找到等于或大于给定数字的最接近的数字

Question

I have a python function in which I need to find the number in a list 'values' that's closest to a give number 'n' but that's greater than or equal to 'n'.

So far I have this:

def nearest_largest_value2 (n, values):
      closest = []
      for i in values:
        if i == n:
          closest = i
        elif (i > n) and (i-n < 2):
          closest = i
       return closest

print(nearest_largest_value2(5, [1,3,4,6,7]))
print(nearest_largest_value2(5, [7,6,4,3,1]))
print(nearest_largest_value2(5, [1,3,4,5,6,7]))

Problem is that I'm getting the answer I want for the first two print statements (6) but I'm getting '6' for the last print statement when I want to get 5.

I'm new to Python but I thought once the first if clause was satisfied the code would stop.

Answer 1

The condition in < 2 is wrong. Instead, you should be checking if the current i is closer than the current closest . Eg:

def nearest_largest_value2 (n, values):
  closest = None
  for i in values:
    if (i >= n) and (closest is None or i < closest):
      closest = i
  return closest

EDIT:
An alternative way to describe the problem is to find the minimal value that's larger or equal than n . Describing the problem like that allows for a more "pythonic" oneliner:

def nearest_largest_value2 (n, values):
    return min(v for v in values if v >= n)

EDIT2:
As @ekhumoro pointed out, the alternative solution presented in the previous edit will break if values does not contain any element that is equal to or greater than n . He also graciously offered a fix for it:

def nearest_largest_value2 (n, values):
    min([v for v in values if v >= n] or [None])

Answer 2

If you have multiple queries with different numbers, but to the same list , then it's more efficient to first sort the list and then do a binary search on it.

from bisect import bisect_left

values = [7, 6, 4, 3, 1]

values.sort()

def nearest_largest_value2 (n, values):
    try:
        result = values [bisect_left (values, n)]
    except IndexError: # when number is > every element of list 
        result = None
    return result

print (nearest_largest_value2 (5, values)) # 6
print (nearest_largest_value2 (3, values)) # 3
print (nearest_largest_value2 (8, values)) # None

Let p correspond to length of 'values' list and q correspond to the number of inputs 'n' (or # times nearest_largest_value2 is invoked).

In the stated case , Time Complexity of the previous solution is O(pq) whereas Time Complexity of the current solution is O((p + q)logp) .