python 列表汇总

Question

有一组列表
[1, 1, Decimal('1000.00'), 3, 3, Decimal('3000.00'), 0, 0, 0, 7, 7, Decimal('6200.00'), 11, 11, Decimal('10200.00')]
[28, 29, Decimal('25300.00'), 30, 30, Decimal('29500.00'), 0, 0, 0, 99, 100, Decimal('94600.00'), 157, 159, Decimal('149400.00')]
[3, 3, Decimal('2500.00'), 8, 8, Decimal('8000.00'), 0, 0, 0, 17, 17, Decimal('16500.00'), 28, 28, Decimal('27000.00')]
现在需要将每个列表对应索引上的值求和取出组成新的列表
该如何解答

Answer 1

是这样吗？取出来直接加不就行了。。。还是我理解错了，有点不敢相信你问的是这么简单的问题。

from decimal import *
a=[1, 1, Decimal('1000.00'), 3, 3, Decimal('3000.00'), 0, 0, 0, 7, 7, Decimal('6200.00'), 11, 11, Decimal('10200.00')]
b=[28, 29, Decimal('25300.00'), 30, 30, Decimal('29500.00'), 0, 0, 0, 99, 100, Decimal('94600.00'), 157, 159, Decimal('149400.00')]
c=[3, 3, Decimal('2500.00'), 8, 8, Decimal('8000.00'), 0, 0, 0, 17, 17, Decimal('16500.00'), 28, 28, Decimal('27000.00')]
d=[]
for i in range(len(a)):
    d.append(a[i]+b[i]+c[i])
    
print d

Answer 2

zip 是干“对应索引位组合”这事的。

# -*- coding: utf-8 -*-

from decimal import Decimal

a = [1, 1, Decimal('1000.00'), 3, 3, Decimal('3000.00'), 0, 0, 0, 7, 7, Decimal('6200.00'), 11, 11, Decimal('10200.00')]
b = [28, 29, Decimal('25300.00'), 30, 30, Decimal('29500.00'), 0, 0, 0, 99, 100, Decimal('94600.00'), 157, 159, Decimal('149400.00')]
c = [3, 3, Decimal('2500.00'), 8, 8, Decimal('8000.00'), 0, 0, 0, 17, 17, Decimal('16500.00'), 28, 28, Decimal('27000.00')]

print [sum(x) for x in zip(a, b, c)]