有效地查找数量最小的索引，该索引大于大型排序列表中的某个值

Question

If I have a long list of sorted numbers, and I want to find the index of the smallest element larger than some value, is there a way to implement it more efficiently than using binary search on the entire list?

For example:

import random
c = 0
x = [0 for x in range(50000)]

for n in range(50000):
    c += random.randint(1,100)
    x[n] = c

What would be the most efficient way of finding the location of the largest element in x smaller than some number, z

I know that you can already do:

import bisect
idx = bisect.bisect(x, z)

But assuming that this would be performed many times, would there be an even more efficient way than binary search? Since the range of the list is large, creating a dict of all possible integers uses too much memory. Would it be possible to create a smaller list of say every 5000 numbers and use that to speed up the lookup to a specific portion of the large list?

Answer 1

Can you try if this can be a solution? It takes long to generate the list, but seems fast to report the result.

Given the list:

import random
limit = 50 # to set the number of elements
c = 0
x = [0 for x in range(limit)]

for n in range(limit):
    c += random.randint(1,100)
    x[n] = c
print(x)

Since it is a sorted list, you could retrieve the value using a for loop:

z = 1600 # reference for lookup

res = ()
for i, n in enumerate(x):
  if n > z:
    res = (i, n)
    break

print (res) # this is the index and value of the elements that match the condition to break
print(x[res[0]-1]) # this is the element just before