问题：Python dict如何创建密钥或向密钥添加元素？
我有一本空字典。名称：dict_x
将具有其值为列表的键。
从一个单独的迭代中，我获得一个键（例如：）key_123和一个项目（一个元组），将其放置在dict_xvalue 的列表中key_123。
如果该键已经存在，我想添加此项。如果此键不存在，我想用一个空列表创建它，然后追加到它或只在其中添加一个元组。
将来再次出现此键时，由于它存在，我希望再次添加该值。
我的代码包含以下内容：
获取关键和价值。
看看中是否存在NOT键dict_x。
如果没有创建它： dict_x[key] == []
之后： dict_x[key].append(value)
这是这样做的方式吗？我应该尝试使用try/except积木吗？
I have an empty dictionary. Name: dict_x
It is to have keys of which values are lists.
From a separate iteration, I obtain a key (ex: key_123), and an item (a tuple) to place in the list of dict_x‘s value key_123.
If this key already exists, I want to append this item.
If this key does not exist, I want to create it with an empty list and then append to it or just create it with a tuple in it.
In future when again this key comes up, since it exists, I want the value to be appended again.
My code consists of this:
Get key and value.
See if NOT key exists in dict_x.
and if not create it: dict_x[key] == []
Afterwards: dict_x[key].append(value)
Is this the way to do it? Shall I try to use try/except blocks?
回答 0
用途dict.setdefault()：
dic.setdefault(key,[]).append(value)
help（dict.setdefault）：
setdefault(...)
D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D
Use dict.setdefault():
dic.setdefault(key,[]).append(value)
help(dict.setdefault):
setdefault(...)
D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D
回答 1
这是执行此操作的各种方法，因此您可以比较它的外观并选择所需的内容。我以我认为最“ pythonic”的方式对它们进行了排序，并评论了乍一看可能并不明显的利弊：
使用collections.defaultdict：
import collections
dict_x = collections.defaultdict(list)
...
dict_x[key].append(value)
优点：可能是最佳性能。缺点：在Python 2.4.x中不可用。
使用dict().setdefault()：
dict_x = {}
...
dict_x.setdefault(key, []).append(value)
缺点：未使用list()s的创建效率低。
使用try ... except：
dict_x = {}
...
try:
values = dict_x[key]
except KeyError:
values = dict_x[key] = []
values.append(value)
要么：
try:
dict_x[key].append(value)
except KeyError:
dict_x[key] = [value]
Here are the various ways to do this so you can compare how it looks and choose what you like. I’ve ordered them in a way that I think is most “pythonic”, and commented the pros and cons that might not be obvious at first glance:
Using collections.defaultdict:
import collections
dict_x = collections.defaultdict(list)
...
dict_x[key].append(value)
Pros: Probably best performance. Cons: Not available in Python 2.4.x.
Using dict().setdefault():
dict_x = {}
...
dict_x.setdefault(key, []).append(value)
Cons: Inefficient creation of unused list()s.
Using try ... except:
dict_x = {}
...
try:
values = dict_x[key]
except KeyError:
values = dict_x[key] = []
values.append(value)
Or:
try:
dict_x[key].append(value)
except KeyError:
dict_x[key] = [value]
回答 2
您可以为此使用defaultdict。
from collections import defaultdict
d = defaultdict(list)
d['key'].append('mykey')
这比setdefault没有创建最终不会使用的新列表要有效得多。每次调用setdefault都会创建一个新列表，即使该条目已存在于字典中也是如此。
You can use a defaultdict for this.
from collections import defaultdict
d = defaultdict(list)
d['key'].append('mykey')
This is slightly more efficient than setdefault since you don’t end up creating new lists that you don’t end up using. Every call to setdefault is going to create a new list, even if the item already exists in the dictionary.
回答 3
您可以使用defaultdict在collections。
来自doc的示例：
s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
d = defaultdict(list)
for k, v in s:
d[k].append(v)
You can use defaultdict in collections.
An example from doc:
s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
d = defaultdict(list)
for k, v in s:
d[k].append(v)
回答 4
dictionary['key'] = dictionary.get('key', []) + list_to_append
dictionary['key'] = dictionary.get('key', []) + list_to_append
