Question

我们假设我有以下列表（实际上他们有很多子列表）：

list_1 = [['Hi my name is anon'],
                 ['Hi I like #hokey']]


list_2 = [['Hi my name is anon_2'],
                 ['Hi I like #Basketball']]

我想计算所有可能的配对的distance没有重复（没有替换的组合，产品？）。例如：

distance between: ['Hi my name is anon'] and ['Hi my name is anon_2']
distance between: ['Hi my name is anon'] and ['Hi I like #Basketball']
distance between: ['Hi I like #hokey'] and ['Hi my name is anon_2']
distance between: ['Hi I like #hokey'] and ['Hi I like #Basketball']

将分数放入这样的列表中：

[distance_1,distance_2,distance_3,distance_4]

为此，我考虑使用itertools product或combination。这就是我试过的：

strings_1 = [i[0] for i in list_1]
strings_2 = [i[0] for i in list_2]

import itertools

scores_list = [dis.jaccard(i,j) for i,j in zip(itertools.combinations(strings_1, strings_2))]

问题是我得到了这个追溯：

    scores_list = [dis.jaccard(i,j) for i,j in zip(itertools.combinations(strings_1, strings_2))]
TypeError: an integer is required

如何才能有效地完成这项任务？如何计算这种类似产品组合的操作呢？

Answer 1

您需要使用itertools.product来获取笛卡尔积，如此

[dis.jaccrd(string1, string2) for string1, string2 in product(list_1, list_2)]

该产品会对这些项目进行分组，例如

>>> from pprint import pprint
>>> pprint(list(product(list_1, list_2)))
[(['Hi my name is anon'], ['Hi my name is anon_2']),
 (['Hi my name is anon'], ['Hi I like #Basketball']),
 (['Hi I like #hokey'], ['Hi my name is anon_2']),
 (['Hi I like #hokey'], ['Hi I like #Basketball'])]

如果您只想将jaccrd函数应用于列表中的字符串，那么您可能需要预处理列表，例如

>>> list_11 = [item for items in list_1 for item in items]
>>> list_21 = [item for items in list_2 for item in items]
>>> pprint([str1 + " " + str2 for str1, str2 in product(list_11, list_21)])
['Hi my name is anon Hi my name is anon_2',
 'Hi my name is anon Hi I like #Basketball',
 'Hi I like #hokey Hi my name is anon_2',
 'Hi I like #hokey Hi I like #Basketball']
>>> pprint([dis.jaccard(str1, str2) for str1, str2 in product(list_11, list_21)])
...
...

根据Ashwini在评论中的建议，对于您的情况，您可以直接使用itertools.starmap，就像这样

>>> from itertools import product, starmap
>>> list(starmap(dis.jaccrd, product(list_11, list_21)))

例如，

>>> list_1 = ["a1", "a2", "a3"]
>>> list_2 = ["b1", "b2", "b3"]
>>> from itertools import product, starmap
>>> list(starmap(lambda x, y: x + " " + y, product(list_1, list_2)))
['a1 b1', 'a1 b2', 'a1 b3', 'a2 b1', 'a2 b2', 'a2 b3', 'a3 b1', 'a3 b2', 'a3 b3']

Answer 2

product有效，但由于你只有配对，这也有效：

[dis.jaccard(string1, string2) for string1 in list_1 for string2 in list_2]

也就是说，当然starmap + product组合获胜。

是否以迭代方式计算成对列表的得分？

2 个答案: