这并不是直接回答这个问题，但对我来说，看起来您可能需要collections.Counter的功能。

from collections import Counter

to_count = ["foo", "foo", "bar", "baz", "foo", "bar"]

count = Counter(to_count)

print(count)

print("acts just like the desired dictionary:")
print("bar occurs {} times".format(count["bar"]))

print("any item that does not occur in the list is set to 0:")
print("dog occurs {} times".format(count["dog"]))

print("can iterate over items from most frequent to least:")
for item, times in count.most_common():
    print("{} occurs {} times".format(item, times))

这就产生了输出

Counter({'foo': 3, 'bar': 2, 'baz': 1})
acts just like the desired dictionary:
bar occurs 2 times
any item that does not occur in the list is set to 0:
dog occurs 0 times
can iterate over items from most frequent to least:
foo occurs 3 times
bar occurs 2 times
baz occurs 1 times

有点晚了，但应该有用。

my_dict = {}
my_dict[key] = my_dict[key] + 1 if key in my_dict else 1

要回答“我如何才能知道字典中给定的索引是否已经被设置为非none值”的问题，我更喜欢这样:

try:
  nonNone = my_dict[key] is not None
except KeyError:
  nonNone = False

这符合已经引用的EAFP概念(请求原谅比请求许可更容易)。它还避免了字典中的重复键查找，因为它会在my_dict和my_dict[key]不是None的key中查找，如果查找是昂贵的，这是有趣的。

对于您提出的实际问题，即增加int值(如果存在)，或将其设置为默认值(否则)，我还建议使用

my_dict[key] = my_dict.get(key, default) + 1

正如安德鲁·威尔金森的回答。

如果要在字典中存储可修改的对象，还有第三种解决方案。一个常见的例子是multimap，在其中存储键的元素列表。在这种情况下，你可以使用:

my_dict.setdefault(key, []).append(item)

如果字典中不存在forkey的值，setdefault方法将其设置为setdefault的第二个参数。它的行为就像一个标准的my_dict[key]，返回键的值(可能是新设置的值)。

您正在寻找collections.defaultdict(适用于Python 2.5+)。这

from collections import defaultdict

my_dict = defaultdict(int)
my_dict[key] += 1

会做你想做的事。

对于常规的Python字典，如果给定的键没有值，则在访问字典时不会得到None——会引发KeyError。所以如果你想使用一个普通的字典，而不是你的代码，你会使用

if key in my_dict:
    my_dict[key] += 1
else:
    my_dict[key] = 1

我个人喜欢使用setdefault()

my_dict = {}

my_dict.setdefault(some_key, 0)
my_dict[some_key] += 1

这并不是直接回答这个问题，但对我来说，看起来您可能需要collections.Counter的功能。

from collections import Counter

to_count = ["foo", "foo", "bar", "baz", "foo", "bar"]

count = Counter(to_count)

print(count)

print("acts just like the desired dictionary:")
print("bar occurs {} times".format(count["bar"]))

print("any item that does not occur in the list is set to 0:")
print("dog occurs {} times".format(count["dog"]))

print("can iterate over items from most frequent to least:")
for item, times in count.most_common():
    print("{} occurs {} times".format(item, times))

这就产生了输出

Counter({'foo': 3, 'bar': 2, 'baz': 1})
acts just like the desired dictionary:
bar occurs 2 times
any item that does not occur in the list is set to 0:
dog occurs 0 times
can iterate over items from most frequent to least:
foo occurs 3 times
bar occurs 2 times
baz occurs 1 times