更新值时是否可以使用python dict理解？

Question

counts = defaultdict(int)
for elem in sets:         #list of sets
    for _ in elem:        #each set contains elements that may be duplicates among the sets
        counts[_] += 1

有没有办法对这样的代码使用字典理解？

Answer 1

Don't use a list comprehension for side effects（即更新counts）。

相反，如果将defaultdict(int) 替换为Counter，则可以将解决方案写在一行中。 （我假设你的最终目标是让你的代码更短/更干净。）

sets = [
    {1, 2, 3},
    {1, 2},
    {3},
    {3}]
# ---
from collections import Counter
counts = Counter(x for s in sets for x in s)
print(counts)  # -> Counter({3: 3, 1: 2, 2: 2})

• 注意：您可能更喜欢使用itertools.chain.from_iterable(sets)，而不是嵌套的生成器表达式。

Answer 2

如果你坚持使用列表推导，你可以这样做：

>>> counts = {1:4, 2:5, 3:8, 4:2, 5:9}
>>> sets = [{2,3}, {3,1}, {2}]
>>> [[counts.update({e: counts[e]+1}) for e in si] for si in sets]
[[None, None], [None, None], [None]]
>>> counts
{1: 5, 2: 7, 3: 10, 4: 2, 5: 9}

但是是的，正如其他人所建议的，您可以使用 Counter：

#1：使用 Counter 从集合中获取新计数

#2：合并新的和默认的计数器

#3：让 Counter 回到你的字典中

带样本输入：

>>> counts = {1:4, 2:5, 3:8, 4:2, 5:9}
>>> sets = [{2,3}, {3,1}, {2}]
>>> from collections import Counter
>>> cntr_new = Counter(x for s in sets for x in s)
>>> cntr_new
Counter({2: 2, 3: 2, 1: 1})
>>> cntr_orig = Counter(counts)
>>> cntr_final = cntr_orig + cntr_new

最终结果：

>>> dict(cntr_final)
{1: 5, 2: 7, 3: 10, 4: 2, 5: 9}