Python - 用频率和项目的乘积替换列表中相同的连续元素

给定一个列表，任务是编写一个Python程序，用频率和 Item 的乘积替换连续元素的分组。

Input : test_list = [3, 3, 3, 3, 6, 7, 5, 5, 5, 8, 8, 6, 6, 6, 6, 6, 1, 1, 1, 2, 2]

Output : [12, 6, 7, 15, 16, 30, 3, 4]

Explanation : 3 occurs 4 times in consecution hence, 3*4 = 12, the result.

Input : test_list = [3, 3, 3, 3, 6, 7, 5, 5, 5]

Output : [12, 6, 7, 15]

Explanation : 5 occurs 3 times in consecution hence, 5*3 = 15, the result.

方法一：使用循环。

在此，将元素与前一个元素进行比较以决定组结束，如果元素不同，则将计数与元素的乘积作为结果附加。

Python3

# Python3 code to demonstrate working of
# Equal Consecution Product
# Using loop
 
# initializing list
test_list = [3, 3, 3, 3, 6, 7, 5, 5, 5, 8,
             8, 6, 6, 6, 6, 6, 1, 1, 1, 2, 2]
              
# printing original list
print("The original list is : " + str(test_list))
 
res = []
count = 1
for idx in range(1, len(test_list)):
     
    # checking with prev element
    if test_list[idx - 1] != test_list[idx]:
         
        # appending product
        res.append((test_list[idx - 1] * count))
        count = 0
    count += 1
res.append((test_list[-1] * count))
 
# printing result
print("Elements after equal Consecution product : " + str(res))

输出：

The original list is : [3, 3, 3, 3, 6, 7, 5, 5, 5, 8, 8, 6, 6, 6, 6, 6, 1, 1, 1, 2, 2]

Elements after equal Consecution product : [12, 6, 7, 15, 16, 30, 3, 4]

方法二：使用groupby() + sum()

在这种情况下，使用 groupby() 形成组，并且分组元素的总和给出所需的产品。

蟒蛇3

# Python3 code to demonstrate working of
# Equal Consecution Product
# Using groupby() + sum()
from itertools import groupby
 
# initializing list
test_list = [3, 3, 3, 3, 6, 7, 5, 5, 5, 8, 8, 6, 6, 6, 6, 6, 1, 1, 1, 2, 2]
              
# printing original list
print("The original list is : " + str(test_list))
 
# creating Consecution groups and summing for required values
res = [sum(group) for k, group in groupby(test_list)]
 
# printing result
print("Elements after equal Consecution product : " + str(res))

输出：

The original list is : [3, 3, 3, 3, 6, 7, 5, 5, 5, 8, 8, 6, 6, 6, 6, 6, 1, 1, 1, 2, 2]

Elements after equal Consecution product : [12, 6, 7, 15, 16, 30, 3, 4]