从众多答案中可以看出，有几种解决方案。LBYL的一个实例(三思而后行)还没有提到，has_key()方法:

my_dict = {}

def add (key):
    if my_dict.has_key(key):
        my_dict[key] += 1
    else:
        my_dict[key] = 1

if __name__ == '__main__':
    add("foo")
    add("bar")
    add("foo")
    print my_dict

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

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

我正在寻找它，没有在网上找到它，然后尝试我的运气与尝试/错误，找到了它

my_dict = {}

if my_dict.__contains__(some_key):
  my_dict[some_key] += 1
else:
  my_dict[some_key] = 1