Python - 连续元素功率的总和

给定一个列表，任务是编写一个Python程序来计算连续元素出现频率的幂的总和。

例子：

Input : test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5] 
Output : 110 
Explanation : 2^3 + 3^4 + 4^2 + 5 = 110
 

Input : test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4] 
Output : 105 
Explanation : 2^3 + 3^4 + 4^2 = 105 

方法#1：使用循环

在此，我们对每个元素进行迭代并测试下一个元素，如果发现不同，则对连续元素的幂进行求和，否则添加计数器以获取频率以增加数量。

Python3

# Python3 code to demonstrate working of
# Summation of consecutive elements power
# Using loop
  
# initializing list
test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
  
# printing original lists
print("The original list is : " + str(test_list))
  
freq = 1
res = 0
for idx in range(0, len(test_list) - 1):
  
    # adding powers
    if test_list[idx] != test_list[idx + 1]:
        res = res + test_list[idx] ** freq
        freq = 1
    else:
        freq += 1
  
# catering for last element
res = res + test_list[len(test_list) - 1] ** freq
  
# printing result
print("Computed summation of powers : " + str(res))

输出

The original list is : [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Computed summation of powers : 110

方法 #2：使用defaultdict() + loop + sum()

在这里，我们使用键值对使用 defautldict 捕获频率，并使用循环来遍历每个元素。接下来，使用 sum() 执行求和。

蟒蛇3

# Python3 code to demonstrate working of
# Summation of consecutive elements power
# Using defaultdict() + loop + sum()
from collections import defaultdict
  
# initializing list
test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
  
# printing original lists
print("The original list is : " + str(test_list))
  
# getting frequency
temp = defaultdict(int)
for ele in test_list:
    temp[ele] += 1
  
temp = dict(temp)
  
# computing summation
res = sum([key ** temp[key] for key in temp])
  
# printing result
print("Computed summation of powers : " + str(res))

输出

The original list is : [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Computed summation of powers : 110