如何比较字典中的值

Question

如果我有这样的字典

times = {'personA': [9,10,11,15,16,18,19],'personB': [12,14,15,16,19]}

我试图找到这两个人都可用的时间。 这是两个人不可用的“阻塞”时间。 因此，9,10,11将是上午9点至11点，15,16将是下午3点至4点，依此类推。 我想找出两个人都可以使用的1-24之间的所有数字。 是否有捷径可寻？

Answer 1

使用集合而不是列表

例如

>>> times = {'personA': {9,10,11,15,16,18,19},'personB': {12,14,15,16,19}}
>>> hours = set(range(1,25))
>>> both = hours - (times['personA'] | times['personB'])
>>> both
{1, 2, 3, 4, 5, 6, 7, 8, 13, 17, 20, 21, 22, 23, 24}

Answer 2

times = {'personA': [9,10,11,15,16,18,19],'personB': [12,14,15,16,19]}
# Just a set with all 24 hours because
whole_day = {x for x in range(25)}

# Getting available time for both since it's what we need in fact, not busy time
free_times = {key: whole_day - set(value) for key, value in times.items()}

# Getting intersection of those free hours -> result
common_free_time = free_times['personA'] & free_times['personB']

Answer 3

times = {'personA': [9,10,11,15,16,18,19],'personB': [12,14,15,16,19]}
combined = times['personA'] + times['personB']
all_times = [i for i in range(1, 25)]
result = [x for x in all_times if x not in combined]

Answer 4

我将创建一组所有可用时间，并删除其中一个人不可用的所有插槽：

available = set(range(12,25))
for v in times.values():
    available = available - set(v)