问题描述
我正在从 codingbat.com 进行以下 Python 列表练习:
I am working on the following Python list exercise from codingbat.com:
给定一个整数数组,返回数组中前两个元素的和大批.如果数组长度小于2,只需将元素相加即可存在,如果数组长度为 0,则返回 0.示例:
Given an array of ints, return the sum of the first 2 elements in the array. If the array length is less than 2, just sum up the elements that exist, returning 0 if the array is length 0. Examples:
sum2([1, 2, 3]) → 3 sum2([1, 1]) → 2 sum2([1, 1, 1, 1]) → 2
我的解决方案如下:
def sum2(nums): if len(nums)>=2: return nums[0] + nums[1] elif len(nums)==1: return nums[0] return 0
但我想知道有没有什么办法可以用更少的条件语句来解决这个问题.
But I wonder if there's any way to solve the problem with fewer conditional statements.
推荐答案
有.解决方案的两个要素 - 内置函数 sum 和列表的 切片:
There is. Two elements of the solution - builtin function sum and lists's slices:
>>> sum([1,2,3][:2]) 3 >>> sum([1,1,1,1][:2]) 2 >>> sum([1,1][:2]) 2 >>> sum([1][:2]) 1 >>> sum([][:2]) 0