对这种方法不是很兴奋，但是它将带您到那里。使用lst对字典defaultdict进行迭代，以mid的值进行分组，然后对那个进行迭代以产生输出，并结合msg和antivirus键。

from collections import defaultdict

lst = [{'mid': 123,'msg': 'sometext','antivirus': 'positive'},{'mid': 123,'msg': 'sometext2',{'mid': 456,'msg': 'sometext3','msg': 'sometext4',{'mid': 789,'msg': 'sometext5','antivirus': 'positive'}]

dd = defaultdict(list)
for d in lst:
    key = d['mid']
    dd[key].append(d)

output = []
for (k,v) in dd.items():
    output.append({
        'mid':       k,'msg':       ','.join(x['msg']       for x in v),'antivirus': ','.join(x['antivirus'] for x in v),})

print(output)

[
  {'mid': 123,'msg': 'sometext,sometext2','antivirus': 'positive,positive'},'msg': 'sometext3,sometext4','antivirus': 'positive'}
]

您可以只使用pandas dataFrame：

import pandas as pd

lst  = [{'mid': 123,'antivirus': 'positive'}]

d = (pd.DataFrame(lst)
       .groupby(['mid'])
       .agg(','.join)
       .reset_index()
       .to_dict('r'))

print (d)

输出：

[{'mid': 123,positive',sometext2'},sometext4'},'antivirus': 'positive','msg': 'sometext5'}]

将您的一个变量（list）与内置变量相同是一个不好的主意，因此我在此处使用l

使用中间defaultdict：

from collections import defaultdict


intermediate = defaultdict(lambda: defaultdict(list))
for record in l:
    mid = record["mid"]
    for key,value in record.items():
        if key == "mid":
            continue
        intermediate[mid][key].append(value)

result = [
    {"mid": mid,**{key: ",".join(value) for key,value in attributes.items()}}
    for mid,attributes in intermediate.items()
]
result

（list是python中的关键字，因此我将名称更改为mylist） 这是您必须提供的一线服务：

import itertools; map(lambda sub: reduce(lambda a,b: { key : ",".join(set(filter(lambda x: x!='',[str(a.get(key,''))] + [str(b.get(key,''))]))) for key in set(a.keys() + b.keys()) },sub,{}),map(lambda sub: list(sub[1]),itertools.groupby(mylist,lambda lst: lst['mid'])))

不那么令人讨厌：

import itertools
groups = map(lambda sub: list(sub[1]),lambda lst: lst['mid'])) # get the dicts organized into groups on key 'mid'

def joindicts(a,b):
    result = dict()
    for key in set(a.keys() + b.keys()): # get union of keys for both dicts
        val_a = str(a.get(key,''))
        val_b = str(b.get(key,''))
        val = ','.join([x for x in [val_a] + [val_b] if x != ''])
        result.update({key:val})
    return result

map(lambda sub: reduce(joindicts,groups)