我有以下示例数据框。让我们假设每个字母实际上都是一个单词。例如,a = 'ant'和b = 'boy'。id words1 [a, b, c, d, e, f, g]1 [h, I, o]1 1 [a, b, c]2 [e, f, g, m, n, q, r, s]2 [w, j, f]3 [l, t, m, n, q, s, a]3 [c, d, e, f, g]4 4 [f, g, z]创建上述示例数据框的代码:import pandas as pd d = {'id': [1, 1, 1, 1, 2, 2, 3, 3, 4, 4], 'words': [['a', 'b', 'c', 'd', 'e', 'f', 'g'], ['h', 'I', 'o'], '', ['a', 'b', 'c'], ['e', 'f', 'g', 'm', 'n', 'q', 'r', 's'], ['w', 'j', 'f'], ['l', 't', 'm', 'n', 'q', 's', 'a'], ['c', 'd', 'e', 'f', 'g'], '', ['f', 'g', 'z']]}df = pd.DataFrame(data=d)我在其上运行以下 NLP 代码以执行以下操作: 给我一个从“单词”字段并置在一起的各种 3 词组合的计数。from nltk.collocations import *from nltk import ngramsfrom collections import Countertrigram_measures = nltk.collocations.BigramAssocMeasures()finder = BigramCollocationFinder.from_documents(df['words'])finder.nbest(trigram_measures.pmi, 100) s = pd.Series(df['words'])ngram_list = [pair for row in s for pair in ngrams(row, 3)]counts = Counter(ngram_list).most_common()df = pd.DataFrame.from_records(counts, columns=['gram', 'count'])示例结果假设输出如下(数据值是假的):gram count a, b, c 13c, d, e 9g, h, i 6q, r, s 1问题是我希望将结果输出按“id”字段拆分。 我想要的样本输出如下(数据是假的和随机的):id gram count 1 a, b, c 131 c, d, e 91 g, h, i 61 q, r, s 12 a, b, c 62 w, j, f 33 l, t, m 43 e, f, g 24 f, g, z 1我如何实现这一目标?...通过“id”字段获取结果?
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