我有一个项目列表,例如:
my_list = [ {'id':100, 'location':'A'}, {'id':100, 'location':'B'}, {'id':100, 'location':'C'}, {'id':101, 'location':'A'}, {'id':101, 'location':'G'}, {'id':100, 'location':'F'},{'id':100, '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','B','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。
答案 0 :(得分:3)
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'},
{'id': 100, 'location': 'C'},
{'id': 101, 'location': 'A'},
{'id': 101, 'location': 'G'},
{'id': 100, 'location': 'F'},
{'id': 100, '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。
答案 1 :(得分:2)
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'}, {'id':100, 'location':'C'}, {'id':101, 'location':'A'}, {'id':101, 'location':'G'}, {'id':100, 'location':'F'},{'id':100, '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', 'B', 'C']
101-round#1 ['A', 'G']
100-round#2 ['F', 'R']