Question

我正在尝试创建具有多个字典（键，值）对的列表。例如，如果有一个学生并且他在不同的时间学习，那么我必须将不同的时间表合并为一个列表。我正在从数据库中提取数据。

获取的值如下：

{'name':'a', 'study_time':{'start':9,'end':10}}
{'name':'a', 'study_time':{'start':11,'end':12}}
{'name':'a', 'study_time':{'start':13,'end':15}}
{'name':'a', 'study_time':{'start':16,'end':19}}
{'name':'a', 'study_time':{'start':20,'end':21}}

我需要的是

{
    'name':'a',
    'study_time':[
        {'start':9,'end':10},
        {'start':11,'end':12},
        {'start':13,'end':15},
        {'start':16,'start':19},
    ],
}

我得到的是

{
    'name':'a',
    'study_time': [
        [
            [
                {'start':9,'end':10},
                {'start':11,'end':12},
            ],
            {'start':13,'end':15},
            {'start':16,'start':19},
        ],
        {'start':20,'end':21},
    ],
}

我尝试过的是

parsed_user_details=[]
if len(parsed_user_details) == 0:
    parsed_user_details.append(parsed_user_detail)
    print("I am inside if", parsed_user_details,file=sys.stderr)
else:
    parsed_user_details[0]['time'] = [
        parsed_user_details[0]['time'],
        parsed_user_detail['time'],
    ]
    print("I am inside else", parsed_user_details)

总体上我尝试过的是，如果时间列表为0，只需将值添加到列表中，否则将值添加到列表中。我解析的用户详细信息是 parsed_user_detail = {'name'：'a'，'study_time'：{'start'：9，'end'：10}} 每当我循环播放时，我都会一一得到下面的parsed_user_detail。

{'name':'a', 'study_time':{'start':9,'end':10}}
{'name':'a', 'study_time':{'start':11,'end':12}}
{'name':'a', 'study_time':{'start':13,'end':15}}
{'name':'a', 'study_time':{'start':16,'end':19}}
{'name':'a', 'study_time':{'start':20,'end':21}}

Answer 1

您很近：

list_dict = [{'name':'a',"study_time":{'start':9,'end':10}},
{'name':'a',"study_time":{'start':11,'end':12}},
{'name':'a',"study_time":{'start':13,'end':15}},
{'name':'a',"study_time":{'start':16,'end':19}}]

name = list_dict[0]["name"]

parsed_user_details = []

for i in list_dict : 
  if len(parsed_user_details) == 0:
    parsed_user_details.append(i)
    parsed_user_details[0]["study_time"] = [parsed_user_details[0]["study_time"]]
  else : 
    parsed_user_details[0]['study_time'].append(i['study_time'])



print(parsed_user_details)

您需要添加它，使第一个study_time成为一个数组，然后追加到它：

parsed_user_details[0]["study_time"] = [parsed_user_details[0]["study_time"]]

输出：

{'name': 'a',
  'study_time': [{'end': 10, 'start': 9},
   {'end': 12, 'start': 11},
   {'end': 15, 'start': 13},
   {'end': 19, 'start': 16}]}

Answer 2

您可以使用[Python 3.Docs]: itertools.groupby(iterable, key=None)。我还添加了其他用户记录以增加复杂性：

public Bitmap mergeMultipleImages(Bitmap[] img_bitmaps){

    Bitmap result = Bitmap.createBitmap(img_bitmaps[0].getWidth() * 2, img_bitmaps[0].getHeight() * 2, Bitmap.Config.ARGB_8888);
    Canvas canvas = new Canvas(result);
    Paint paint = new Paint();
    for (int i = 0; i < img_bitmaps.length; i++) {
        canvas.drawBitmap(img_bitmaps[i], img_bitmaps[i].getWidth() * (i % 2), img_bitmaps[i].getHeight() * (i / 2), paint);
    }
        return result;
    }

Answer 3

所以我有一种解决方案，有点round回，但可以得到您的结果。我假设您实际上可能有不同的名称，并且每个名称都想要一个对象？

from collections import defaultdict
l1 = [{'name':'a', 'study_time':{'start':9,'end':10}},
{'name':'a', 'study_time':{'start':11,'end':12}},
{'name':'a', 'study_time':{'start':13,'end':15}},
{'name':'a', 'study_time':{'start':16,'end':19}},
{'name':'a', 'study_time':{'start':20,'end':21}},]

dict_holder = defaultdict(list)
for d in l1:
    dict_holder[d['name']].append(d['study_time'])

new_dicts = []
for name, study_list in dict_holder.items():
    new_dicts.append({'name': name, 'study_time': study_list})

new_dicts
[{
     'name': 'a', 
     'study_time': [
         {'start': 9, 'end': 10}, 
         {'start': 11, 'end': 12}, 
         {'start': 13, 'end': 15}, 
         {'start': 16, 'end': 19}, 
         {'start': 20, 'end': 21}]
 }]

Answer 4

对于仅内置解决方案，请尝试以下操作：

data = [
    {'name':'a', 'study_time':{'start':9,'end':10}},
    {'name':'a', 'study_time':{'start':11,'end':12}},
    {'name':'a', 'study_time':{'start':13,'end':15}},
    {'name':'a', 'study_time':{'start':16,'end':19}},
    {'name':'a', 'study_time':{'start':20,'end':21}}
]

result = []
for entry in data:
    item = next(
        (x for x in result if x.get('name') == entry.get('name')),
        None
    )
    if not item:
        item = {
            'name': entry.get('name'),
            'study_time': []
        }
        result.append(item)
    item['study_time'].append(entry.get('study_time'))

Answer 5

如果预期结果是多个词典的列表，则可以执行此操作。

from itertools import groupby

def get_studetails(student):
    res = {}
    for i in student:
        if not res.get('name'):
            res['name'] = i.get('name')
        if not res.get('study_time') or res.get('study_time') == []:
            res['study_time'] = []
            res['study_time'].append(i.get('study_time'))
        else:
            temp_study = res.get('study_time') 
            temp_study.append(i.get('study_time'))
            res['study_time'] = temp_study
    return res

total_list = []

studentdetailsl = [{'name':'a','study_time':{'start':9,'end':10}},{'name':'a','study_time':{'start':11,'end':12}},{'name':'a','study_time':{'start':13,'end':15}},{'name':'a','study_time':{'start':16,'end':19}},{'name':'a','study_time':{'start':20,'end':21}}, {'name':'b','study_time':{'start':20,'end':21}},{'name':'b','study_time':{'start':11,'end':12}}]

for k,v in groupby(studentdetailsl, lambda x: x['name']):
    total_list.append(get_studetails(list(v)))
from pprint import pprint
pprint(total_list)

输出：

[{'name': 'a',
  'study_time': [{'end': 10, 'start': 9},
                 {'end': 12, 'start': 11},
                 {'end': 15, 'start': 13},
                 {'end': 19, 'start': 16},
                 {'end': 21, 'start': 20}]},
 {'name': 'b',
  'study_time': [{'end': 21, 'start': 20}, {'end': 12, 'start': 11}]}]

将值作为多个列表添加到列表中

5 个答案: