创建奇数索引python的总和 [英] creating sum of odd indexes python
本文介绍了创建奇数索引python的总和的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我试图创建一个函数等于列表中每个其他数字的总和。例如,如果列表是[0,1,2,3,4,5],则函数应该等于5 + 3 + 1。我怎么能这样做?我的Python知识并没有比while和for循环更远。谢谢。
解决方案
以下是简单的一行:
In [37]:L
Out [37]:[0,1,2,3,4,5]
In [38]:sum L [1 :: 2])
Out [38]:9
代码 L [1 :: 2]
表示从第一个索引开始,获取 L
/ b>
这是一种完成所有繁重工作的方法:
L = [0,1,2,3,4,5]
total = 0
在范围内(len(L)):
if i%2:#if this奇数索引
total + = L [i]
这是另一种方法,使用枚举
:
L = [0,1,2,3,4, 5]
total = 0
for i,枚举数(L):
if i%2:
total + = num
I'm trying to create a function equal to the sum of every other digit in a list. For example, if the list is [0,1,2,3,4,5], the function should equal 5+3+1. How could I do this? My knowledge of Python does not extend much farther than while and for loops. Thanks.
解决方案
Here is a simple one-liner:
In [37]: L
Out[37]: [0, 1, 2, 3, 4, 5]
In [38]: sum(L[1::2])
Out[38]: 9
In the above code, L[1::2]
says "get ever second element in L
, starting at index 1"
Here is a way to do all the heavy lifting yourself:
L = [0, 1, 2, 3, 4, 5]
total = 0
for i in range(len(L)):
if i%2: # if this is an odd index
total += L[i]
Here's another way, using enumerate
:
L = [0, 1, 2, 3, 4, 5]
total = 0
for i,num in enumerate(L):
if i%2:
total += num
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