Quick implementation of character n-grams for word

To generate bigrams:

In [8]: b='student'

In [9]: [b[i:i+2] for i in range(len(b)-1)]
Out[9]: ['st', 'tu', 'ud', 'de', 'en', 'nt']

To generalize to a different n:

In [10]: n=4

In [11]: [b[i:i+n] for i in range(len(b)-n+1)]
Out[11]: ['stud', 'tude', 'uden', 'dent']

Try zip:

>>> def word2ngrams(text, n=3, exact=True):
...   """ Convert text into character ngrams. """
...   return ["".join(j) for j in zip(*[text[i:] for i in range(n)])]
... 
>>> word2ngrams('foobarbarblacksheep')
['foo', 'oob', 'oba', 'bar', 'arb', 'rba', 'bar', 'arb', 'rbl', 'bla', 'lac', 'ack', 'cks', 'ksh', 'she', 'hee', 'eep']

but do note that it's slower:

import string, random, time

def zip_ngrams(text, n=3, exact=True):
  return ["".join(j) for j in zip(*[text[i:] for i in range(n)])]

def nozip_ngrams(text, n=3):
    return [text[i:i+n] for i in range(len(text)-n+1)]

# Generate 10000 random strings of length 100.
words = [''.join(random.choice(string.ascii_uppercase) for j in range(100)) for i in range(10000)]

start = time.time()
x = [zip_ngrams(w) for w in words]
print time.time() - start

start = time.time()
y = [nozip_ngrams(w) for w in words]
print time.time() - start        

print x==y

[out]:

0.314492940903
0.197558879852
True

Although late, NLTK has an inbuilt function that implements ngrams

# python 3
from nltk import ngrams
["".join(k1) for k1 in list(ngrams("hello world",n=3))]

['hel', 'ell', 'llo', 'lo ', 'o w', ' wo', 'wor', 'orl', 'rld']