Reputation: 551
I am currently using a dictionary keyed on seconds since epoch combined with a nested dictionary and list to store and look up events. The strength is that I can quickly look up a value with good performance (think hash). The weakness is that it is clumsy to search and manipulate the subsequent contents. I can't help but think my approach is not very Pythonic so I am looking for suggestions.
Here is a simple example using integers rather than seconds since epoch:
D = {}
D[1] = {"995" : ["20905", "200101"]}
D[2] = {"991" : ["20901"], "995" : ["20905"]}
eventCode = '995'
error = '90900'
# find entry
if 1 in D:
# if eventCode exists then append error code
if D[1][eventCode]:
D[1][eventCode].append(error)
I can look up D[1] quickly, however the remainder of the code doesn't seem very pythonic. Any suggestions or am I paranoid?
I should check to see if "error" is already in the list. However, I am unsure how to check for membership in this construct. This snippet does not work for me:
if error not in D[1][eventCode]
Upvotes: 5
Views: 1164
Reputation: 7598
Apart from the other suggestions, you could use defaultdict and sets in this way:
from collections import defaultdict
D = defaultdict(lambda: defaultdict( lambda: None))
D[1] = {"995" : {"20905", "200101"}}
D[2] = {"991" : {"20901"}, "995" : {"20905"}}
eventCode = '995'
error = '90900'
tmp = D[1][eventCode]
if tmp is None:
D[1][eventCode]={error}
else:
tmp.add(error)
Upvotes: 1
Reputation: 39546
I'm not sure if that's you want, but you can change lists with sets to handle duplicates and use dict.get to skip checking if key exists:
D = {"995" : {"20905", "200101"}} # dict: str -> set
# 1) event code not exists, nothing changes:
D.get('111', set()).add('7777')
print(D) # {'995': {'20905', '200101'}}
# 2) event code exists, error already present, nothing changes:
D.get('995', set()).add('20905')
print(D) # {'995': {'20905', '200101'}}
# 3) event code exists, error not present, error will be added:
D.get('995', set()).add('7777')
print(D) # {'995': {'20905', '7777', '200101'}}
Upvotes: 2