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
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