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從python中的單詞列表中查找最長(zhǎng)的常用單詞序列

從python中的單詞列表中查找最長(zhǎng)的常用單詞序列

慕斯王 2022-12-20 14:32:43
我搜索了很多解決方案,我確實(shí)發(fā)現(xiàn)了類似的問(wèn)題。此答案返回可能不屬于輸入列表中所有字符串的最長(zhǎng)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(zhǎng)公共 WORDS 序列。我正在尋找上述解決方案的組合。也就是說(shuō),我想要可能不會(huì)出現(xiàn)在輸入列表的所有單詞/短語(yǔ)中的最長(zhǎng)的常見(jiàn)單詞序列。以下是預(yù)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
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SMILET

TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超4個(gè)贊

此代碼還將按列表中最常見(jiàn)的單詞對(duì)所需列表進(jìn)行排序。它會(huì)計(jì)算列表中每個(gè)單詞的數(shù)量,然后剪切只出現(xiàn)一次的單詞并對(duì)其進(jìn)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)


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波斯汪

TA貢獻(xiàn)1811條經(jīng)驗(yàn) 獲得超4個(gè)贊

一個(gè)相當(dāng)粗略的解決方案,但我認(rèn)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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