在python中排序小写字母

Question

I am trying to convince myself that a counting sort performs faster than the sorted method in Python. However calling the sorted builtin seems to be faster even for large inputs like 10 million elements. What can I do to make the counting sort faster?

I generate a list of lowercase letters to simplify the example to 26 unique values:

letters = [random.choice(string.ascii_lowercase) for i in range(10000000)]

I then do the following variation on counting sort:

def sorted_count(letters):
 counts = [0] * 26
 for letter in letters:
     counts[ord(letter) - 97] += 1
 out = [None] * len(letters)
 j = 0
 for i in range(len(counts)):
     while counts[i] > 0:
         out[j] = chr(i + 97)
         counts[i] -= 1
         j += 1
 return out

Even on 10,000,000 elements the call to sorted(letters) is ~4x faster. How can I improve the speed of my sort?

Answer 1

Instead of using a while loop inside the forloop at the end . you could simple use

for i in range(len(counts)):
 if counts[i]>0:
     out[j] =counts[i]*chr(i + 97)
 j+=1
return out

Answer 2

Here's a modified function, which is 3 times faster than the proposed one and twice as fast as sorted :

import random
import string
import timeit
N = 1000000
letters = [random.choice(string.ascii_lowercase) for i in range(N)]


def original_sorted_count(letters):
    counts = [0] * 26
    for letter in letters:
        counts[ord(letter) - 97] += 1
    out = [None] * len(letters)
    j = 0
    for i in range(len(counts)):
        while counts[i] > 0:
            out[j] = chr(i + 97)
            counts[i] -= 1
            j += 1
    return out

def eric(letters):
    counts = [0] * 26
    for letter in letters:
        counts[ord(letter) - 97] += 1
    out = []
    for i in range(len(counts)):
        out += [chr(i+97)] * counts[i]
    return out

print('Original : %.3fs' %timeit.timeit(lambda: original_sorted_count(letters), number=20))
print('Sorted   : %.3fs' %timeit.timeit(lambda: sorted(letters), number=20))
print('Eric     : %.3fs' %timeit.timeit(lambda: eric(letters), number=20))

print(eric(letters) == sorted(letters))

It outputs:

Original : 9.616s
Sorted   : 6.367s
Eric     : 3.604s
True