CODEWITHSUNDEEP
pythonperformancepython-2.7comparisonstring-comparison
lzc
lzc

Reputation: 1705

Fastest way to compare one string of two lengths

I'm looking for the fastest way to compare if a string has one of two lengths. This string must be either one letter longer or shorter, and I feel that the if statement below might not be the fastest and I don't see much of an improvement in time with it, or without it.

I'm comparing between two string, if my_word length is more than 1 character less or more than compare_word then I shall continue my loop.

if (len(compare_word) > (len(my_word)+1)) and (len(compare_word) < (len(my_word)-1)): 
    continue

Upvotes: 2

Views: 1222

Answers (2)

miku
miku

Reputation: 188164

  • you do not need to call len twice,
  • you can utilize abs

Example:

s = "Hello"
t = "Worl"

if abs(len(s) - len(t)) > 1:
    print("string lengths differ by more than 1")

Update: With ipython's timeit there are almost no speed gain, however:

In [10]: s = str(range(100000))

In [11]: t = str(range(100001))

In [12]: %timeit len(s) > len(t) + 1 and len(s) < len(t) - 1
10000000 loops, best of 3: 106 ns per loop

In [13]: %timeit abs(len(s) - len(t)) > 1
10000000 loops, best of 3: 115 ns per loop

In [14]: %timeit 1 >= len(s) - len(t) >= -1
10000000 loops, best of 3: 113 ns per loop

Here's another run with shorter strings, but about the same result: https://gist.github.com/miku/6904419.

Nevertheless, in the context of OPs code, abs(len(s) - len(t)) > 1 really is faster.

Upvotes: 4

hivert
hivert

Reputation: 10667

It seems that the fastest way is

if 1 >= len(s) - len(t) >= -1:
    print("string lengths differ by more than 1")

Indeed:

>>> %timeit abs(a) <= 1
1000000 loops, best of 3: 283 ns per loop
>>> %timeit  1 >= a >= -1
10000000 loops, best of 3: 198 ns per loop

Upvotes: 1

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