如何從默認字典（串列）中提取頂部專案？

我不熟悉與defaultdicts 合作。我有一個匹配腳本，它將唯一識別符號作為“鍵”放置，然后使用defaultdict(list). 匹配是公司名稱、地址和匹配分數（基于匹配演算法）。有時是 1-1 匹配，這意味著有 1 個鍵與匹配相關聯，但有時演算法會捕獲接近匹配，因此有時會有多個匹配。對于那些想選擇這個得分最高的比賽的人。

目標： 從 defaultdict(list) 中提取每個唯一識別符號的資料。如果唯一識別符號的值大于 1，則精確計算具有最高 Lev Score、Fuzzy Score 和 Jaro 分數的資料。

這是資料的預覽：

#imports
from collections import defaultdict
test_dic_stack = defaultdict(list)

#testing data (unique1 has a 1-1 match &  unique2 has a 1-5 match)
test_dic_stack['unique1'].append({'Account Name': 'company1', 'Matching Account': 'company1', 'Account_Address': '123 Road', 'Address_match': '123 Road',  'Lev_score': 98.0, 'Fuzzy_score': 100, 'Jaro_Score': 99.0})
test_dic_stack['unique2'].append({'Account Name': 'company1', 'Matching Account': 'company1', 'Account_Address': '1 awesome street', 'Address_match': '1 awesome street',  'Lev_score': 91.0, 'Fuzzy_score': 89, 'Jaro_Score': 99.0})
test_dic_stack['unique2'].append({'Account Name': 'company2', 'Matching Account': 'company2', 'Account_Address': '1 awesome street', 'Address_match': '1 awesome st',  'Lev_score': 71.0, 'Fuzzy_score': 82, 'Jaro_Score': 84.0})
test_dic_stack['unique2'].append({'Account Name': 'company3', 'Matching Account': 'company3', 'Account_Address': '1 awesome street', 'Address_match': '1 awesome street suite 1',  'Lev_score': 88.0, 'Fuzzy_score': 89, 'Jaro_Score': 90.0})
test_dic_stack['unique2'].append({'Account Name': 'company4', 'Matching Account': 'company4', 'Account_Address': '1 awesome street', 'Address_match': '1 awe street',  'Lev_score': 81.0, 'Fuzzy_score': 90, 'Jaro_Score': 86.0})
test_dic_stack['unique2'].append({'Account Name': 'company5', 'Matching Account': 'company5', 'Account_Address': '1 awesome street', 'Address_match': '1 awe st',  'Lev_score': 70.0, 'Fuzzy_score': 86, 'Jaro_Score': 89.0})

#defaultdict preview
defaultdict(list,
            {'unique1': [{'Account Name': 'company1',
               'Matching Account': 'company1',
               'Account_Address': '123 Road',
               'Address_match': '123 Road',
               'Lev_score': 98.0,
               'Fuzzy_score': 100,
               'Jaro_Score': 99.0}],
             'unique2': [{'Account Name': 'company1',
               'Matching Account': 'company1',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awesome street',
               'Lev_score': 91.0,
               'Fuzzy_score': 89,
               'Jaro_Score': 99.0},
              {'Account Name': 'company2',
               'Matching Account': 'company2',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awesome st',
               'Lev_score': 71.0,
               'Fuzzy_score': 82,
               'Jaro_Score': 84.0},
              {'Account Name': 'company3',
               'Matching Account': 'company3',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awesome street suite 1',
               'Lev_score': 88.0,
               'Fuzzy_score': 89,
               'Jaro_Score': 90.0},
              {'Account Name': 'company4',
               'Matching Account': 'company4',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awe street',
               'Lev_score': 81.0,
               'Fuzzy_score': 90,
               'Jaro_Score': 86.0},
              {'Account Name': 'company5',
               'Matching Account': 'company5',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awe st',
               'Lev_score': 70.0,
               'Fuzzy_score': 86,
               'Jaro_Score': 89.0}]})

這是我要求的結果：
提取 unique1 資料并提取 unique2 “最佳匹配”資料。請注意，有時最好的匹配并不總是第一個

results = [{'unique1': {'Account Name': 'company1',
               'Matching Account': 'company1',
               'Account_Address': '123 Road',
               'Address_match': '123 Road',
               'Lev_score': 98.0,
               'Fuzzy_score': 100,
               'Jaro_Score': 99.0},

          'unique2': {'Account Name': 'company1',
               'Matching Account': 'company1',
               'Account_Address': '1 awesome street',
               'Address_match': '1 awesome street',
               'Lev_score': 91.0,
               'Fuzzy_score': 89,
               'Jaro_Score': 99.0}]

uj5u.com熱心網友回復：

您可以使用字典理解，max將三個分數的總和用作鍵。

假設d輸入字典。

out = {k:max(v, key=lambda x: sum((x['Fuzzy_score'], x['Lev_score'], x['Jaro_Score'])))
       for k,v in d.items()}

輸出：

{'unique1': {'Account Name': 'company1',
  'Matching Account': 'company1',
  'Account_Address': '123 Road',
  'Address_match': '123 Road',
  'Lev_score': 98.0,
  'Fuzzy_score': 100,
  'Jaro_Score': 99.0},
 'unique2': {'Account Name': 'company1',
  'Matching Account': 'company1',
  'Account_Address': '1 awesome street',
  'Address_match': '1 awesome street',
  'Lev_score': 91.0,
  'Fuzzy_score': 89,
  'Jaro_Score': 99.0}}

標籤：Python 列表 字典 默认字典

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