熊猫：仅当另一列中的值匹配时才计算行之间的重叠单词

如何解决熊猫：仅当另一列中的值匹配时才计算行之间的重叠单词

我有一个如下所示的数据框，但有很多行：

import pandas as pd

data = {'intent':  ['order_food','order_food','order_taxi','order_call','order_taxi'],'Sent': ['i need hamburger','she wants sushi','i need a cab','call me at 6','she called me','i would like a new taxi' ],'key_words': [['need','hamburger'],['want','sushi'],['need','cab'],['call','6'],['call'],['new','taxi']]}

df = pd.DataFrame (data,columns = ['intent','Sent','key_words'])

我使用下面的代码（不是我的解决方案）计算了 jaccard 相似度：

def lexical_overlap(doc1,doc2): 
    words_doc1 = set(doc1) 
    words_doc2 = set(doc2)

    intersection = words_doc1.intersection(words_doc2)


    return intersection

并修改 @Amit Amola 给出的代码以比较每两行之间重叠的单词并从中创建一个数据框：

overlapping_word_list=[]

for val in list(combinations(range(len(data_new)),2)):
     overlapping_word_list.append(f"the shared keywords between {data_new.iloc[val[0],0]} and {data_new.iloc[val[1],0]} sentences are: {lexical_overlap(data_new.iloc[val[0],1],data_new.iloc[val[1],1])}")
#creating an overlap dataframe
banking_overlapping_words_per_sent = DataFrame(overlapping_word_list,columns=['overlapping_list'])

由于我的数据集很大，当我运行此代码来比较所有行时，它需要很长时间。所以我想只比较具有相同意图的句子，而不比较具有不同意图的句子。我不确定如何继续这样做

解决方法

IIUC 您只需要遍历 intent 列中的唯一值，然后使用 loc 来获取与之对应的行。如果您有两行以上，您仍然需要使用 combinations 来获取相似意图之间的唯一 combinations。

from itertools import combinations

for intent in df.intent.unique():
    # loc returns a DataFrame but we need just the column
    rows = df.loc[df.intent == intent,["Sent"]].Sent.to_list()
    combos = combinations(rows,2)
    for combo in combos:
        x,y = rows
        overlap = lexical_overlap(x,y)
        print(f"Overlap for ({x}) and ({y}) is {overlap}")

#  Overlap for (i need hamburger) and (she wants sushi) is 46.666666666666664
#  Overlap for (i need a cab) and (i would like a new taxi) is 40.0
#  Overlap for (call me at 6) and (she called me) is 54.54545454545454

,

好的，所以我想出了如何根据@gold_cy 的回答在评论中提到我想要的输出：

for intent in df.intent.unique():
# loc returns a DataFrame but we need just the column
   rows = df.loc[df.intent == intent,['intent','key_words','Sent']].values.tolist()
   combos = combinations(rows,2)
   for combo in combos:
       x,y = rows
       overlap = lexical_overlap(x[1],y[1])
       print(f"Overlap of intent ({x[0]}) for ({x[2]}) and ({y[2]}) is {overlap}")