构建字典框架，用not in进行判断分类的思想

#将list1转化为list4形式
list1 = [
    {'name': 'alex', 'hobby': '喝酒'},
    {'name': 'alex', 'hobby': '抽烟'},
    {'name': 'alex', 'hobby': '烫头'},
    {'name': 'alex', 'hobby': 'Massage'},
    {'name': 'wusir', 'hobby': '喊麦'},
    {'name': 'wusir', 'hobby': '街舞'},
]
list4 = [
    {'name': 'alex', 'hobby_list': ['喝酒', '抽烟', '烫头', 'Massage']},
    {'name': 'wusir', 'hobby_list': ['喊麦', '街舞']}
]
#通过构建字典框架，再对键进行not in判断，实际上是对不确定具体值的键进行了分类
dic = {}
for d in list1:
    if d['name'] not in dic:
        dic[d['name']] = {'name': d['name'], 'hobby_list': [d['hobby']]}
    else:
        dic[d['name']]['hobby_list'].append(d['hobby'])
list_new = list(dic.values())
print(dic)
print(list_new)

标签：alex,name,框架,list1,list,dic,构建,hobby,字典
来源： https://www.cnblogs.com/hlzzhq/p/15878591.html