Python 如果键存在，则删除字典项
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Delete a dictionary item if the key exists
提问by Simon Hughes
Is there any other way to delete an item in a dictionary only if the given key exists, other than:
仅当给定的键存在时，是否还有其他方法可以删除字典中的项目，除了：
if key in mydict:
del mydict[key]
The scenario is that I'm given a collection of keys to be removed from a given dictionary, but I am not certain if all of them exist in the dictionary. Just in case I miss a more efficient solution.
场景是我得到了一组要从给定字典中删除的键，但我不确定字典中是否存在所有键。以防万一我错过了更有效的解决方案。
采纳答案by Adem ?zta?
回答by mgilson
There is also:
还有：
try:
del mydict[key]
except KeyError:
pass
This only does 1 lookup instead of 2. However, exceptclauses are expensive, so if you end up hitting the except clause frequently, this will probably be less efficient than what you already have.
这只会执行 1 次查找，而不是 2 次。但是，except子句的开销很大，因此如果您最终经常使用 except 子句，这可能会比您已有的效率低。
回答by hughdbrown
Approach: calculate keys to remove, mutate dict
方法：计算要删除的键，变异 dict
Let's call keysthe list/iterator of keys that you are given to remove. I'd do this:
让我们调用keys您要删除的键的列表/迭代器。我会这样做：
keys_to_remove = set(keys).intersection(set(mydict.keys()))
for key in keys_to_remove:
del mydict[key]
You calculate up front all the affected items and operate on them.
您预先计算所有受影响的项目并对其进行操作。
Approach: calculate keys to keep, make new dict with those keys
方法：计算要保留的键，用这些键创建新的字典
I prefer to create a new dictionary over mutating an existing one, so I would probably also consider this:
我更喜欢创建一个新的字典而不是改变一个现有的字典，所以我可能也会考虑这个：
keys_to_keep = set(mydict.keys()) - set(keys)
new_dict = {k: v for k, v in mydict.iteritems() if k in keys_to_keep}
or:
或者：
keys_to_keep = set(mydict.keys()) - set(keys)
new_dict = {k: mydict[k] for k in keys_to_keep}
