[英]Handling a dictionary with multiple values per key
我正在遍历一个列表并尝试使用key/pair
值创建一个字典,问题是当我遍历列表时,我可能会发现一个键的重复实例,在这种情况下我想添加另一对。 我知道我需要的是某种列表列表,虽然我让它适用于一个键有两个值的情况,但当我达到 3 个或更多时它会失败。
这是我当前的代码:
team2goals = {}
<loop through list>
timestamp = xValue
if name in team2goals:
temp = team2goals[name]
temp = {temp,timestamp}
team2goals[name] = temp
else:
team2goals[name] = timestamp
其中team2goals
是<str>,<str>
的字典。 它检查键 ( name
) 的实例是否存在,如果存在,它存储当前键值并创建值的字典,否则它只是添加一个新的键/对值。
我试过类似的东西:
tempT = team1goals[name]
temp = []
temp.append(tempT)
temp.append(timestamp)
team1goals[name] = temp
但这似乎只是开始嵌套字典,即[[timestamp,timestamp2],timestamp3]
有没有办法做我正在尝试的事情?
您几乎拥有它,您需要为每个键创建一个列表,然后只需 append 时间戳。
尝试这个:
team2goals = {}
<loop through list>
timestamp = xValue
if name not in team2goals:
team2goals[name] = []
team2goals[name].append(timestamp)
那么Dict[str, List[str]]
呢? 这很容易使用defaultdict
from collections import defaultdict
team_to_goals = defaultdict(list)
<loop>:
team_to_goals[name].append(x_value)
你可以做几件事。 如果同一键值有多个条目,则应使用一个键值,但保留结果列表。 如果每个结果都有多个组件,您可以将它们保存在子列表或元组中。 对于您的应用程序,我建议:
键:列表(元组(数据,时间戳))
基本上保存您提到的 2 个字符串对的列表。
此外,默认字典对此很有用,因为它将在字典中创建新的列表项作为“默认项”。
代码:
from collections import defaultdict
# use default dictionary here, because it is easier to initialize new items
team_goals = defaultdict(list) # just list in here to refer to the class, not 'list()''
# just for testing
inputs = [ ['win big', 'work harder', '2 May'],
['be efficient', 'clean up', '3 may'],
['win big', 'be happy', '15 may']]
for key, idea, timestamp in inputs:
team_goals[key].append((idea, timestamp))
print(team_goals)
for key in team_goals:
values = team_goals[key]
for v in values:
print(f'key value: {key} is paired with {v[0]} at timestamp {v[1]}')
产量:
defaultdict(<class 'list'>, {'win big': [('work harder', '2 May'), ('be happy', '15 may')], 'be efficient': [('clean up', '3 may')]})
key value: win big is paired with work harder at timestamp 2 May
key value: win big is paired with be happy at timestamp 15 may
key value: be efficient is paired with clean up at timestamp 3 may
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