python - python 中的快速/优化 N-gram 实现

Question

哪个 ngram 实现在 python 中最快？

我尝试分析 nltk 与 scott 的 zip (http://locallyoptimal.com/blog/2013/01/20/elegant-n-gram-generation-in-python/):

from nltk.util import ngrams as nltkngram
import this, time

def zipngram(text,n=2):
  return zip(*[text.split()[i:] for i in range(n)])

text = this.s

start = time.time()
nltkngram(text.split(), n=2)
print time.time() - start

start = time.time()
zipngram(text, n=2)
print time.time() - start

[输出]

0.000213146209717
6.50882720947e-05

在 python 中生成 ngram 有没有更快的实现？

Answer 1

一些分析的尝试。我认为使用生成器可以提高这里的速度。但与原来的轻微修改相比，改进并不明显。但是，如果您不需要同时使用完整列表，生成器函数应该会更快。

import timeit
from itertools import tee, izip, islice

def isplit(source, sep):
    sepsize = len(sep)
    start = 0
    while True:
        idx = source.find(sep, start)
        if idx == -1:
            yield source[start:]
            return
        yield source[start:idx]
        start = idx + sepsize

def pairwise(iterable, n=2):
    return izip(*(islice(it, pos, None) for pos, it in enumerate(tee(iterable, n))))

def zipngram(text, n=2):
    return zip(*[text.split()[i:] for i in range(n)])

def zipngram2(text, n=2):
    words = text.split()
    return pairwise(words, n)


def zipngram3(text, n=2):
    words = text.split()
    return zip(*[words[i:] for i in range(n)])

def zipngram4(text, n=2):
    words = isplit(text, ' ')
    return pairwise(words, n)


s = "Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum."
s = s * 10 ** 3

res = []
for n in range(15):

    a = timeit.timeit('zipngram(s, n)', 'from __main__ import zipngram, s, n', number=100)
    b = timeit.timeit('list(zipngram2(s, n))', 'from __main__ import zipngram2, s, n', number=100)
    c = timeit.timeit('zipngram3(s, n)', 'from __main__ import zipngram3, s, n', number=100)
    d = timeit.timeit('list(zipngram4(s, n))', 'from __main__ import zipngram4, s, n', number=100)

    res.append((a, b, c, d))

a, b, c, d = zip(*res)

import matplotlib.pyplot as plt

plt.plot(a, label="zipngram")
plt.plot(b, label="zipngram2")
plt.plot(c, label="zipngram3")
plt.plot(d, label="zipngram4")
plt.legend(loc=0)
plt.show()

对于这个测试数据，zipngram2 和 zipngram3 似乎是最快的。