字典中唯一嵌套值的计数

Question

我有一个看起来像这样的模式，我想获得提到唯一drugs的最大数量的journal。

my_list = [{'atccode': 'A04AD',
'drug': 'DIPHENHYDRAMINE',
'mentioned_in': [{'date': '01/01/2019',
'journal': 'Journal of emergency nursing'},
{'date': '01/01/2019',
'journal': 'Journal of emergency nursing'
},
{'date': '1 January 2020',
'journal': 'Journal of emergency nursing'
},
{'date': '1 January 2020',
'journal': 'Journal of emergency nursing'},
{'date': '1 January 2020',
'journal': 'Journal of emergency nursing'
}]},
{'atccode': 'S03AA',
'drug': 'TETRACYCLINE',
'mentioned_in': [{'date': '02/01/2020',
'journal': 'American journal of veterinary research'
},
{'date': '2020-01-01',
'journal': 'Psychopharmacology'}]},
{'atccode': 'V03AB',
'drug': 'ETHANOL',
'mentioned_in': [{'date': '2020-01-01',
'journal': 'Psychopharmacology'
}]},
{'atccode': 'A01AD',
'drug': 'EPINEPHRINE',
'mentioned_in': [{'date': '01/02/2020',
'journal': 'The journal of allergy and clinical '
'immunology. In practice'},
{'date': '01/03/2020',
'journal': 'The journal of allergy and clinical '
'immunology. In practice'
},
{'date': '27 April 2020',
'journal': 'Journal of emergency nursing'
}]}]

所以结果看起来像这样:

{
"journal":"Psychopharmacology",
"unique_drug_mentions" : 2
},
{
"journal" : "Psychopharmacology",
"unique_drug_mentions":2
}

到目前为止我一直在尝试的是

from collections import Counter
mentions_counts = Counter(d['journal'] for d in my_list)
most_common = {'unique_drug_mentions': mentions_counts.most_common(1)[0][0], "journal" :d["journal"]}

但是没有成功。

Answer 1

我要循环遍历list:

# store counts of unique drugs here
counts = {}
# loop through your dicts in the list
for d in my_list:
# look in each journal mention
for d2 in d['mentioned_in']:
# if we haven't seen this journal before
if d2['journal'] not in counts:
counts[d2['journal']] = set()
counts[d2['journal']].add(d['drug'])
# this would have all your verbose info as you want it
unique_drug_counts = [
{
"journal": journal,
"unique_drug_mentions": len(drugs)
}
for journal, drugs in counts.items()
]
# max value (the answer to your question
max(counts.items(), key=lambda x: len(x[1]))