python遍历列表并基于“新到货”添加特殊键

如何解决python遍历列表并基于“新到货”添加特殊键

我有一个项目列表，例如：

my_list = [ {'id':100,'location':'A'},{'id':100,'location':'B'},'location':'C'},{'id':101,'location':'G'},'location':'F'},'location':'R'}]

如果将所有内容添加到字典中，我将有100：['A'，'B'，'C'，'F'，'R']和101：['A'，'G]，但我想要显示100个有2个“回合”，例如：

100-round#1:['A','B','C'] #note the key is 100-round#1
101-round#1:['A','G']
...
100-round#2:['F','R']

我这样做的目的是获得一个全局词典（部分需要）

location_dict={}
for r in my_list:
   id = r['id']
   location = r['location']
   id_locations = location_dict.get(id,[])
   id_locations.append(location)
   location_dict[id]=id_locations 

结果：

{100:['A','C','F','R'],101:['A','G]}

我如何遍历列表并创建一个词典，其中的键指示ID（数字）以及“回合”，所以我可以说列表中有100个回合。

解决方法

我假设“舍入”是指列表中连续id个值的条纹。我将使用defaultdict来避免为新键分配一个空列表，但这并不会改变与您所质疑的逻辑相关的内容。试试

from collections import defaultdict

my_list = [
    {'id': 100,'location': 'A'},{'id': 100,'location': 'B'},'location': 'C'},{'id': 101,'location': 'G'},'location': 'F'},'location': 'R'}
]

# This keeps track of how many times we've seen each
# id,and defaults to 0.
rounds = defaultdict(int)

locations = defaultdict(list)
current_id = ''

for data in my_list:
    id = data['id']

    if id != current_id:
        # If this is true,we start a new round.
        round = rounds[id] + 1
        rounds[id] += 1
        current_id = id

    locations[f'{id}-round{round}'].append(data['location'])

print(dict(locations))

产生

{'100-round1': ['A','B','C'],'101-round1': ['A','G'],'100-round2': ['F','R']}

当然，如果locations适合您，则不必将defaultdict强制转换为列表末尾。

,

此脚本将创建一个有序词典，其中键是字符串<id>-round#<round number>，值是实际回合的位置：

from itertools import groupby,count
from collections import defaultdict,OrderedDict


my_list = [ {'id':100,'location':'A'},{'id':100,'location':'B'},'location':'C'},{'id':101,'location':'G'},'location':'F'},'location':'R'}]

out,c = OrderedDict(),defaultdict(lambda: count(1))
for id_,g in groupby(my_list,lambda k: k['id']):
    out['{}-round#{}'.format(id_,next(c[id_]))] = [d['location'] for d in g]


for k,v in out.items():
    print(k,v)

打印：

100-round#1 ['A','C']
101-round#1 ['A','G']
100-round#2 ['F','R']