Python，检查数字是否在列表中的多个范围内。

Question

If there are integer lists like these:

a_list = [2501, 2783, 3088, 3980, 465, 1001, 39392911, 39394382, 488955,489087, ......]
b_list = [474, 498, 47478821, 47479800, 3774, 8970, 484000, 486000......]

every 2 numbers indicate a range of natural numbers, for example, the ranges of a_list would be:

2501     2783      # 2501，2502，2503，2504，2505，2506,......,2783
3088     3980 
465      1001 
39392911 39394382 
488955   489087
......

For a given number, search for the range where it belongs to, and with priority of a_list > b_list ie if a range is found in a_list, stop searching and move on to the next number for searching.

I had test run for searching around 50 numbers which took about 7 minutes. I have a big dataset which could be 20 million numbers need to be searched in his way.

How to code this to do it faster?

============= more conditions and information =============

• could be more than 10 thousand numbers in each list.
• could be up to 30 million numbers for searching.
• the size of list is always n * 2
• a_list: [1st < 2nd, 3rd < 4th, ......]
• the numbers in the lists might occur more than once.
• the priority: a_list > b_list.

I have code as following:

hasFound = 0

if hasFound == 0:
    for x, y in izip(*[iter(a_list)]*2):   # gives every 2 numbers
        if aNumber in range(x,y):
            a_list_counter +=1 
            hasFound = 1
            break

if hasFound == 0:       
    for x, y in izip(*[iter(b_list)]*2):
        if aNumber in range(x,y):
            b_list_counter += 1
            hasFound = 1
            break

Many thanks in advance.

Answer 1

Toss them all in one big dictionary:

a_list = [2501, 2783, 3088, 3980, 465, 1001, 39392911, 39394382, 488955,489087, ......]
b_list = [474, 498, 47478821, 47479800, 3774, 8970, 484000, 486000......]
# into
ranges = {'a': [2501, 2783, 3088, 3980, 465, 1001, 39392911, 39394382, 488955,489087, ......],
          'b': [474, 498, 47478821, 47479800, 3774, 8970, 484000, 486000......]}

Then go through each list in order, mostly the way you were doing it before:

numbers = [list of your target numbers]
scores = {} # dict to store results in

for number in numbers:

    for range_name in sorted(ranges):
        range_list = ranges[range_name]
        groups = zip(*[iter(range_list)] * 2)
        if any(start <= number < end for start,end in groups):
            scores.setdefault(range_name, 0) += 1

Alternatively (and I'm not sure if this is faster or not) you could do:

for number in numbers:
    for range_name in sorted(ranges):
        range = ranges[range_name]
        if sorted(range + [number]).index(number) % 2:
            scores.setdefault(range, 0) += 1

In this case you're throwing a new number into a sorted list, re-sorting it (which is fast using TimSort), and seeing if it falls between two existing numbers.