如何创建一个输出列表，它是输入 number_list 中交替元素的总和

Question

I'm trying to sum the even and odd-indexed elements of a list together without the use of the sum() method, but something is going wrong. I'm new to Python so help would be appreciated.

An example would be input number_list as [1, 2, 3, 4] then the output list will be [4, 6], which is from the sum of the even-numbered elements 1 and 3, and the odd-numbered elements 2 and 4 from the list number_list . Similarly, [5, 8, 3, 2, 6, 7, 1] would have an output list of [8, 10, 9, 9, 5].

This is my code:

number_list = [2, 6, 3, 5, 9, 1, 7]

res=[0, 0, 0, 0, 0, 0]
for i in range(0, len(number_list)): 
    if(i % 2): 
        res[1] += number_list[i] 
    else : 
        res[0] += number_list[i] 
print("Input List: {0}\n".format(number_list))
print("Output List: {0}\n".format(res))

Answer 1

I think the following is what you were going for; your first attempt was definitely on target:

!/usr/bin/env python3

number_list = [2, 6, 3, 5, 9, 1, 7]

i = 0
r = [0,0]
for n in number_list:
  if (i % 2) == 0:
    # Even
    r[0] += n
  else :
    # Odd
    r[1] += n
  i += 1

print(" Input List: {0}". format(number_list))
print("Output List: {0}". format(r))

The result is:

 Input List: [2, 6, 3, 5, 9, 1, 7]
Output List: [21, 12]

There are some simplifications here, namely throwing out Python nuances and just looking at iterating over a list of numbers, determining whether they are in the even or odd position (the i index), and adding (summing) them to the base r array. Is this the most compact and optimized Python code you will encounter? No. But is it clear as to what you're trying to do - I think so.

Answer 2

Joe's answer is absolutely correct. However more compact answer would be..

r = [0,0]
for n, val in enumerate(number_list):
   j = n % 2
   r[j] += val

print(" Input List: {0}". format(number_list))
print("Output List: {0}". format(r))