Question

我有一套清单：

a = [{'foo','cpu','phone'},{'foo','mouse'}, {'dog','cat'}, {'cpu'}]

预期结果：

我想查看每个单独的字符串，进行计数并以原始格式返回所有内容x >= 2：

a = [{'foo','cpu','phone'}, {'foo','mouse'}, {'cpu'}]

我尝试使用列表理解进行循环，但不适用于集合列表：

a = [k for k in a if a.count(k) >= 2]

Answer 1

from collections import Counter

a = [{'foo','cpu','phone'},{'foo','mouse'}, {'dog','cat'}, {'cpu'}]
x = Counter(sum([list(i) for i in a], []))
z = [a for a, b in dict(x).items() if b >=2]
res = []
for i in a:
    for j in i:
        if j in z: 
            res.append(i)
            break 
print(res)

输出

[{'foo', 'phone', 'cpu'}, {'foo', 'mouse'}, {'cpu'}]

Answer 2

from collections import Counter
counter = Counter()
for a_set in a:
    # First we create a counter with counts for all words
    counter.update(a_set)
result = []
for a_set in a:
    for word in a_set:
        if counter[word] >= 2:
            # If any word in any set has a count of 2 or more we append the set
            result.append(a_set)
            break
print(result)

通过一组列表过滤出现次数

2 个答案: