我有這個字典串列，我想將那些具有相同“姓名”和“學校”值的人放入一個新串列中，并將他們的“年齡”也合并到一個串列中，而字典的其余部分與像往常一樣添加到串列中不同..
這是字典串列的示例
[{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
我想讓它變成這樣的東西..
[{'name': 'Jane', 'age': [12,14,16], 'school': 'SIT'}, {'name': 'John', 'age': 13, 'school': 'SMU'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
使用Python ..請幫忙！
嘗試使用計數器，回圈但仍然無法使其作業..
uj5u.com熱心網友回復：
你應該使用itertools.groupby().
例子：
import itertools
from pprint import pprint
data = [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
keyfunc = lambda x: (x["name"], x["school"])
# needs to be sorted to use groupby
data.sort(key=keyfunc)
output = []
for k,v in itertools.groupby(data, key=keyfunc):
this_group = {
"name": k[0],
"school": k[1],
"age": [i["age"] for i in v],
}
output.append(this_group)
pprint(output)
輸出是：
[{'age': [12, 14, 16], 'name': 'Jane', 'school': 'SIT'},
{'age': [13], 'name': 'John', 'school': 'NUS'},
{'age': [13], 'name': 'John', 'school': 'SMU'}]
以供參考：
- https://docs.python.org/3/library/itertools.html#itertools.groupby
uj5u.com熱心網友回復：
x = [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
new_x = {}
for r in x:
if r['name'] in new_x.keys():
if not isinstance(new_x[r['name']]['age'], list):
new_x[r['name']]['age'] = [new_x[r['name']]['age']]
if r['age'] not in new_x[r['name']]['age']:
new_x[r['name']]['age'].append(r['age'])
else:
new_x[r['name']] = {'age': r['age'], 'school': r['school']}
z = [v.update(name=k) for k, v in new_x.items()]
z = [v for k, v in new_x.items()]
uj5u.com熱心網友回復：
這是您問題的通用解決方案。只有name和school被認為是“特殊的”。age當必須添加新值時，所有其他鍵都將轉換為串列。
l = [
{"name": "Jane", "age": 12, "school": "SIT"},
{"name": "John", "age": 13, "school": "SMU"},
{"name": "Jane", "age": 14, "school": "SIT"},
{"name": "Jane", "age": 16, "school": "SIT"},
{"name": "John", "age": 13, "school": "NUS"},
]
r = {}
for x in l:
id = f"{x['name']}-{x['school']}"
if id in r:
for k,v in x.items():
if k not in ["name", "school"]:
if k in r[id]:
if isinstance(r[id][k], list):
r[id][k].append(v)
else:
r[id][k] = [r[id][k], v]
else:
r[id][k] = v
else:
r[id] = x
result = [x for x in r.values()]
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標籤：Python列表循环字典
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