我有以下清單：
asset_list =
[
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-1', 'state': 'Alabama', 'rid': 103, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'},
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-2', 'state': 'Alaska', 'rid': 104, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-3', 'state': 'Alabama', 'rid': 105, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-3', 'state': 'Alabama', 'rid': 105, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-4', 'state': 'Arizona', 'rid': 106, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-5', 'state': 'California', 'rid': 107, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-7', 'state': 'Alabama', 'rid': 108, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'}
]
我必須從上面的串列中創建另一個串列，其中producer和state是唯一的，因此新串列將如下所示：
[
{'producer': 'HIK', 'state': 'Alabama'},
{'producer': 'HIK', 'state': 'Alaska'},
{'producer': 'TF Inc.', 'state': 'Alabama'},
{'producer': 'TF Inc.', 'state': 'Arizona'},
{'producer': 'TF Inc.', 'state': 'California'},
]
為了解決這個問題，我首先獲取所有唯一的生產者名稱：
producer_unique_list = []
for producer in asset_list:
if producer['producer'] not in producer_unique_list:
producer_unique_list.append(producer['producer'])
然后將每個資產的生產者與 比較producer_unique_list，只添加 (producer, state) 唯一的條目：
producer_list = []
run_once = True
for ass in asset_list:
p_d = dict()
if run_once:
p_d['producer'] = ass['producer']
p_d['state'] = ass['state']
producer_list.append(p_d)
run_once = False
else:
for y in producer_list:
if ass['producer'] != y['producer'] or ass['state'] != y['state']:
p_d = dict()
p_d['producer'] = ass['producer']
p_d['state'] = ass['state']
producer_list.append(y)
但看起來這個邏輯行不通。任何人都可以提出一些解決這個問題的好方法。謝謝
uj5u.com熱心網友回復：
您可以從基于producer和state構建一組元組開始。使用集合將處理重復問題。您可以基于該集合構建字典串列，如下所示：
asset_list = [
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-1', 'state': 'Alabama',
'rid': 103, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'},
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-2', 'state': 'Alaska',
'rid': 104, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-3', 'state': 'Alabama',
'rid': 105, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-3', 'state': 'Alabama',
'rid': 105, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-4', 'state': 'Arizona',
'rid': 106, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 64, 'client_id': '232-00000008', 'prem_id': 'PID-5', 'state': 'California',
'rid': 107, 'total_premid': 0, 'total_site': 0, 'producer': 'TF Inc.'},
{'cid': 60, 'client_id': '232-00000004', 'prem_id': 'PID-7', 'state': 'Alabama',
'rid': 108, 'total_premid': 0, 'total_site': 0, 'producer': 'HIK'}
]
s = set()
for a in asset_list:
s.add((a['producer'], a['state']))
out = [{'producer': i[0], 'state': i[1]} for i in s]
print(out)
uj5u.com熱心網友回復：
使用以下代碼，我能夠獲得所需的結果：
unique_values_set = set([tuple((entry["producer"], entry["state"])) for entry in asset_list])
unique_dict_list = [{'producer': value[0], 'state': value[1]} for value in unique_values_set]
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/382235.html