比较 Python 列表中的字典并附加结果答案

【问题标题】：Compare dictionaries in a Python list and append results比较 Python 列表中的字典并附加结果
【发布时间】：2020-06-26 06:34:17
【问题描述】：

我正在尝试遍历字典数组，比较它们之间的每个值，然后将结果附加到字典中。

例如，如果我们有以下数组：

epidemics = [{"Disease": "B", "year": "1980"},
             {"Disease": "C", "year": "1975"},
             {"Disease": "B", "year": "1996"},
             {"Disease": "E", "year": "2000"},
             {"Disease": "B", "year": "2020"}]

如果其他年份也发生了同一种疾病的流行，我想要达到的结果是：

 epidemics = [{"Disease": "B", "year": "1980", "occurredIn": ["1996", "2020"]},
             {"Disease": "C", "year": "1975"},
             {"Disease": "B", "year": "1996", "occurredIn": ["1980", "2020"]},
             {"Disease": "E", "year": "2000"},
             {"Disease": "B", "year": "2020", "occurredIn": ["1980", "1996"]}]

这是我到目前为止的地方：

for index, a in enumerate(epidemics):
  v_b = []

  for b in epidemics[index+1:]:
    if a['Disease'] == b['Disease']:
        v_b.append(b["year"])
        a['occurredIn'] = v_b

它打印我：

[{'Disease': 'B', 'year': '1980', 'occurredIn': ['1996', '2020']},
 {'Disease': 'C', 'year': '1975'},
 {'Disease': 'B', 'year': '1996', 'occurredIn': ['2020']},
 {'Disease': 'E', 'year': '2000'}, 
 {'Disease': 'B', 'year': '2020'}]

提前致谢

【问题讨论】：

如果可以的话，我不认为你想要的结果实际上是一个好的结构。为什么不只存储疾病 B 一次，将所有年份都放在同一个地方？如果您每年需要一些东西，您总是可以相应地访问该信息。这个结构是一个简单的字典，键是疾病，值是年份列表
@ParitoshSingh 感谢您的提示....此数据只是一个示例，我正在使用的数据更加可靠。无论如何，我会记住这一点。欢呼

标签： python dictionary append compare

【解决方案1】：

您可以建立一个临时字典并为每种疾病建立一组年份。然后使用这本字典改造epidemics 列表：

from collections import defaultdict

epidemics = [{"Disease": "B", "year": "1980"},
             {"Disease": "C", "year": "1975"},
             {"Disease": "B", "year": "1996"},
             {"Disease": "E", "year": "2000"},
             {"Disease": "B", "year": "2020"}]

d = defaultdict(set)
for dd in epidemics:
    disease = dd['Disease']
    d[disease].add(dd['year'])

for i,dd in enumerate(epidemics):
    years = d[dd['Disease']]-set([dd['year']])
    if years:
        epidemics[i]['occurredIn'] = years

print(epidemics)

输出：

[{'Disease': 'B', 'year': '1980', 'occurredIn': {'2020', '1996'}},
 {'Disease': 'C', 'year': '1975'},
 {'Disease': 'B', 'year': '1996', 'occurredIn': {'1980', '2020'}},
 {'Disease': 'E', 'year': '2000'},
 {'Disease': 'B', 'year': '2020', 'occurredIn': {'1980', '1996'}}]

【讨论】：

改用defaultdict而不是setdefault()以使其更具可读性。
谢谢，非常感谢。
@Helder'hp'Pinto 不用担心。很高兴为您提供帮助。

【解决方案2】：

您可以根据疾病发生的年份构建新词典，例如：

epidemics = [{"Disease": "B", "year": "1980"},
             {"Disease": "C", "year": "1975"},
             {"Disease": "B", "year": "1996"},
             {"Disease": "E", "year": "2000"},
             {"Disease": "B", "year": "2020"}]

output = []
for ep in epidemics:
    other_years = set(
        other_ep['year'] for other_ep in epidemics if (
            ep['Disease'] == other_ep['Disease'] and ep['year'] != other_ep['year']
            )
    )

    output.append(
        ep if not other_years else {**ep, 'occurredIn': list(other_years)}
    )

print(output)
>>> {'Disease': 'B', 'year': '1980', 'occurredIn': ['2020', '1996']}
>>> {'Disease': 'C', 'year': '1975'}
>>> {'Disease': 'B', 'year': '1996', 'occurredIn': ['1980', '2020']}
>>> {'Disease': 'E', 'year': '2000'}
>>> {'Disease': 'B', 'year': '2020', 'occurredIn': ['1980', '1996']}

【讨论】：

穆哈斯·格拉西亚斯·马科斯