Python - 双元组交替求和

给定双元组，交替执行交替元素的求和，即索引的求和。

Input : test_list = [(4, 1), (5, 6), (3, 5), (7, 5)] 
Output : 18 
Explanation : 4 + 6 + 3 + 5 = 18, Alternatively are added to sum. 

Input : test_list = [(4, 1), (5, 6), (3, 5)] 
Output : 13 
Explanation : 4 + 6 + 3 = 183, Alternatively are added to sum. 

方法#1：使用循环

在这里，我们遍历每个元组，并检查列表中的索引，如果偶数，则对元组的第一个元素求和，否则将第二个元素相加以计算总和。

Python3

# Python3 code to demonstrate working of
# Dual Tuple Alternate summation
# Using loop
  
# initializing list
test_list = [(4, 1), (5, 6), (3, 5), (7, 5), (1, 10)]
  
# printing original list
print("The original list is : " + str(test_list))
  
res = 0
for idx in range(len(test_list)):
  
    # checking for Alternate element
    if idx % 2 == 0:
        res += test_list[idx][0]
    else:
        res += test_list[idx][1]
  
# printing result
print("Summation of Alternate elements of tuples : " + str(res))

输出

The original list is : [(4, 1), (5, 6), (3, 5), (7, 5), (1, 10)]
Summation of Alternate elements of tuples : 19

方法 #2：使用列表理解 + sum()

在这里，我们使用sum()执行求和任务，并且使用列表理解为列表中的元组迭代提供紧凑的解决方案。

蟒蛇3

# Python3 code to demonstrate working of 
# Dual Tuple Alternate summation
# Using list comprehension + sum()
  
# initializing list
test_list = [(4, 1), (5, 6), (3, 5), (7, 5), (1, 10)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# summation using sum(), list comprehension offers shorthand
res = sum([test_list[idx][0] if idx % 2 == 0 else test_list[idx][1] for idx in range(len(test_list))])
  
# printing result 
print("Summation of Alternate elements of tuples : " + str(res))

输出

The original list is : [(4, 1), (5, 6), (3, 5), (7, 5), (1, 10)]
Summation of Alternate elements of tuples : 19