python - 在其他内部字典键的每个组合以及外部字典键的每个组合中搜索内部字典键的每个组合

Question

我不确定标题是否很好地描述了我的问题，但如果有问题我会稍后编辑。我已经检查了许多与此相关的问题，但由于代码是如此嵌套，我在编程方面不是很有经验，我需要使用我无法处理的组合。

我有一个嵌套字典，与此类似:

example_dictionary = {'I want to eat peach and egg.':{'apple':3, 'orange':2, 'banana':5},\
                   'Peach juice is so delicious.':{'apple':3, 'orange':5, 'banana':2}, \
'Goddamn monkey ate my banana.':{'rice':4, 'apple':6, 'monkey':2}, \
'They say apple is good for health.':{'grape':10, 'monkey':5, 'peach':5, 'egg':8}}

我想做的是通过遵循一些规则构建邻接矩阵。 规则是:

1)如果任何一个句子(外部字典键)中存在任何一个内部字典中的单词，则在相关句子之间添加一个权重作为该单词的值。

2)如果两个句子中的任何一个具有相同的内部字典键(单词)但值不同，则将单词的值相乘并添加相关句子之间的权重。

额外注意:内部字典可以有不同的长度，相同的内部字典键(单词)可能有不同的值。我希望它们仅在这种情况下相乘，如果它们具有我不想考虑的相同值。

示例:

Sentence1(0): I want to eat peach and egg. {'apple':3, 'orange':2, 'banana':5}

Sentence2(1): Peach juice is so delicious. {'apple':3, 'orange':5, 'banana':2}

Sentence3(2): Goddamn monkey ate my banana.{'rice':4, 'apple':6, 'monkey':2}

Sentence4(3): They say apple is good for health. {'grape':10, 'monkey':5, 'peach':5, 'egg':8}

Between 0 and 1: 5*2+5*2=20 (because, their apple's has the same value, just multiplied the values for orange and banana. And none of the words exists in any sentence.)

Between 2 and 3: (2*5=10 (monkey is the same key with different value) +

6 (the key of sentence3 'apple' exists in sentence4) +

5 (the key of sentence4 'monkey' exists in sentence3)= 21

Between 0 and 3: 3+5+8=16 (sentence1 key 'apple' exists in sentence4, and sentence4 keys 'egg' and 'peach' exist in sentence1.

我希望这些例子能够清楚地说明这一点。

我尝试过的(由于嵌套结构和组合，这对我来说非常困惑):

from itertools import combinations, zip_longest
import networkx as nx

def compare_inner_dicts(d1,d2):
#this is for comparing the inner dict keys and multiplying them
#if they have the same key but different value
    values = []
    inner_values = 0
    for common_key in d1.keys() & d2.keys():
        if d1[common_key]!= d2[common_key]:
            _value = d1[common_key]*d2[common_key]
            values.append(_value)
            inner_values = sum([p for p in values])

    inner_dict_values = inner_values
    del inner_values  

    return inner_dict_values


def build_adj_mat(a_dict):
    gr = nx.Graph()
    for sentence, words in a_dict.items():

        sentences = list(a_dict.keys())
        gr.add_nodes_from(sentences)
        sentence_pairs = combinations(gr.nodes, 2)
        dict_pairs = combinations(a_dict.values(), 2)
        for pair, _pair in zip_longest(sentence_pairs, dict_pairs):
            numbers = []
            x_numbers = []
            #y_numbers = []
            sentence1 = pair[0]
            sentence2 = pair[1]
            dict1 = _pair[0]
            dict2 = _pair[1]

            inner_dict_numbers = compare_inner_dicts(dict1, dict2)
            numbers.append(inner_dict_numbers)

            for word, num in words.items():
                if sentence2.find(word)>-1:
                    x = words[word]
                    x_numbers.append(x)
                    numbers.extend(x_numbers)
#                if sentence1.find(word)>-1: #reverse case
#                    y = words[word]
#                    y_numbers.append(y)
#                    numbers.extend(y_numbers)

                    total = sum([p for p in numbers if len(numbers)>0])

                    if total>0:
                        gr.add_edge(sentence1, sentence2, weight=total)
                        del total
                    else: del total
                else: 
                    continue
                    numbers.clear()
                    x_numbers.clear()
                   #y_numbers.clear()

    return gr

G = build_adj_mat(example_dictionary)
print(nx.adjacency_matrix(G))

