearances = np.unique(big_list, return_counts=True)
index = np.flip(np.argsort(number_of_appearances))
print(unique_strings[index], number_of_appearances[index])

这会使您的列表变得平坦，找到唯一的字符串，并按它们出现的次数(大多数至少至少)对它们进行排序。第一个字符串将是“最找到的元素”，任何计数超过1的字符串都会在多个列表中重复。

Answer 2

我认为这会有所帮助；

from functools import reduce
reduce(numpy.intersect1d, (my_lists))

来源：https://numpy.org/doc/stable/reference/generated/numpy.intersect1d.html