如何比较 Pandas Dataframe 中的两列以查找匹配百分比并根据该逻辑返回一个值？

Question

I need to compare two columns in a Pandas data frame and fuzzy match.

If the fuzzy match is above a certain percentage (eg 85), I need to return that percentage, or a string saying "Partial Match"

If it matches fully, return "Full Match"

If it doesn't match, return "No Match"

Solutions I've tried:

Attempt #1

 conditions = [
     (df['one'] == df['two']),fuzz.ratio((df['one'],df['two'])) > 80, 
      fuzz.ratio((df['one'],df['two'])) <= 80]

  choices = ["FULL Match", fuzz.ratio((df['one'],df['two'])),"NO MATCH"]

df['result'] = np.select(condition,choices, default = np.nan)

====================================================================

Attempt #2

df['result'] = np.where(fuzz.ratio(df['one'], df['two']) >= 85, "Partial Match", 'No Match')

 import pandas as pd
 import numpy as np
 from fuzzywuzzy import fuzz
 import os


 df = pd.read_csv('data.csv')

 >x = fuzz.ratio(df['one'], df['two']) >= 85

 df['result'] = np.where(x, "Match", 'No Match')'''

Expected Result

         one          two    result
 0    apple        Apple     Partial Match
 1  banana       bannana     Partial Match
 2     kiwi  dragonfruit     No Match
 3    mango        mango     Full Match

===================================================================

Error Message:

Attempt #1

IndexError: tuple index out of range

Attempt #2

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Answer 1

尝试将最后两个命令合二为一

df['result'] = np.where(fuzz.ratio(df['one'], df['two']) >= 85, "Match", 'No Match')

Answer 2

I think this does the trick:

from difflib import SequenceMatcher

def similar(a, b):
    match_score = SequenceMatcher(None, a, b).ratio()
    if match_score == 1.0:
        result = "Full Match"
    elif match_score >= .85:
        result = "Partial Match"
    else:
        result = "No Match"
    return result

df["result"]=df[['one','two']].apply(lambda df: similar(df.one, df.two), axis=1)