预期结果:

(0, 1) 5*2+5*2=20
(0, 2) 3*6=18+5=23
(0, 3) 3+5+8=16
(1, 0) 20
(1, 2) 3*6=18+2=20
(1, 3) 3+5=8
(2, 0) 23
(2, 1) 20
(2, 3) 2*5=10+5+6=21
(3, 0) 16
(3, 1) 8
(3, 2) 21

输出:

  (0, 2)        23
  (0, 3)        6
  (1, 2)        23
  (1, 3)        6
  (2, 0)        23
  (2, 1)        23
  (2, 3)        16
  (3, 0)        6
  (3, 1)        6
  (3, 2)        16

通过比较预期输出和比较输出，我可以理解其中一个问题，即我的代码只是检查 sentence1 中的单词是否存在于 sentence2 中，但不检查不要做相反的事情。我尝试使用注释掉的部分来解决它，但它返回了更多无意义的结果。另外我不确定是否还有其他问题。我不知道如何得到正确的结果，这两种组合和嵌套结构让我完全迷失了。抱歉问了这么长的问题，为了清楚起见，我描述了一切。任何帮助将不胜感激，提前致谢。

Answer 1

您可以使用以下功能:

from collections import defaultdict
import itertools as it
import re


def compute_scores(sentence_dict):
    scores = defaultdict(int)
    for (j, (s1, d1)), (k, (s2, d2)) in it.combinations(enumerate(sentence_dict.items()), 2):
        shared_keys = d1.keys() & d2.keys()
        scores[j, k] += sum(d1[k]*d2[k] for k in shared_keys if d1[k] != d2[k])
        scores[j, k] += sum(d1[k] for k in d1.keys() & get_words(s2))
        scores[j, k] += sum(d2[k] for k in d2.keys() & get_words(s1))
    return scores


def get_words(sentence):
    return set(map(str.lower, re.findall(r'(?<=\b)\w+(?=\b)', sentence)))

结果当然取决于您对单词的定义，因此您需要在函数 get_words 中填写您自己的定义。默认实现似乎适合您的示例数据。由于根据您的定义，句子对的分数是对称的，因此也无需考虑反向配对(它具有相同的分数)；即 (0, 1) 与 (1, 0) 具有相同的分数。这就是代码使用 itertools.combinations 的原因.

运行示例数据:

from pprint import pprint

example_dictionary = {
    'I want to eat peach and egg.': {'apple':3, 'orange':2, 'banana':5},
    'Peach juice is so delicious.': {'apple':3, 'orange':5, 'banana':2},
    'Goddamn monkey ate my banana.': {'rice':4, 'apple':6, 'monkey':2},
    'They say apple is good for health.': {'grape':10, 'monkey':5, 'peach':5, 'egg':8}}

pprint(compute_scores(example_dictionary))

给出以下分数:

defaultdict(<class 'int'>,
            {(0, 1): 20,
             (0, 2): 23,
             (0, 3): 16,
             (1, 2): 20,
             (1, 3): 8,
             (2, 3): 21})

如果字典不仅可以包含单词，还可以包含短语(即多个单词)，只需对原始实现进行轻微修改即可(也适用于单个单词):

scores[j, k] += sum(weight for phrase, weight in d1.items() if phrase in s2.lower())
scores[j, k] += sum(weight for phrase, weight in d2.items() if phrase in s1.lower())