我正在為大學做一個專案，我想在其中使用兩個資料框之間的關聯。
import pandas as pd
from fuzzywuzzy import fuzz
from fuzzywuzzy import process
dataframe_one = pd.DataFrame({'city_name': ['LISBOA', 'LISBOA', 'RIO DE JANEIRO',
'BARCELONA', 'PARIS'],
'address_name':['ESPLANADA DOM CARLOS',
'AVENIDA BRASILIA',
'PARQUE NACIONAL DA TIJUCA',
'C D ARISTIDES MAILLOL',
'CHAMP DE MARS 5 AV ANATOLE']})
dataframe_two = pd.DataFrame({'city': ['RIO DE JANEIRO', 'RIO DE JANEIRO', 'LISBOA'],
'population': [6748, 6748, 504],
'address':['PQ NACIONAL TIJUCA', 'URCA',
'ESPLANADAA DOM CARLO']})
我想要做的是比較兩個資料框。所以如果城市相同，則分析地址列之間的相似度。如果地址相似度值大于一個 THRESHOLD，那么我在新列中賦值。我做了以下事情：
THRESHOLD = 80
dataframe_two['NewColumn1'] = None
dataframe_two['NewColumn2'] = 0
for i in range(0, len(dataframe_one)):
for j in range(0, len(dataframe_two)):
if(dataframe_one['city_name'].iloc[i] == dataframe_two['city'].iloc[j]):
adress1 = dataframe_one['address_name'].iloc[i]
adress2 = dataframe_two['address'].iloc[j]
value_fuzzy = fuzz.ratio(adress1, adress2)
if(value_fuzzy > THRESHOLD):
dataframe_two['NewColumn1'].loc[j] = \
dataframe_one['address_name'].iloc[i]
dataframe_two['NewColumn2'].loc[j] = value_fuzzyvalue_fuzzy
實作按預期輸出到第二個資料幀，但我想用更高效的代碼替換它。
輸出：
city population address NewColumn1 NewColumn2
RIO DE JANEIRO 6748 PQ NACIONAL TIJUCA PARQUE NACIONAL DA TIJUCA 84
RIO DE JANEIRO 6748 URCA None 0
LISBOA 504 ESPLANADAA DOM CARLO ESPLANADA DOM CARLOS 95
uj5u.com熱心網友回復：
這是另一種在您的初始資料框中找到最佳匹配的實作。
l = []
TRESHOLD = 80
# ITERATE OVER THE SECOND DATAFRAME
for _ ,row in dataframe_two.iterrows():
# CREATE NEW ROW FOR THE RESULT
d = {**row, 'NewColumn1':None, 'NewColumn2':None}
# FIND MATCHES
d1_subset = dataframe_one.loc[dataframe_one['city_name'] == row['city']]
# ITERATE OVER THE MATCHES
if d1_subset.shape[0] > 0:
best_adress = None
best_score = 0
adresses = d1_subset['address_name'].tolist()
# FIND BEST MATCH
for adress in adresses:
value_fuzzy = fuzz.ratio(adress, row['address'])
if value_fuzzy > best_score:
best_adress = adress
best_score = value_fuzzy
# IF BEST SCORE IS ABOVE THRESHOLD
if best_score > TRESHOLD:
d['NewColumn1'] = best_adress
d['NewColumn2'] = best_score
l.append(d)
# CREATE DATAFRAME FROM THE RESULT
df = pd.DataFrame(l)
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