Question

我正在尝试将winning_numbers与较大的previous_winners列表进行比较，然后得出某个数字出现多少次的计数。

winning_numbers = [20, 49, 47, 40, 36, 4, 2]
previous_winners = [
    [1, 8, 11, 25, 28, 4, 6],
    [13, 16, 34, 35, 45, 10, 12],
    [4, 5, 8, 31, 43, 2, 9],
    [2, 12, 15, 34, 50, 3, 4]
]

我一直在尝试以下

compare = set(winning_numbers) & set(previous_winners)
print(compare)

但是它会产生错误TypeError: unhashable type: 'list'，除非我在previous_winners上使用单个列表，例如给出{4, 2}但...如何计数这些数字的次数出现在previous_winners列表中？

我想最终打印出类似“我们匹配4个，它被匹配8次”之类的内容

Answer 1

您可以展平previous_winners列表，并使用collections.Counter来计算每个数字的出现次数，以便您可以迭代winning_numbers来生成中奖号码与其号码的对应关系在previous_winner中计数：

from collections import Counter
counts = Counter(n for l in previous_winners for n in l)
print({n: counts.get(n, 0) for n in winning_numbers})

这将输出：

{20: 0, 49: 0, 47: 0, 40: 0, 36: 0, 4: 3, 2: 2}

或者，如果您只希望以前出现的数字，则将winning_numbers首先设为集合，并仅对集合中的那些数字从previous_winners中产生计数会稍微有效率：< / p>

from collections import Counter
winning_set = set(winning_numbers)
print(Counter(n for l in previous_winners for n in l if n in winning_set))

这将输出：

Counter({4: 3, 2: 2})

Answer 2

您可以使用chain.from_iterable展平previous_winners并使用list comprehension创建字典，并使用.count来计数previous_winners中的每次出现：

l = list(chain.from_iterable(previous_winners))
{i:l.count(i) for i in winning_numbers}
{20: 0, 49: 0, 47: 0, 40: 0, 36: 0, 4: 3, 2: 2}

Answer 3

您可以遍历previous_winners的每个列表并调用集合的交集，然后遍历compare的每个交集并在每次发现匹配项时计数。

matches = {}
for prev in previous_winners:
    compare = set(winning_numbers) & set(prev)
    for match in compare:
        matches[match] = matches.get(match, 0) + 1

for k, v in matches.items():
    print(k, 'occurred', v, 'times')

输出：

4 occurred 3 times
2 occurred 2 times

比较两个列表，然后说一个列表中的一个项目匹配多少次

3 个答案: