我正在寻找一种将文本拆分为n-gram的方法。 通常我会做类似的事情：
import nltk
from nltk import bigrams
string = "I really like python, it's pretty awesome."
string_bigrams = bigrams(string)
print string_bigrams
我知道nltk只提供bigrams和trigrams，但有没有办法将文本分成4克，5克甚至100克？
谢谢！
答案 0 :(得分：162)
其他用户提供的基于本机python的优秀答案。但这是nltk方法（以防万一，OP因重新发明nltk库中已有的内容而受到惩罚）。
人们很少在nltk中使用ngram module。这不是因为它很难阅读ngrams，而是训练基于ngrams的模型，其中n> 3将导致大量数据稀疏。
from nltk import ngrams
sentence = 'this is a foo bar sentences and i want to ngramize it'
n = 6
sixgrams = ngrams(sentence.split(), n)
for grams in sixgrams:
print grams
答案 1 :(得分：53)
我很惊讶这还没有出现：
In [34]: sentence = "I really like python, it's pretty awesome.".split()
In [35]: N = 4
In [36]: grams = [sentence[i:i+N] for i in xrange(len(sentence)-N+1)]
In [37]: for gram in grams: print gram
['I', 'really', 'like', 'python,']
['really', 'like', 'python,', "it's"]
['like', 'python,', "it's", 'pretty']
['python,', "it's", 'pretty', 'awesome.']
答案 2 :(得分：10)
这是做n-gram的另一种简单方法
>>> from nltk.util import ngrams
>>> text = "I am aware that nltk only offers bigrams and trigrams, but is there a way to split my text in four-grams, five-grams or even hundred-grams"
>>> tokenize = nltk.word_tokenize(text)
>>> tokenize
['I', 'am', 'aware', 'that', 'nltk', 'only', 'offers', 'bigrams', 'and', 'trigrams', ',', 'but', 'is', 'there', 'a', 'way', 'to', 'split', 'my', 'text', 'in', 'four-grams', ',', 'five-grams', 'or', 'even', 'hundred-grams']
>>> bigrams = ngrams(tokenize,2)
>>> bigrams
[('I', 'am'), ('am', 'aware'), ('aware', 'that'), ('that', 'nltk'), ('nltk', 'only'), ('only', 'offers'), ('offers', 'bigrams'), ('bigrams', 'and'), ('and', 'trigrams'), ('trigrams', ','), (',', 'but'), ('but', 'is'), ('is', 'there'), ('there', 'a'), ('a', 'way'), ('way', 'to'), ('to', 'split'), ('split', 'my'), ('my', 'text'), ('text', 'in'), ('in', 'four-grams'), ('four-grams', ','), (',', 'five-grams'), ('five-grams', 'or'), ('or', 'even'), ('even', 'hundred-grams')]
>>> trigrams=ngrams(tokenize,3)
>>> trigrams
[('I', 'am', 'aware'), ('am', 'aware', 'that'), ('aware', 'that', 'nltk'), ('that', 'nltk', 'only'), ('nltk', 'only', 'offers'), ('only', 'offers', 'bigrams'), ('offers', 'bigrams', 'and'), ('bigrams', 'and', 'trigrams'), ('and', 'trigrams', ','), ('trigrams', ',', 'but'), (',', 'but', 'is'), ('but', 'is', 'there'), ('is', 'there', 'a'), ('there', 'a', 'way'), ('a', 'way', 'to'), ('way', 'to', 'split'), ('to', 'split', 'my'), ('split', 'my', 'text'), ('my', 'text', 'in'), ('text', 'in', 'four-grams'), ('in', 'four-grams', ','), ('four-grams', ',', 'five-grams'), (',', 'five-grams', 'or'), ('five-grams', 'or', 'even'), ('or', 'even', 'hundred-grams')]
>>> fourgrams=ngrams(tokenize,4)
>>> fourgrams
[('I', 'am', 'aware', 'that'), ('am', 'aware', 'that', 'nltk'), ('aware', 'that', 'nltk', 'only'), ('that', 'nltk', 'only', 'offers'), ('nltk', 'only', 'offers', 'bigrams'), ('only', 'offers', 'bigrams', 'and'), ('offers', 'bigrams', 'and', 'trigrams'), ('bigrams', 'and', 'trigrams', ','), ('and', 'trigrams', ',', 'but'), ('trigrams', ',', 'but', 'is'), (',', 'but', 'is', 'there'), ('but', 'is', 'there', 'a'), ('is', 'there', 'a', 'way'), ('there', 'a', 'way', 'to'), ('a', 'way', 'to', 'split'), ('way', 'to', 'split', 'my'), ('to', 'split', 'my', 'text'), ('split', 'my', 'text', 'in'), ('my', 'text', 'in', 'four-grams'), ('text', 'in', 'four-grams', ','), ('in', 'four-grams', ',', 'five-grams'), ('four-grams', ',', 'five-grams', 'or'), (',', 'five-grams', 'or', 'even'), ('five-grams', 'or', 'even', 'hundred-grams')]
答案 3 :(得分：10)
仅使用nltk工具
from nltk.tokenize import word_tokenize
from nltk.util import ngrams
def get_ngrams(text, n ):
n_grams = ngrams(word_tokenize(text), n)
return [ ' '.join(grams) for grams in n_grams]
示例输出
get_ngrams('This is the simplest text i could think of', 3 )
['This is the', 'is the simplest', 'the simplest text', 'simplest text i', 'text i could', 'i could think', 'could think of']
为了保持数组格式的ngrams，只需删除' '.join
答案 4 :(得分：4)
您可以使用itertools：
from itertools import izip, islice, tee
s = 'spam and eggs'
N = 3
trigrams = izip(*(islice(seq, index, None) for index, seq in enumerate(tee(s, N))))
list(trigrams)
# [('s', 'p', 'a'), ('p', 'a', 'm'), ('a', 'm', ' '),
# ('m', ' ', 'a'), (' ', 'a', 'n'), ('a', 'n', 'd'),
# ('n', 'd', ' '), ('d', ' ', 'e'), (' ', 'e', 'g'),
# ('e', 'g', 'g'), ('g', 'g', 's')]
答案 5 :(得分：2)
我从未处理过nltk，而是将N-gram作为一些小班项目的一部分。如果你想找到字符串中出现的所有N-gram的频率，这是一种方法。 D将为您提供N字的直方图。
D = dict()
string = 'whatever string...'
strparts = string.split()
for i in range(len(strparts)-N): # N-grams
try:
D[tuple(strparts[i:i+N])] += 1
except:
D[tuple(strparts[i:i+N])] = 1
答案 6 :(得分：2)
对于four_grams，它已经在NLTK，这里有一段代码可以帮助你解决这个问题：
from nltk.collocations import *
import nltk
#You should tokenize your text
text = "I do not like green eggs and ham, I do not like them Sam I am!"
tokens = nltk.wordpunct_tokenize(text)
fourgrams=nltk.collocations.QuadgramCollocationFinder.from_words(tokens)
for fourgram, freq in fourgrams.ngram_fd.items():
print fourgram, freq
我希望它有所帮助。
答案 7 :(得分：2)
使用python的内置zip()构建bigrams的更优雅的方法。
只需通过split()将原始字符串转换为列表，然后通常将列表传递一次，然后将一个元素偏移一次。
string = "I really like python, it's pretty awesome."
def find_bigrams(s):
input_list = s.split(" ")
return zip(input_list, input_list[1:])
def find_ngrams(s, n):
input_list = s.split(" ")
return zip(*[input_list[i:] for i in range(n)])
find_bigrams(string)
[('I', 'really'), ('really', 'like'), ('like', 'python,'), ('python,', "it's"), ("it's", 'pretty'), ('pretty', 'awesome.')]
答案 8 :(得分：2)
对于需要双字母组或三字母组的情况，人们已经很好地回答了，但是在这种情况下，如果您需要每个字母组作为句子，则可以使用nltk.util.everygrams
>>> from nltk.util import everygrams
>>> message = "who let the dogs out"
>>> msg_split = message.split()
>>> list(everygrams(msg_split))
[('who',), ('let',), ('the',), ('dogs',), ('out',), ('who', 'let'), ('let', 'the'), ('the', 'dogs'), ('dogs', 'out'), ('who', 'let', 'the'), ('let', 'the', 'dogs'), ('the', 'dogs', 'out'), ('who', 'let', 'the', 'dogs'), ('let', 'the', 'dogs', 'out'), ('who', 'let', 'the', 'dogs', 'out')]
如果您有一个限制，例如三字组的最大长度应为3，则可以使用max_len参数来指定它。
>>> list(everygrams(msg_split, max_len=2))
[('who',), ('let',), ('the',), ('dogs',), ('out',), ('who', 'let'), ('let', 'the'), ('the', 'dogs'), ('dogs', 'out')]
您可以修改max_len参数以实现任意克，即4克，5克，6甚至100克。
可以修改前面提到的解决方案以实现上面提到的解决方案，但是此解决方案比这要简单得多。
要进一步阅读，请单击here
并且当您只需要特定的语法（例如双字母组或三字母组等）时，可以使用M.A. Hassan的答案中提到的 nltk.util.ngrams 。
答案 9 :(得分：1)
如果效率是一个问题，你必须建立多个不同的n-gram（如你所说的那样多达一百），但你想使用纯python我会这样做：
from itertools import chain
def n_grams(seq, n=1):
"""Returns an itirator over the n-grams given a listTokens"""
shiftToken = lambda i: (el for j,el in enumerate(seq) if j>=i)
shiftedTokens = (shiftToken(i) for i in range(n))
tupleNGrams = zip(*shiftedTokens)
return tupleNGrams # if join in generator : (" ".join(i) for i in tupleNGrams)
def range_ngrams(listTokens, ngramRange=(1,2)):
"""Returns an itirator over all n-grams for n in range(ngramRange) given a listTokens."""
return chain(*(n_grams(listTokens, i) for i in range(*ngramRange)))
用法：
>>> input_list = input_list = 'test the ngrams generator'.split()
>>> list(range_ngrams(input_list, ngramRange=(1,3)))
[('test',), ('the',), ('ngrams',), ('generator',), ('test', 'the'), ('the', 'ngrams'), ('ngrams', 'generator'), ('test', 'the', 'ngrams'), ('the', 'ngrams', 'generator')]
〜与NLTK相同的速度：
import nltk
%%timeit
input_list = 'test the ngrams interator vs nltk '*10**6
nltk.ngrams(input_list,n=5)
# 7.02 ms ± 79 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
input_list = 'test the ngrams interator vs nltk '*10**6
n_grams(input_list,n=5)
# 7.01 ms ± 103 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
input_list = 'test the ngrams interator vs nltk '*10**6
nltk.ngrams(input_list,n=1)
nltk.ngrams(input_list,n=2)
nltk.ngrams(input_list,n=3)
nltk.ngrams(input_list,n=4)
nltk.ngrams(input_list,n=5)
# 7.32 ms ± 241 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
input_list = 'test the ngrams interator vs nltk '*10**6
range_ngrams(input_list, ngramRange=(1,6))
# 7.13 ms ± 165 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
从previous answer重新发布。
答案 10 :(得分：0)
您可以使用sklearn.feature_extraction.text.CountVectorizer：
import sklearn.feature_extraction.text # FYI http://scikit-learn.org/stable/install.html
ngram_size = 4
string = ["I really like python, it's pretty awesome."]
