Summing first 2 elements in a Python list when the

问题:

I am working on the following Python list exercise from codingbat.com:

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

My solution below works:

def sum2(nums):
  if len(nums)>=2:
    return nums[0] + nums[1]
  elif len(nums)==1:
    return nums[0]
  return 0

But I wonder if there's any way to solve the problem with fewer conditional statements.

回答1:

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

回答2:

If you can't use sum, one possible solution uses exceptions:

totalsum = 0
try:
  totalsum += nums[0]
  totalsum += nums[1]
except IndexError:
  pass
return totalsum

Catch the error and short-circuit the summation if an element doesn't exist. Easier to ask forgiveness than permission, as they say.

回答3:

try this:

def sum2(nums):
  if len(nums) == 1:
    return nums[0]
  else:
    return sum(nums [:2])