我有一个如下所示的DataFrame。
DF1:
A
Any Match
Credit
I need a debit card.
Logging
Awesome
我还有另一个DataFrame,如下所示:
DF2:
B
I did not find any match.
I want a credit card.
I need a debit card.
I do not know.
I am logging into credit portal.
我需要输出为:
B A
I did not find any match. Any Match
I want a credit card. Credit
I need a debit card. I need a debit card.
I am logging into credit portal. logging,credit
如果DF1中存在的短语在DF2中存在的任何文本中,则为
。 将o / p打印为“文本”和重要短语。
答案 0 :(得分:0)
尝试Fuzzywuzzy
:
import pandas as pd
from fuzzywuzzy import fuzz
matched_entities = []
for row in df1.index:
name1 = vendor_df.get_value(row,"A")
for columns in df2.index:
name2=df2.get_value(columns,"B")
matched_token=fuzz.partial_ratio(name1,name2)
if matched_token> 80:
matched_vendors.append([A,B])
df_partial_ratio = pd.DataFrame(columns=['A', 'B'], data=matched_entities)
在您的数据库中,如果fuzz.partial_ratio
不起作用,请尝试fuzz.ratio
或fuzz.token_sort_ratio
。可以通过将以下代码中的一行更改为以下代码来实现这两项:
matched_token=fuzz.ratio(name1,name2)
OR
matched_token=fuzz.token_sort_ratio(name1,name2)
答案 1 :(得分:0)
您可以执行以下操作。首先,定义与“规范化”文本匹配的查找功能,例如小写:
def lookup(x, values):
for value in values:
if value.lower() in x.lower():
return value
然后将此功能应用于DF2:
dfB['A'] = dfB['B'].apply(lambda x: lookup(x, dfA['A']))
哪些应该给您:
B A
0 I did not find any match. Any Match
1 I want a credit card. Credit
2 I need a debit card. Debit
3 I do not know. None
答案 2 :(得分:0)
尝试
df1['B'] = float('nan')
pos = 0
for i in range(len(df1)):
for j in range(len(df2)):
if df1['A'][i].lower() in df2['B'][j].lower():
df1['B'].iloc[pos] = df2['B'][j]
pos+=1
break
df1.dropna(axis=0)
输出
A B
0 Any Match I did not find any match.
1 Credit I want a credit card.
2 I need a debit card I need a debit card.