Question

I have a Python list of strings, e.g. initialized as follows:

l = ['aardvark', 'cat', 'dog', 'fish', 'tiger', 'zebra']

I would like to test an input string against this list, and find the “closest string below it” and the “closest string above it”, alphabetically and case-insensitively (i.e. no phonetics, just a<b etc). If the input exists in the list, both the “below” and “above” should return the input.

Several examples:

Input  | Below    |  Above   
-------------------------------
bat    | aardvark | cat      
aaa    | None     | aardvark 
ferret | dog      | fish     
dog    | dog      | dog

What’s the neatest way to achieve this in Python? (currently I’m iterating over a sorted list using a for loop)

To further clarify: I’m interested in simple dictionary alphabetical comparison, not anything fancy like Levenshtein or phonetics.

Thanks

Answer 1

This is exactly what the bisect module is for. It will be much faster than just iterating through large lists.

import bisect

def closest(haystack, needle):
    if len(haystack) == 0: return None, None

    index = bisect.bisect_left(haystack, needle)
    if index == 0:
        return None, haystack[0]
    if index == len(haystack):
        return haystack[index], None
    if haystack[index] == needle:
        return haystack[index], haystack[index]        
    return haystack[index-1], haystack[index]

The above code assumes you’ve sanitized the input and list to be all upper or lower case. Also, I wrote this on my iPhone, so please do check for typos.