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

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

其他回答

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

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)。

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

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

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