如何查找字典中的键是否已设置为非none值?

如果已经有一个值,我想增加值,否则设置为1:

my_dict = {}

if my_dict[key] is not None:
  my_dict[key] = 1
else:
  my_dict[key] += 1

当前回答

同意cgoldberg的观点。我是这样做的:

try:
    dict[key] += 1
except KeyError:
    dict[key] = 1

所以,要么像上面那样做,要么像其他人建议的那样使用默认字典。不要使用if语句。这不是Pythonic。

其他回答

我个人喜欢使用setdefault()

my_dict = {}

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

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

下面是我最近想出的解决这个问题的一句话。它基于setdefault字典方法:

my_dict = {}
my_dict[key] = my_dict.setdefault(key, 0) + 1

你需要字典成语中的关键字。

if key in my_dict and not (my_dict[key] is None):
  # do something
else:
  # do something else

但是,您可能应该考虑使用defaultdict(正如dF建议的那样)。

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