问题
Which ngram implementation is fastest in python?
I've tried to profile nltk's vs scott's 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
[out]
0.000213146209717
6.50882720947e-05
Is there any faster implementation for generating ngrams in python?
回答1:
Some attempts with some profiling. I thought using generators could improve the speed here. But the improvement was not noticeable compared to a slight modification of the original. But if you don't need the full list at the same time, the generator functions should be faster.
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()
For this test data, zipngram2 and zipngram3 seems to be the fastest by a good margin.
回答2:
Extending M4rtini's code, I made three additional versions with a hardcoded n=2
parameter:
def bigram1(text):
words = iter(text.split())
last = words.next()
for piece in words:
yield (last, piece)
last = piece
def bigram2(text):
words = text.split()
return zip(words, islice(words, 1, None))
def bigram3(text):
words = text.split()
return izip(words, islice(words, 1, None))
Using timeit
, I get these results:
zipngram(s, 2): 3.854871988296509
list(zipngram2(s, 2)): 2.0733611583709717
zipngram3(s, 2): 2.6574149131774902
list(zipngram4(s, 2)): 4.668303966522217
list(bigram1(s)): 2.2748169898986816
bigram2(s): 1.979405164718628
list(bigram3(s)): 1.891601800918579
bigram3
is the fastest for my tests. There does seem to be a slight benefit to hardcoding and from using iterators if they're used throughout (at least for this parameter value). We see the benefit from iterators throughout in the bigger difference between zipngram2
and zipngram3
for n=2
.
I also tried getting a boost from using PyPy, but it seemed to actually make things slower here (this included attempts to warm up the JIT by calling it 10k times on functions before doing the timing test). Still, I'm very new to PyPy so I may be doing something wrong. Possibly using Pyrex or Cython would enable greater speedups.
来源:https://stackoverflow.com/questions/21883108/fast-optimize-n-gram-implementations-in-python