我搜索了很多解決方案，我確實發現了類似的問題。此答案返回可能不屬于輸入列表中所有字符串的最長字符序列。此答案返回必須屬于輸入列表中所有字符串的最長公共 WORDS 序列。我正在尋找上述解決方案的組合。也就是說，我想要可能不會出現在輸入列表的所有單詞/短語中的最長的常見單詞序列。以下是預期的一些示例：['exterior lighting', 'interior lighting']-->'lighting'['ambient lighting', 'ambient light']-->'ambient'['led turn signal lamp', 'turn signal lamp', 'signal and ambient lamp', 'turn signal light']-->'turn signal lamp'['ambient lighting', 'infrared light']-->''謝謝
2 回答
SMILET
TA貢獻1796條經驗 獲得超4個贊
此代碼還將按列表中最常見的單詞對所需列表進行排序。它會計算列表中每個單詞的數量，然后剪切只出現一次的單詞并對其進行排序。
lst=['led turn signal lamp', 'turn signal lamp', 'signal and ambient lamp', 'turn signal light']
d = {}
d_words={}
for i in lst:
for j in i.split():
if j in d:
d[j] = d[j]+1
else:
d[j]= 1
for k,v in d.items():
if v!=1:
d_words[k] = v
sorted_words = sorted(d_words,key= d_words.get,reverse = True)
print(sorted_words)
波斯汪
TA貢獻1811條經驗 獲得超4個贊
一個相當粗略的解決方案，但我認為它有效：
from nltk.util import everygrams
import pandas as pd
def get_word_sequence(phrases):
ngrams = []
for phrase in phrases:
phrase_split = [ token for token in phrase.split()]
ngrams.append(list(everygrams(phrase_split)))
ngrams = [i for j in ngrams for i in j] # unpack it
counts_per_ngram_series = pd.Series(ngrams).value_counts()
counts_per_ngram_df = pd.DataFrame({'ngram':counts_per_ngram_series.index, 'count':counts_per_ngram_series.values})
# discard the pandas Series
del(counts_per_ngram_series)
# filter out the ngrams that appear only once
counts_per_ngram_df = counts_per_ngram_df[counts_per_ngram_df['count'] > 1]
if not counts_per_ngram_df.empty:
# populate the ngramsize column
counts_per_ngram_df['ngramsize'] = counts_per_ngram_df['ngram'].str.len()
# sort by ngramsize, ngram_char_length and then by count
counts_per_ngram_df.sort_values(['ngramsize', 'count'], inplace = True, ascending = [False, False])
# get the top ngram
top_ngram = " ".join(*counts_per_ngram_df.head(1).ngram.values)
return top_ngram
return ''
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