Question

我正在尝试获取匹配的ID并将数据存储到一个列表中。我有一个词典列表：

list = [
            {'id':'123','name':'Jason','location': 'McHale'},
            {'id':'432','name':'Tom','location': 'Sydney'},
            {'id':'123','name':'Jason','location':'Tompson Hall'}
       ]

预期输出类似于

# {'id':'123','name':'Jason','location': ['McHale', 'Tompson Hall']},
# {'id':'432','name':'Tom','location': 'Sydney'},

如何根据dict ID值获取匹配数据？我试过了：

for item in mylist:
    list2 = []
    row = any(list['id'] == list.id for id in list)
    list2.append(row)

这不起作用（抛出：TypeError: tuple indices must be integers or slices, not str）。如何获得具有相同ID的所有项目并存储到一个词典中？

Answer 1

鉴于此输入

mylist = [
            {'id':'123','name':'Jason','location': 'McHale'},
            {'id':'432','name':'Tom','location': 'Sydney'},
            {'id':'123','name':'Jason','location':'Tompson Hall'}
       ]

你可以用理解来提取它

matched = [d for d in mylist if d['id'] == '123']

然后你想要合并这些位置。假设matched不为空

final = matched[0]
final['location'] = [d['location'] for d in matched]

这是解释器

In [1]: mylist = [
   ...:                 {'id':'123','name':'Jason','location': 'McHale'},
   ...:                 {'id':'432','name':'Tom','location': 'Sydney'},
   ...:                 {'id':'123','name':'Jason','location':'Tompson Hall'}
   ...:            ]

In [2]: matched = [d for d in mylist if d['id'] == '123']    
In [3]: final=matched[0]
In [4]: final['location'] = [d['location'] for d in matched]
In [5]: final
Out[5]: {'id': '123', 'location': ['McHale', 'Tompson Hall'], 'name': 'Jason'}

显然，您希望将'123'替换为包含所需id值的变量。

将其全部包含在函数中：

def merge_all(df):
    ids = {d['id'] for d in df}
    result = []
    for id in ids:
        matches = [d for d in df if d['id'] == id]
        combined = matches[0]
        combined['location'] = [d['location'] for d in matches]
        result.append(combined)
    return result

另外，请不要将list用作变量名。它会隐藏内置list类。

Answer 2

首先，您在for循环中遍历词典列表，但从不引用您在item中存储的词典。我想当你写list[id]时，你的意思是item[id]。

其次，any()返回一个布尔值（true或false），这不是你想要的。相反，也许尝试row = [dic for dic in list if dic['id'] == item['id']]

第三，如果你在for循环中定义了list2，它将在每次迭代时消失。在for循环之前移动list2 = []。

这应该会给你一个良好的开端。请记住，row只是具有相同ID的所有词典的列表。

Answer 3

在将我期望成为列表的字典条目转换为列表之后，我会使用kdopen的方法和合并方法。当然，如果你想避免冗余，那就制作它们。

mylist = [
            {'id':'123','name':['Jason'],'location': ['McHale']},
            {'id':'432','name':['Tom'],'location': ['Sydney']},
            {'id':'123','name':['Jason'],'location':['Tompson Hall']}
       ]


def merge(mylist,ID):
    matches = [d for d in mylist if d['id']== ID]

    shell = {'id':ID,'name':[],'location':[]}

    for m in matches:
        shell['name']+=m['name']    
        shell['location']+=m['location']
        mylist.remove(m)

    mylist.append(shell)
    return mylist

updated_list = merge(mylist,'123')

词典列表：获取基于id的匹配词典列表

3 个答案: