Python - 如何撰寫 Python 程式來接受 List,并提取所有頻率大于 K 的元素?請指教?
示例輸入/輸出:
輸入串列:
[4, 6, 4, 3, 3, 4, 3, 4, 6, 6]
輸入 K:
2
所需輸出:[4, 3, 6]
uj5u.com熱心網友回復:
使用串列理解:
a = [4, 6, 4, 3, 3, 4, 3, 4, 6, 6]
k = 2
out = [i for i in set(a) if a.count(i) > k]
uj5u.com熱心網友回復:
在python中使用計數器
list1 = [4, 6, 4, 3, 3, 4, 3, 4, 6, 6]
d = Counter(list1)
new_list = list([item for item in d if d[item] > 1])
print(new_list) #output: [4, 6, 3]
uj5u.com熱心網友回復:
inputs = 0
user_list = []
while inputs < 8:
user_input = int(input("Enter a number: "))
user_list.append(user_input)
inputs = 1
input_list = [4, 6, 4, 3, 3, 4, 6, 6]
input_k = 2
def extract(x, y):
product = []
for i in x:
list_element = i
if list_element > y:
product.append(i)
else:
break
product = list(set(product))
return product
print(extract(user_list, input_k))
print(extract(input_list, input_k))
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