1 
import pandas as pd
from difflib import SequenceMatcher

df = pd.DataFrame({"id":[9,12,13,14],
                   "text":["Error number 609 at line 10", "Error number 609 at line 22", "Error string 'foo' at line 11", "Error string 'bar' at line 14"]})

输出：

   id                           text
0   9    Error number 609 at line 10
1  12    Error number 609 at line 22
2  13  Error string 'foo' at line 11
3  14  Error string 'bar' at line 14

我想用来difflib.SequenceMatcher删除低于80行的相似度分数，只保留一个。

a = "Error number 609 at line 10"
b = "Error number 609 at line 22"
c = "Error string 'foo' at line 11"
d = "Error string 'bar' at line 14"
print(SequenceMatcher(None, a, b).ratio()*100) #92.5925925925926
print(SequenceMatcher(None, b, c).ratio()*100) #60.71428571428571
print(SequenceMatcher(None, c, d).ratio()*100) #86.20689655172413
print(SequenceMatcher(None, a, c).ratio()*100) #64.28571428571429

如何在 Python 中获得如下预期结果？您可以使用difflib或其他 python 包。谢谢你。

   id                           text
0   9    Error number 609 at line 10
2  13   Error string 'foo' at line 11
4

1 回答 1

2

您可以使用：

#cross join with filter onl text column
df = df.assign(a=1).merge(df[['text']].assign(a=1), on='a')
#filter out same columns per rows
df = df[df['text_x'] != df['text_y']]
#sort columns per rows
df[['text_x','text_y']] = pd.DataFrame(np.sort(df[['text_x','text_y']],axis=1), index=df.index)
#remove duplicates
df = df.drop_duplicates(subset=['text_x','text_y'])
#get similarity
df['r'] = df.apply(lambda x: SequenceMatcher(None, x.text_x, x.text_y).ratio(), axis=1)
#filtering
df = df[df['r'] > 0.8].drop(['a','r'], axis=1)
print (df)
    id                         text_x                         text_y
1    9    Error number 609 at line 10    Error number 609 at line 22
11  13  Error string 'bar' at line 14  Error string 'foo' at line 11
于 2020-01-07T09:02:52.947 回答