vect = sklearn.feature_extraction.text.CountVectorizer(ngram_range=(ngram_size,ngram_size))
vect.fit(string)
print('{1}-grams: {0}'.format(vect.get_feature_names(), ngram_size))
输出：
4-grams: [u'like python it pretty', u'python it pretty awesome', u'really like python it']
您可以将ngram_size设置为任何正整数。即你可以分成4克，5克甚至100克的文字。
答案 11 :(得分：0)
Nltk很棒，但有时候是一些项目的开销：
import re
def tokenize(text, ngrams=1):
text = re.sub(r'[\b\(\)\\\"\'\/\[\]\s+\,\.:\?;]', ' ', text)
text = re.sub(r'\s+', ' ', text)
tokens = text.split()
return [tuple(tokens[i:i+ngrams]) for i in xrange(len(tokens)-ngrams+1)]
使用示例：
>> text = "This is an example text"
>> tokenize(text, 2)
[('This', 'is'), ('is', 'an'), ('an', 'example'), ('example', 'text')]
>> tokenize(text, 3)
[('This', 'is', 'an'), ('is', 'an', 'example'), ('an', 'example', 'text')]
答案 12 :(得分：0)
你可以使用下面没有其他包的代码获得所有4-6gram：
from itertools import chain
def get_m_2_ngrams(input_list, min, max):
for s in chain(*[get_ngrams(input_list, k) for k in range(min, max+1)]):
yield ' '.join(s)
def get_ngrams(input_list, n):
return zip(*[input_list[i:] for i in range(n)])
if __name__ == '__main__':
input_list = ['I', 'am', 'aware', 'that', 'nltk', 'only', 'offers', 'bigrams', 'and', 'trigrams', ',', 'but', 'is', 'there', 'a', 'way', 'to', 'split', 'my', 'text', 'in', 'four-grams', ',', 'five-grams', 'or', 'even', 'hundred-grams']
for s in get_m_2_ngrams(input_list, 4, 6):
print(s)
输出如下：
I am aware that
am aware that nltk
aware that nltk only
that nltk only offers
nltk only offers bigrams
only offers bigrams and
offers bigrams and trigrams
bigrams and trigrams ,
and trigrams , but
trigrams , but is
, but is there
but is there a
is there a way
there a way to
a way to split
way to split my
to split my text
split my text in
my text in four-grams
text in four-grams ,
in four-grams , five-grams
four-grams , five-grams or
, five-grams or even
five-grams or even hundred-grams
I am aware that nltk
am aware that nltk only
aware that nltk only offers
that nltk only offers bigrams
nltk only offers bigrams and
only offers bigrams and trigrams
offers bigrams and trigrams ,
bigrams and trigrams , but
and trigrams , but is
trigrams , but is there
, but is there a
but is there a way
is there a way to
there a way to split
a way to split my
way to split my text
to split my text in
split my text in four-grams
my text in four-grams ,
text in four-grams , five-grams
in four-grams , five-grams or
four-grams , five-grams or even
, five-grams or even hundred-grams
I am aware that nltk only
am aware that nltk only offers
aware that nltk only offers bigrams
that nltk only offers bigrams and
nltk only offers bigrams and trigrams
only offers bigrams and trigrams ,
offers bigrams and trigrams , but
bigrams and trigrams , but is
and trigrams , but is there
trigrams , but is there a
, but is there a way
but is there a way to
is there a way to split
there a way to split my
a way to split my text
way to split my text in
to split my text in four-grams
split my text in four-grams ,
my text in four-grams , five-grams
text in four-grams , five-grams or
in four-grams , five-grams or even
four-grams , five-grams or even hundred-grams
您可以在blog
上找到更多详细信息答案 13 :(得分：0)
大约7年后，使用collections.deque是一个更优雅的答案：
def ngrams(words, n):
d = collections.deque(maxlen=n)
d.extend(words[:n])
words = words[n:]
for window, word in zip(itertools.cycle((d,)), words):
print(' '.join(window))
d.append(word)
words = ['I', 'am', 'become', 'death,', 'the', 'destroyer', 'of', 'worlds']
输出：
In [15]: ngrams(words, 3)
I am become
am become death,
become death, the
death, the destroyer
the destroyer of
In [16]: ngrams(words, 4)
I am become death,
am become death, the
become death, the destroyer
death, the destroyer of
In [17]: ngrams(words, 1)
I
am
become
death,
the
destroyer
of
In [18]: ngrams(words, 2)
I am
am become
become death,
death, the
the destroyer
destroyer of
答案 14 :(得分：0)
如果您想为具有恒定内存使用量的大字符串提供纯迭代器解决方案：
from typing import Iterable
import itertools
def ngrams_iter(input: str, ngram_size: int, token_regex=r"[^\s]+") -> Iterable[str]:
input_iters = [
map(lambda m: m.group(0), re.finditer(token_regex, input))
for n in range(ngram_size)
]
# Skip first words
for n in range(1, ngram_size): list(map(next, input_iters[n:]))
output_iter = itertools.starmap(
lambda *args: " ".join(args),
zip(*input_iters)
)
return output_iter
测试：
input = "If you want a pure iterator solution for large strings with constant memory usage"
list(ngrams_iter(input, 5))
输出：
['If you want a pure',
'you want a pure iterator',
'want a pure iterator solution',
'a pure iterator solution for',
'pure iterator solution for large',
'iterator solution for large strings',
'solution for large strings with',
'for large strings with constant',
'large strings with constant memory',
'strings with constant memory usage']